calculus laboratory(with derive)

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A CALCULUS LABORATORY MANUAL

USING DERIVE

Gabriela R. Sanchis

Elizabethtown College

September 15, 1996



Contents

1 Functions and Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 An Introduction to Limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28

4 Average and Instantaneous Velocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

5 Derivatives of Some Elementary Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

6 Newton’s Method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

7 Three Important Theorems About Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . 69

8 An Introduction to Riemann Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82

9 The Fundamental Theorem of Calculus . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 96

10 Approximate Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 107

11 Volumes of Revolution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 120

12 Exponential and Logarithmic Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 134

13 Inverse Trigonometric Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 143

14 Integration of Rational Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 160

15 Improper Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 172

16 Sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 185

17 P-series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 195

18 Taylor Polynomials . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 208

i



ii CONTENTS

19 Parametric Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221

20 Polar Coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 232

21 An Introduction to Vectors . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 244

22 Cylinders and Quadric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 255

23 Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 272

24 Curvature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 284

25 Functions of 2 Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293

26 Directional Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307

27 Relative Extrema for Functions of Two Variables . . . . . . . . . . . . . . . . . . . . . . . . . 323



Laboratory # 1

Functions and Graphs

Teaches:

• functions – domain, range, x and y intercepts, symmetry, odd, even

• using Derive to find zeroes

1. Author

sin(x) − sin(3x)

3 +

sin(5x)

5

2. Choose Plot Beside <Enter> to open a graphics window, then choose Plot to plot the highlighted

expression.

3. Use F10 to zoom out. This allows you to see more of the graph. If you look at the bottom of the

screen, you’ll see the x and y scales increase each time you hit F10 . The x scale is the number of

units that each tick on the x axis represents; likewise for the y scale. F9 zooms in, so each time you

hit F9 you’ll see the scales decrease. You can also use the Scale command to set your scales manually.

Definition 1.1 The domain of a function y = f (x) is the set of all xs for which the function is

defined. The range of a function y = f (x) is the set of all y s such that y = f (x) for some x.

1



2 LABORATORY # 1. FUNCTIONS AND GRAPHS

4. Answer Worksheet Question #1.

Definition 1.2 The x-intercepts of the graph of f are the x coordinates of the points where the graph

crosses the x axis. The y-intercept of the graph of f is the y coordinate of the point where the graph

crosses the y axis.


6. One way to evaluate a function at a particular x is to use the command Manage Substitute. For

example, to evaluate expression #1 at x = π/2:

(a) make sure the Algebra window is active ( F1 toggles between windows);

(b) make sure the expression you wish to evaluate is highlighted (i.e. expression #1);

(c) choose Manage Substitute <Enter>;

(d) type in the value of x, i.e. π/2 (alt-p for π),then <Enter>;

(e) finally, choose approXimate <Enter>.

7. Answer Worsheet Question #3.

8. To examine the graph of f (x − 1), highlight expression #1, choose Manage Substitute <Enter> and

type x − 1 <Enter>. Then choose Plot Plot.

9. Answer Worksheet Questions #4 and #5.

Definition 1.3 A zero or root of a function f is any number a such that f (a) = 0.

10. In terms of the graph of f , the zeroes are the x- intercepts. So once you Plot a function, you can

approximate the zeroes by moving the cross onto the x-intercepts and reading off the x coordinate of

the point from the bottom of the screen.




3

12. Factor (d) in Worksheet Question #6. Because one of the factors is cubic, it is difficult to find the roots

by hand. However, Derive is still able to find the roots exactly, using the soLve command. Highlight

this expression and choose soLve <Enter>. Besides x = 0, you should get 3 more roots. ApproXimate

each of them to express them as decimals.


14. Finally, Author 2x − x2 and Plot. What happens when you try to find the roots using the soLve

command? Derive does not seem able to find the exact roots. From the graph it appears that x = 2

and x = 4 are roots, and plugging in these values for x confirms this.


16. Now make sure the Algebra window is active, then choose Options Precision Approximate <Enter>.

Try the soLve command again on 2x − x2. Since we can see from the graph that the root in question

is between −1 and 0, enter −1 for the lower limit and 0 for the upper limit.




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WORKSHEET FOR LABORATORY #1

Functions and Graphs

NAME: DATE:

LAB PARTNER: INSTRUCTOR:

Questions 1–4 refer to the function f (x) = sin x − sin(3x)3 + sin(5x)

5

1. (a) What is the domain of f (x)?

(b) What is the range of f (x)?

[Hint: Use the cross to locate the smallest and biggest y values on the graph; use F9 to zoom

in, Center to center the viewing window at the cross, and the arrow keys to adjust the position of

the cross; repeat until the graph on the screen is a horizontal line; with your cross on the graph,

read off the y coordinate from the bottom of the screen.

To reset, use the Scale command to reset your x and y scales both to 1, Move the cross to (0, 0),

and Center.]

2. (a) What are the x-intercepts of f (x)? Find all x-intercepts, not just the ones in your viewing window.

5



6 Worksheet # 1. FUNCTIONS AND GRAPHS

[Hint: Set your x scale to π/2 (π is entered by typing alt-p). ]

(b) What is the y-intercept of f (x)?

3. (a) Evaluate f (x) at the given values of x:

x f (x) x f (x)

0.4 −0.4

0.8 −0.8

1.2 −1.2

1.6 −1.6

2.0 −2.0

(b) What relationship appears to exist between f (x) and f (−x)?

(c) Complete the following definition:

Definition 1.4 The graph of f is symmetric with respect to the origin (in which case f is

said to be an odd function) if for all x, .

4. To answer (a)–(d), use Manage Substitute on the expression sin x − sin(3x)3 + sin(5x)

5 to obtain the given

transformation, then Plot to obtain the graph.



Name: 7

(a) How does the graph of f (x − 1) compare to that of f (x)?

(b) How does the graph of f (x + 2) compare to that of f (x)?

(c) How does the graph of f (2x) compare to that of f (x)?

(d) How does the graph of f (x/2) compare to that of f (x)?

(e) How does the graph of f (x) + 1 compare to that of f (x)?

Questions 5 and 6 refer to the function f (x) = x2

−1x2+1 . With the graphics window active, choose Delete

All.

5. Author and Plot

x2 − 1

x2 + 1




(a) Evaluate this function at the given values of x:

x f (x) x f (x)

0.5 −0.5

1.0 −1.0

1.5 −1.5

2.0 −2.0

(b) What relationship appears to exist between f (x) and f (−x)?

(c) Complete the following definition:

Definition 1.5 The graph of f is symmetric with respect to the y axis (in which case f is

said to be an even function) if for all x, .

6. Find all the zeroes of the following functions by following the technique just outlined. Note: not all the

zeroes may be in your viewing window, so you may have to zoom out ( F10 ) to see them all. On the

other hand, once your cross is on or near the zero, you should Center and zoom in ( F9 ) in order to

get a closer look at that section of the graph. Keep zooming in, Centering, and adjusting the position

of the cross, until you are satisfied that the coordinate of the zero is correct to three decimal places.

(a) y = x3 − 2x2 − x + 2



Name: 9

(b) y = x3 − x2 + x − 1

(c) y = 2x4 + x3 − 3x2 + x

(d) y = 16x4 − 20x3 + 5x

(e) y = 2x − x2

7. Use Derive’s Factor Rational command to factor (a), (b), and (c) in question #6, then find the roots

exactly by setting each factor equal to 0. You will need the quadratic formula

x = −b ± √

b2 − 4ac

2a

to find the zeroes of any quadratic factor ax2 + bx + c. Express all your answers as decimals.

(a) x3 − 2x2 − x + 2 =




(b) x3 − x2 + x − 1 =

(c) 2x4 + x3 − 3x2 + x =

8. What are the real roots (i.e. those not involving the imaginary number i) of y = 16x4 − 20x3 + 5x?

9. In question #6e, you found an approximation to the negative root of y = 2x − x2. Use Manage

Substitute to plug this value in for x. Is your approximation correct, or is it too big, or is it too small?

[Hint: Look at the graph of y = 2x − x2; what is the sign of y when x is a little to the right of the

negative root, and when x is a little to the left of the negative root?]

10. (a) What does Derive give you for the negative root of 2x − x2?



Name: 11

(b) Use Manage Substitute to plug in this value for x, then approXimate. What do you get?

(c) Is Derive’s approximation too big or too small? Explain.

(d) Choose Options Precision TAB to change the number of digits to 10. Then try soLve again to

find the negative root of 2x − x2. Do you get something bigger or smaller than before? In view

of your answer to (c), is this what you expected?



Laboratory # 2

An Introduction to Limits

Teaches:

• limits

• one-sided limits

• limits involving infinity – vertical and horizontal asymptotes

1. Open a graphics window by choosing Plot Beside <ENTER>.

2. In Calculus, we often deal with problems in which the answer cannot be obtained directly but must be

approached by a series of approximations. Consider the function f (x) = sinxx

; f (x) cannot be evaluated

at x = 0; however, f (x) can be evaluated at values of x as close to 0 as we might want.

3. Recall that in order to use Derive to evaluate a function such as f (x) = sinx

x

at, say, x = 1, you need

to Author sin(x)x

, choose Manage Substitute to replace x with 1, and then choose approXimate (this is

the method you used in the previous lab).

4. If your graphics window is active, choose Algebra, then Choose Options Precision, TAB over to Digits,

and change the number of digits to 11 <ENTER>. Now answer Worksheet Question #1.

12



13

5. If F (x) “gets close to” some number L as x “gets close to” a number a, we write

limx→aF (x) = L.

We read this as “The limit of F (x) as x approaches a is L”.

Your calculations in Worksheet Question #1 should suggest that limx→0

sin x

x = 1.


7. Let us explore the concept of a limit graphically. Plot sinxx

. Our conjecture is that limx→0sinxx

= 1.

To obtain a better sense of this graphically, do the following:

(a) Move your cross to (0, 1) and Center. Notice how the graph seems to go through this point (it

doesn’t really since sinxx is not defined at 0 so the graph really has a tiny hole at (0, 1); what you

are noticing is that for x’s that are really close to 0, the y’s are really close to 1).

(b) Suppose “y close to 1” means that 0.99 < y < 1.01. Author and Plot the following two expressions:

y = .99 y = 1.01

(c) Then “y close to 1” means that y is between these two horizontal lines. Change your y scale to

.01. (This will “zoom you in” on the y axis.)

(d) Use Shift- F7 to “zoom in” on the x axis. Keep zooming in until the entire graph on your screen

lies between the two horizontal lines. (What you’ve done is taken a small box around the point

(0, 1) and blown it up:

6

-

?

6

6

-−0.2 −0.1 0.1

1.01

0.99

1.00

−2 −1 1



14 LABORATORY # 2. AN INTRODUCTION TO LIMITS

The fact that you can do this no matter how close together your two horizontal lines are is what

we mean by saying that the limit of sinxx as x approaches 0 is 1.

8. We are now ready to state the definition of a limit. As you read it, keep in mind the previous example

where f (x) = sinxx , a = 0, L = 1, and = .01:

Definition 2.1 We say limx→a

f (x) = L if for any 2 horizontal lines y = L + and y = L − , we can

find a viewing window centered at (a, L) such that for all x’s in our viewing window (except possibly

x = a) the graph of f (x) lies between the two horizontal lines.

9. Now let’s look at the function y = sin

1x

. Reset your graphics window by making it active ( F1

makes the next window active), then choosing Window Designate 2D-Plot. Then Author and Plot

sin

1

x

I want you to disprove the following claim:

limx→0

sin1

x = 0 (2.1)

Do this as follows:

(a) Author and Plot:

y = .01

y = −.01

(b) set your y scale to .01.

(c) Zoom in horizontally using Shift- F7 . Can you zoom in enough so that the entire graph in your

viewing window lies between the two horizontal lines? The fact that you can’t indicates that (2.1)

is false.




15

11. The table that you filled in in Worksheet Question #4 suggests that limx→1|x2−x|x−1

does not exist, since

there is no unique number that |x2−x|x

−1

approaches as x gets close to 1. However if we restrict ourselves

to values of x that are close to but greater than 1, then we see that |x2−x|x−1

approaches 1; similarly, if

we restrict ourselves to values of x that are close to but less than 1, then we see that |x2−x|x−1 approaches

−1. We call 1 the right limit of |x2−x|x−1 as x approaches 1; we call −1 the left limit of

|x2−x|x−1 as x

approaches 1. We write:

limx→1+

|x2 − x|x − 1

= 1 and limx→1−

|x2 − x|x − 1

= −1

12. Reset your graphics window and Plot |x2−x|x−1 . Notice how the y values for the part of the graph to the

right of x = 1 are close to 1 when x is close to 1; similarly, notice how the y values for the part of the

graph to the left of x = 1 are close to −1 when x is close to 1.

a b c


14. As you can see, as x gets closer and closer to (but greater than) -1, 1x+1

gets larger and larger. We say

that the limit of 1x+1 as x approaches −1 from the right is infinity and we write

limx→−1+

1

x + 1 = ∞.



16 LABORATORY # 2. AN INTRODUCTION TO LIMITS

15. Similarly, as x gets closer and closer to (but less than) −1, 1x+1

decreases without bound. We say that

the limit of 1x+1 as x approaches −1 from the left is minus infinity and we write

limx→−1−

1

x + 1 = −∞.

16. Let’s now examine the graph of y = 1x+1

near x = −1. Reset your graphics window if necessary, set

your x scale to 2 and your y scale to 10, then Plot 1x+1

.


18. The unbounded behavior of a function f as x approaches a (from the right or from the left) is reflected

in the graph via vertical asymptotes. The following picture illustrates the different possibilities, where

the dotted lines x = a, x = b, and x = c are vertical asymptotes:

-a b c

We write

limx→a−

f (x) = ∞; limx→b+

f (x) = −∞;

limx→a+

f (x) = ∞; limx→c−

f (x) = −∞;

limx→b−

f (x) = ∞; limx→c+

f (x) = ∞.

19. Answer Worksheet Questions #8–#10.



17

20. Your calculation should suggest that

limx→∞x

√ x2 + 1 = 1 and limx→−∞x

√ x2 + 1 = −1

Definition 2.2 If limx→∞

f (x) = L or limx→−∞

f (x) = L , then the line y = L is a horizontal asymp-

tote of f (x).

Thus y = 1 and y = −1 are horizontal asymptotes for the function f (x) = x√ x2+1

.




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An Introduction to Limits

NAME: DATE:


1. (a) Fill in the following table:

x sinxx x sinx

x

0.1 −0.1

0.01 −0.01

0.001

−0.001

0.0001 −0.0001

(b) What number does sinxx approach as x gets closer and closer to 0?

2. Fill in the following tables, then conjecture what the value of the given limit is; if there appears that

there is no limit, i.e. the values of f (x) do not seem to be approaching a unique number, then write

“DNE” (for Does Not Exist).

19



20 Worksheet # 2. AN INTRODUCTION TO LIMITS

(a)

x x2 x x2

2.1 1.9

2.01 1.99

2.001 1.999

2.0001 1.9999

limx→2

x2 =

(b)

x sin 1x x sin 1

x0.1 −0.1

0.01 −0.01

0.001 −0.001

0.0001 −2.0001

limx→0

sin

1

x

=

(c)

x (1 + x)1/x x (1 + x)1/x

0.1 −0.1

0.01 −0.01

0.001 −0.001

0.0001 −0.0001

limx→0

(1 + x)1/x =

[Hint: The above calculations should indicate that limx→0(1 + x)1/x is approximately 2.7. The

actual limit turns out to be an irrational number which we generally denote by e; we will see more

of this number in Calculus II. You can obtain a decimal approximation to e in Derive by Authoring



Name: 21

“Alt-e”, then using the approXimate command. Use this approximation for your answer above.]

(d)

x x3−1|x−1| x x3−1

|x−1|

0.9 1.1

0.99 1.01

0.999 1.001

0.9999 1.0001

lim

x→1

x3 − 1

|x − 1| =

3. Determine whether the following statements of the form limx→a f (x) = L are true or false. In each

case follow the procedure outlined in the two examples, i.e.

• Reset your graphics window by making it active, then choosing Window Designate 2D-Plot.

• Author and Plot the given function f (x) as well as the horizontal lines y = L +.01 and y = L−.01

where L is the conjectured limit – e.g. for (a) you would Author and Plot the following three

expressions:

1 − √ x + 1

x y = −.5 − .01 y = −.5 + .01

(recall that the square root sign is obtained by typing alt-q)

• Move the cross and to the given point (a, L) and Center; (e.g. to (0, −0.5) for (a)); S=set your y

scale to .01, and Center again.

• Use Shift- F7 to zoom in until the entire graph in your viewing window lies between the two

horizontal lines. If you can do it, that’s evidence that the given statement is true; if not, conclude

that the statement is false.




(a) limx→0

1 − √ x + 1

x = −0.5 [Does the graph of y = 1−√ x+1

x initially appear to go through the point

(0,

−.5)?]

(b) limx→0

(1 + x)1/x = 2.7[Does the graph of y = (1 + x)(1/x) initially appear to go through the point

(0, 2.7)?]

(c) limx→0

x sin(1/x) = 0[Does the graph of y = x sin(1/x) initially appear to go through the point

(0, 0)?]

4. (a) Fill in the following table:

x |x2−x|x−

1 x |x2−x|x−

1

1.1 0.9

1.01 0.99

1.001 0.999

1.0001 0.9999

(b) What number does |x2−x|x−1

approach as x gets closer and closer to 1, if we restrict ourselves to

x > 1?

(c) What number does |x2−x|x−1 approach as x gets closer and closer to 1, if we restrict ourselves to



Name: 23

x < 1?

5. Use Derive to obtain plots of the following functions; then use the plot to conjecture the value of the

given limits:

(a) f (x) = x2−1|x−1|

limx→1+ f (x) = limx→1− f (x) =

(b) f (x) = x3−x|x2−x|

limx→1+ f (x) = limx→1− f (x) =

limx→0+ f (x) = limx→0− f (x) =

(c) f (x) = x2

|x|

limx→0+ f (x) = limx→0− f (x) =

6. (a) Investigate the limit limx→−11

x+1 by filling in the following table:

x 1x+1 x 1

x+1

−0.9 −1.1

−0.99 −1.01

−0.999 −1.001

−0.9999 −1.0001

(b) What happens to 1x+1 as x approaches −1 if we restrict ourselves to x > −1?




(c) What happens to 1x+1

as x approaches −1 if we restrict ourselves to x < −1?

7. Copy the graph of y = 1x+1

below; draw the vertical asymptote exhibited by the graph as a dotted

line.

−10

−5

5

10

−4 −2 2 4

8. Plot each of the following functions to find the vertical asymptotes. Use the graph to determine whether

the limit as x approaches a from the right/left is ∞ or −∞.

(a) f (x) = 1(x+1)2

(b) f (x) = x3

(x+2)2(x−3)



Name: 25

(c) f (x) = x−2x2−4

9. Use the results of the previous question to formulate a theorem about vertical asymptotes:

Theorem 2.1 Let P and Q be polynomials. Then the line x = a is a vertical asymptote of y = P (x)Q(x)

if Q(a) and P (a) .

[Hint: In 8c, what can you say about P (2) and Q(2)? Is x = 2 a vertical asymptote?]

10. (a) Fill in the following table.

x x√ x2+1

x x√ x2 +1

10 −10

100 −100

103 −104

104 −105

105 −106

(b) What happens to x√ x2+1

as x increases without bound?

(c) What happens to x√ x2+1

as x decreases without bound?

(d) Reset your graphics window, then use Derive to obtain a plot of y = x√ x2+1

and copy the graph

below; draw the horizontal asymptotes exhibited by the graph as dotted lines.




−2

−1

1

2

−2 −1 1 2

11. What is the maximum number of horizontal asymptotes that a function can have?

12. For each of the following rational functions, determine the horizontal asymptotes by inspecting the

graphs:

F (x) limx

→∞F (x) lim

x

→−∞F (x) Horizontal Asymptotes

7x5+4x3+2x+1−x8+2x2+1

x99+1x100+1

2x2+3x+14x2−2x+3

6x3+4x2−20x+2−2x3+6x2−x+7

13. Based on the above table, what conjecture can you make about the horizontal asymptotes of a rational

function (i.e a polynomial divided by a polynomial) when

(a) the degree of the numerator < the degree of the denominator:



Name: 27

(b) the degree of the numerator = the degree of the denominator:

[Hint: look at the highest powers of x in the numerator and denominator]



Laboratory # 3

Continuity

Teaches:

• types of discontinuities – jump, removable, infinite

• intermediate value theorem

• bisection-type method for finding roots of functions

1. Consider again the function f (x) defined by f (x) =

x2, x ≤ 1;

2x, x > 1.

We say that this function is defined piecewise.

2. Transfer Load the Utility file LABS.MTH; this will load a function that will make it easier for you to

author and plot piecewise-defined functions.

3. To enter the above function, Author and Plot

PIECES

[x2, 2x], [1]

.


28



29

5. The function g(x) defined by

g(x) = x2, x

≤1;

2x − 1, x > 1.

can be graphed by Authoring and Plotting

PIECES

[x2, 2x − 1], [1]


7. Let h(x) be defined by

h(x) =

x, x < −1;

−x + 1, −1 ≤ x < 1;

2x, x ≥ 1.

(3.1)

To graph this function Author and Plot

PIECES ([x, −x + 1, 2x], [−1, 1])

I hope that by now you are getting the idea of how to use PIECES to graph piecewise-defined functions.

8. Anwer Worksheet Question #3.

9. The original function h defined by (3.1) had jumps or discontinuities at x = −1 and x = 1. More gen-

erally, the graph of a function f (x) will have a discontinuity at x = a if limx→a+ f (x) = limx→a− f (x):

-

6

?

a

limx→a+

f (x)

limx→a−

f (x)



30 LABORATORY # 3. CONTINUITY

10. Delete All graphs in your graphics window, then Author and Plot

1

x − 1

As you can see, this graph also has a “break” in it.


12. Author and Plot

F (x) := x2 + 4x + 3

x + 1

This function appears to be continuous (i.e. no breaks). However, in actual fact, we know the graph

must have a break or “hole” at x = −1 since the function is not defined at x = −1.


14. The discontinuity of F (x) = x2+4x+3x+1

at x = −1 is called a “removable discontinuity” because it consists

only of a little hole that can be filled in by defining F (−1) in the right way.


16. The following pictures illustrate the different types of discontinuities that can occur. All of them

involve some kind of “break” in the graph:

6

-

?

-

?

b 6

-

?

b6

The discontinuity exhibited by the first graph at x = −1 is called an infinite discontinuity; the

second graph has a removable discontinuity at x = 0; the second graph has a jump discontinuity

at x = 1.



31

Intermediate Value Theorem

Definition 3.1 We say the function f is continuous on the open interval (a, b) if it has no

discontinuities on (a, b). We say f is continuous on the closed interval [a, b] if it is continuous on

the open interval (a, b) and

limx→a+

f (x) = f (a) and limx→b−

f (x) = f (b).

17. As you’ve seen, at discontinuity at x = a means that the graph has a “gap” or “break” or “hole” at

x = a. It follows then that a function that is continuous on an interval [a, b], loosely speaking, is one

whose graph can be drawn without lifting pencil from paper.


19. Your answer to Worksheet Question #8 should suggest the following:

Theorem 3.1 (Intermediate Value Theorem) If f is a function continuous on the closed interval

[a, b] for which f (a) = f (b), and if N is any number between f (a) and f (b), then there exists at least

one number c between a and b such that f (c) = N .

The following picture illustrates the theorem:

N N

f (a)

f (b)

a c c b

6

-



32 LABORATORY # 3. CONTINUITY

20. By letting N = 0, we obtain the following corollary:

Corollary 3.1 If f is a function continuous on [a, b] and if f (a) and f (b) are of opposite algebraic

signs, then f has a zero between a and b, i.e. there exists at least one number c between a and b such

that f (c) = 0.

A Method for Finding Zeroes

21. The Intermediate Value Theorem allows us to approximate the zeroes of a continuous function to any

degree of accuracy. The method, which we will illustrate with an example, is similar to the bisection

method discussed in the text.

22. Author F (x) := x6 − 3x − 1. The corollary above tells us that F (c) = 0 for at least one c between 1

and 2, since F (1) = −3 < 0 and F (2) = 57 > 0. Plotting this function reveals that there is only one

zero in the interval (1, 2).

23. Author

VECTOR([x, F (x)], x, 1, 2, 0.1)

and approXimate. The above expression tells DERIVE to evaluate the points (x, F (x)) where x steps

from 1 to 2 in increments of size 0.1. By noting where F changes sign, we can determine which of the

intervals [1.1.1], [1.1, 1.2],.. . ,[1.9, 2] contains the zero c. Since F (1.3) < 0 and F (1.4) > 0, we conclude

that 1.3 < c < 1.4.

24. Next we wish to evaluate F (x) at points between 1.3 and 1.4, so Author

VECTOR([x, F (x)], x, 1.3, 1.4, .01)

and approXimate. Now we note that F (1.3) < 0 and F (1.31) > 0 so we conclude that 1.3 < c < 1.31.

25. Answer Worksheet Questions#9 and #10.



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Continuity

NAME: DATE:


1. (a) At x = 1, the graph jumps from y= to y = .

(b) limx→1+

f (x)=

(c) limx→1−

f (x) =

2. (a) How does this graph differ from the previous one?

(b) limx→1+

g(x) =

(c) limx→1−

g(x) =

3. (a) At x = −1 the graph jumps from y = to y = .

(b) limx→−1+

h(x) =

(c) limx→−1−

h(x) =

(d) Suppose I want to “join” the left and middle pieces of this graph by shifting up the left piece.

This can be done by redefining h(x) to be x + a for x < −1 where a is the number of units by

33



34 Worksheet # 3. CONTINUITY

which the left piece of the graph will be shifted up. What should a be so that the jump at x = −1

disappears?

(e) Author and Plot

PIECES ([x + a, −x + 1, 2x], [−1, 1])

where you’ve replaced a with the value you came up with in (d). Has the jump disappeared?

(f) For this redefined function h(x),

limx→−1+

h(x) =

limx→−1−

h(x) =

(g) At x = 1 the graph jumps from y = to y = .

(h) limx

→1+

h(x) =

(i) limx→1−

h(x) =

(j) Find a number b so that redefining h(x) to be 2x + b for x ≥ 1 makes the second jump disappear.

[Hint: since you want to shift the right piece down, what should be the sign of b?]

(k) Author and Plot

PIECES ([x + a, −x + 1, 2x + b], [−1, 1])

where you’ve replaced a and b with the values you came up with in (d) and (j). Have both jumps

disappeared?



Name: 35

(l) For this redefined function h(x),

limx→

1+h(x) =

limx→1−

h(x) =

4. (a) At what x value does the break occur?

(b) Describe the behavior of the graph near this x value.

5. Move the cross along the graph until the x coordinate is as close to −1 as possible.

(a) What is the y coordinate of the cross?

(b) This graph, then, actually has a tiny hole (too small to see) at what point?

(c) lim

x→−1

x2 + 4x + 3

x + 1

=

(d) Show that x2+4x+3x+1 equals x +3 when x = −1. Is this consistent with your answer to (c)? Explain.




[Hint: factor the numerator.]

(e) With the cross on the “hole”, Center, then zoom in ( F9 ) until the graph begins to break up and

you can see the hole. Author and Plot

y = x + 3

What is (or should be) the difference between the graph of y = x2+4x+3x+1 and the graph of y = x+3?

(f) The hole in the graph of F (x) can be “filled in” by defining F (−1) to be .

6. Reset the graphics window, then Author and Plot: y = sin x

x

(a) The function f (x) = sinxx

has a removable discontinuity (i.e. a tiny hole) at x = .

(b) How can this discontinuity be removed?

7. Author and Plot: y =

√ x − 2

x − 4

(a) The function f (x) =√ x−2x−4 has a removable discontinuity at x = .



Name: 37

(b) How can this discontinuity be removed?

8. Suppose f is a function continuous on [a, b] and such that f (a) = f (b). Let N be any number between

f (a) and f (b), as in the following picture:

6

-

a b

f (b)

f (a)

N

r

r

A

B

Draw a continuous curve from A = (a, f (a)) to B = (b, f (b)).

(a) What must be true about the curve you drew as it relates to the line y = N ?

(b) Explain why there must be a point c between a and b such that f (c) = N . Label this point c on




your picture.

9. Continuing in this manner, do three more iterations; after each one, write down the interval that

contains c.

(a) < c <

(b) < c <

(c) < c <

10. Use the method we just discussed to find the root c of

F (x) = x5 + x + 1.

Begin by verifying that there is a root between −1 and 0. Do four iterations.



Laboratory # 4

Average and Instantaneous Velocity

Teaches:

• average and instantaneous velocity

• rectilinear motion

1. Laura lives in Hamilton, New York; her parents live 60 miles away in Syracuse. She recently took a

car trip to visit her parents. She left Hamilton at 12:00 p.m. and arrived at her parents house at 1:30

p.m. Laura claims that she never once exceeded the 55 miles per hour speed limit during the entire

trip. In this lab, we shall try to discover whether Laura was telling the truth.

2. Transfer Load the Utility file LABS.MTH, then Author

S (t) := t2

(36t2

− 428t + 801)9

3. Choose Plot Beside <ENTER> to open a graphics window; then set your x scale to 0.4 and your y

scale to 15; Move the cross to x = 0.75 and y = 30; finally Center and Plot.

39



40 LABORATORY # 4. AVERAGE AND INSTANTANEOUS VELOCITY

4. The graph in front of you is the plot of Laura’s trip; time (in hours, starting at 12:00 p.m.) is plotted

on the horizontal axis, while distance travelled (in miles) is plotted on the vertical axis. So for example,

the graph begins at the origin (0, 0) indicating that at noon (x = 0) she had travelled 0 miles, and it

ends at (1.5, 60) indicating that 1 12 hours later (x = 1.5) she had travelled 60 miles. By moving the

cross along the graph and reading off the coordinates of the cross from the bottom of the screen, you

can estimate how far Laura had travelled at any time x ,0 ≤ x ≤ 1.5.


6. Recall that the average velocity of a moving object over a time interval is computed by dividing the

distance travelled during that time interval by the length of the time interval. Answer Worksheet

Questions #2–#4.

7. Your answers to the previous questions should indicate that sometime during the second half hour of

the trip Laura must have exceeded the speed limit. Was she speeding during the whole half hour, or

only for a couple of minutes, perhaps to pass another vehicle? In order to answer these question, we

need to keep calculating her average velocity over smaller and smaller time intervals. Since this can be

tedious, I’ve defined some functions to help you out.



41

8. The function VELOCITY will calculate average velocities for you. In your Algebra window, Author

VELOCITY(S (t), 0, 1.5, 3)

and approXimate. You will obtain the vector

[33.1111, 57.7777, 29.1111]. (4.1)

These numbers represent the average velocities during the first, second, and third half hours of the

trip. They should agree with your answer to Worksheet Question #4.

9. To calculate average velocities over 15-minute intervals, you must divide the original time interval

[0, 1.5] into six subintervals, each of length 1/4 hour. To do this, Author

VELOCITY(S (t), 0, 1.5, 6)

and approXimate. (In general, VELOCITY(S (t), a, b, n) calculates the average velocities over the time

intervals obtained by dividing the interval [a, b] into n equal subintervals.)


11. The function VELPLOT will give you a plot of the average velocities. Author VELPLOT(S (t), 0, 1.5, 3),

then approXimate and Plot. As you can see, the graph you obtain consists of 3 horizontal lines whose

heights are given precisely by (4.1).


13. The graph that you’ve obtained with VELPLOT is a graph of average velocities over tiny (one-minute)

intervals. You will learn later on that the instantaneous velocity is given by a function called the

derivative of S (t). Just for fun, highlight the expression for S (t) (it should be expression #1) then

choose Calculus Differentiate, then <Enter> three times. Simplify and Plot.




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Average and InstantaneousVelocity

NAME: DATE:


1. Fill in the following table by reading the coordinates of the cross from the bottom of the screen.

Remember that time t is the x coordinate (in hours) and distance travelled d is the y coordinate (in

miles). Give your answer to the nearest mile.

Time Distance travelled

(to the nearest mile)

12:00

12:20

12:40

1:00

1:20

1:30

43



44 Worksheet # 4. AVERAGE AND INSTANTANEOUS VELOCITY

2. (a) What was Laura’s average velocity during her trip?

(b) Does this seem to support Laura’s claim that she did not exceed the speed limit? Explain.

3. (a) Calculate Laura’s average velocity during the first 45 minutes of the trip. Use the function S (t)

to evaluate this exactly (instead of reading off an approximation of S (t) from the graph).

Distance travelled= miles

Length of time interval= hours

Average velocity= ÷ = miles per hour

(b) Calculate Laura’s average velocity during the last 45 minutes of the trip.




(c) On average, during which half of the trip was she travelling faster?

(d) Do your answers support Laura’s claim? Explain.



Name: 45

4. (a) Calculate Laura’s average velocity during the first half hour of the trip.




(b) Calculate Laura’s average velocity during the second half hour of the trip.



Average velocity= ÷

= miles per hour

(c) Calculate Laura’s average velocity during the last half hour of the trip.




(d) On average, on which third of the trip was she travelling the fastest?

(e) Do your answers support Laura’s claim?

5. During which 15 minute interval (i.e. from what time to what time) was Laura travelling the fastest?




What was her average velocity during this time interval?

6. Use the VELOCITY function to fill in the following table:

Time Interval Average Velocity Time Interval Average Velocity

12:00 to 12:05 12:45 to 12:50

12:05 to 12:10 12:50 to 12:55

12:10 to 12:15 12:55 to 1:00

12:15 to 12:20 1:00 to 1:05

12:20 to 12:25 1:05 to 1:10

12:25 to 12:30 1:10 to 1:15

12:30 to 12:35 1:15 to 1:20

12:35 to 12:40 1:20 to 1:25

12:40 to 12:45 1:25 to 1:30

7. Use the VELPLOT function to obtain a graph of Laura’s average velocities over 1-minute intervals.

(The plotting may take a while.) Copy the graph onto the coordinate axes below:



Name: 47

30

60

0.4 0.8 1.2 1.5 1.6

Velocity vs Time

8. (a) Use the velocity graph to answer the following questions as accurately as you can. During which

time interval(s) (i.e. from what time to what time) did Laura exceed the 55 mph limit? (Give

your answer in real time to the nearest minute.)

(b) During which time interval(s) was Laura’s velocity below 20 mph?

(c) What was Laura’s approximate velocity

i. from 12:45 to 12:46?

ii. From 1:20 to 1:21? p.m.?




9. Suppose a fly is moving along a horizontal coordinate line, so that its position, or coordinate, on the

line is given by the function s(t) = −4t2 + 10t + 6, 0 ≤ t ≤ 3, where t represents time measured in

seconds. The motion of the fly can be represented by the following graph:

-

6 8 10 12420

t = 0

t = 3

-s

s

(a) I have indicated on the graph the position of the fly at times t = 0 (s(0) = 6) and t = 3 (s(3) = 0).

Compute s(1) and s(2) and label the graph accordingly.

(b) According to your calculations as illustrated by the graph, the fly turns around sometime between

t = and t =

(c) Author

S (t) := −4t2 + 10t + 6

VELOCITY (S (t), 0, 3, 3)

and approXimate the last expression. This will give you the average velocities of the fly over

each of the three time intervals [0, 1], [1, 2], and [2, 3]. Verify these numbers, and explain why

the last two are negative. [Hint: how far has the fly moved over the interval [1 , 2] and in which



Name: 49

direction?]

(d) Reset the graphics window, then Plot S (t); set the x scale to 0.8, the y scale to 5; Move the cross

to x = 1.5 and y = 0, then Center. These settings should give you a good view of the graph

for 0 ≤ x ≤ 3. On this graph, time t is measured on the x axis, and the position of the fly is

measured on the y axis. Copy the graph below and label the points (both x and y coordinates)

for x = 1, 2, 3.

6

-

1 2 3

0

4

8

12

−12

−8

−4

(e) What can you say about the motion of the fly for values of t for which the graph of S (t) is rising?




(f) What can you say about the motion of the fly for values of t for which the graph of S (t) is falling?

(g) What then do the x and y coordinates of the peak of this graph represent?

(h) To obtain an approximate graph of the velocity of the fly, Author, approXimate, and Plot:

VELPLOT (S (t), 0, 3, 100)

What can you say about the motion of the fly for values of t for which the velocity is positive?

(i) What can you say about the motion of the fly for values of t for which the velocity is negative?

(j) So what does the x (or t) intercept of the velocity graph represent?

(k) Again highlight the expression for S (t) and choose Calculus Differentiate,<Enter> three times,

Simplify, to obtain the derivative of S (t). This function gives the instantaneous velocity of the



Name: 51

fly at any time t. Plot it to see how close the VELPLOT approximation was. According to your

answer to the previous question, at what exact time does the fly turn around?



Laboratory # 5

Derivatives of Some Elementary

Functions

Teaches:

• definition of derivative• power rule for derivatives

1. Recall that the derivative of the function f (x) is denoted by f (x) and is defined by

f (x) = limh→0

f (x + h) − f (x)

h . (5.1)

2. In this lab, we will investigate the derivatives of some elementary functions.

3. Author and Plot

F (x) := sin x

4. Now Author

M (x, h) := F (x + h) − F (x)

h (compare with (5.1))

52



53

D(x) := M (x, .01)


6. Plot D(x). Answer Worksheet Questions #2 and #3.

7. Verify your answer to Worksheet Question #2 by plotting that function with DERIVE and seeing

whether it coincides with the graph of D(x). Does it?

8. Delete all graphs. Author and Plot

F (x) := −x2 + 2x + 1

then replot D(x). Answer Worksheet Questions #4 and #5.

9. Verify your answer to Worksheet Question #4 by plotting that function with DERIVE and seeing

whether the graph coincides with the graph of D(x). Does it?

10. Reset the graphics window, then zoom out once ( F10 ). Then choose Options Color Auto No. Next

choose Options Color Plot and pick your favorite color for the next plot.

11. Author

VECTOR (F (a) + D(a)(x − a), a, −4, 4, .2)

This will calculate the (approximate) tangent lines to F (x) at x = −4, −3.8, −3.6, . . . , 3.6, 3.8, 4 (these

are only approximate because we are using D for the slope rather than the actual derivative). Now

simplify and Plot‘ 1. Finally choose Options Color Auto Yes and Plot F (x). Do you see now how these

tangent lines can give us a lot of information about the graph of y = F (x)?

12. Author and Plot

F (x) := 2x



54 LABORATORY # 5. DERIVATIVES OF SOME ELEMENTARY FUNCTIONS

then replot expression D(x). Answer Worksheet Questions #6–#10.

Many students have a difficult time using the definition (5.1) to evaluate derivatives, mainly because

the algebra involved can get a bit messy. So we are going to “cheat” a little bit and use DERIVE to

help with the algebra.

13. Close the graphics window. Recall that M (x, h) = F (x+h)−F (x)h and that F (x) = limh→0 M (x, h).

14. For our first example, author F (x) := 1. Simplify the expression for M (x, h) and enter the result in

the second column of Table 5.1 on the appropriate line. In column 3 you will enter the limit as h → 0

of the entries in column 2.

[Hint: most of the time, to evaluate the limit as h → 0, simply “plug in” 0 for h.]

15. Now author F (x) := x and again simplify M (x, h). Then complete the second line of Table 5.1.

16. Repeat the above with F (x) := x2 and then with F (x) := x5.


18. The derivatives that we have explored so far are for functions of the form xn where n is a positive

integer. See what happens when n is negative by repeating the above with F (x) := 1x , F (x) := 1

x2 ,

and F (x) := 1x4 . Enter your results in Table 5.1. Then answer Worksheet Question #12.

19. Now author F (x) := x1/2 and again simplify M (x, h). You should get:

√ x + h − √

x

h

You can’t evaluate the limit as h → 0 yet, since plugging in 0 for h yields 0

0 . You will need to do a

little algebra. Answer Worksheet Questions #13 and #14.

20. Author F (x) := x−1/2 and again simplify M (x, h). Answer Worksheet Questions #15 and #16.




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Derivatives of Some ElementaryFunctions

NAME: DATE:


1. Explain why D(x) can be used as an approximation for F (x).

2. What familiar function does the graph of D(x) look like?

3. Copy the graphs of F and its derived function D onto the coordinate axes below. Also fill in the blank:

55



56 Worksheet # 5. DERIVATIVES OF SOME ELEMENTARY FUNCTIONS

?

-

6

F (x) = sin x

?

-

6

D(x) ≈

4. Find an equation for D(x), then Author and Plot your equation to verify that it coincides with the

graph of D(x).

5. Copy the graphs of F and its derived function D onto the coordinate axes below. Also fill in the blank:

?

-

6

F (x) = x2 + 2x + 1

?

-

6

D(x) =

6. Copy the graphs of F (x) = 2x and its derived function D onto the coordinate axes below.

?

-

6

F (x) = 2x versus its derived function D(x)



Name: 57

7. By comparing the graphs of F (x) and D(x) above, you might conjecture that D(x) = aF (x) for some

constant a. Approximate a (to the nearest tenth) and then verify your answer by plotting aF (x) and

checking that the plot coincides with the graph of D(x).

[Hint: If D(x) = aF (x) for all x, what’s D(0)?]

a =

8. Author F (x) := 3x. Replot F and D and copy the graphs onto the coordinate axes below:

?

-

6

F (x) = 3x versus its derived function D(x)

9. By comparing the graphs of F (x) and D(x) above, you might again conjecture that D(x) = aF (x) for

some constant a. Approximate a (to the nearest tenth) and then verify your answer by plotting aF (x)

and checking whether the plot coincides with the graph of D(x).

a =

10. EXTRA CREDIT: You’ve compared the graphs of 2x and its derivative, as well as the graphs of 3x

and its derivative. By trial and error, find b (to the nearest tenth) such that the graphs of bx and its

derivative coincide exactly.

b =

11. Based on your calculations, can you come up with a formula for the derivative of xn?

d

dxxn =




12. Does the formula that you came up with in question 11 seem to work when n is a negative integer?

Verify that it does for each of the three examples that you have worked out.

13. Find the limit as h → 0 of M (x, h). Enter your results in Table 5.1 on the appropriate line. [Hint:

rationalize the numerator by multiplying both the numerator and the denominator of the expression

by√

x + h +√

x.]

14. Does the formula that you came up with in question 11 seem to work when n is a positive fraction?

Verify that it does for n = 1/2.



Name: 59

15. Find the limit as h → 0 of M (x, h). Enter your results in Table 5.1 on the appropriate line. [Hint:

Again, rationalize the numerator].

16. Does the formula that you came up with in question 11 seem to work when n is a negative fraction?

Verify that it does for n = −1/2.

17. Based on what you learned in this lab, evaluate the following derivatives:

(a) d

dx x100 =

(b) d

dx

1

x342 =

(c) d

dx x4/7 =




(d) d

dx

1

x3/5 =

F (x) M (x, h) limh→0

M (x, h)

1

x

x2

x5

1x

1x2

1x4

√ x

1√ x

Table 5.1: Derivatives of some elementary functions



Laboratory # 6

Newton’s Method

Teaches:

• Derivation of Newton’s method

1. In your Algebra window set Options Precision to Approximate with 10 digits.

2. Author

F (x) := x5 + 2x3 + 1 (6.1)

D(x) := 5x4 + 6x2 (6.2)

y = D(a)(x − a) + F (a) (6.3)

then Plot expression (6.1). Note that D(x) is the derivative of F (x) and so expression (6.3) is the

equation of the line tangent to the graph of y = F (x) at the point (a, F (a)).


4. Set your x and y scales to 0.5; Move the cross to (0, 0) and Center. Choose Algebra Options Precision

61



62 LABORATORY # 6. NEWTON’S METHOD

Exact <Return>. Then answer Worksheet Questions #3-#7.

5. The algorithm that you just performed to approximate the zero of F (x) is called Newton’s method.

The basic idea is as follows:

(a) Choose an initial approximation x0 to the zero (in the example you worked through, this initial

approximation was x0 = −1).

(b) Draw the line tangent to the graph of the function at x0. Find x1, the x intercept of this line.

It should be a number that’s closer to the zero than your first approximation. (This was your

answer to Worksheet Question #3).

(c) Draw the line tangent to the graph of the function at x1, then find x2, the x intercept of this line.

This should be a number that’s even closer to the zero than your previous approximation. (This

was your answer to Worksheet Question #4).

(d) Continue in this manner. At each step you are finding the x intercept of the line tangent to the

graph of the given function at the approximation found in the previous step.

(e) Stop when you’ve attained required accuracy.


7. Author

F (x) :=

D(x) :=

This will “clear” the previous definitions of F and D. Then answer Worksheet Question #9.

8. To automate Newton’s method, we proceed as follows:

(a) Come up with an initial approximation x0 for the root x.



63

(b) Plug this approximation in for a into the function you found in Worksheet Question #9 to come

up with your next approximation x1.

(c) Plug x1 in for a into the function you found in Worksheet Question #9 to come up with your

next approximation x2.

(d) Continue in this manner until two successive approximations agree in as many decimal places as

your desired accuracy.

9. Use this procedure to redo your calculation of √

7. Remember that you must first define F and D. If

your calculations don’t agree with your previous ones, you’ve make a mistake.


11. Finally, Derive’s ITERATES function can be used to automate iteration techniques such as the one

you have studied in this lab. To see how this works, Author

ITERATES (a − F (a)/D(a), a, 1, 4) (6.4)

and approXimate. The resulting string of numbers should agree with your answers to Worksheet

Question #10b. The general syntax for the ITERATES function is

ITERATES(u, x, x0, n)

where x0 is the initial value of x at which expression u will be evaluated, and n is the number of

iterations that will be performed. You may want to rework Worksheet Questions #8 and #10a using

the ITERATES function. For example, to rework Question #10a, Author

F (x) := x3 + x + 1

D(x) := 3x2 + 1

ITERATES(a − F (a)/D(a), a, −1, 4)

then approXimate the last expression.



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Newton’s Method

NAME: DATE:


1. F (x) has one zero. Use the cross to approximate it. Zoom in ( F9 ) to get as much accuracy as you

can.

x = (approximately)

2. Derive’s soLve command internally uses the bisection method which we studied earlier in this course

to find a zero (as long as Options Precision is set to Approximate). Use it to approximate the zero of

F (x). You will need to give Derive an initial interval in which to start searching for the root.

x =

3. Write down the equation of the line tangent to the graph of F (x) at x0 = −1, then Plot it.[Hint:

Manage Substitute −1 for a in expression (6.3), then use the Expand command to write it in the form

y = mx + b (Just hit Return when prompted for the expand variable).] Draw the tangent line on the

graph below, and label x1, its x intercept. Find x1 by highlighting the right side of the equation of the

tangent line and then using the soLve command, followed by approXimate.

Tangent line: y = x+

65



66 Worksheet # 6. NEWTON’S METHOD

−3

−2

−1

1

2

3

−2 x0 1 2

..........

...........

..........

.........

........

........

.......

........

.......

.......

............................................................................................................................................................................................................................................................................................................

................................

....................................................................................

.....................

.................................................................

............................................................................................................

x intercept: x1=

4. Write down the equation of the line tangent to the graph of F (x) at the point x1 that you found in

Question #3, then Plot it. Also find x2, the x intercept of this line.

[Note: always use approX to obtain decimal representations of numbers.]


x intercept: x2=

5. Write down the equation of the line tangent to the graph of F (x) at the point that you found in

Question #4, then Plot it. Also find x3, the x intercept of this line.


x intercept: x3=

6. Continue in this manner until successive calculations are equal in the first 6 decimal places.

x4=

x5=



Name: 67

7. How does your answer to Question #6 compare to your answers to Questions #1 and #2?

8. Use Newton’s method to approximate√

7 to five decimal places (i.e. until successive approximations

agree in the first 5 decimal places). At each step, write down the equation of the tangent line in

y = mx + b form as well as the x intercept. Use x0 = 3 for your initial approximation. [Hint: consider

the function F (x) = x2 − 7 and the appropriate D(x)].

x0 = 3 y = x +

x1 = y = x +

x2 = y = x +

x3 = y = x +

x4 =

9. For any a, find a formula for the x intercept of the line tangent to y = F (x) at (a, F (a)).



68 Worksheet # 6. NEWTON’S METHOD

10. Use Newton’s method (i.e. the formula in Question #9) to find approximations (to five decimal places)

to all real roots of the following functions. In each case, plot the function to determine the number of

roots, and use the closest integer to the root for x0. Remember to redefine F (x) and D(x) each time.

(a) F (x) = x3 + x + 1

x0 =

x1 =

x2 =

x3 =

x4 =

x5 =

(b) F (x) = cos x − x

x0 =

x1 =

x2 =

x3 =

x4 =



Laboratory # 7

Three Important Theorems About

Derivatives

Teaches:

• First derivative at interior extrema• Rolle’s Theorem

• Mean Value Theorem

1. Author and Plot the function y = 1 − x2 and locate the highest point on the graph for −2 < x < 2.

Move your cross to this highest point, then Center and zoom in (use F9 ). Keep zooming in and

centering until the graph in front of you is a straight line.


3. Reset your graphics window by choosing Window Designate 2D-Plot. Author and Plot the function

y = 12x5 −45x4 + 20x3 + 90x2 −120x +43, −2 < x < 2. You should leave your x scale at 1 (so that the

interval [−2, 2] is visible in your graphics window) and set the y scale so that the entire graph is visible.

69



70 LABORATORY # 7. THREE IMPORTANT THEOREMS ABOUT DERIVATIVES

You may want to use F8 to zoom out in the vertical direction; F7 zooms the vertical direction in.

Again locate the highest point with your cross, then Center and zoom in until the graph in front of

you is a straight line. Answer Worksheet Question #2.

4. Repeat the above for the function y = 3−4x2

1+4x2 , −2 < x < 2. Answer Worksheet Question #3.

5. Complete the statement of Theorem 7.3 in Worksheet Question #4. Then answer Worksheet Questions

#5 – #7.

6. Let f be a function that is continuous on [a, b] and differentiable on (a, b). Assume f (a) = f (b). Let

f (c) be the maximum height of f on [a, b]. Let f (d) be the minimum height of f on [a, b]:

6

-

f (d)

f (a) = f (b)

f (c)

a c d b


8. Hopefully, you are now convinced of the following theorem:

Theorem 7.1 (Rolle’s Theorem) Let f be a continuous function on the closed interval [a, b] and

have a derivative for all x in the open interval (a, b). If f (a) = f (b), then there is at least one number

c in (a, b) such that f (c) = 0.




71

10. Finally, suppose f satisfies all the hypotheses of Rolle’s theorem, except that f (a) and f (b) are not

necessarily equal. The Mean Value Theorem says that there must be some c in (a, b) at which the

tangent line is parallel to the line joining (a, f (a)) and (b, f (b)):

6

-

(a, f (a)) r

r

r (b, f (b))

(c, f (c))

a c b

11. Since parallel means having the same slope, and the slope of the line joining (a, f (a)) and (b, f (b)) is

f (b)−f (a)b

−a

, we have:

Theorem 7.2 (Mean Value Theorem) Let f be a continuous function on the closed interval [a, b]

and have a derivative for all x in the open interval (a, b). Then there is at least one number c in (a, b)

such that f (c) = f (b)−f (a)b−a .




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Three Important Theorems AboutDerivatives

NAME: DATE:


1. (a) What are the coordinates of the highest point of the graph of y = 1 − x2, −2 < x < 2?

(b) What is the slope of the line that you see (i.e. of the tangent line at this highest point)?

2. (a) What are the coordinates of the highest point of the graph of y = 12x5 − 45x4 + 20x3 + 90x2 −

120x + 43, −2 < x < 2?


73



74 Worksheet # 7. THREE IMPORTANT THEOREMS ABOUT DERIVATIVES

3. (a) What are the coordinates of the highest point of the graph of y = 3−4x2

1+4x2 , −2 < x < 2?


4. Theorem 7.3 (Theorem of the Interior Extremum) Let f be a function defined at least on the

open interval (a, b). If f takes on a maximum value (or a minimum value) at a number c in this

interval and if f (c) exists, then f (c) = .

5. Give an example of a function which is continuous on a closed interval for which the theorem fails

(i.e. for which the tangent line at the highest point is not horizontal). [Hint: look for a function whose

highest point occurs at one of the endpoints of the interval.]

6. Give an example of a function f defined on some interval [a, b] that has a horizontal tangent at some

point (c, f (c)) but such that (c, f (c)) is not the highest (or lowest) point on the graph. [Hint: look at

y = x3 on some appropriate interval.]

7. For each function f in Questions #1–#3, verify (show your work) that f (c) = 0 for the maximum



Name: 75

point (c, f (c)) that you found.

8. (a) If a < c < b, what can we conclude about f (c) by Theorem 7.3?

(b) If a < d < b, what can we conclude about f (d) by Theorem 7.3?

(c) Suppose both c and d are endpoints of [a, b]. This means that both the highest and lowest points

of the graph occur at the endpoints.

What kind of function must f be on [a, b]? (Hint: Recall that we are assuming that f (a) = f (b),

so that the height at both endpoints is the same.]

So what must f (x) be for any x in (a, b)?




9. Let f (x) = 32x6 − 48x4 + 18x2 − 1.

(a) Does f (x) satisfy the hypotheses of Rolle’s theorem on the interval [−

1, 1]? Explain.

(b) Plot f (x). How many points c are there in (−1, 1) that satisfy the conclusion of Rolle’s theorem

(i.e. for which f (c) = 0)? Find them all, correct to 6 decimal places.[Hint: evaluate f (x), then

Author it; use Derive’s Factor Rational command to factor f (x); so you should then be able to

find its zeroes exactly, then approXimate them.]



Name: 77

The following instructions pertain to questions 10–13.

For each of the following functions f and intervals [a, b]:

(a) Plot f on [a, b] and copy the graph onto the given coordinate axes.

(b) Find the slope of the line through (a, f (a)) and (b, f (b)).

(c) Find the equation of the line through (a, f (a)) and (b, f (b)). Write your equation in slope-intercept

form. [Note: Figure out the equation in point-slope form first, then soLve for y, Expand, and

approXimate to obtain the equation in slope-intercept form.] Plot the line and copy the plot onto

the coordinate axes.

(d) If the hypotheses of the Mean Value Theorem are satisfied:

i. Write down the equation that you must solve to find the points c that satisfy the conclusion

of the Mean Value Theorem. This equation will be of the form:

f (c) = m

where m is the slope that you found in (a), and f (c) is the derivative of f (evaluated at c)

that you must evaluate.

ii. Author this equation and soLve for c to find all points c that satisfy the conclusion of the

mean value theorem; label these points (both x and y coordinates) on your graph.

Otherwise, explain which hypothesis the Mean Value Theorem does not hold.

(e) Find the equation of the tangent line to the graph of f at each point c found above. Write it in

slope-intercept form. [Note: in point-slope form, the equation is y − f (c) = m(x − c), where m is

the number that you found in (a). See (c) to convert to slope-intercept form.]

Plot each line and copy it onto the coordinate axes.




10. f (x) = x3 − 12x on [−2, 2]

-

6

?

4

8

12

16

−4

−8

−12

−16

1 2−1−2

(b) Slope of line through (a, f (a)) and (b, f (b)) is :

.

(c) Equation of line through (a, f (a)) and (b, f (b)): x+

(d) If MVT applies:

(i) Equation to solve for c:

(ii) c =

If MVT doesn’t apply, explain why:

(e) If MVT applies, write down equations of tangent lines:

y = x+

y = x+



Name: 79

11. f (x) = 8x4 − 8x2 + 1 on [−1, 0.5]

-

6

?

(b) Slope of line through (a, f (a)) and (b, f (b)) is:


(d) If MVT applies:


(ii) c =



y = x+

y = x+




12. f (x) = 1(x−2)2 on [0, 3]

-

6

?



(d) If MVT applies:


(ii) c =



y = x+

y = x+



Name: 81

13. f (x) = x2/3 on [−2, 2] (Choose Manage Branch Any before Plotting)

-

6

?



(d) If MVT applies:


(ii) c =



y = x+

y = x+



Laboratory # 8

An Introduction to Riemann Sums

Teaches:

• summation formulas

• approximating areas via step functions

• Riemann sums

1. There are two ways in which you can have Derive evaluate a sum for you, such as5

k=1

(2k − 1).

First way : Author 2k − 1, then choose Calculus Sum <Enter>; type k for the variable of summation,

1 for the lower limit, and 5 for the upper limit. Then Simplify the resulting expression. Try this.

Second way : Author SUM(2k − 1, k, 1, 5) and Simplify. Try this as well; you should get the same

answer.


3. Summation Formulas. As we get into Riemann sums and integration, we will find ourselves often

needing to evaluate sums of 1,000 or 10,000 terms. Clearly, most of the time this won’t be possible

without a computer. However, sometimes these sums can be evaluated using summation formulas. For

82



83

example,

1,000k=1

2 = 2 + 2 + · · · + 2 1, 000 2’s

= 1, 000 × 2 = 2, 000;

and in general,

nk=1

c = c + c + · · · + c n c’s

= n × c = nc.

This is our first summation formula.


5. The functions defined in the previous exercises are called step functions.

Definition 8.1 A step function is a function whose domain is a union of nonoverlapping intervals

and such that the function is constant on each interval.

Derive has an easy way of expressing step functions using the CHI function. For example, the function

in Worksheet Question #11 would be written as

CHI(−2, x, 1) + 3 CHI(1, x, 2) + 2 CHI(2, x, 4).

6. Author the above and plot and compare with the sketch you made in Worksheet Question #11.

7. It should be clear to you now that if f is a step function, then the region under f is the union of

rectangles and so the area under f can easily be computed using basic geometry.

8. Let f (x) = −12

x3 + x2 + 1 and suppose we want to compute the area above the x axis and below this

curve between x = −1 and x = 2.

9. Author

F (x) := −1

2x3 + x2 + 1



84 LABORATORY # 8. AN INTRODUCTION TO RIEMANN SUMS

10. Author and Plot

F (x) CHI(−1, x, 2) (8.1)

Also set the x and y scales to 2. (Multiplying by the CHI function has the effect of “zeroing out” f (x)

outside the interval [−1, 2]).

11. We will try to approximate the area under the graph by finding a step function whose graph is “close”

to that of f ; then the area under the step function should be “close” to the area under f .

12. Author and Plot:

CHI(−1, x, 0)F (0) + CHI(0, x, 1)F (1) + CHI(1, x, 2)F (2) (8.2)

13. The step function (8.2) was obtained by dividing the interval [−1, 2] into three equal subintervals [−1, 0],

[0, 1], and [1, 2], and then defining a step function so that the height on each of these subintervals was

the same as the height of F at the right-hand endpoint of the subinterval (0 for [−1, 0], 1 for [0, 1], and

2 for [1, 2]). Answer Worksheet Question #13.

14. In your graphics window choose Delete Last to remove the graph of (8.2). Then Author and Plot

CHI(−1, x, −.5)F (−0.5) + CHI(−.5, x, 0)F (0) + CHI(0, x , .5)F (0.5)

+CHI(.5, x, 1)F (1) + CHI(1, x, 1.5)F (1.5)+ CHI(1.5, x, 2)F (2) (8.3)

15. The step function (8.3) was obtained by dividing the interval [-1,2] into 6 equal subintervals and then

defining a step function whose height on each subinterval was equal to the height of F at the right-hand

endpoint of the subinterval.




85

17. We can generalize this process by using n subintervals. In this case, the area under the graph of

y = F (x) (where F is a positive function) for a ≤ x ≤ b, will be approximated by

nk=1

b − a

n F

a +

k (b − a)

n

(8.4)

Definition 8.2 Any expression of the form

nk=1

b − a

n f (x∗k)

where a + (k − 1) b−an ≤ x∗k ≤ a + k b−an

is called a Riemann sum.

NOTES:

(a) In Worksheet Questions #13–#15, you evaluated some Riemann sums where a = −1, b = 2, x∗k

is the righthand endpoint of the kth interval, and n = 3, 6, and 12, respectively.

(b) In general, if f ≥ 0, then a Riemann sum is used to approximate the area under f on the interval

[a, b].

(c) Other common choices for x∗k are the left-hand endpoint of the kth subinterval or the midpoint

of the k th subinterval.

18. In our previous example, we had F (x) = −12 x3 + x2 + 1, a = −1, and b = 2. Author the sum (8.4)

using Calculus Sum. Then verify (using Manage Substitute and approXimate) that when n = 3, you

get 3.5. Also verify that:

(a) When n = 6 you obtain the answer to Worksheet Question #14.

(b) When n = 12 you obtain the answer to Worksheet Question #15.

19. Answer Worksheet Question #16 and #17.



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An Introduction to Riemann Sums

NAME: DATE:


1. Use Derive to calculate the following sums; express your answers as decimals:

(a)100k=1

k2

(b)7

j=−3

1

j2 + 1

2. Express the following sum using sigma notation, then use the first summation formula (i.e. don’t

use Derive) to evaluate:

3 + 3 + 3 + · · · + 3 25 times

3. Use sigma notation to express the sum 1 + 2 + · · · + n, then use Calculus Sum to obtain our second

summation formula.

4. Express each of the following sums using sigma notation, then use the second summation formula

to evaluate; show your work; don’t use Derive, except to check your answer:

87



88 Worksheet # 8. AN INTRODUCTION TO RIEMANN SUMS

(a) 1 + 2 + 3 + 4 + 5

(b) 1 + 2 + · · · + 100

5. Use sigma notation to express the sum 12 + 22 + · · · + n2, then use Calculus Sum to obtain our third

summation formula.

6. Express each of the following sums using sigma notation, then use the third summation formula to

evaluate; show your work; don’t use Derive except to check your answer:

(a) 12 + 22 + 32 + 42 + 52 + 62

(b) 12 + 22 +

· · ·+ 1002

7. Use sigma notation to express the sum 13 + 23 + · · ·+ n3, then use Calculus Sum to obtain our fourth

summation formula.

8. Express each of the following sums using sigma notation, then use the fourth summation formula to

evaluate:

(a) 13 + 23 + 33



Name: 89

(b) 13 + 23 + · · · + 103

9. Use summation formulas and the properties of sigma notation to evaluate the following sum:

7j=1

(2 j2 − 3 j + 4)

10. a. Sketch a graph of y = 2 on the coordinate axes below and shade in the region that lies above the

x axis and below the graph of y = 2 between x = −1 and x = 3.

-

6

?

1 2 3−1−2−3

1

2

3

−1

−2

−3

b. Calculate the area of the shaded region.

11. a. Sketch the graph of f (x) defined by f (x) =

1, −2 ≤ x < 1;

3, 1 ≤ x < 2;

2, 2 ≤ x ≤ 4.




-

6

?

1 2 3 4−1−2−3

1

2

3

−1

−2

−3

b. Shade in the region above the x axis that lies under the graph of f (x) between x = −2 and x = 4

and calculate the area of this region.

12. For each of the following, find the area of the region above the x axis that lies below the function f .

In (a) and (b) sketch the region:

a. f (x) =

2, 1 ≤ x < 3;

3, 3 ≤ x < 4;

1, 4 ≤ x < 4.5;

5, 4.5 ≤ x ≤ 5.



Name: 91

-

6

?

2 4 6−2−4−6

2

4

6

−2

−4

−6

b. f (x) =

3, −3 ≤ x ≤ −2;

2, 0 < x ≤ 1;

1.5, 1 < x ≤ 2.




-

6

?

1 2 3−1−2−3

1

2

3

−1

−2

−3

c. f (x) =

A, a ≤ x < b;

B, b ≤ x < c;

C, c ≤ x < d;

D, d ≤ x ≤ e.

13. Below is the graph of f (x) = −12

x3 + x2 + 1. Copy the graph of the step function (8.2) onto the same

axes; shade in the area of the region under the step function and calculate the area of the shaded

region.



Name: 93

−1

1

2

3

4

−2 −1 1 2 3

.............................................................................................................................................................................................................................................................................

.............................

..................................

..............................................................................................................................................................................................................................

Area= × F( ) + × F( ) + × F( )

=3

k=1

× F( )

= × + × + ×

=

14. Copy the graph of the step function (8.3) onto the coordinate axes below. Shade in the area of the

region under the step function and calculate the area of the shaded region.

−1

1

2

3

4

−2 −1 1 2 3

.............................................................................................................................................................................................................................................................................

.............................

..................................

................................................................................................................................................................................

..............................................

Area= ×F( ) + ×F( ) + ×F( ) + ×F( ) + ×F( ) + ×F( )




=6

k=1

× F( )

=

15. Now draw the step function that you get when you use 12 subintervals, then shade in the region under

the step function and calculate the area of the shaded region.

−1

1

2

3

4

−2 −1 1 2 3

........................................................................................................................................................................................................................................................

....................................

.............................

.........................................

........................................................................................................................................................................................................

Area=

12

k=1

× F( ) =

[Note: Use Derive’s Calculus Sum and approXimate commands to evaluate the sum.]

16. Use Manage Substitute and approXimate to approximate the area under F (x) = −12

x3 + x2 + 1 when



Name: 95

a. n = 20

b. n = 100

c. n = 1000

17. Author and Plot

F (x) :=

1 − x2

then use the formula (8.4) (with a = −1 and b = 1 to approximate the area under F (x)

(a) when n = 100

(b) when n = 1000

(c) Use geometry to determine the exact area under the curve y =√

1 − x2 from x = −1 to x = 1.

Explain.[Note: you may want to approXimate the exact answer so that you can compare it to

your answers above.]



Laboratory # 9

The Fundamental Theorem of

Calculus

Teaches:

• definite and indefinite integrals• The Fundamental Theorem of Calculus – derivative form

• The Fundamental Theorem of Calculus – antiderivative form

1. Recall that a Riemann sum is any expression of the form

nk=1

b − a

n f (x∗k), (9.1)

where a + b−an (k − 1) ≤ x∗k ≤ a + b−a

n k.

2. Also recall that if f ≥ 0 on [a, b], then (9.1) represents the areas of rectangles which approximate the

area under f between x = a and x = b; the following picture illustrates this for n = 5:

96



97

...........................

................................

................................................

.............................................................................................................................................................................................................................................................................................................................................................................................................

...........................................................

..........

a x∗1 x∗2 x∗3 x∗4 x∗5 b

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Definition 9.1 The definite integral of f from a to b, denoted by ba f (x) dx, is defined by

ba

f (x) dx = limn→∞

nk=1

b − an

f (x∗k). (9.2)

x is called the variable of integration.

3. Note that the variable of integration makes no difference to the value of the integral, so that

ba

f (x) dx =

ba

f (u) du =

ba

f (y) dy

4. In the special case when a = 0, b = x. and x∗k = a + b−an k = kxn , (9.2) becomes:

x0

f (u) du = limn→∞

nk=1

x

nf

kx

n

5. In this lab we will investigate the above limit for different functions f . Author

F (x) := x

nk=1

x

n F kx

n (9.3)

and Simplify expression (9.3).


7. Author F (x) := x2 and simplify expression (9.3) again; also evaluate the limit as n → ∞.



98 LABORATORY # 9. THE FUNDAMENTAL THEOREM OF CALCULUS

8. Repeat the above for each of the following functions; you may use Calculus Limit to evaluate the limits:

(a) F (x) := x3

(b) F (x) := x4

(c) F (x) := x5

9. Anwer questions #2 and #3.

10. I hope that you have discovered the following wonderful result:

ddx

x0

f (u) du = f (x).

Now as a matter of fact, it turns out that the above result is still true if we replace the lower limit 0

with any other number a:

Theorem 9.1 (Fundamental Theorem of Calculus–derivative form) Let f (x) be continuous on

an interval [a, b]. Then

d

dx

xa

f (u) du = f (x). (9.4)

This theorem will be crucial in helping us evaluate definite integrals. The only way that we know how

to do this at the moment is to evaluate limits of Riemann sums.

11. Suppose we wish to evaluate ba f (u) du for a given function f and a given interval [a, b]. We first make

the following definition:

Definition 9.2 An antiderivative of f is a function g whose derivative is f . We denote an an-

tiderivative of f by

f (x) dx. We say that

f (x) dx is the indefinite integral of f .

For example, an antiderivative of x is 12

x2 since the derivative of 12

x2 is x. So is 12

x2 + 3 since it has

the same derivative.



99


13. Even though a function can have many antiderivatives, once you have found one antiderivative g you’ve

essentially found them all, for any other antiderivative is of the form g + C where C is some constant.

We would then write, for example,

x dx =

1

2x2 + C,

to denote that every function whose derivative is x “looks like” 12

x2 +C for some number C ; C is called

the constant of integration and must always be included when evaluating indefinite integrals.


15. You have (hopefully) now discovered the second formulation of the Fundamental Theorem of Calculus:

Theorem 9.2 (Fundamental Theorem of Calculus–antiderivative form) Let f be continuous

on [a, b] and let G be any antiderivative of f . Then

b

a

f (u) du = G(b)−

G(a).

Example: Since G(x) = 13

x3 is an antiderivative of x2, we have that 2

1 x2 dx = G(2) − G(1) =

13

(23) − 13

(13) = 83 − 1

3 = 7

3.


17. If you let x∗k = a + b−an k inn (9.1), then you obtain the Riemann sum

nk=1

b − an

f a + b − an

k (9.5)

ba

f (x) dx is the limit as n → ∞ of the above. To Author expression (9.5), first Author b−an F

a + b−an k

,

then use the Calculus Sum command to obtain the above sum.




100 LABORATORY # 9. THE FUNDAMENTAL THEOREM OF CALCULUS

19. When f ≥ 0 on [a, b], we’ve seen that ba

f (x) dx represents the area bounded by the graph of f and

the x axis between x = a and x = b.If f (x) < 0 for some x ∈ [a, b], then

b

a f (x) dx still represents the

area bounded by f and the x axis, except that the area of the regions that lie below the x axis get

subtracted. So in the following picture, ba

f (x) dx = area A − area B + area C .

-A

B

C

a b

6

?




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The Fundamental Theorem of Calculus

NAME: DATE:


1. (a) What do you get when you simplify expression (9.3)?

(b) What is the limit of expression (9.3) as n → ∞?

(c) Explain the result geometrically (i.e. what area does x

0 u,du represent) ?

101



102 Worksheet # 9. THE FUNDAMENTAL THEOREM OF CALCULUS

2. Complete the following table:

F (x)n

k=1

x

n

F (kx/n) limn→∞

n

k=1

x

n

F (kx/n) = x

0

F (u) du

x x2(n+1)2n

x2

2

x2

x3

x4

x5

3. What is the relationship between the entries in column 1 and column 3 of the above table?

4. Find antiderivatives for the following functions:

(a) f (x) = x2

(b) f (x) = sin x

(c) f (x) = sin3x



Name: 103

5. Explain why if g is an antiderivative of f , then so is g + C for any constant C .

6. Consider now the problem of evaluating ba

f (u) du, where f is continuous on [a, b].

(a) Use the Fundamental Theorem of Calculus (9.4) to show that if F (x) = xa f (u) du, then F is an

antiderivative of f .

(b) Suppose that G is any antiderivative of f . In view of the previous remarks, there must be a

constant C such that for any x ∈ [a, b],

G(x) = F (x) + C (9.6)

Show that C = G(a). [Hint: plug in x = a in (9.6)]

(c) Show that F (b) = G(b) − G(a). [Hint: plug in x = b in (9.6)]

7. (a) Show geometrically that 2

1 x dx = 3

2.




(b) Show that 2

1 x dx = 3

2 by using the Fundamental Theorem of Calculus–antiderivative form (The-

orem 9.2).

8. (a) Use the Riemann sum (9.5) with n = 100 to approximate the integral 51 (x2 + 3x + 1) dx. [Hint:

Author F (x) := x2 + 3x + 1 then Manage Substitute a = 1, b = 5 and n = 100 into (9.5);

approXimate.

(b) Use the Riemann sum (9.5) with n = 100 to approximate the integral 2

0

√ 4 − x2 dx.

(c) Use the Riemann sum (9.5) with n = 100 to approximate the integral 3

11√ x3+1

dx.

(d) Evaluate the integral 5

1 (x2 + 3x + 1) dx exactly by using the Fundamental Theorem of Calcu-

lus.(Check your answer by Authoring x2 + 3x + 1 then choosing Calculus Integrate <Return>

<Return < <Return>; then Simplify; compare to your answer to (a).)



Name: 105

(e) An antiderivative for√

4 − x2 will not be available to us until we have studied inverse trig func-

tions. (Use the Calculus Integrate command to see what the antiderivative looks like.) So you

can’t (at least not yet) use the Fundamental Theorem of Calculus to evaluate 2

0

√ 4 − x2 exactly.

However, find an exact value for 2

0

√ 4 − x2 dx by using geometry. (Compare to your answer to

(b).)

(f) Finding an antiderivative for 1√ x3+1

is again difficult; even Derive can’t do it (See what happens

when you use Calculus Integrate). In this instance, it is not possible to obtain an exact value for

31

1√ 1+x3

dx. The best that we can do is approximate it as you did above. We will study some

more effective approximation techniques later on; for now, use Derive’s built-in approximation

technique as follows: choose Calculus Integrate as before, but type in 1 and 3 for the lower and

upper limits, respectively. Then approXimate the resulting definite integral. (Compare to your

answer to (c).)

9. Use the fundamental theorem of calculus to evaluate 3

0 (2x−1) dx, then check your result geometrically.

10. Give geometric arguments for the following. Draw a picture to illustrate your argument:




(a) If f is an even function (i.e. symmetric with respect to the y axis), then

a

−af (x) dx = 2

a

0

f (x) dx.

(b) If f is an odd function (i.e. symmetric with respect to the origin) then a−a

f (x) dx = 0.



Laboratory # 10

Approximate Integration

Teaches:

• numerical integration techniques – Midpoint, Trapezoidal, and Simpson’s

rules

• error bounds

1. Transfer Load the Utility file LABS.MTH.

2. The fundamental theorem of calculus gives us a very easy way of evaluating ba

f (x) dx provided

we can find an antiderivative for f . Many times this isn’t possible and the best we can do is

approximate the definite integral.

3. Let

{a = x0 < x1 <

· · ·< xn = b

}be the regular partition of the interval [a, b] into n equal subintervals,

so that xk = a + k(b−a)n

. One way of approximating ba

f (x) dx is to calculate the Riemann sum:

nk=1

b − a

n f (x∗k)

where xk−1 ≤ x∗k ≤ xk. When x∗k is the midpoint of [xk−1, xk], we call this approximation technique

the midpoint rule.

107



108 LABORATORY # 10. APPROXIMATE INTEGRATION

4. Set Options Precision to Exact with 10 digits. Author and Plot

F (x) := x sin(πx)

5. In your graphics window, choose Options State Rectangular Connected Large <Enter>. Also set the

x and y scales to .5, Move the cross to (.5, .5) and Center. Then Author , Simplify, and Plot

GMIDPOINT(x sin(πx), 0, 1, 4)

to see a graphical display of the 4 rectangles whose areas approximate the integral 1

0 x sin(πx) dx.


7. Now Author and approXimate

MIDPOINT(x sin(πx), 0, 1, 4)

Your answer should agree with your answer to Worksheet Question #1. In general, Author, Simplify,

and Plot GMIDPOINT(F (x), a, b, n) to obtain a graphical display of the n rectangles whose areas ap-

proximate ba

F (x) dx using the midpoint rule; then Author and approXimate MIDPOINT(F (x), a , b , n)

to evaluate the combined area of these rectangles.


9. Some more popular approximation techniques are based on the fact that since polynomials are easily

integrated, we should try to approximate f on each subinterval by a suitably chosen polynomial and

then integrate the polynomial. The simplest polynomial is a straight line, and this is the basis for the

Trapezoidal Rule.

10. Delete All graphs and replot F (x) = x sin(πx). Then Author, Simplify,

and Plot

GTRAP(F (x), 0, 1, 4)



109

This will give you a graphical display of the trapezoids used to approximate 1

0 x sin(πx) dx using the

trapezoidal rule with n = 4.

11. Recall that the area of a the trapezoid depicted below is 12

(w)(u + v).

u

v

w



TRAP(F (x), 0, 1, 4)


and Plot GTRAP(F (x), a, b, n) to obtain a graphical display of the n trapezoids whose areas approx-

imate ba F (x) dx using the trapezoidal rule; then Author and approXimate TRAP(F (x), a , b , n) to

evaluate the combined area of these trapezoids.


15. Simpson’s Rule is derived by approximating the function f over every two subintervals [x2k−2, x2k] by

the parabola (i.e. second degree polynomial) that goes through [x2k−

2, f (x2k−

2)], [x2k−

1, f (x2k−

1)],

and [x2k, f (x2k)]. So for Simpson’s rule, n must be even.

16. Delete All graphs and replot F (x) = x sin(πx). Then Author, Simplify,

and Plot

GSIMPSON(F (x), 0, 1, 4)




This will give you a graphical display of the parabolic regions used to approximate 1

0 x sin(πx) dx

using Simpson’s rule with n = 4.

17. For example, the first parabola has to go through the points (0, F (0)), (.25, F (.25)), and (.5, F (.5)).

To evaluate the equation of this parabola, Author and Simplify, and Plot

FIT

[x,ax2 + bx + c], [[0, F (0)], [.25, F (.25)], [.5, F (.5)]]

The graph should coincide with the previous one for 0 ≤ x ≤ .5.



SIMPSON(F (x), 0, 1, 4)


and Plot GSIMPSON(F (x), a , b , n) to obtain a graphical display of the n parabolic regions whose areas

approximate b

a F (x) dx using Simpson’s rule; then Author and approXimate SIMPSON(F (x), a , b , n)

to evaluate the combined area of these regions.


21. In all of our approximations to 1

0 x sin(πx) dx, we were able to determine the accuracy of the approx-

imation since we knew what the exact answer was. Usually, we do not know the exact value of the

integral that we are approximating, but we still need to know how accurate our approximation is.

Theorem 10.1 (Error Bound for Midpoint Rule) Suppose I = ba

f (x) dx and M n is an approx-

imation to I using the midpoint rule with n rectangles. If there exists a number M > 0 such that

|f (x)| ≤ M for all x in [a, b], then

|I − M n| ≤ M (b − a)3

24n2



111

22. As an example, consider the integral 3

1

√ 1 + x3 dx. Delete All graphs, and set both x and y scales to

1; then Author

1 + x3

and choose Calculus Integrate <Enter> <Enter> <Enter>; Simplify. As you can see, even Derive

cannot come up with an antiderivative for√

1 + x3, so we cannot evaluate 3

1

√ 1 + x3 dx exactly.

23. Suppose we wish to use the midpoint rule to approximate this integral accurate to 3 decimal places

(i.e. so that the error

|I

−M n

| < 5

×10−4). In order to use the above theorem, we need to find M .

The number M is an upper bound for |f (x)| where f (x) =√

1 + x3. So highlight√

1 + x3 and use

the Calculus Differentiate command with order 2 to find the second derivative. Simplify and Plot.


Theorem 10.2 (Error Bound for Trapezoidal Rule) Suppose I = ba

f (x) dx and T n is an ap-

proximation to I using the trapezoidal rule with n trapezoids. If there exists a number M > 0 such that

|f (x)| ≤ M for all x in [a, b], then

|I − T n| ≤ M (b − a)3

12n2


Theorem 10.3 (Error Bound for Simpson’s Rule) Suppose I = ba

f (x) dx and S n is an approx-

imation to I using Simpson’s rule with n subintervals. If there exists a number M > 0 such that

|f (4)(x)| ≤ M for all x in [a, b], then

|I − S n| ≤ M (b − a)5

180n4

26. The number M that comes up in the error estimate for Simpson’s rule is not necessarily the same one

as the one used for the Midpoint and Trapezoidal rules, as it relates to the fourth derivative of f (x).




Use Derive to find the fourth derivative of √

1 + x3. Then Plot it. The graph should convince you that

the same value of M that was used for our error estimates using the Midpoint and Trapezoidal rules

will also work (in this case) for Simpson’s rule.


28. Finally we note that approximate integration can be used to find decimal approximations to irrational

numbers such as π . Answer Worksheet Question #10.



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Approximate Integration

NAME: DATE:


Note: Give decimal approximations for your answers unless you are asked specifically for an exact answer.

1. On the axes below is the graph of y = x sin(πx); sketch and shade the rectangles whose area represents

the midpoint approximation with n = 4 to 1

0 x sin(πx) dx. Then evaluate the area of the shaded

rectangles.

.5

1

.5 1

.......................................................................................................................................

....................

................

......................

..................

.............

.....................

................

.............................

..........................

.....................................

....................

...............................................................................................................................................................................................................................................................................................................................................................................................

.......................................

113



114 Worksheet # 10. APPROXIMATE INTEGRATION

Area of rectangle #1: ×F ( ) =

Area of rectangle #2: ×

F ( ) =



Total area:

2. (a) Use the midpoint rule with n = 10 to approximate 1

0 x sin(πx) dx (i.e. Author and approXimate

MIDPOINT(F (x), 0, 1, 10)).

(b) Use the midpoint rule with n = 100 to approximate 1

0 x sin(πx) dx.

(c) To use the Fundamental Theorem of Calculus to evaluate 1

0 x sin(πx) dx exactly, we need to find

an antiderivative for x sin(πx). Later on, you will be able to use the technique of integration

by parts to do this; for now, use Derive: Author

x sin(πx)

then choose Calculus Integrate <Enter> <Enter> <Enter>, then Simplify. Use this antiderivative

to evaluate 1

0 x sin(πx) dx exactly; then approXimate it to obtain a decimal expansion:

(d) To how many decimal places was your midpoint approximation with n = 100 accurate?



Name: 115

3. On the axes below is the graph of y = x sin(πx); sketch and shade the trapezoids whose area represents

the trapezoidal rule approximation with n = 4 to

1

0 x sin(πx) dx. Then evaluate the area of the shaded

trapezoids.

.5

1

.5 1

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Area of trapezoid #1: 12 × ×

F( ) + F( )

=


F( ) + F( )

=


F( ) + F( )

=


F( ) + F( )

=

Total area:

4. (a) Use the trapezoidal rule with n = 10 to approximate 1


TRAP(F (x), 0, 1, 10)).

(b) Use the trapezoidal rule with n = 100 to approximate 1

0 x sin(πx) dx.

(c) To how many decimal places was your trapezoidal rule approximation with n = 100 accurate?




(Recall that the exact value of this integral was obtained in Question 2c)

5. (a) Write down the equation of the parabola that intersects the graph of F (x) = x sin(πx) at x = 0,

x = .25, and x = .5.

(b) Find the area under this parabola from x = 0 to x = .5 by integrating this parabola from x = 0

to x = 0.5. (You may use Calculus Integrate, entering 0 and .5 for the lower and upper limits,

respectively; then Simplify and approXimate.)

(c) Find the equation of the parabola that intersects the graph of F (x) = x sin(πx) at x = .5, x = .75,

and x = 1 (i.e. Author and Simplify

FIT

[x,ax2 + bx + c], [[0.5, F (0.5)], [.75, F (.75)], [1, F (1)]]

and Plot to make sure it coincides with the previous graph for .5 ≤ x ≤ 1.

(d) Find the area under this parabola from x = .5 to x = 1 by integrating this parabola from x = .5

to x = 1.

(e) Find Simpson’s rule’s approximation of 1

0 x sin(πx) dx with n = 4 by adding your answers to (b)



Name: 117

and (d).

6. (a) Use Simpson’s rule with n = 10 to approximate 1


SIMPSON(F (x), 0, 1, 10)).

(b) Use Simpson’s rule with n = 100 to approximate 1

0 x sin(πx) dx.

(c) To how many decimal places was Simpson’s rule’s approximation with n = 100 accurate?

7. (a) M has to be a number such that |f (x)| ≤ M for all x in [1, 3] i.e. the graph of f (x) must lie

entirely between the horizontal lines y = M and y = −M for 1 ≤ x ≤ 3. Looking at the graph of

f (x), what’s the smallest integer that we can let M be?

(b) Author the inequality

M (b − a)3

24n2 < 0.0005

then use Manage Substitute to plug in a = 1, b = 3, and M = the value that you found in (a).

Then choose soLve to solve the inequality. What is the smallest positive integer n that satisfies




the inequality?

(c) Use the midpoint rule with this value of n to approximate 3

1

√ 1 + x3 dx to three decimal places

(i.e. Author and approXimate MIDPOINT(√

1 + x3, 1, 3, n) where n is the number you found in

(b)).

8. (a) In order to figure out the value of n needed to insure accuracy to 4 decimal places in the trapezoidal

rule, Author the inequality

M (b − a)3

12n2 < 0.00005

and use Manage Substitute to plug in the same values of a, b, and M as before. Then soLve the

inequality. What is the smallest positive integer n that satisfies the inequality?

(b) Use the trapezoidal rule with this value of n to approximate 3

1

√ 1 + x3 dx to four decimal places.

9. (a) In order to figure out the value of n needed to insure accuracy to 5 decimal places in Simpson’s

rule, Author the inequality

M (b − a)5

180n4 < 0.000005



Name: 119

and use Manage Substitute to plug in the same values of a, b, and M as before. Then soLve the

inequality. What is the smallest positive even integer n that satisfies the inequality?

(b) Use Simpson’s rule with this value of n to approximate 3

1

√ 1 + x3 dx to 5 decimal places.

(c) Which of the three rules (midpoint, trapezoidal, and Simpson’s) seems to be the most efficient

(i.e. the one that gives the most accuracy with the fewest subintervals n)?

10. In Calculus II you will learn that 1

04

1+x2 dx = π.

(a) Find the smallest integer M such that |f (4)(x)| ≤ M for 0 ≤ x ≤ 1 where f (x) = 41+x2 . [Hint:

Use Derive to find the fourth derivative of f (x) then Plot it with x scale 1 and y scale 40.]

(b) Author and soLve the inequality

M (b − a)5

180n4 < 0.00000005

where you have plugged in a = 0, b = 1 and M = the value found in (a). What is the smallest

positive even integer that satisfies the inequality?

(c) Use Simpson’s rule with this value of n to approximate π = 1

04

1+x2 dx to 7 decimal places.



Laboratory # 11

Volumes of Revolution

Teaches:

• solids of revolution

• approximating volumes using disks

• approximating volumes using shells

1. We’ve seen that in calculating the area of an irregular region, it was possible to approximate the region

by rectangles for which we already know how to calculate the area.

2. We will see that we can extend this idea to volumes, by approximating a solid by solids whose volumes

we already know how to calculate.

3. Below, I remind you of some formulas for volumes that you may have forgotten. You will want to refer

to them as you work through this lab.

120



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Volumes of Revolution

NAME: DATE:


1. For each of the following, state what kind of solid is generated by revolving the given region R about

the given axis L. Then use your knowledge of geometry to write down the volume of the solid of

revolution.

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-

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h

r

R

x

y

(a) axis L = y axis

solid = volume =

123



124 Worksheet # 11. VOLUMES OF REVOLUTION

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a−a

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solid = volume =

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h

r

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(c) axis L = y axis

solid = volume =

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r1 r2

R

x

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(d) axis L = x axis

solid = volume =



Name: 125

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(e) axis L = y axis

solid = volume =

2. Consider the triangular region below, bounded by the x axis, the line y = x/2, and the line x = 8; also

consider the solid generated by revolving this region about the x axis:

(a) The volume of this solid of revolution can be approximated by the volume of the thin cylinders one

obtains by revolving the shaded region indicated in the picture below. We call these thin cylinders

disks. Use (1d) to find the volume of each disk, then sum them all up to find an approximation for

the desired volume. Give your final answer both in terms of π and as a decimal approximation.




Disk # r2 − r1 h volume

1

2

3

4

Total volume=

(b) Repeat using 8 subdivisions:



Name: 127


1

2

3

4

5

6

7

8

Total volume=

(c) Transfer Load the Utility file LABS.MTH. You can now use the command VOLUME X(F (x), a, b, n)

to approximate the volume of the solid of revolution obtained by revolving the region under the

graph of the (nonnegative) function F (x) between x = a and x = b about the x axis. For example,

Author and approXimate each of the following two expressions:

VOLUME X(x/2, 0, 8, 4)

VOLUME X(x/2, 0, 8, 8)

the approximations should agree with your answers to (a) and (b) above. Now Author

VOLUME X(x/2, 0, 8, n)

and use Manage Substitute to plug in each given value of n, then approXimate to find the

approximate volume of revolution using n disks:




n Approximate Volume n Approximate Volume

4 300

8 400

100 500

200 1, 000

(d) Use the formula for the volume of a cone to find the exact volume of the solid of revolution.

3. Consider the half-circle below, bounded by the x axis and the the curve y =√

4 − x2; also consider

the solid obtained by revolving this region about the x axis:

(a) The volume of this solid of revolution can be approximated by the volume of the disks one obtains

by revolving the shaded region indicated in the picture below. Use (1d) to find the volume of

each disk, then sum them all up to find an approximation for the desired volume. Give your final

answer both in terms of π and as a decimal approximation.



Name: 129


1

2

3

4

Total volume=






1

2

3

4

5

6

7

8

Total volume=

(c) Now Author

VOLUME X(

4 − x2, −2, 2, n)

and use Manage Substitute to plug in each given value of n, and approXimate to estimate the

volume of the solid of revolution using n disks; the approximations for n = 4 and n = 8 should

agree with your answers to (a) and (b) above:


4 300

8 400

100 500

200 1, 000

(d) Use the formula for the volume of a sphere to find the exact volume of the solid of revolution.



Name: 131

4. Consider the triangular region below, bounded by the x and y axis, and the line y = 4 − x/2; also

consider the solid generated by revolving this region about the y axis:

(a) The volume of this solid of revolution can be approximated by the volume of the shells one obtains

by revolving the shaded region indicated in the picture below. Use (1e) to find the volume of each

shell, then sum them all up to find an approximation for the desired volume. Give your final

answer both in terms of π and as a decimal approximation.




Shell # r1 r2 h volume

1

2

3

4

Total volume=


Shell # r1 r2 h volume

1

2

3

4

5

6

7

8



Name: 133

Total volume=

(c) To approximate the volume of the solid of revolution obtained by revolving the region under the

graph of the (nonnegative) function F (x) between x = a and x = b about the y axis using n

shells, you can Author and approXimate VOLUME Y(F (x), a , b , n). For example, Author and

approXimate each of the following two expressions:

VOLUME Y(4 − x/2, 0, 8, 4)

VOLUME Y(4 − x/2, 0, 8, 8)

the approximations should agree with your answers to (a) and (b) above.

(d) Now Author

VOLUME Y(4 − x/2, 0, 8, n)

and use Manage Substitute to plug in each given value of n, and approXimate to estimate the

volume of the solid of revolution using n shells:


4 300

8 400

100 500

200 1, 000

(e) Use the formula for the volume of a cone to find the exact volume of the solid of revolution.



Laboratory # 12

Exponential and Logarithmic

Functions

Teaches:

• natural logarithm function -definition, graph, derivative• natural exponential function -definition, graph, derivative

1. By the power rule, we can find antiderivatives for almost all functions of the form xr. Answer Worksheet

Questions #1.

2. Since we do not yet have an antiderivative for 1x

, we cannot use the fundamental theorem of calculus

to evaluate the integral ba 1x dx. However, we have several approximation techniques at our disposal.

3. Transfer Load the Utility file LABS.MTH. Then Author and Plot 1/x; Move the cross to (2, 1) and

Center; choose Options State Rectangular Connected Large <Return>; also choose Options Color

Auto No. We will use Simpson’s rule with n = 20 to approximate integrals. To obtain a graphical

134



135

display of the approximate area represented by the integral 2

11x dx, Author, Simplify, and Plot

GSIMPSON(1/x, 1, 2, 20)

then to approximate this area, Author and approXimate

SIMPSON(1/x, 1, 2, 20)

4. To obtain a graphical display of the approximate area represented by 3

11x dx, in your graphics window

choose Options Color Plot to change the next plot color, then Author, Simplify, and Plot

GSIMPSON(1/x, 1, 3, 20)

then to approximate this area, Author and approXimate

SIMPSON(1/x, 1, 3, 20)


6. The properties that you have discovered about F (x) should remind you of some important functions

which you’ve studied in high school called logarithms .

Definition 12.1 The function F (x) = x

11t dt, x > 0, is called the natural logarithm function.

We write F (x) = ln x.

Some properties of the natural logarithm:

ln(xy) = ln(x) + ln(y)

ln

x

y

= ln(x) − ln(y)

In other words, logs turn products into sums, and logs turn quotients into differences. Answer Work-

sheet Questions #6 and #7.



136 LABORATORY # 12. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

7. The number G(1) is a very important irrational number which is denoted by e. Because G(2) = e2

and G(3) = e3, and in general G(r) = er for rational numbers r , we use the notation G(x) = ex.

Definition 12.2 The function G(x) = ex, defined as the inverse of the natural logarithm function

F (x) = ln x, is called the exponential function with natural base e. The number e is an irrational

number which is approximately equal to 2.7.

8. Analogous to the properties of the natural logarithm, we obtain some properties of the exponential

function:

ex+y = ex · ey

ex−y = ex

ey

In other words, exponential functions turn sums into products, and differences into quotients.




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Exponential and LogarithmicFunctions

NAME: DATE:


1. (a) What is an antiderivative of xr, r = −1?

(b) Can this rule be used when r = −1? Explain.

2. (a) Below are two graphs of y = 1/x; in the first graph, shade the region whose area is represented

by the definite integral 2

11x

dx. In the second graph, shade the region whose area is represented

by the definite integral 31 1x dx.

137



138 Worksheet # 12. EXPONENTIAL AND LOGARITHMIC FUNCTIONS

−4

−2

2

4

−4 −2 2

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4

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(b) Explain why 2

11x dx should be smaller than

31

1x dx.

(c) In general, if 1 < a < b, which will be larger, a

11x dx, or

b1

1x dx? Explain.

3. So if we define F (x) = x1

1t dt, x > 0, use Simpson’s rule with n = 20 to approximate the following:

(a) F (2) = 2

11t dt =

21

1x dx ≈

(b) F (3) = 3

11t dt =

31

1x

dx ≈

(c) F (1/2) = 1/2

11t dt = − 1

1/21x

dx ≈



Name: 139

(d) F (1/3) = 1/3

11t dt = − 1

1/31x dx ≈

(e) Conjecture a property about F (x) based on this calculations.

(f) Why is F (x) not defined when x < 0?

4. Use Simpson’s rule with n = 20 to approximate F (x) = x

11t dt at the given values of x. Round your

answers to the nearest hundredth.

x F (x) x F (x) x F (x)

1 0.1 4

0.75 0.01 6

0.5 2 7

0.25 3 8

5. (a) Compare F (6) = F (2 × 3) to F (2)+ F (3); then compare F (0.75) = F (3 × 0.25) to F (3)+ F (.25).

What do you notice? Conjecture a general property about F (x) based on your observations.

(b) Compare F (3) = F

0.750.25

to F (0.75) − F (0.25); then compare F (4) = F

82

to F (8) − F (2).




What do you notice? Conjecture a general property about F (x) based on your observations.

6. (a) Graph the points (x, F (x)) that you calculated in Question #4 above, then connect them with

a smooth curve to obtain a rough graph of the function F (x). Then on the second set of axes,

graph the points (F (x), x) to obtain a rough graph of the inverse of F (x), which we will call G(x).

-

6

?

y = F (x)

5−5

5

−5

-

6

?

y = G(x)

5−5

5

−5

(b) Use the definition of F (x) (as an integral) and the Fundamental Theorem of Calculus to find

F (x). For x > 0, is F increasing or decreasing? Explain.



Name: 141

(c) Find F (x). Is this consistent with the concavity exhibited by your graph? Explain.

(d) Your graph of F (x) should suggest that limx→0+ F (x) = −∞ and limx→∞ F (x) = ∞. What then

are the following limits?

limx→−∞

G(x) =

limx→∞G(x) =

7. (a) Approximate G(1) to one decimal place,i.e, you want to find b such that F (b) = 1. [Hint: you

want to find b such that F (b) = b

11x

dx = 1. Since you know (approximately) the values of F (2)

and F (3), explain (in terms of areas) why b must be between 2 and 3, then use Simpson’s rule to

approximate F (2.1), F (2.2), etc. to find the number b for which F (b) is closest to 1.

(b) Use the same technique as above to find G(2) to one decimal place.(i.e. to find b such that

F (b) = b

11x dx = 2). Start by explaining why G(2) must be between 7 and 8.




(c) Verify that G(2) ≈ [G(1)]2.

(d) Approximate G(3) to one decimal place. Start by explaining why G(3) must be between 20 and

21. You will need to approximate F (20) and F (21) using Simpson’s rule with n = 50.

(e) Verify that G(3) ≈ [G(1)]3.

8. Author G(x) := ex; (note: to enter the number e in Derive, you must use Alt-e; it will appear on the

screen as e); Plot to verify that your graph of G(x) above is correct. Then Author

G(x + .01) − G(x)

0.01

This last expression should be approximately equal to G(x). Plot it. What seems to be the relationship

between G(x) = ex and G(x)?



Laboratory # 13

Inverse Trigonometric Functions

Teaches:

• the six inverse trigonometric functions – graphs, domains, ranges, and derivatives

1. Author and Plot

y = sin x

then zoom out once ( F10 ) and answer Worksheet Question #1.

2. The function which you just graphed is called the inverse sine or arcsine. We generally write

y = arcsin(x) or y = sin−1(x). Derive’s notation for this function is ASIN, so Author and Plot

y = ASIN(x)

This graph should coincide with your previous one.

3. Each time we introduce a new function, we are interested in evaluating its derivative. Answer Worksheet

Question #2.

143



144 LABORATORY # 13. INVERSE TRIGONOMETRIC FUNCTIONS

4. Delete All graphs, then Author and Plot

y = cos x

and answer Worksheet Question #3.

5. The function which you just graphed is called the inverse cosine or arccosine. We generally write

y = arccos(x) or y = cos−1(x). Derive’s notation for this function is ACOS, so Author and Plot

y = ACOS(x)


6. Again, we are interested in evaluating the derivative of the inverse cosine function, so answer Worksheet

Question #4.


y = tan x


8. The function which you just graphed is called the inverse tangent or arctangent. We generally

write y = arctan(x) or y = tan−1(x). Derive’s notation for this function is ATAN, so Author and Plot

y = ATAN(x)


9. Again, we are interested in evaluating the derivative of the inverse tangent function, so answer Work-

sheet Question #6.


y = cot x




145

11. The function which you just graphed is called the inverse cotangent or arccotangent. We generally

write y = arccot(x) or y = cot−1(x). Derive’s notation for this function is ACOT, so Author and Plot

y = ACOT(x)


12. Again, we are interested in evaluating the derivative of the inverse cotangent function, so answer

Worksheet Questions #8 – #10.



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Inverse Trigonometric Functions

NAME: DATE:


1. (a) The function y = sin x does not have an inverse. Why not?

(b) What is the biggest number a such that, if you restrict the domain of y = sin x to the interval

[−a, a], then it will have an inverse?

(c) Author and Plot

[x, sin x]

(this will plot points (x, y) such that y = sin x and x is in the interval specified by you); use your

answer to (b) to enter the min and max values. Copy the graph below (recall that π is intered as

Alt-p), and write down the domain and range of this function.

147



148 Worksheet # 13. INVERSE TRIGONOMETRIC FUNCTIONS

?

6

-

2 4−2−4

2

4

−2

−4

domain=

range=

(d) To obtain a graph of the inverse of this restricted sine function, we need to interchange x and y

values from the previous graph. To do this, Author [sin x, x] and Plot; the min and max values

should remain the same as before. Copy the graph below and write down the domain and range.

?

6

-

2 4−2−4

2

4

−2

−4

domain=

range=



Name: 149

2. To obtain an approximate graph of ddx sin−1(x), Author and Plot

ASIN(x + .01)−

ASIN(x)

.01 (13.1)

(a) Copy the graph below:

?

6

-

2 4−

2−

4

2

4

−2

−4

(b) This graph doesn’t really look like any “familiar” functions; for example it’s too fat to be a

parabola. However, Author and Plot

1ASIN(x+.01)−ASIN(x)

.01

2

Note: recall that F3 can be used to bring down a highlighted expression to the Author line; F4

brings it down with parentheses around it. Copy this graph below:

?

6

-

2 4−2−4

2

4

−2

−4




(c) This graph looks much more like a parabola; in fact it is one. Find its equation. [Hint: compare

this graph to the graph of y = x2 and use your knowledge of shifts and/or reflections of functions.]

(d) Verify that your answer above is correct by Plotting the equation that you found to make sure

that its graph coincides witht the previous one. Then since

“answer to (c)” = 1

ddx sin−1 x

2,

you should now be able to solve for ddx

sin−1 x.

3. (a) The function y = cos x does not have an inverse. Why not?

(b) What is the biggest number a such that, if you restrict the domain of y = cos x to the interval

[0, a], then it will have an inverse?

(c) Author and Plot

[x, cos x]



Name: 151

use your answer to (b) to enter the min and max values. Copy the graph below, and write down

the domain and range of this function.

?

6

-

2 4−2−4

2

4

−2

−4

domain=

range=

(d) To obtain a graph of the inverse of this restricted cosine function, we need to interchange x and

y values from the previous graph. To do this, Author [cos x, x] and Plot; the min and max values

should remain the same as before. Copy the graph below and write down the domain and range.

?

6

-

2 4−2−4

2

4

−2

−4

domain=

range=




4. Author and Plot ASIN(x) + ACOS(x).

(a) What do you notice? Write down an identity.

(b) Use your identity and the derivative of sin−1 x which we have already calculated to obtain the

derivative of cos−1 x.

5. (a) The function y = tan x does not have an inverse. Why not?

(b) What is the biggest number a such that, if you restrict the domain of y = tan x to the interval

(−a, a), then it will have an inverse?

(c) Author and Plot

[x, tan x]

for the min and max values enter an interval slightly smaller than the restricted domain that you

found in (b), e.g. [−1.5, 1.5] (if you plug in the exact endpoints the plotting will take much longer

because the function becomes infinite at the endpoints). Copy the graph below, and write down

the domain and range of this function.



Name: 153

?

6

-

2 4−2−4

2

4

−2

−4

domain=

range=

(d) To obtain a graph of the inverse of this restricted tangent function, we need to interchange x and

y values from the previous graph. To do this, Author and Plot

[tan x, x]

the min and max values should remain the same as before. Copy the graph below and write down

the domain and range.

?

6

-

2 4−2−4

2

4

−2

−4

domain=

range=




6. To obtain an approximate graph of ddx

tan−

1(x), Author and Plot

ATAN(x + .01) − ATAN(x)

.01 (13.2)

(a) Copy the graph below:

?

6

-

2 4−2−4

2

4

−2

−4

(b) Again, this graph doesn’t really look like any “familiar” functions; so Author and Plot

1ATAN(x+.01)−ATAN(x)

.01

again using F3 or F4 to save typing. Copy this graph below:

?

6

-2 4−2−4

2

4

−2

−4



Name: 155

(c) This graph looks much more like a parabola; in fact it is one. Find its equation. [Hint: again

compare this graph to the graph of y = x2.]

(d) Verify that your answer above is correct by Plotting the equation that you found to make sure

that its graph coincides witht the previous one. Then since

“answer to (c)” = 1ddx tan−1 x

,

you should now be able to solve for ddx

tan−1 x.

7. (a) The function y = cot x does not have an inverse. Why not?

(b) What is the biggest number a such that, if you restrict the domain of y = cot x to the interval

(0, a), then it will have an inverse?

(c) Author and Plot

[x, cot x]




for the min and max values enter an interval slightly smaller than the restricted domain that you

found in (b), e.g. [.1, 3.1]. Copy the graph below, and write down the domain and range of this

function.

?

6

-

2 4−2−4

2

4

−2

−4

domain=

range=

(d) To obtain a graph of the inverse of this restricted cotangent function, we need to interchange x

and y values from the previous graph. To do this, Author and Plot

[cot x, x]

the min and max values should remain the same as before. Copy the graph below and write down

the domain and range.



Name: 157

?

6

-

2 4−2−4

2

4

−2

−4

domain=

range=

8. Author and Plot ATAN(x) + ACOT(x).

(a) What do you notice? Write down an identity.

(b) Use your identity and the derivative of tan−1 x which we have already calculated to obtain the

derivative of cot−1 x.

9. Finally, let’s look at the inverse secant and cosecant functions. Delete All graphs, then Author and

Plot

y = sec x




y = ASEC(x)

(a) Copy the graphs below:

-

6

?

y = sec(x)

5−5

5

−5

-

6

?

y = sec−1(x)

5−5

5

−5

(b) Recall that the graph of the inverse of a function can be obtained by reflecting the graph of the

original function across the diagonal y = x. With a darker color pen, go over the part of the

graph of y = sec x that corresponds to the part of the graph that was “inverted” to obtain the

graph of y = ASEC(x). What are the domain and range of this restricted secant function? What

are the domain and range of the inverse secant function?

domain of restricted secant=

range of restricted secant=

domain of inverse secant=

range of inverse secant=

(c) Use Calculus Differentiate to obtain the derivative of the inverse secant.


y = csc x



Name: 159

y = ACSC(x)

(a) Copy the graphs below:

-

6

?

y = csc(x)

5−5

5

−5

-

6

?

y = csc−1(x)

5−5

5

−5

(b) With a darker color pen, go over the part of the graph of y = csc x that corresponds to the part

of the graph that was “inverted” to obtain the graph of y = ACSC(x). What are the domain and

range of this restricted cosecant function? What are the domain and range of the inverse cosecant

function?

domain of restricted cosecant=

range of restricted cosecant=

domain of inverse cosecant=

range of inverse cosecant=

(c) Use Calculus Differentiate to obtain the derivative of the inverse cosecant.



Laboratory # 14

Integration of Rational Functions

Teaches:

• uisng partial fraction decompositions to integrate rational functions


2. If we wanted to find antiderivative of 3x+1 − 1

x−2, we could easily use the integration formula in

Worksheet Question #1(a) to obtain 3 ln |x + 1| − ln |x − 2| + C . However, if instead we were given

the (equivalent) expression 2x−7x2−x−2

, the problem of finding an antiderivative is not simple, unless we

realize that the expression can be “decomposed” into the partial fractions 3x+1 − 1

x−2.

3. In this lab we will look at complicated rational expressions of the form F (x)G(x)

, where F (x) and G(x) are

polynomials such that the degree of the numerator is strictly less than the degree of the numerator.

(If the degree of the numerator is greater than or equal to that of the denominator, long division

will reduce the expression to the desired form.) We will learn that such complicated expressions can

be “decomposed” into simpler partial fractions, each of which can be integrated by using one of the

formulas from Worksheet Question #1.

160



161


5. As you have seen, if the denominator of your rational expression factors completely into linear factors

so that each factor is of the form (x − a)n,then the expression can always be decomposed into terms

of the form A(x−a)k

, and these can be integrated by using formula (a) (if k = 1) or formula (b) if k > 1

in Worksheet Question #1.

6. However, not all polynomials factor this nicely. In particular, a quadratic expression of the form

ax2 + bx + c will not factor into linear factors if b2 − 4ac < 0 (recall the quadratic formula).

7. Answer Worksheet Questions #4 and #5 to see how to handle these kinds of expressions.



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Integration of Rational Functions

NAME: DATE:


1. In this lab, we will need a small table of integrals which we will develop here. Some of these integrals

you should be able to do on your own (in fact, all except (e) are straight forward; (e) requires a trig

substitution); however, if you need help you may use Derive to do the integration; for example for (a),

you can Author 1u−a , then choose Calculus Integrate, enter u for the integration variable, and leave

the upper and lower limits blank; then Simplify.

Notes:

• Derive does not use absolute values for the first integral, but you should.

• Derive does not include the “+C” in the integration, but you should.

(a)

du

u − a =

(b)

du

(u − a)n =

(c) du

u2 + a2 =

(d)

u du

u2 + a2 =

(e)

du

(u2 + a2)2 =

(f)

u du

(u2 + a2)2 =

163



164 Worksheet # 14. INTEGRATION OF RATIONAL FUNCTIONS

2. Author

8x5

−81x4 + 298x3

−456x2 + 222x + 41

x6 − 10x5 + 35x4 − 44x3 − 9x2 + 54x − 27

Complicated as this expression appears, you will be able to integrate it using only the short table that

you developed in Question #1 (and some clever algebra which we will let Derive take care of).

(a) The first thing to notice is that the denominator factors. Highlight the denominator only of

this expression (use the right arrow key) and choose Factor raDical to factor it. Write down the

factored denominator:

(b) Now Expand this last expression and write down the partial fraction decomposition that you

obtain.

(c) Now integrate each term of the partial fraction decomposition (use the appropriate formulas from

Worksheet Question #1 if you need to).

(d) Notice that each factor of the form (x − a)n in the denominator of the original expression gave

rise to n terms in the partial fraction decomposition of the form A(x−a)k

, k = 1, 2, . . . , n:



Name: 165

Factor n Terms in expansion

(x−

3)3 3 1

x−3

,−

1

(x−3)2 , 2

(x−3)3

(x − 1)2 2 3x−1

,− 2(x−1)2

x + 1 1 4x+1

So if I changed the denominator to (x − 3)3(x − 1)2(x +1)(x + 2)4, the partial fraction decompo-

sition would now contain more terms of the form (use A, B, C,.. . for unknown

constants):

3. Author

x

x3 − 3x2 + x + 1

(a) Highlight the denominator and choose Factor raDical. This time the denominator factors into

(x − a)(x − b)(x − c), only the numbers a, b , and c are not all rational.(Be careful with your

signs!)

a =

b =

c =

(b) Expand the last expression; if you collect terms that have the same denominator, the resulting

expression can be put into the form

A

x − a +

B

x − b +

C

x − c

where a, b, c, are as above and

A =

B =




C =

(c) Now you should be able to to the integration (use (a) from the table in Question #1 if you really

need to).

4. Author

3x4 + x3 + 5x2 + 3x + 18

x5 + 2x4 − 8x2 − 12x − 8

(a) Highlight the denominator, then choose Factor raDical to factor it. Write down the factored

denominator:

(b) Now Expand the last expression; write down the expansion, but collect the terms with the same

denominator:

(c) Only one of the three terms that you obtained above can be integrated easily. Do the integration

for this term (using (a) from the table in Question #1 if you need to):



Name: 167

(d) Let’s look at the term in the expansion of the form Ax+Bax2+bx+c . Author this term by itself. The

goal now is to find a u substitution that will allow us to integrate this term using formulas (c)

and/or (d) from the table. To do this, you need to use the technique of “completing the square”

to express the denominator as u2 + a2. Do this now and identify u and a; also solve for x in terms

of u:

i. u =

ii. a =

iii. x =

(e) To do the the u substitution, use Manage Substitute on the expression Ax+Bax2+bx+c that you Authored

above to replace x by your answer to (4(d)iii) above. Then Expand. Write down what you get.

(f) You should now be able to integrate your answer to (e) (use formulas (c) and (d) from the table

in Question #1 if you need to) then rewrite in terms of x.

(g) Now, Author the term in the expansion that you obtained in (4b) that looks like Ax+B(ax2+bx+c)2 by

itself. Again choose Manage Substitute to write in terms of u using the same u substitution that




you came up with in (4d); Expand.

(h) Now integrate your answer to (g) (you will need formulas (e) and (f ) from the table in Question

#1); then rewrite in terms of x.

(i) Finally, put everything together to obtain an antiderivative for the original expression 3x4+x3+5x2+3x+18x5+2x4−8x2−12x−8

:

(j) Notice that the factor of the form (ax + bx + c)n in the denominator of the original expression

gave rise to n terms in the partial fraction decomposition of the form Ax+B(ax2+bx+c)k

, k = 1, 2, . . . , n:

Factor n Terms in expansion

(x2 + 2x + 2)2 2 2x−3x2+2x+2 , x−1

(x2+2x+2)2

So if I changed the denominator to (x − 2)(x2 + 2x + 2)2(x2 − 2x + 5)3, the partial fraction

decomposition would now contain more terms of the form (use A, B, C,.. .for



Name: 169

unknown constants):

5. Author

8x7 + 50x6 + 241x5 + 477x4 + 763x3 − 361x2 + 273x + 169

x8 + 7x7 + 34x6 + 62x5 + 65x4 − 169x3

then follow the same procedure as above to integrate, i.e.

(a) Highlight the denominator, then choose Factor raDical to factor it:

(b) Expand the entire expression, then write down the expansion, collecting terms with like denomi-

nator.

(c) Integrate each of the four term of the form A(x−a)n

(you may use the integration formulas (a) and

(b) in Worksheet Question #1 if you need to).




(d) For the two terms of the form Ax+B(ax2+bx+c)n

, you need to complete the square to write the denomi-

nator ax2 + bx + c in the form u2 + k2. Identify u and k ; also solve for x in terms of u:

i. u =

ii. k =

iii. x =

(e) Author the term in (5b) that has the form Ax+Bax2+bx+c by itself and use Manage Substitute to do

the u substitution in (5d(iii)) above. Then Expand. Write down the expansion:

(f) Now integrate (use formulas (c) and (d) in Question #1 if you need to). Then rewrite in terms

of x.

(g) Finally Author the term in (5b) that has the form Ax+B(ax2+bx+c)2 by itself and use Manage Substitute

to do the same u substitution as before; Expand and write down the expansion:

(h) Integrate the above (you will need formulas (e) and (f) from Question #1); then rewrite in terms



Name: 171

of x.

(i) Now, put everything together to obtain the integral of the original expression.



Laboratory # 15

Improper Integrals

Teaches:

• convergence and divergence of improper integrals

1. Transfer Load the Utility File LABS.MTH.

Author and Plot 1x2 , then use Calculus Integrate, setting the lower limit at −1 and the upper limit at

1, to obtain the expression

1

−1

1

x2 dx


3. As you see, the integral 1−11x2 dx should not be evaluated using the fundamental theorem of calculus.

Let us now focus our attention on half of this integral, i.e. on 1

01x2 dx; if we can figure out the value

of this integral (if indeed the area of this unbounded region can be assigned a finite value?) then, by

symmetry, the original integral must be twice that value.

4. To obtain a graphical display of the approximate area represented by 1

01x2 dx, set your x scale to

172



173

.25 and the y scale to 50, Move the cross to (.5, 75) and Center; choose Options State Rectangular

Connected Large <Enter>; then choose Options Color Auto No; then choose Options Color Plot if

you wish to change the Next plot color; finally Author, Simplify, and Plot

GTRAP

1

x2, 0, 1, 15


6. The integrals 1−1

1x2 dx,

10

1x2 dx, and

10

1√ x

dx are examples of improper integrals. These integrals

are improper because the function being integrated has an infinite discontinuity somewhere in the

interval of integration. Both 1/x2 and 1/√ x have infinite discontinuities at x = 0.

7. In terms of areas, the integrals 1

0 f (x) dx, where f (x) = 1

x2 or f (x) = 1√ x

, represent areas of “un-

bounded” regions. We try to get around this problem by approximating 1t f (x) dx and so ignoring

the area from 0 to t where t is a number very close to 0:

..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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0 t 1

ignore -

1t

f (x) dx

Definition 15.1 Let f be a function continuous on (a, b) (in the above picture, a = 0 and b = 1).

(a) Suppose f has an infinite discontinuity at a (but is continuous at b). Then the integral ba

f (x) dx

is called an improper integral. If

limt→a+

bt

f (x) dx = L



174 LABORATORY # 15. IMPROPER INTEGRALS

exists, then we say this improper integral converges to L. If this limit does not exist (or is infinite),

we say that this improper integral diverges.

(b) If the infinite discontinuity occurs at b, then the convergence or divergence of the improper integral

ba

f (x) dx is determined by evaluating

limt→b−

ta

f (x) dx

8. Your calculations should suggest that 1

01x2 dx diverges and

10

1√ x

dx converges. Answer Worksheet

Question #8.

9. Another kind of improper integral occurs when the interval of integration is infinite, as in ∞a f (x) dx.

The convergence/divergence of such an integral is determined by whether the limit limt→∞

ta

f (x) dx

exists.




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Improper Integrals

NAME: DATE:


1. (a) On the axes below, sketch the graph of y = 1x2 ; then shade the (unbounded) region whose area is

represented by the integral 1−1

1x2 dx.

?

6

-

2 4−2−4

2

4

−2

−4

(b) Simplify the expression 1−1

1x2 dx. Explain why the answer that you obtain cannot possibly be

correct.

(c) Explain how Derive obtained this answer by “blindly” applying the fundamental theorem of

175



176 Worksheet # 15. IMPROPER INTEGRALS

calculus.

(d) Explain why the fundamental theorem of calculus should not be used to evaluate this integral.

[Hint: what’s wrong with f (x) = 1x2 on the interval [−1, 1]?]

2. (a) Why do you only see 14 trapezoids when you asked for 15?

(b) Try to evaluate the total area of these trapezoids by Authoring and approXimating

TRAP

1

x2, 0, 1, 15

Explain Derive’s answer. [Hint: Recall that the area of a trapezoid is equal to its width w times

the average of its two parallel sides l1 and l2: w(l1+l2 )2 ]

(c) Perhaps we should just ignore that first “trapezoid” that’s causing so many problems. To evaluate

only the area of the other 14 trapezoids, you need to Author and approXimate

TRAP

1

x2, t, 1, 14



Name: 177

where t =

[Hint: to verify your answer, Author the above using GTRAP instead of TRAP, then Simplify

and Plot using a different color; you should obtain a graph of the same 14 trapezoids.]

(d) What is the combined area of the 14 “real” trapezoids?

3. Repeat the above exercise with 30 trapezoids. Again you should find that the first trapezoid causes

problems.

(a) So to see a graphical display of only the last 29 trapezoids and evaluate their combined areas, you

must Author, Simplify, and Plot

GTRAP

1

x2, t, 1, 29

then Author and approXimate

TRAP

1

x2, t, 1, 29

where t =

(b) What is the combined area of these 29 trapezoids?

4. Repeat using 45 trapezoids.

(a) Ignoring the first (unbounded) one, you should Author, Simplify, and Plot

GTRAP

1

x2, t, 1, 44

then Author and approXimate

TRAP

1

x2, t, 1, 44

where t =




(b) What is the combined area of these 44 trapezoids?

5. Was it legitimate to ignore the first trapezoid in each approximation – i.e. do the approximations of

10

1x2 dx that we obtained by neglecting the first (thin and unbounded) trapezoid appear to approach

a limit?

6. We saw that when Derive Simplifies a definite integral, it uses the fundamental theorem of calculus

(whether it’s appropriate or not). When approXimating a definite integral, Derive uses a modified

Simpson’s rule.What happens when you approXimate the expression 1−1

1x2 dx?

7. Delete All Graphs, reset the x and y scales both to 1, Move the cross to (0 , 0) and Center; then Author

and Plot y = 1√ x

Copy the graph below, then shade the (unbounded) region whose area is given by

the definite integral 1

01√ x

dx.

?

6

-

2 4−2−4

2

4

−2

−4



Name: 179

(a) Set your x scale to 0.25 and your y scale to 1; then Move the cross to (.5, 1.5) and Center. Author,

Simplify, and Plot

GTRAP

1√

x, 0, 1, 15

then use the TRAP command to evaluate the combined area of the last 14 of these trapezoids:

(b) Find the combined area of the 29 trapezoids used to approximate 1

01√ x

dx obtained by dividing

the interval [0, 1] into thirty equal subintervals and ignoring the first trapezoid.

(c) Find the combined area of the 99 trapezoids used to approximate 1

01√ x


the interval [0, 1] into 100 equal subintervals and ignoring the first trapezoid.

(d) Find the combined area of the 999 trapezoids used to approximate 10 1√ x dx obtained by dividing

the interval [0, 1] into 1000 equal subintervals and ignoring the first trapezoid.

(e) Find the combined area of the 9,999 trapezoids used to approximate 1

01√ x


the interval [0, 1] into 10,000 equal subintervals and ignoring the first trapezoid.

(f) Was it legitimate to ignore the first trapezoid in each approximation – i.e. do the approximations

of 1

01√ x

dx that we obtained by neglecting the first (thin and unbounded) trapezoid appear to




approach a limit?

(g) Finally, approXimate the integral 1

01√ x

dx. Is Derive’s approximation consistent with your pre-

vious approximations?

8. (a) Evaluate 1

0.011√ x

dx by using the fundamental theorem of calculus; give both an exact answer

and an approXimated answer (compare to 7c).

(b) Evaluate 1

0.0011√ x

dx by using the fundamental theorem of calculus; again give both an exact and

an approXimated answer (compare to 7d).

(c) Evaluate 1t

1√ x

dx for general t by using the fundamental theorem of calculus.



Name: 181

(d) Evaluate 1

01√ x

dx exactly by taking the limit of 1t

1√ x

dx as t → 0+.

(e) Does this agree with our earlier analysis of the integral 1

01√ x

dx in terms of trapezoidal rule

approximations? Explain.

9. (a) Author and Plot

F (x) := 1√ 1 + x3

Copy the graph below and shade the (unbounded) region whose area is given by the improper

integral ∞1 1√ 1+x3 dx.

-

6

2 4

2

4

(b) Use Calculus Integrate (with no limits of integration) to try to find an antiderivative for F (x). Is

Derive able to find one?




(c) Without an antiderivate for F (x), we cannot use the fundamental theorem of calculus to calculate

c

11√

1+x3 dx, so we again must resort to approximation techniques. ApproXimate

100

11√

1+x3 dx

(Derive uses a modified Simpson’s rule to approXimate definite integrals, as opposed to trying to

use the Fundamental Theorem which is what it does when you Simplify a definite integral).

(d) ApproXimate 1000

11√

1+x3 dx.

(e) ApproXimate 5000

11√

1+x3 dx.

(f) What do you think, does the improper integral ∞

11√

1+x3 dx converge or diverge? Explain.

(g) Use Calculus Integrate with lower limit 1 and upper limit INF to obtain the expression ∞

11√

1+x3 dx.

Then approXimate. Write down what you get, and explain why you think this is or is not a good

approximation to the improper integral.

10. (a) Author and Plot

F (x) := 1

ln x

Copy the graph below and shade the (unbounded) region whose area is given by the improper

integral ∞

21

lnx dx.



Name: 183

6

?

-

2 4 6 8

2

4

−4

−2

(b) ApproXimate 100

21

ln x dx.

(c) ApproXimate 1000

21

lnx dx.

(d) ApproXimate

5000

21

lnx dx.

(e) What do you think, does the improper integral ∞

21

lnx dx converge or diverge? Explain.

(f) Use Calculus Integrate with lower limit 2 and upper limit INF to obtain the expression ∞2 1lnx dx.

Then approXimate.




(g) Choose Options Precision and increase the number of digits to 12; then approXimate ∞

21

lnx dx

again.



Laboratory # 16

Sequences

Teaches:

• Graphs and limits of sequences

• Recursively defined sequences

• Repeating decimals as sequences

Definition 16.1 A sequence is a function whose domain is the set of positive integers. Instead of using

the usual functional notation a(n), it is customary to write an for the value of the function a at n, and to

refer to an as the nth term of the sequence.

1. Just like any other function, you can graph sequences and evaluate the limit as n → ∞. For example,

for the sequence an = (

−1)n + n

2 we have

a(1) = a1 = (−1)1 + 1

2 = −1 +

1

2 = −1

2

a(2) = a2 = (−1)2 + 2

2 = 1 + 1 = 2

185



186 LABORATORY # 16. SEQUENCES

a(3) = a3 = (−1)3 + 3

2 = −1 +

3

2 =

1

2

a(4) = a4 = (−

1)4 + 4

2 = 1 + 2 = 3

so we graph the points

1, −12

, (2, 2),

3, 1

2

, and (4, 3).

2. Choose Plot Under <Enter>. Author and Plot each of the following 4 expressions

[1, −12 ]

[2, 2]

[3, 12 ]

[4, 3]

to Plot the first four terms of the sequence an = (−1)n + n2 .


4. Delete All graphs, then choose Options State, tab over to Size and choose Small <Return>. Also,

change your x scale to 25 and the y scale to 1, Move your cross to (100 , 1), and Center.

5. Author and approXimate

VECTOR

n,

2n

n + 1

, n, 1, 200

to generate all the points

n, 2nn+1

for n = 1, 2, . . . , 200; then Plot to obtain a graph of the first 200

terms of the sequence an = 2nn+1

. Answer Worksheet Questions #2 and #3.

6. Sequences are sometimes defined recursively . This means that each term is defined in terms of previous

terms. For example, given a1 = 4 and an+1 = 12 an + 1, then

a2 = 1

2a1 + 1 =

1

2 × 4 + 1 = 3



187

a3 = 1

2a2 + 1 =

1

2 × 3 + 1 =

5

2

a4 = 1

2a3 + 1 =

1

2 × 5

2 + 1 =

9

4

and so on. You can use Derive’s ITERATES command to evaluate recursively defined functions: to

evaluate the first 200 terms of the above sequence, Author and approXimate

ITERATES (0.5x + 1, x, 4, 200)

the first 4 terms should match the above calculations of a1, a2, a3, and a4. To graph this sequence,

Author

a := #“expression number of the approximated ITERATES expression”

(e.g. a := #24); (note: this expression is too long to be able to use F3 to bring it down to the Author

line); then Author, approXimate, and Plot

VECTOR ([n, ELEMENT(a, n)] , n, 1, 200)

Answer Worksheet Questions #4–#7.

7. Finally we consider a sequence obtained from an infinite decimal expansion by truncating. For example

the decimal 0.23232323 . . . can be viewed as the limit of the following sequence:

a1 = 0.23 = 23100

a2 = 0.2323 = 0.23+ 0.0023 = 23

100 +

23

10, 000

a3 = 0.232323 = 0.23+ 0.0023 + 0.000023 = 23

100 +

23

10, 000 +

23

1, 000, 000



188 LABORATORY # 16. SEQUENCES

and in general

an =

23

102 +

23

104 +

23

106 + . . . +

23

102n =

n

k=1

23

102k

8. Author

SUM

23

102k, k, 1, n

(16.1)

then Simplify. Answer Worksheet Questions #8 and #9.



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Sequences

NAME: DATE:


1. Below is the graph of the first 4 terms of the sequence an = (−1)n + n2 ; evaluate the next four terms

and graph them:

1

1

2 3 4

2

3

−1

q

q

q

q

?

4

5

5 6 7 8

6

-

2. At what height does the graph of the sequence an = 2nn+1

level off (i.e. what is limn→∞ 2nn+1

)?

189



190 Worksheet # 16. SEQUENCES

3. Examine the graph of each of the following sequences; then determine the limit of the sequence by

noting at what height the graph levels off. [Hint: to obtain each graph, Author, approXimate, and

Plot VECTOR ([n, an], n, 1, 200) where an is the given expression; you may need to increase the y scale

in order to see the entire graph in your viewing window]

(a) an = n sin(1/n)

(b) an = cos(1/n)

(c) an =

1 + 1n

n

(d) an =√

n + 1 − √ n

4. At what height does the graph of the sequence defined by

a1 = 4

an+1 = 1

2an + 1

level off, i.e. what is limn→∞ an?



Name: 191

5. For each of the following recursively defined sequences, calculate a2 and a3; your answers should be

accurate to 4 decimal places. Then examine the graph of the sequence to determine the limit of the

sequence by noting at what height the graph levels off.

(a) a1 =√

6, an+1 =√

6 + an

[Hint: You need to

i. Author and approXimate ITERATES(√

6 + x,x,√

6, 200);

ii. then Author a := #“expression number of approximation”;

iii. then Simplify VECTOR ([n, ELEMENT(a, n)] , n, 1, 200)]

(b) a1 = 0.1, an+1 = 2.9an − 1.9a2n

[Hint: first Author and approXimate ITERATES 2.9x − 1.9x2

, x , .1, 200, then proceed as in (ii)

and (iii) above]

6. Examine the graphs of each of the following recursively defined sequences to determine the limit.

(a) a1 = 1, an+1 = 12

an + 4

an

[Hint: first Author and approXimate ITERATES

12

x + 4

x

, x, 1, 200

, then proceed as in (ii)




and (iii) in the hint to Question #5a]

(b) a1 = 1, an+1 = 12

an + 9

an

[Hint: same as above except change the 4x to 9

x ]

(c) a1 = 1, an+1 = 12

an + 16

an

(d) Based on your observations, conjecture what the limit of the sequence defined by a1 = 1, an+1 =

12

an + a

an

is for any number a > 0.

(e) Use your conjecture to estimate√

2 using only the +, × and ÷ keys on your calculator (i.e. don’t

use the √

x key); stop when successive approximations agree in the first 3 decimal places.

7. The following sequences exhibit a different behavior than the ones that you have looked at so far. In

each case, look at the graph, then describe the behavior of the sequence. [Hint: for (a)–(e), see hint to

Question #3]



Name: 193

(a) an = n ln(n)

(b) an = (−1)n

(c) an = sin(n)

(d) an = sin(πn)n

(e) an = cos(πn)n

(f) a1 = 0.1, an+1 = 3.1an − 2.1a2n

[Hint: first Author and approXimate ITERATES

3.1x − 2.1x2, x, 0.1, 200; then proceed as in

(ii) and (iii) of hint to Question #5a]




(g) a1 = 0.1, an+1 = 3.5an − 2.5a2n

8. (a) What does the sequence an = n

k=123

102k simplify to?

(b) Use the simplified expression for an obtained above to evaluate limn→∞an as a fraction:

9. Consider the repeating decimal 0.5454 . . .

(a) Write the first three terms, a1, a2, and a3, both as decimals and as a sum of fractions, and then

the general term an, of the sequence of sums that converges to 0 .54545454 . . ..

(b) Use Derive to simplify the general term of this sequence.

(c) Use the simplified expression for an obtained above to evaluate limn→∞

an as a fraction:



Laboratory # 17

P-series

Teaches:

• convergence and divergence of p-series – the integral test

1. Consider the following sequence

{an

}: a1 = 1√

1; a2 = 1√

1 + 1√

2, a3 = 1√

1 + 1√

2 + 1√

3, and in general,

an = 1 + 1√

2+

1√ 3

+ . . . + 1√

n =

nk=1

1√ k


3. To have Derive help you evaluate an for larger values of n, Author the following:

A(n) := SUM(1/√

k,k, 1, n)

4. Verify your answer to Question #1 by Authoring and then approXimating A(1), A(2), A(3), and A(4).


6. Choose Plot Beside <Enter> to open a graphics window; choose Options State, tab over to Size and

choose Small <Enter>. Also, change your x scale to 10 and your y scale to 5, Move your cross to

(20, 10) and Center.

195



196 LABORATORY # 17. P-SERIES

7. To graph the first 50 terms of the sequence {an}, Author, approXimate, and Plot

VECTOR ([n, A(n)], n, 1, 50)

Answer Worksheet Question #3.

8. You will have noticed that each calculation takes longer than the previous one.(Remember that a5000

is the sum of 5000 numbers!) We need a quicker way of calculating an for large n, or at least to get an

idea of how big an is.


10. Let us now consider the sequence {bn} defined as follows: b1 = 112 ; b2 = 1

12 + 122 ; b3 = 1

12 + 122 + 1

32 ;

and in general,

bn = 1

12 +

1

22 +

1

32 + . . . +

1

n2 =

nk=1

1

k2


12. To have Derive help you evaluate bn for larger values of n, Author the following:

B(n) := SUM(1/k2, k, 1, n)

13. Verify your answer to Question #5 by Authoring and then approXimating B(1), B(2), B(3), and B(4).


15. Again, because it takes so long to evaluate bn for large values of n, we need to find another way of

determining how large bn gets. Answer Worksheet Question #7.

16. Finally, let’s consider the sequence cn defined as follows: c1 = 11 ; c2 = 1

1 + 12 ; c3 = 1

1 + 12 + 1

3 ; and in

general,

cn = 1

1 +

1

2 +

1

3 + . . . +

1

n =

nk=1

1

k

The sequence cn is called the harmonic series. Answer Worksheet Question #8.



197

17. To have Derive help you evaluate cn for larger values of n, Author the following:

C (n) := SUM(1/k,k, 1, n)

18. Verify your answer to Question #8 by Authoring and then approXimating C (1), C (2), C (3), and C (4).


20. Again, we need a quicker way of determining how large cn gets. Answer Worksheet Questions #10 and

#11.



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P-Series

NAME: DATE:


1. Let an = 1 + 1√

2+

1√ 3

+ . . . + 1√

n =

nk=1

1√ k

.

(a) Evaluate a1, a2, a3, and a4. Express your answers as decimals. (You may use either Derive or a

calculator).

(b) What kind of a sequence is an? (i.e. is it increasing, decreasing, nonincreasing, nondecreasing, or

none of these?)

2. (a) Use Derive to calculate a100, a1000, and a5000.

a100 =

a1000 =

a5000 =

199



200 Worksheet # 17. P-SERIES

(b) Based on your calculations, do you think an converges or diverges?

3. Does the graph of A(n) seem to indicate that an converges? i.e. does the graph seem to level off in

height or does it seem to be heading toward infinity?

4. Consider the function F (x) = 1√ x

.

(a) Below is a graph of y = F (x); express the shaded area in terms of the function A(n) by first

expressing it as the sum of the areas of rectangles. [Hint: the width of each rectangle is 1; the

height is given by F (an endpoint of subinterval)]. Then write down an inequality by comparing

the shaded area to the area under the curve y = 1√ x

on an appropriate interval. Write down

the integral involved, then perform the integration (you may use Derive’s Calculus Integrate

command if you wish) and express your answer exactly (i.e. in terms of square roots, not as a

decimal approximation).

1 2 3 4 5 6

1

2

F (x) = 1√ x

......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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Name: 201

Shaded region = + + + +

= A( ) >

=

(b) Repeat the above for the picture below:

1 2 3 4 5 6 7 8 9 10

1

2

F (x) = 1√ x

...................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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Shaded region = + + + . . . + +

= A( ) > =

(c) Based on your calculations above, fill in the blank:

an = A(n) >

(d) Use the above inequality to find an integer n so that an > 1000:

(e) Find an integer n such that an > 50, 000:




(f) limn→∞ an ≥ limn→∞ =

(g) Is an bounded? Does it converge? Explain.

5. Let bn = 1

12 +

1

22 +

1

32 + . . . +

1

n2 =

nk=1

1

k2

(a) Evaluate b1, b2, b3, and b4. Express your answers as decimals. (You may use either Derive or a

calculator).

(b) What kind of a sequence is bn? (i.e. is it increasing, decreasing, nonincreasing, nondecreasing, or

none of these?)

6. (a) Use Derive to calculate b100, b1000, and b5000.

b100 =

b1000 =

b5000 =



Name: 203

(b) Based on your calculations, do you think bn converges or diverges?

7. Consider the function F (x) = 1x2 .

(a) Below is a graph of y = F (x); express the shaded area in terms of the function B(n) by first

expressing it as the sum of the areas of rectangles. Then write down an inequality by comparing

the shaded area to the area under the curve y = 1x2 on an appropriate interval. Write down the

integral involved, then perform the integration and express your answer exactly (i.e. as a fraction,

not as a decimal approximation).

1 2 3 4 5 6

1

F (x) = 1x2

..................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .

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. . . . . .. . . . . . .. . . . . .. . . . . . .. . . . . . . . . . . . . .


= B( ) − < =





1 2 3 4 5 6 7 8

1

F (x) = 1x2

.................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................... . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .

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. . . . . .. . . . . . .. . . . . .. . . . . . .. . . . . . . . . . . . . . . . . . . . . . . . . . . .

Shaded region = + + + . . . + +

= B( ) − <

=


bn = B(n) < < (a number)

(d) Is bn bounded? Does it converge? Explain.

8. Let cn = 1

1 +

1

2 +

1

3 + . . . +

1

n =

nk=1

1

k.

(a) Evaluate c1, c2, c3, and c4. Express your answers as decimals. (You may use either Derive or a



Name: 205

calculator).

(b) What kind of a sequence is cn? (i.e. is it increasing, decreasing, nonincreasing, nondecreasing, or

none of these?)

9. (a) Use Derive to calculate c100, c1000, and c5000.

c100 =

c1000 =

c5000 =

(b) Based on your calculations, do you think cn converges or diverges?

10. Consider the function F (x) = 1x .

(a) Below is a graph of y = F (x); express the shaded area in terms of the function C (n) by first

expressing it as the sum of the areas of rectangles. Then write down an inequality by comparing

the shaded area to the area under the curve y = 1x on an appropriate interval. Write down the

integral involved, then perform the integration and express your answer exactly (i.e. in terms of

logarithms, not as a decimal approximation).




1 2 3 4 5 6

1

F (x) = 1x

.......................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .

. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .


= C ( ) >

=


1 2 3 4 5 6 7 8 9 10

1

F (x) = 1x

....................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .

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. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .. . . . . . .

Shaded region = + + + . . . + +

= C ( ) >

=



Name: 207


cn = C (n) >

(d) Use the above inequality to find an integer n so that cn > 20.

(e) Is cn bounded? Does it converge? Explain.

11. In this lab we looked at various sequences of the form: an =n

k=1

1

k p

For which values of p did we obtain a convergent sequence? For which values of p did we obtain a

divergent sequence?



Laboratory # 18

Taylor Polynomials

Teaches:

• evaluating Taylor polynomials

• using Taylor polynomials to approximate functions

• differentiation and integration of Taylor polynomials

1. We have seen that the line tangent to the graph of a function y = F (x) can be thought of as the best

linear approximation of the function at and near the point of tangency. In this lab you will learn how

to find better approximations. Answer Worksheet Question #1.

2. Choose Plot Beside <Enter> to open a graphics window; then split the graphics window horizontally

by choosing Window Split Horizontal <Enter>; then go to the Algebra window and split it horizontally;

finally Window Designate window #1 as a 2D-Plot window. You should have 4 windows, 3 of which

are graphics windows.

3. Author

F (x) := ln(x + 1)

208



209

and Plot in window #1, and also in window #3. Also Plot the tangent line whose equation you just

found in both windows.

4. As we calculate successive approximations to the graph of y = ln(x + 1), we will use window #1 to

plot them, window #3 to “zoom in” on the graphs, and window #4 to do some error analysis.

5. Let’s call G(x) a “good” approximation to F (x) if F and G differ by less that 0.0005. Author

0.0005

then go to window #4, set the y scale to 0.0005, and Plot. This gives us a horizontal line at our

tolerance level of y = 0.0005. Now Author

G(x) := ax + b

replacing a and b by the values that you found in Worksheet Question #1.

6. The difference between F and G is our error. So Author

|F (x) − G(x)|

and Plot in window #4. The interval for which the graph of the error lies beneath the line y = 0.0005

is very small; use Shift- F7 to blow it up (Shift- F8 shrinks it back down); once you have blown up

the interval as large as you can (while still keeping the endpoints where the two graphs intersect in

the viewing window), change the x scale in window #3 to the same value as the x scale in window

#4, and the y scale to twice this value. Note: we will always keep the x scales in windows #3 and #4

the same; in this way both windows will show the same interval ( u, v) where |F (x) − G(x)| < 0.0005;

window #3 illustrates the closeness of the two graphs on that interval:



210 LABORATORY # 18. TAYLOR POLYNOMIALS


8. Note that the tangent line of the graph of F (x) = ln(x + 1) is the unique linear function G(x) such

that F (0) = G(0) and F (0) = G

(0) (i.e. the graphs of F and G intersect at x = 0 and have the same

slope at x = 0).

9. We will now try to find the best quadratic approximation to F (x) = ln(x + 1) near (0, 0) by finding

the unique function G(x) = a + bx + cx2 for which F (0) = G(0), F (0) = G(0) and F (0) = G (0).

You already found a and b in Worksheet Question #1. To find c, Author the following , replacing a

and b by the values that you found in Worksheet Question #1 (make sure you use := for the first one,

but only = for the second one):

G(x) := a + bx + cx2

F (x) = G(x)

use Calculus Differentiate to find the second derivative of the last expression (i.e. enter 2 for the order).



211

Simplify; what you obtain is an equation that sets F (x) equal to G(x). Since we want to find c so

that F (0) = G(0), use Manage Substitute to plug in 0 for x, then soLve for c. Answer Worksheet

Question #3.

10. Now we want to find the best cubic approximation to F (x) = ln(1 + x) near (0, 0) by finding the

unique function G(x) = a + bx + cx2 + dx3 for which F (0) = G(0), F (0) = G(0), F (0) = G(0), and

F (0) = G(0). You’ve already found a, b, and c in Worksheet Question #3. To find d, Author

G(x) := a + bx + cx2 + dx3

where you have replaced a, b, and c by their known values; then highlight the expression F (x) = G(x)

and use Calculus Differentiate to find the third derivative of this expression. Simplify; what you obtain

is an equation that sets F equal to G. Since we want to find d so that F (0) = G(0), use Manage

Substitute to plug in 0 for x, then soLve for d.


12. The polynomials that you have been calculating are called Taylor polynomials (at the point x = 0).

Derive has an easy way of calculating these polynomials. Author and Simplify

TAYLOR (ln(x + 1), x, 0, 10)

The polynomial that you obtain should be the same as the one that you wrote down in Worksheet

Question #6b.

13. Answer Worksheet Question #7–#10.



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Taylor Polynomials

NAME: DATE:


1. Find the equation of the line G(x) = a + bx tangent to the graph of y = ln(x + 1) at(0, 0).

a = b=

equation of line: G(x) =

2. Use the cross in window #4 to determine the interval (u, v) on which the tangent line G(x) at x = 0

that you found above is a “good” approximation to F (x) = ln(x + 1). [Hint: remember that “good”

means that |F (x) − G(x)| < 0.0005 which happens when the graph in window #4 lies beneath the

line y = 0.0005; so what you want are the x coordinates of the points of intersection of the graph in

window #4 with the line y = 0.0005] Round your answers to two decimal places.

213



214 Worksheet # 18. TAYLOR POLYNOMIALS

3. (a) What is is the equation of G(x) = a + bx + cx2, the best quadratic approximation of F (x) near

(0, 0)?

c =

equation of parabola: G(x) =

(b) Author G(x) := a + bx + cx2, where you have replaced a, b, and c by their known values (or

use Manage Substitute on a previous expression to replace c); then Plot in window #1 and in

window #3. Again we want to find the interval on which G is a good approximation to F , so

Plot |F (x) − G(x)| in window #4. Use Shift- F8 to shrink down the interval in your viewing

window until you can see where the graph of the error intersects the line y = 0.0005; also change

the x scale in window #3 so that it’s the same as the x scale in window #4, and set the y scale

in window #3 to be twice the x scale. Use the cross in window #4 to find the interval (u, v) for

which the parabola G(x) is a good approximation to F (x). Again, round to two decimal places.

(c) Is the interval that you found bigger or smaller than the one that you found in Question #2?



Name: 215

4. (a) What is the equation of G(x) = a+bx+cx2+dx3, the best cubic approximation of F (x) = ln(1+x)

near (0, 0)?

d =

equation of cubic: G(x) =

(b) Author G(x) := a + bx + cx2 + dx3, where you have replaced a, b, c, and d by their now known

values (or use Manage Substitute on a previous expression to replace d) then Plot in window

#1 and in window #3. To find the interval on which G is a good approximation to F , Plot

|F (x) − G(x)| in window #4. Use Shift- F8 to shrink down the interval in your viewing window

until you can see where the graph of the error intersects the line y = 0.0005; then change the x

and y scales in window #3 accordingly. Use the cross in window #4 to find the interval ( u, v) for

which the cubic G(x) is a good approximation to F (x). Again, round to two decimal places.

(c) Is the interval that you found bigger or smaller than the one that you found in Question #3b?

5. (a) Proceed as before to find the best fourth degree polynomial than approximates F (x) near (0, 0).

This time, you will need to Author G(x) := a + bx + cx2 + dx3 + ex4, replacing a, b, c, and d

with their known values, then take the fourth derivative of the equation F (x) = G(x), Simplify,

Manage Substitute 0 for x, and soLve for e.

e =

equation of 4th degree polynomial: G(x) =




(b) Author G(x) := a + bx + cx2 + dx3 + ex4, where you have replaced a, b, c, d, and e by their known

values (or use Manage Substitute on a previous expression to replace e) then Plot in window #1

and in window #3. Plot |F (x) − G(x)| in window #4. Use the cross to find the interval (u, v)

for which the 4th degree polynomial G(x) is a good approximation to F (x). Again, round to two

decimal places.

(c) Is the interval that you found bigger or smaller than the one that you found in Question #4b?

6. Hopefully, you are now able to discern a pattern in these polynomials.

(a) So what is the best fifth degree polynomial that approximate F (x) near (0, 0)?

(b) What is the best tenth degree polynomial that approximate F (x) near (0, 0)?

7. (a) Fill in the following table. [Hint: you have already found the polynomials G(x) and the cor-

responding intervals for n = 1 through n = 4; for the fifth degree Taylor polynomial, you can

Author and Simplify G(x) := TAYLOR (ln(x + 1), x, 0, 5) then Plot |F (x) − G(x)| in window #4

and proceed as before.]



Name: 217

degree n nth degree Taylor polynomial G(x) interval where F (x) ≈ G(x)

1

2

3

4

5

6

7

10

(b) What degree Taylor polynomial G(x) should you use to approximate F (.3) = ln(1.3) if you want

the error to be less that 0.0005? Choose the shortest polynomial that will work and evaluate

G(.3) to approximate ln(1.3). Compare the approximation to Derive’s approximation of ln(1.3).

8. (a) Delete All graphs from all three graphics windows; also set the x and y scales in window #1 to 2;

use Derive to find the third, fifth, and seventh degree Taylor polynomials for f (x) = sin x. Plot

both f (x) and the polynomials, then copy the graphs onto the axes below. Make sure you label

all your graphs.

P 3(x) =

P 5(x) =

P 7(x) =




-

?

6

1 2 3 4−4 −3 −2 −1

1

2

−2

−1

(b) Find the interval on which the third degree Taylor polynomial approximates sin x with an error

< 0.0005. [Hint: Author and Plot | sin x − TAYLOR(sin x,x, 0, 3)| in window #4; you want the

interval on which the graph of this error lies below the line y = 0.0005 (which you should rePlot)]

(c) Find the shortest Taylor polynomial that can be used to approximate sin(2) so that the error

is < 0.0005.Then use it to to approximate sin(2) and compare the approximation to Derive’s

approximation to sin(2). [ Hint: set the x scale in window #4 to 2, then Plot the error

|sin x

−TAYLOR(sin x,x, 0, n)| for n = 1, 2, 3, . . . until the interval on which the graph lies below the line

y = 0.0005 includes 2.]

9. (a) Find the derivatives of the polynomials that you found in Question #8a above.

P 3(x) =

P 5(x) =



Name: 219

P 7(x) =

(b) Use Derive to find the second, fourth, and sixth degree Taylor polynomials for f (x) = cos x.

P 2(x) =

P 4(x) =

P 6(x) =

(c) What do you notice?

10. (a) Use Derive to find the third, fifth, and seventh degree Taylor polynomials for f (x) = 11+x2 .

P 3(x) =

P 5(x) =

P 7(x) =

(b) Find antiderivatives of the polynomials that you found above.

P 3(x) dx = P 5(x) dx = P 7(x) dx =

(c) Use Derive to find the fourth, sixth, and eighth degree Taylor polynomials for f (x) = tan−1 x




(recall that tan−1 x is entered as ATAN(x) in Derive)

P 4(x) =

P 6(x) =

P 8(x) =

(d) What do you notice?



Laboratory # 19

Parametric Equations

Teaches:

• Parametrizations of plane curves

• Curvilinear motion

1. Suppose a fly is moving on the plane so that its position at time t (measured in seconds) is given by

the coordinates (2t, t2 − 2t). For example, at time t = 0, the fly is at the origin, and at time t = 2, its

coordinates are (2 · 2, 22 − 2 · 2) = (4, 0). By computing its position at different times, we can sketch a

rough graph of the motion of the fly:

2 4 6 8 10

2

4

6

8

rr

r

r

r

t=0

t=1

t=2

t=3

t=46

-

221



222 LABORATORY # 19. PARAMETRIC EQUATIONS


3. You will have noticed that the path of both flies in Worksheet Question #1 are the same, but their

rates of travel are different. Indeed, it took the second fly half as long (2 seconds vs 4 seconds for the

first fly) to cover the same ground as the first fly.

Definition 19.1 In general, if a moving particle (or fly) is moving on the plane so that its coordinates

at time t are given by (f (t), g(t)) for some continuous functions f and g defined on an interval I , we

call the set C of points (f (t), g(t)) a plane curve. The equations x = f (t) and y = g(t) are called

parametric equations for C . The variable t is called the parameter.

4. Derive can graph plane curves. For example, to graph the motion of the first fly in Worksheet Question

#1, Author

[t, t2]

then Plot. Enter 0 and 4 for the min and max values. Also set your x scale to 2 and your y scale to 8

so that you are able to see the entire graph. Then Author

[2t, 4t2]

and Plot. Enter 0 and 2 for the min and max values. Both these graphs should coincide with the

sketches that you made in Worksheet Question #1.

5. Suppose that a ball is thrown from the origin at an angle θ0 with an initial velocity v0. It can be shown

that if the only force acting on the ball is gravity, then the coordinates of the ball at any time t are

given by

x = (v0 cos θ0)t y = −1

2gt2 + (v0 sin θ0)t (19.1)

where g is the acceleration due to gravity, approximately 9.8 meters/second2 .



223

-

6

s (x, y)

7 6

v0 sin θ0

v0 cos θ0

v0

wθ0




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Parametric Equations

NAME: DATE:


1. (a) Suppose the fly is moving so that its position at time t is given by the coordinates (t, t2), 0 ≤ t ≤ 4.

Fill in the table below by computing the position of the fly at the given times, then make a rough

graph depicting the motion of the fly.

time t x coordinate t y coordinate t2 time t x coordinate t y coordinate t2

0 2

0.5 2.5

1 3

1.5 4

4

8

12

16

1 2 3 4

225



226 Worksheet # 19. PARAMETRIC EQUATIONS

(b) Suppose a second fly is moving so that at time t its position is given by (2t, 4t2), 0 ≤ t ≤ 2. Again,

fill in the table below by computing the position of the fly at the given times, then make a rough

graph depicting the motion of the fly.

time t x coordinate= 2t y coordinate= 4t2 time t x coordinate = 2t y coordinate= 4t2

0 1

0.25 1.25

0.5 1.5

0.75 2

4

8

12

16

1 2 3 4

(c) Compare the motion of the two flies. What is the same? What is different?

(d) Estimate the first fly’s velocity at t = 1 by computing its approximate average velocity over the

time interval [1, 1.01], [Hint: average velocity = distance travelledelapsed time = distance travelled

0.01 . The

distance travelled is approximately equal to the distance between the fly’s position at t = 1:

(1, 12) and the fly’s position at t = 1.01: (1.01, 1.012). So use the distance formula: Distance from



Name: 227

(x1, y1) to (x2, y2) is

(x2 − x1)2 + (y2 − y1)2.]

(e) The second fly reaches the point (1, 1) at time t = 1/2. Estimate the second fly’s velocity at t = 1/2

by computing its approximate average velocity over the interval [1 /2, 1/2 + .01] = [0.5, 0.51].

(f) How do the two flies’ velocities at the times they pass through the point (1, 1) compare?

2. Suppose you throw a ball at an angle of θ0 = 60 deg with an initial velocity of v0 = 20 meters/second.

(a) Use (19.1) to write down the parametric equations for the position (x1, y1) of the ball at any time

t.

(b) At what time will the ball hit the ground? Use the parametric equations that you found in (a)

(which give the position (x1, y1) of the ball at any time t) to find an exact answer, then also give




a decimal approximation.

(c) Use Derive to obtain a plot of the motion of the ball.

[Hint: Author and Plot [x1, y1] where x1 and y1 are the parametric equations that you found in

(a). Use your answer to (b) to determine the min and max values of t that you need to enter.

Change your scales so that the entire path of the ball from when it’s thrown to when it hits the

ground is visible in your viewing window.]

5 10 15 20 25 30 35

5

10

15

(d) How far away from you will the ball hit the ground? Use the parametric equations and your

answer to (b) to obtain your answer. Check that your answer is consistent with your plot.

(e) What is the maximum height reached by the ball? Use the parametric equations and calculus

(what are you trying to maximize, so what derivative should you set to 0?) to obtain your answer.

Show your calculations; approximate your answer as a decimal.Again, make sure that your answer



Name: 229

is consistent with your plot.

3. Suppose that the bottom of a ferris wheel of radius 7 meters is located 25 meters away from where

you are standing ready to throw your ball as in the previous question. The bottom of the ferris wheel

is moving toward you. Your friend is riding on the ferris wheel, and at time t = 0 (which is when you

throw the ball) she is half way up of the ferris wheel on your side.

5 10 15 20 25 30 35

5

10

15

.................................................................................................

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y

ball

? •

•........................................................................................................................................................................................................

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............

..............friendstarts here

(t = 0)

friend is herea little later(at time t)

?

7θb(x1, y1)

(x2, y2)

←− a −→

ferris wheel

.......................

.......

.............

.............................

(a) Express a and b as functions of θ.




(b) Express the position (x2, y2) of your friend as functions of θ.

(c) If the wheel makes one revolution (i.e. 2π radians) every 12 seconds, then

dθdt

= radians per second, so

θ = t.

(d) Use your answers above to find parametric equations for x2 and y2 in terms of the parameter t.

4. In the previous two questions, you found parametric equations for both the position (x1, y1) of the

ball and the position (x2, y2) of your friend. Recall that the distance between two points (x1, y1) and

(x2, y2) is

(x2 − x1)2 + (y2 − y1)2.

(a) Find an expression for the distance between the ball and your friend as a function of time.

(b) Use Derive’s calculus commands to find the minimum distance between the ball and your friend

(what are you trying to minimize, so what derivative should you set to 0?) Recall that when

Derive is unable to find the zeroes of a function in exact mode (using the soLve command),



Name: 231

you must set Options Precision to Approximate so that Derive will use the bisection method to

approximate the zero.

(c) Your friend will catch the ball if it comes within one third of a meter of her. Does she catch the

ball? Explain.



Laboratory # 20

Polar Coordinates

Teaches:

• graphs of polar equations – cardioids, limacons, rose curves, conics

• area in polar coordinates – approximations using circular sectors


2. Derive is able to plot polar graphs. Author

r = 2 + 2 cosθ

(θ is entered as alt-h) then in your graphics window choose Options State Polar Connected Small

<Enter>. Now Plot from θ = 0 to θ = 2π. Your plot should match the one from Question #2.


4. Given r = f (θ), α ≤ θ ≤ β , we are often interested in finding the area of the region bounded by the

graphs of f and the rays θ = α and θ = β :

232



233

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...... .

θ = α

θ = β r = f (θ)

5. Transfer Load the Utility file LABS.MTH.

6. Just as for rectangular coordinates we approximated the area under a curve y = f (x) by the areas

of rectangles, we use circular sectors to approximate areas of regions defined by polar equations. A

circular sector with radius r and central angle θ has area 12

θr2:

.......

.......

.......

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........................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................

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θ r

7. Author, Simplify, and Plot

PAREA(2 sin θ, 0, π, 20)

from θ = 0 to θ = π. This will give you a graphical display of the 20 circular sectors we may use to

approximate the area of this circle. The area of each sector of radius r and central angle θ can then

be evaluated using the formula 12 r2θ.



234 LABORATORY # 20. POLAR COORDINATES

8. To evaluate the sum of the areas of the sectors, Author and approXimate

PSUM(2 sin θ, 0, π, 20)

Answer Worksheet Question #7.

9. More generally, suppose you want to approximate the area bounded by the graph of r = f (θ) between

θ = α and θ = β :

......................................................................................................

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.......... .

θ = α

θ = β

r = f (θ)

Then what you will do (in Worksheet Question #8) is Author, Simplify, and Plot PAREA(f (θ), α , β , n)

to obtain a graphical display of the n circular sectors used to approximate the area; then Author and

approXimate PSUM(f (θ), α , β , n) to approximate the sum.




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Polar Coordinates

NAME: DATE:


1. Recall that the location of a point P can either be specified by its rectangular coordinates (x, y) or its

polar coordinates (r, θ). Use the following picture to establish relationships between rectangular (x, y)

and polar (r, θ) coordinates:

6

-

rP

y

x

rθ

(a) sin θ = or y =

(b) cos θ = or x =

(c) tan θ = or θ =

(d) x2 + y2 = or r =

235



236 Worksheet # 20. POLAR COORDINATES

2. Consider the polar equation

r = 2 + 2 cosθ

(a) Find some points on the graph by filling in the following table; use decimal approximations

rounded to two decimal places:

θ r θ r

0 5π/4

π/4 3π/2

π/2 7π/4

3π/4 2π

π

(b) Use your table to sketch the graph of r = 2 + 2 cosθ below:

−3 0 3

−3

0

3

...

.......

.............

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............. ............. ........

..... .............

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................

r = 2 + 2cos θ

3. The family of curves r = a ± a sin θ and r = a ± a cos θ, where a is any positive real number, are called

cardioids. Use Derive to plot some cardioids using different values for a.(e.g. a = 1, 2, 3, . . .). Then

sketch the different shapes below:



Name: 237

6

?

-

r = a + a cos θ

6

?

-

r = a + a sin θ

6

?

-

r = a − a cos θ

6

?

-

r = a − a sin θ

4. Cardioids are special cases of polar curves known as limacons, whose equations are of the form

r = a ± b sin θ or r = a ± b cos θ. Use Derive to sketch different limacons; you get different shapes

depending on your choice of a and b; sketch the different shapes below:




6

?

-

r = a + b sin θ, 0 < a < b

6

?

-

r = a + b sin θ,b < a < 2b

6

?

-

ra + b sin θ, a ≥ 2b

6

?

-

r = a − b sin θ, 0 < a < b

6

?

-

r = a − b sin θ,b < a < 2b

6

?

-

ra − b sin θ, a ≥ 2b

5. If n is a positive integer, n ≥ 2, the graphs of r = a sin(nθ) or r = a cos(nθ) are called rose curves.

(a) Use Derive to sketch some rose curves:



Name: 239

6

?

-

r = 2 cos(2θ)

6

?

-

r = 2 cos(3θ)

6

?

-

r = 2 cos(4θ)

6

?

-

r = 2 cos(5θ)

(b) When n is odd, how is the number of petals related to n?

(c) When n is even, how is the number of petals related to n?

Just for fun, try graphing r = 2 cos(2.1θ), 0 ≤ θ ≤ 50.




(d) What kind of graphs do r = a sin θ and r = a cos θ produce (a > 0)?

6. Any polar equation of the form

r = a

1 + b cos θ or r =

a

1 + b sin θ

is a conic section with focus at the origin and axis along a coordinate axis. Experiment by graphing

some of these.

(a) What shape do these graphs have when −1 < b < 1?

(b) What kind of graph do you get when |b| > 1? [Note: Derive will take a long time to graph if

you give it a domain that results in a denominator that’s zero. For example, to graph 21+2 cosθ ,

0 ≤ θ ≤ 2π, note that the denominator is zero when θ = 2π/3, 4π/3 (Derive’s soLve command

can help you find these values), so plot this in three steps, first from 0 to 2 π/3 − .1, then from

2π/3 + .1 to 4π/3 − .1, and finally from 4π/3 + .1 to 2π.]

(c) What kind of graph do you get when b = ±1? [Again, to plot 11+cosθ

, for instance, plot first from

θ = 0 to θ = π − .1, and then from θ = π + .1 to θ = 2π, to avoid points near θ = π which is

where the denominator is 0.]



Name: 241

(d) What’s the difference between the graphs involving sin θ and the graphs involving cos θ?

7. We want to approximate the area inside one petal of the rose curve r = sin5θ. To see a graphical

display of the circular sectors used in this approximation, Delete All graphs then Author, Simplify and

Plot

PAREA (sin(5θ), 0, π/5, 10)

from θ = 0 to θ = 2π; this will give you a graph of the entire rose curve (since you graphed from 0 to

2π) as well as the 10 sectors used to approximate the area of the one petal with domain 0 ≤ θ ≤ π/5.

Author the above with PAREA replaced by PSUM then approXimate in order to find the sum of the

areas of these sectors.

8. Delete All graphs, then approximate the area of the region in the first quadrant that is outside the

circle r = 1 and inside the rose curve r = 2 sin(2θ) as follows:

(a) Plot both curves r = 1 and r = 2sin(2θ) and copy the graphs below. Shade the region whose

area we want to approximate.




?

-

6

(b) Draw the two rays θ = α and θ = β that go from the origin to the points of intersection in the

first quadrant. Find α and β by solving 1 = 2 sin(2θ) for θ.

α =

β =

(c) Use PAREA (to obtain a graphical display) and PSUM to approximate (to two decimal places)

the area inside the petal and between the two rays. [Hint: keep approximating using more and

more circular sectors until successive approximations agree in the first three decimal places, so

you can round to the second place.]

(d) Use PAREA (to see a graphical display of the circular sectors – make sure you simplify before

you Plot) and PSUM to approximate (to two decimal places) the area inside the circle r = 1 and

between the two rays.

(e) Subtract your answer to (d) from your answer to (c) to obtain an approximation of the desired

area.

(f) In general, the area bounded by the graphs of r = f (θ), r = g(θ), θ = α, and θ = β , can be found



Name: 243

by evaluating the following integral:

β

α

1

2 [f (θ)]2

−[g(θ)]

2 dθ

With f (θ) = 2 sin(2θ) and g(θ) = 1, use Calculus Integrate to evaluate the area of this region

exactly. Give both the exact answer (using Simplify) and the decimal approximation obtained

using approXimate.



Laboratory # 21

An Introduction to Vectors

Teaches:

• vectors in two dimensions

• vector operations–addition, scalar multiplication, dot product.

1. Transfer Load the Utililty file LABS.MTH. Choose Plot Beside <Enter> to open a graphics window,

then choose Options State Rectangular Connected Small <Enter>.

Definition 21.1 A vector is a quantity that has both magnitude and direction. Geometrically, a

vector is represented by a directed line segment – i.e. an arrow. The vector below is denoted by −→AB.

The point A is called the initial point of −→AB and the point B is called the terminal point.

A

B

2. To graph vectors in the xy plane, we need to specify the initial and terminal points. Often, we will

want the initial point to be the origin. In Derive, use V 0(a, b) to obtain a graph of the vector whose

initial point is at the origin (0, 0) and whose terminal point is (a, b).

244



245

3. Author and Plot

V 0(1, 2)

4. This vector is called the position vector < 1, 2 >. The term position vector just means that the

initial point is at the origin.



V 0(1, 2)

V 1(1, 2, 2, −1)

The first vector is the position vector < 1, 2 >; the second vector is the position vector < 1, 2 >

translated so that its initial point is (2, −1). In general, use V 1(a,b,c,d) to obtain the position vector

< a,b > translated so that it’s initial point is (c, d). Thus, V 0(a, b) is the same as V 1(a,b, 0, 0).


8. We now need to learn how to add vectors v1 and v2. We do so as follows: move the vector v2 so that

its initial point coincides with the terminal point of v1; the vector v1 + v2 is the vector whose initial

point is the origin and whose terminal point is the new terminal point of the translated vector v2. The

following picture indicates how we obtain v1 + v2:

6

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*

J

JJ

JJJ

v1

v2

v2 (translated)

v1 + v2




246 LABORATORY # 21. AN INTRODUCTION TO VECTORS

10. We would now like to learn how to multiply vectors. There are two important kinds of vectors products;

in this lab we shall focus on the dot product.

Definition 21.2 The dot product of two vectors u =< a, b > and v =< c, d > is denoted by u · v

and is defined by

u · v ==< a,b > · < c, d >= ac + bd

11. Notice that u · v is a number , not a vector. For example, < 2, 3 > · < −1, 2 >= (2)(−1) + (3)(2) =

−2 + 6 = 4.




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An Introduction to Vectors

NAME: DATE:


1. (a) Use Derive to obtain graphs of the following position vectors; then draw and label the vectors on

the axes below: v1 =< −3, 1 >, v2 =< 2, −2 > and v3 =< 0, 4 >.

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6

?

(b) Every vector v has a length or magnitude, denoted by |v|, which can be computed using the

Pythagorean theorem, if you think of the vector as the hypotenuse of a certain right triangle.

Calculate the length of each vector in (a).

|v1| =

|v2| =

|v3| =

247



248 Worksheet # 21. AN INTRODUCTION TO VECTORS

(c) What’s a general formula for evaluating the magnitude of the position vector < a,b >?

2. Multiplication of a scalar (i.e. a number) c with a vector v =< a,b > is defined by cv =< ca,cb >.

Delete All graphs, then Author and Plot V 0(1, 2) to obtain the graph of v =< 1, 2 >;

(a) Author and Plot V 0(2, 4) to obtain a plot of 2v =< 2, 4 >. What is the relationship between the

vectors v and 2v?

(b) Author and Plot V 0(.5, 1) to obtain a plot of .5v =< .5, 1 >. What is the relationship between

the vectors v and .5v?

(c) Author and Plot V 0(−2, −4) to obtain a plot of −2v =< −2, −4 >. What is the relationship

between the vectors v and −2v?

(d) In general, what appears to be the relationship between any vector v =< a, b > and the vector

cv =< ca, cb >, if c > 0?

(e) In general, what appears to be the relationship between any vector v and the vector cv, if c < 0?



Name: 249

3. For each of the following, use Derive to obtain a graph of the given position vector (using V 0) as well

as the translated vector (using V 1) so that you can locate the terminal point with the cross. Draw

both on the given axes, and label both initial and terminal points.

(a) What will be the terminal point of the vector < 1, 2 > if it’s translated so that its initial point is

(2, −1)?

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6

?

(b) What will be the terminal point of the vector < −2, 3 > if it’s translated so that it’s initial point

is (2, 1)?

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6

?

(c) What will be the terminal point of the vector < −1, −1 > if it’s translated so that it’s initial

point is (−1, 0)?




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6

?

(d) In general, what is the terminal point of the vector < a,b > if it’s translated so that its initial

point is (c, d)?

4. Delete All graphs; for each of the following, use Derive to graph the vectors v1 and v2 (i.e. use V 0);

then obtain a graph of v2 translated so that it’s initial point is at the terminal point of v1 (i.e. use

V 1). The vector v1 + v2 is the vector whose initial point is (0, 0) and whose terminal point is the

terminal point of the translated vector v2. Locate the terminal point with the cross, then use V 0 to

graph v1 + v2. Copy and label all these vectors on the given coordinate axes; make sure you label the

initial and terminal points of all the vectors.

(a) v1 =< −2, 3 >; v2 =< 1, 1 >;



Name: 251

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v1 + v2 =< , >

(b) v1 =< 1, 1 >; v2 =< −2, 3 >;

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v1 + v2 =< , >

(c) v1 =< 1, −2 >; v2 =< 2, −1 >;

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v1 + v2 =< , >




(d) v1 =< 2, −1 >; v2 =< 1, −2 >;

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6

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v1 + v2 =< , >

5. Compare your answers to (a) and (b) above ; then compare your answers to (c) and (d). In view of

your answers above, what property does vector addition have?

6. In view of your answers to Question #4, give a general rule for adding vectors:

< a,b > + < c,d >=< , >

7. Given two vectors u and v, we can talk about the angle θ between the two vectors, as in the following

picture:

-

u

v

θ

In the following table, you are given the vector u and the angle θ; you have to come up with a unit

vector v (assume, as in the picture, that the direction from u to v is counterclockwise; unit vector



Name: 253

means a vector of length 1); for example if u =< 1, −1 > and θ = π4

, then v could be < 1, 0 > (since

we want v to have length 1):

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@@@R

-v

u

π4

(a) Fill in the table (to find v, draw a picture of u and θ as above). You should use the V 0 function

to graph both u and v to verify your answers.

[Hint: recall that |u| is the magnitudeof u =< a, b >, given by√

a2 + b2; for example, | < 1, −1 > | =

(1)

u |u| v |v| θ u · v |u||v| cos θ

< 1, −1 >√

2 < 1, 0 > 1 π4

1√

2 · 1 =√

2√

22

< 1, 0 > 1 π

2

< 1,√

3 > 1 π6

< 1,√

3 > 1 2π3

< 1, 1 > 1 3π4

< 0, 1 > 1 π

(b) Formulate a theorem about dot products based on the relationship among the last three columns

of your table.

Theorem 21.1 For any two vectors u and v, = × .

(c) We say two vectors are orthogonal if the angle between them is a right angle, i.e. is 90 deg or π2

radians. Use the above theorem to formulate a theorem about orthogonal vectors.[Hint: What is

cosπ2

?]




Theorem 21.2 The two vectors u and v are orthogonal if and only if u · v = .

(d) Use the above theorem to determine if the following pair of vectors is orthogonal; you should use

Derive to graph the vectors to verify your answer.

u =< 1, 2 > v =< 4, −2 >

(e) Use the theorem to find a number a so that the vectors u =< 3, −2 > and v =< a, 4 > are

orthogonal. Again, you should graph the vectors with Derive to verify your answer.

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Laboratory # 22

Cylinders and Quadric Surfaces

Teaches:

• how to graph some important surfaces

1. In two dimensions, the graph of x2

+ y2

= 4 is a circle centered at the origin with radius 2. However

in 3-space, the set of all points (x, y, z) for which x2 + y2 = 4 (z arbitrary) is a cylinder.

2. Choose Plot Beside <Enter> to open a graphics window, set the x and y scales both to 5, then split

the graphics window horizontally into two graphics windows by choosing Window Split Horizontal

<Enter>.


4. Next, we need to set up window #3 as follows:

• First, Transfer Load the Utility file LABS.MTH.

• Next, go to window #3 ( F1 takes you to the next window), and “hide” the xy axes by choosing

Options Color Plot and changing the axes and cross colors both to 0.

255



256 LABORATORY # 22. CYLINDERS AND QUADRIC SURFACES

• Then choose Options Color Auto No.

• Finally, Author AXES, then go to window #3 and Plot from 0 to 20 <Ctrl-Enter>.

5. To plot the circle x2 +y2 = 4, z = 0, in window #3, we need to use the parametric equations x = 2 cos t,

y = 2 sin t, z = 0, 0 ≤ t ≤ 2π. Additionally, in order to plot space curves in a 2D-plot window, we need

to use the function ISOMETRIC. So Author and Simplify

ISOMETRIC[2cost, 2sin t, 0]

then go to window #3 and Plot from t = 0 to t = 2π.

6. Simirlarly, we can plot all the points (x, y, z) for which x2 + y2 = 4 and z = 2 by using the parametric

equations x = 2cos t, y = 2sin t, and z = 2, 0 ≤ t ≤ 2π. So Author and Simplify



7. Both these circles lie on the surface x2 + y2 = 4. The union of all the circles with parametric equations

x = 2cos t, y = 2sin t, z = c (where c is any real number) make up the cylinder x2 + y2 = 4. To

generate several of these circles simultaneously, we can use Derive’s VECTOR command. For example,

to generate all the circles from c = −4 to c = 4 in increments of .5, Author and Simplify

VECTOR (ISOMETRIC[2 cos t, 2sin t, c], c, −4, 4, .5)

then go to window #3 and Plot from t = 0 to t = 2π <Ctrl-Enter>.

8. Plotting a lot of space curves that lie on the surface helps us to visualize the surface. You will be using

this technique throughout this lab.




257

10. We use the term cylinder to denote not just a right circular cylinder, but any surface for which there is

a plane such that all cross-sections by planes parallel to that plane are the same. For the right circular

cylinder x2 + y2 = 4, all cross-sections obtained by slicing the cylinder with a plane parallel to the xy

plane are circles of radius 2.

11. Any surface whose equation involves only two of the variables x, y, and z will be a cylinder. Consider,

for example, the surface z = cos y (which does not involve x).

12. Delete All graphs from windows #2 and #3, then if necessary, “unhide” the axes in window #2 (choose

Options Color Axes to change the axes color to 15) and “hide” the axes in window #3 (choose Options

Color Axes to change the axes color to 0). Then redraw the xyz axes by rePlotting AXES in window

#3 from t = 0 to t = 20 <Ctrl-Enter>.


14. To plot the space curve z = cos y, x = 0, that lies on the surface z = cos y (x arbitrary), Author and

Simplify

ISOMETRIC[0, y, cos y]

then go to window #3 and Plot from y = −10 to y = 10.

15. To plot the space curve z = cos y, x = −2, which also lies on the surface z = cos y (x arbitrary),

Author and Simplify

ISOMETRIC[−2, y, cos y]

then go to window #3 and Plot from y = −10 to y = 10.

16. Finally, we can generate the space curves z = cos y, x = c, for several values of c at once by Authoring

and Simplifying

VECTOR (ISOMETRIC[c,y, cos y], c, −4, 4, .5)




then Plotting in window #3 from y = −10 to y = 10 <Ctrl-Enter>.


18. Delete All graphs from windows #2 and #3, and reset the axes in both windows as necessary and set

both scales in window #3 to 2.

19. We now shift our attention to quadric surfaces, which are surfaces whose equations are of the form:

Ax2 + By2 + C z2 + Dx + Ey + F z + G = 0. In this lab, we concentrate on quadric surfaces with

equations of the form:

±x2

a2 ± y2

b2 ± z 2

c2 = 1

20. Consider the surface defined by the equation

x2

4 +

y2

9 + z2 = 1

21. The xy trace of the surface is the intersection of the surface with the xy plane. Since on the xy plane,

z = 0, the equation of the xy trace is x

2

4 + y2

9 + 02 = 1 or x

2

4 + y2

9 = 1, which is the equation of an

ellipse. Similarly, the xz trace and the yz trace are obtained by intersecting the surface with the planes

y = 0 and x = 0, respectively, to obtain the ellipses x2

4 + z2 = 1 and y2

9 + z2 = 1.


23. To begin to “build” the surface in window #3, we need to plot the space curves corresponding to

these cross-sections. Again, whenever we plot space curves in a 2D-plot window, we need to use the

ISOMETRIC command. For example, the space curve corresponding to the xy trace has parametric

equations x = 2cos t, y = 3sin t, z = 0, so Author and Simplify





259

24. Similarly, the space curve corresponding to the yz trace has parametric equations x = 0, y = 3cos t,

and z = sin t, so Author and Simplify

ISOMETRIC[0, 3cos t, sin t]

then go to window #3 and Plot from t = 0 to t = 2π. Also, obtain the parametric equations for the

space curve corresponding to the xz trace and Plot in window #3 as before.

25. Finally, we pick a coordinate plane and try to plot the space curves that are the intersection of the given

surface with planes parallel to the chosen coordinate plane. If we pick the xz plane, then planes parallel

to this one are of the form y = c, so that the cross-sections have equations of the form x2

4 + c2

9 + z2 = 1

or x2

4 + z2 = 1 − c2

9 .


27. The surface you just graphed is called an ellipsoid. Any surface whose equation is of the form x2

a2 +

y2

b2 + z2

c2 = 1 is an ellipsoid and has this basic shape.

28. Again Delete All graphs and reset axes as necessary. Also, set both scales in window #3 to 6.

29. The next surface we will look at is

x2

4 +

y2

9 − z2

9 = 1


31. Again, to plot these traces as space curves, we need to come up with parametric equations. For

example, the xz trace is the hyperbola x2

4 − z2

9 = 1. Solving for z , we obtain

z = ±3

x2

4 − 1

so parametric equations for the top part of the hyperbola are given by x = t, y = 0, and z = 3

t2

4 − 1.

So Author and Simplify

ISOMETRIC

t, 0, 3

t2/4 − 1




then go to window #3 and Plot from t = −10 to t = 10. Then to obtain the graph of the bottom part

of this hyperbola, Author and Simplify

ISOMETRIC

t, 0, −3

t2/4 − 1

then go to window #3 and Plot from t = −10 to t = 10.

32. Use this technique to find parametric equations for the yz trace (which is also a hyperbola) and graph

it in window #3.

33. The xy trace is an ellipse so you should be able to find parametric equations for it using the techniques

used for graphing the ellipsoid. Again, graph it in window #3.


35. Finally, we pick one of the coordinate planes and try to graph the cross-sections along planes parallel

to the chosen coordinate plane. In this case, it’s convenient to choose the xy plane, because the cross-

sections along the planes z = c are ellipses (as opposed to cross-sections along planes y = c or x = c

which would be hyperbolas and so harder to graph). The equation of the cross-section along the plane

z = c has equation

x2

4 +

y2

9 = 1 +

c2

9


37. The surface you just graphed is called a hyperboloid of one sheet. Any surface whose equation is of

the form x2

a2 + y2

b2 − z2

c2 = 1 is a hyperboloid of one sheet and has this basic shape.

38. Delete All graphs and reset the axes in windows #2 and #3 as necessary.

39. Consider now the surface given by the equation

−x2

4 +

y2

9 − z2 = 1

which is called a hyperboloid of two sheets.



261


41. Use the parametric equations found in Worksheet Question #10 to plot these traces as space curves

by Authoring and Simplifying

ISOMETRIC[x, y, z]

where you’ve replaced x, y, and z by the parametric equations, then going to window #3 and Plotting

from t = −10 to t = 10.

42. Finally, we pick one of the coordinate planes and try to graph the cross-sections along planes parallel

to the chosen coordinate plane. In this case, it’s convenient to choose the xz plane, because the cross-

sections along the planes y = c are ellipses (as opposed to cross-sections along planes x = c or z = c

which would be hyperbolas and so harder to graph). The equation of the cross-section along the plane

y = c has equation

x2

4 + z2 =

c2

9 − 1




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Cylinders and Quadric Surfaces

NAME: DATE:


1. The easiest way to graph the circle x2 +y2 = 4 is to use the parametric equations x = 2 cos t, y = 2 sin t,

0 ≤ t ≤ 2π. So Author [2 cos t, 2sin t] and Plot in Window #2 from t = 0 to t = 2π. Copy the graph

below:

263



264 Worksheet # 22. CYLINDERS AND QUADRIC SURFACES

2. Use the graphs of the space curves that you just plotted to draw a picture of the cylinder with equation

x2 + y2 = 4.

6z

HHHHHHHHHj

x y

3. Author and Plot (in window #2) z = cos y, then copy the graph below.



Name: 265

4. Use the graphs of the space curves that you just plotted to draw a picture of the cylinder with equation

z = cos y.

6z

HHHHHHHHHj

x y

5. Recall that parametric equations for the ellipse x2

a2 + y2

b2 = 1 are given by x = a cos t, y = b sin t,

0 ≤ t ≤ 2π. So for example, to obtain the graph of the xy trace of the given surface (which has

equation x2

4 + y2

9 = 1), Author and Plot

[2 cos t, 3sin t]

from t = 0 to t = 2π. Obtain graphs of each trace (in window #2), then copy the graphs below:

yz tracexz tracexy trace

z

y

zy

xx

6

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?

66

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6. (a) Show that x2

4 + z2 = 1 − c2

9 (where c is a fixed constant) is an equation of an ellipse by writing

in the standard form x2

a2 + z2

b2 = 1 and identifying a and b (which will depend on c):

a =

b =

(b) The parametric equations for the ellipse x2

a2 + z2

b2 = 1, y = c, are given by x = a cos t, y = c,

and z = b sin t. Use this and your answer to (a) to write parametric equations for the ellipse

x2

4 + z2 = 1 − c2

9 , y = c.

x =

y =

z =

(c) To Plot these elliptical space curves, Author

ISOMETRIC[x, y, z]

where you have replaced x, y, and z by the parametric equations found in (b). Then choose

Manage Substitute to replace c by 1, say, then Simplify and Plot in window #3 from t = 0 to

t = 2π to obtain a plot of the cross-section of the surface in the plane y = 1. Finally, to obtain

several of these cross-sectional curves at once, Author and Simplify

VECTOR (ISOMETRIC[x, y, z], c, −3, 3, .5)

(again replacing x, y, and z by the parametric equations found in (b)) and Plot in window #3

from t = 0 to t = 2π <Ctrl-Enter>. You should now have a good idea of what the graph of the

surface x2

4 + y2

9 + z2 = 1 looks like, so graph it below:



Name: 267

6z

HHHHHHHHHj

x y

7. Graph the xy, yz, and xz traces of the surface x2

4 + y2

9 − z2

9 = 1 below (they are all familiar conic

sections):

yz tracexz tracexy trace

z

y

zy

xx

6

?

--

?

66

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8. Write down parametric equations for the space curves corresponding to each of the traces of the surface

x2

4 + y2

9 − z2

9 = 1:




xy trace: x = xz trace: x = t yz trace: x = 0

y = y = 0 y =

z = 0 z = z = t

9. (a) Show that x2

4 + y2

9 = 1 + c2

9 (where c is a fixed constant) is an equation of an ellipse by writing

in the standard form x2

a2 + y2


a =

b =


a2 + y2

b2 = 1, z = c, are given by x = a cos t, y = b sin t, and

z = c. Use this and your answer to (a) to write parametric equations for the ellipse x2

4 + y2

9 = 1+c2,

z = c.

x =

y =

z =


ISOMETRIC[x, y, z]



t = 2π. Finally, to obtain several of these cross-sectional curves at once, Author and Simplify

VECTOR (ISOMETRIC[x, y, z], c, −5, 5, .5)



Name: 269



surface x2

4 + y2

9 − z2

9 = 1 looks like, so graph it below:

6z

HHH

H H HHHHj

x y

10. (a) Graph the xy and yz traces of the surface −x2

4 + y2

9 − z2 = 1 below:

xy trace yz trace

-

?

6

-

?

6

x

y

y

z

(b) Find parametric equations for the space curves corresponding to the yz and xy traces.

yz trace: x = 0 xy trace: x = t

y = y =

z = t z = 0




(c) Explain why there is no xz trace.

11. (a) Show that x2

4 + z2 = c2

9 − 1 (where c is a fixed constant) is an equation of an ellipse (for |c| > 3)

by writing in the standard form x2

a2 + z2


a =

b =


a2 + z2

b2 = 1, y = c, are given by x = a cos t, y = c,

and z = b sin t. Use this and your answer to (a) to write parametric equations for the ellipse

x2

4 + z2 = c2

9 − 1, y = c.

x =

y =

z =


ISOMETRIC[x, y, z]



t = 2π. Finally, to obtain several of these cross-sectional curves at once, Author and Simplify

VECTOR (ISOMETRIC[x, y, z], c, −10, −3, .5)



Name: 271

and

VECTOR (ISOMETRIC[x, y, z], c, 3, 10, .5)



surface −x2

4 + y2

9 − z2 = 1 looks like, so graph it below:

6z

HHHHHHHHHj

x y



Laboratory # 23

Vector-Valued Functions

Teaches:

• plane and space curves

• tangent vectors

• moving trihedral

1. Transfer Load the Utility file LABS.MTH. Choose Plot Beside <Enter> to open a graphics window;

then choose Options State Rectangular Connected Small <Enter>; set both the x and y scales to 3.

2. Consider a curve C in the xy-plane parametrized by the parametric equations

x = f (t), y = g(t), a ≤ t ≤ b

It is convenient to define a vector r with the funtions f and g as components:

r(t) = f (t) i + g(t) j

We call r a vector-valued function.

272



273

3. We graph a vector-valued function exactly the way that we graph parametric equations; for example

to graph the vector-valued function r(t) = 2 sin

π2

t

i + 3 cos

π2

t

j, 0 ≤ t ≤ 2 (i.e. the plane curve

parametrized by x = 2 sinπ2

t

and y = 3 cosπ2

t

), Author

F (t) := 2 sin

πt

2

G(t) := 3 cos

πt

2

then Author and Plot

[F (t), G(t)]

and enter 0 and 2 for the min and max values of t.

4. For each value of t, r(t) is a position vector, so you can envision the curve C being traced out by the

moving arrowhead of r(t). To obtain a visual representation of this, Author, Simplify, and Plot

VECTOR (V 0(F (t), G(t)), t, 0, 2, 0.25)

What you will obtain is a graph of the position vectors r(0), r(.25), r(.5), r(.75), . . .r(2).


6. Consider now the derivative of r(t) = f (t) i + g(t) j which is defined to be r (t) = f (t) i + g(t) j

(assuming, of course, that f and g are both differentiable). If you highlight the expression [ F (t), G(t)]

and choose Calculus Differentiate <Enter><Enter><Enter>, you will obtain r(t) = [F (t), G(t)] =

[π cos(πt/2), −3π2

sin(πt/2)].

7. Assign the expressions for F (t) and G(t) to F 1(t) and G1(t), respectively (i.e. Author F 1(t) :=

π cos(πt/2) and G1(t) := − 3π2 sin(πt/2); you can save yourself some typing by using F3 to bring

down highlighted expressions to the Author line.)



274 LABORATORY # 23. VECTOR-VALUED FUNCTIONS

8. What information does r (t) give us about r(t)? Use Manage Substitute to plug in t = 0, then Simplify

to obtain r (0) =< π, 0 >. To graph this position vector, Author and Plot

V 0(π, 0)

Now move this vector so that its initial point is (F (0), G(0)) = (0, 3) by Authoring and Plotting

V 1(π, 0, 0, 3)

9. To graph the vectors r (t) =< F 1(t), G1(t) >, t = 0, .25, .5, . . . , 2 so that the initial point is at the

terminal point of r(t) = (F (t), G(t)), Author, Simplify, and Plot

VECTOR(V 1(F 1(t), G1(t), F (t), G(t)), t, 0, 2, .25)


11. We can also talk about space curves which can be parametrized by three equations:

x = f (t), y = g(t), z = h(t), a

≤t

≤b

Correspondingly a vector-valued function is given by

r(t) = f (t) i + g(t) j + h(t) k

Delete All graphs, then choose Options Color Plot to change the axes and cross colors both to 0. Now

Author and Plot

AXES

from t = 0 to t = 10 <Ctrl–Enter>. Now you have a 3D coordinate system. To plot the space curve

given by the vector-valued function

r(t) = sin t i + cos t j + t

4 k



275

Author, Simplify, and Plot

ISOMETRIC [sint, cos t,t/4]

enter −15 and 15 for the min and max values of the parameter t. Note that this space curve lies on

the cylinder x2 + y2 = 1.

12. We will use the function V 3 to graph vectors in three dimensions; so to draw the 3D vector < x, y,z >

so that its initial point is at (a,b,c), you Author and Plot V 3(x, y, z, a, b, c). For example, to draw the

vector < 0, 1, 1 > so that its initial point is (0, 0, 2) you Author and Plot V 3(0, 1, 1, 0, 0, 2).

13. Again, we want to plot some tangent vectors r (t) = cos t i − sin t j + 14

k. Since we want the initial

point to be the terminal point of r(t) = (sin t, cos t,t/4), Author, Simplify and Plot

VECTOR

V 3

cos t, − sin t,

1

4, sin t, cos t,

t

4

, t, −10, 10, 5

to obtain graphs of the vectors r (t) for t = −10, −5, 0, 5, 10.


15. Let’s now examine the significance of r (t). For our current example,

r (t) = − sin t i − cos t j

16. To plot some of these vectors (again with initial point r(t)), Author, Simplify, and Plot

VECTOR

(V 3

− sin t, − cos t, 0, sin t, cos t,

t

4

, t, −10, 10, 5


18. The third important vector which completes the moving trihedral of vectors is obtained by crossing the

normal and tangent vectors. The corresponding unit vector is called the binormal .




276 LABORATORY # 23. VECTOR-VALUED FUNCTIONS

20. Finally, graph some binormal vectors by Authoring, Simplifying, and Plotting

VECTOR(V 3(x,y,z, sin t, cos t, t

4), t, −10, 10, 5)

replacing x, y, and z with the components of r × r found in Worksheet Question #6.

21. Delete All graphs and rePlot AXES from t = 0 to t = 10 <Ctrl-Enter>.




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Vector-Valued Functions

NAME: DATE:


1. Copy the graph of the half-ellipse traced out by the vector-valued function

r(t) = 2sinπ

2t

i + 3 cosπ

2t

j, 0 ≤ t ≤ 2

as well as the vectors r(t), t = 0, .25, .5, . . . , 2.

6

-6

?

2 4−4 −2

2

4

−4

−2

2. Copy the graphs of the vectors r (t), t = 0, .25, .5, . . . , 2 onto the graph in Question #1. What can

277



278 Worksheet # 23. VECTOR-VALUED FUNCTIONS

you say about the direction of each vector r (t)?

3. Plot the vector-valued function s(t) = 2 sin(πt) i + 3 cos(πt) j, 0 ≤ t ≤ 1 (i.e. Author

F (t) := 2 sin(πt)

G(t) := 3 cos(πt)

then Author and Plot [F (t), G(t)] entering 0 and 1 for the min and max values of t).

(a) What is the relationship between the curve traced out by s(t) and the curve traced out by r(t) in

Question #1, if we think of the parameter t as time?

(b) To graph the vectors s (t), again using the terminal point of s(t) for the initial point, for

t = 0, .25, .5, .75 and 1, first have Derive evaluate the derivatives of F (t) and G(t) by high-

lighting [F (t), G(t)] and choosing Calculus Differentiate, then Simplify to obtain an expression

for [F (t), G(t)]; then assign these expressions to F 1(t) and G1(t), respectively, as before (i.e.

Author F 1(t) := . . . and G1(t) := . . .. Finally, Author, Simplify, and Plot

VECTOR(V 1(F 1(t), G1(t), F (t), G(t)), t, 0, 1, .25)

How do these vectors compare to the ones you graphed in Question #1, i.e. how are they different

and how are they similar?



Name: 279

4. Let r(t) = sin t i + cos t j + t

4

k so that r (t) = cos t i−

sin t j + 1

4

k.

(a) What is the length |r (t)|, for any t?

(b) If we think of r(t) as tracing out the path of a fly, then |r (t)| gives us the speed of the fly at time

t. What can you say about the speed of the fly?

5. Do the vectors r (t) appear to be orthogonal to curve traced out by r(t) (i.e. perpendicular to the

tangent vectors r (t))? Verify that they are by evaluating the dot product r (t) · r (t) for any t.

6. Evaluate the cross product of r (t) and r (t). [Hint: To cross the vectors < a, b,c > and < d, e, f >

in Derive, use CROSS([a,b,c], [d,e,f ])]




7. Consider now the space curve defined by the vector function

r(t) = cos t i + 2 j + sin t k

(a) Evaluate r (t) and |r (t)|.

(b) Plot the space curve traced out by r(t) by Authoring, Simplifying and Plotting ISOMETRIC[cost, 2, sin t]

from t = 0 to t = 2π. Also plot some of the vectors r (t) by Authoring, Simplifying and Plotting

VECTOR(V 3(x,y,z, cos t, 2, sin t), t, 0, 2π,π/2)

replacing x, y, and z with the components of r (t). Copy these graphs below:

6z

HHHHHHHHHj

x y

(c) Evaluate r (t) and plot some of these vectors by Authoring, Simplifying, and Plotting the above

VECTOR equation with x, y, and z replaced by the components of r (t). Do these vectors appear



Name: 281

to be orthogonal to the curve? Show that they are by computing r (t) · r (t).

(d) Evaluate the binormal vector r (t) × r (t).

8. Our final example is

r(t) = t i + t2 j + sin(2πt) k

Delete All graphs and rePlot AXES from 0 to 10.

(a) Evaluate r (t) and |r (t)|.

(b) Plot the space curve traced out by r(t) by Authoring, Simplifying, and Plotting ISOMETRIC[t, t2, sin(2πt)]

from t = −3 to t = 3. Also Plot r (−3/4) by Authoring

V 3(x, y, z, t , t2, sin(2πt))

where you’ve replaced x, y, and z by the components of r (t), then Manage Substituting t = −3/4,

Simplifying, and Plotting. Copy the graphs below:




P P P P P P P P P P P P PPq

6

(c) Evaluate r (t) and Plot r (−3/4) by Authoring the above expression with x, y, and z replaced

by the components of r (t), then Manage Substituting t = −3/4 and Plotting. Do they appear

to be orthogonal to the curve? Verify that they are not by evaluating r (−3/4) · r (−3/4).

(d) Find a unit vector u that’s tangent to the curve traced out by r(t) at t = −3/4. [Hint: since

r (−3/4) is tangent to the curve, let u = r(−3/4)|r(−3/4)| . Let Derive do the algebra for you by Authoring

R(t) := [t, t2, sin(2πt)]

then Authoring and Simplifying

U (t) := DIF(R(t), t, 1)

|DIF(R(t), t, 1)

|and finally Manage Substituting t = −3/4 and Simplifying again.



Name: 283

(e) Now find N (t) = U (t) = the derivative of U (t).[Hint: you can use Calculus Differentiate on

U (t).]. Evaluate N (−3/4).

.

(f) Plot N (−3/4) by Authoring

V 3(x, y, z, t , t2, sin(2πt))

with x, y, and z replaced by the components of N (t), then Manage Substituting t = −3/4 and

Plotting. Does this vector appear to be orthogonal to the curve? Verify that it is by evaluating

r (−3/4) · N (−3/4).



Laboratory # 24

Curvature

Teaches:

• circles of curvature

1. The curvature of a plane or space curve is a measure of how much a curve bends or curves. For example,

the following curve has larger curvature at the point P than at the point Q.

•

•

Q

P

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..................

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2. If a plane curve is given by parametric equations x = f (t) and y = g(t), then the curvature at the

284



285

point (f (t), g(t)) can be obtained by Authoring and approXimating CURV(f (t), g(t)).

3. Open a graphics window by choosing Plot Beside <Enter>.


5. Along with the notion of curvature is the circle of curvature, which is the circle that best fits the curve

at a given point, as illustrated by the following picture:

•

•

Q

P

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As you can see, the bigger the curvature κ, the smaller will be the circle of curvature. In fact, the

radius of the circle of curvature (sometimes called the radius of curvature) is given by the reciprocal

of the curvature. The tangent to the curve at the given point is perpendicular to the radius of the

circle, so the center of the circle can be found by measuring a distance 1/κ along the line perpendicular

to the tangent in the direction of the concave side of the curve:



286 LABORATORY # 24. CURVATURE

•Q = (x, y)

•C = (a, b)

r = 1κ

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6. In Worksheet Questions #2–#4, you will be finding circles of curvature and then graphing them.

The easiest way to graph a circle with center (a, b) and radius r is to use the parametric equations

x = a + r cos t, y = b + r sin t, 0 ≤ t ≤ 2π, i.e. Author

[a + r cos t, b + r sin t]

then use Manage Substitute to plug in the values that you have found for the radius r and the center

(a, b); then Plot from t = 0 to t = 2π.




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Curvature

NAME: DATE:


1. Consider the ellipse parametrized by

x = 2 sin t, y = 3 cos t, 0 ≤ t ≤ 2π

(a) Plot the ellipse (i.e. Author and Plot [2 sint, 3cos t] and enter 0 and 2π for the min and max

values of the parameter t.) From the graph, would you say that the curvature is larger at (0 , 3)

or at (2, 0)?

(b) Find a formula for the curvature (as a function of t) by Authoring and SimplifyingCURV(2 sin t, 3cos t).

(c) Find the curvature of the ellipse at the point (0, 3) (i.e. when t = 0).[Hint: you may want to use

Manage Substitute on your answer to (b) to plug in t = 0, then Simplify.]

287



288 Worksheet # 24. CURVATURE

(d) Find the curvature of the ellipse at the point (2, 0) (i.e. when t = π2

).

(e) Are your answers to (c) and (d) consistent with your answer to (a)? Explain.

2. Consider again the ellipse parametrized by x = 2 sin t and y = 3 cos t, 0 ≤ t ≤ 2π.

(a) Graph the ellipse below; label the points (0, 3) and (2, 0).

6

?

-2 4 6−6 −4 −2

2

4

6

−6

−4

−2

(b) Find the radius r of the circle of curvature at (0, 3). [Hint: remember that the radius is the

reciprocal of the curvature, and you found the curvature at (0, 3) in the previous question.]

(c) Find the center (a, b) of the circle of curvature at (0, 3). [Hint: The tangent line to the ellipse at

(0, 3) is horizontal, and the center of the circle must lie on the line perpendicular to the tangent



Name: 289

at (0, 3), i.e. it must lie on the y axis, a distance r down from (0, 3), where r is the radius that

you found above.]

Now that you have the radius r and the center (a, b) of the circle, you should be able to graph

it using the parametric equations x = a + r cos t and y = b + r sin t, 0 ≤ t ≤ 2π as explained

previously. Copy the graph of the circle onto your graph in (a).

(d) Find the radius r and the center (a, b) of the circle of curvature of the ellipse at (0, 2). Plot the

circle and copy it onto the graph in (a). [Hint: the tangent line at (0 , 2) is vertical, so the center

of the circle should lie on the x axis.]

3. Consider the cardioid given by the polar equation r = 1−sin θ. Recall that the relationships x = r cos θ

and y = r sin θ between polar and rectangular coordinates allow us to easily obtain the parametric

equations x = (1 − sin θ)cos θ and y = (1 − sin θ)sin θ. We want to continue to use the parameter t

instead of θ, so Delete All graphs, then Author and Plot

[(1 − sin t)cos t, (1 − sin t)sin t]

entering 0 and 2π for the min and max values of the parameter t.

(a) Copy the graph of the cardioid below:




6

?

-

1 2−2

1

2

−2

−1

−1

(b) Find an expression for the curvature κ(t) of the cardioid as a function of t. [Hint: First choose

Manage Trigonometry Collect <Enter>; then Author and Simplify CURV((1 − sin t)cos t, (1 −

sin t)sin t).]

(c) Find the coordinates of the point on the cardioid corresponding to t = π6 . Label the point on

your graph in (a).

(d) Find the curvature of the cardioid at this point.

(e) Find the radius r of the circle of curvature at this point.



Name: 291

(f) Find the center (a, b) of the circle of curvature at this point. Then use the parametric equations

x = a + r cos t, y = b + r sin t, 0 ≤ t ≤ 2π to obtain a graph of the circle of curvature. Copy the

graph of the circle onto your graph in (a). [Hint: the tangent line at this point is horizontal.]

(g) Find the coordinates of the point on the cardioid corresponding to t = 3π2

. Label this point on

your graph in (a). Then find the radius and center of the circle of curvature at this point. Obtain

a graph of the circle and copy it onto the graph in (a). [Hint: the tangent line at this point is

horizontal.]

4. Finally consider the limacon with polar equation r = 1+ 2 sin θ. Parametric equations for this limacon

are given by

x = (1 + 2sin t)cos t, y = (1 + 2 sint)sin t, 0 ≤ t ≤ 2π

(a) Graph the limacon below; then find the coordinates of the point corresponding to t = 3π2 and

label it on the graph.




1 2

3

4

-

−2 −1

1

2

?

6

(b) Find an expression for the curvature of the limacon as a function of t.

(c) Find the radius r and the center (a, b) of the circle of curvature at t = 3π2 . Obtain a graph of the

circle and copy it onto your graph in (a).[Hint: the tangent line at this point is horizontal.]



Laboratory # 25

Functions of 2 Variables

Teaches:

• using level curves to graph functions of two variables

• finding relative extrema of functions of two variables

1. The graph of a function z = f (x, y) is a surface in 3-space. The graph is obtained by plotting all points

(x, y, f (x, y)) where (x, y) is in the domain of f :

-

6

=

domain of z=f (x,y)

r

r

(x,y)

(x,y,f (x,y))

f (x,y)

x

z

y

293



294 LABORATORY # 25. FUNCTIONS OF 2 VARIABLES

2. In trying to obtain a graph of a function z = f (x, y), it is often helpful to first graph the level curves

of f . These are the curves defined by f (x, y) = c, for suitable c. These curves tell us what the

cross-section of the surface z = f (x, y) looks like when we slice the surface with the plane z = c:

-

planez=c

surface

z=f (x,y)

f (x,y)=cxy

z6

3. Choose Plot Beside <Enter> to open a graphics window, set the x and y scales both to 5, then split

the graphics window horizontally into two graphics windows by choosing Window Split Horizontal

<Enter>. We will use Window #2 to graph the (2 dimensional) level curves. In Window #3 we will

put the level curves together to “build” the (3 dimensional) surface.


5. Next, we need to set up window #3 as follows:

• First, Transfer Load the Utility file LABS.MTH.

• Next, go to window #3 ( F1 takes you to the next window), choose Options Color Plot and

change the axes and cross colors both to 0.


• Author AXES, then go to window #3 and Plot from 0 to 20 <Ctrl-Enter>.

6. Now we can start “building” our surface. We would like to graph these level curves in window #2 at

heights z = 1, z = 4, and z = 9, respectively in window #3. The first level curve is a circle centered



295

at the origin with radius 1, so it has parametric equations x = cos t, y = sin t, z = 1, 0 ≤ t ≤ 2π. To

graph, Author

ISOMETRIC[cos t, sin t, 1]


Remark: Note that Window #3 is really a 2d-plot window, but we are using it to draw three-

dimensional objects; the function ISOMETRIC calculates the 2D isometric projection of the 3D space

point (x, y, z) so that we can graph it in this two-dimensional window.

7. The second level curve that you graphed, corresponding to c = 4, is a circle centered at the origin with

radius 2; its parametric equations are x = 2 cos t, y = 2 sin t, z = 4. So Author



8. More generally, the level curve x2

+ y2

= c is a circle centered at the origin with radius √ c; to graph

it at height z = c in window #3, we use the parametric equations x =√

c cos t, y =√

c sin t, z = c. To

graph all the level curves from c = 0 to c = 5 in increments of 0.2, Author and Simplify the following:

VECTOR

ISOMETRIC[√

c cos t,√

c sin t, c], c, 0, 5, .2

go to window #3 and Plot from t = 0 to t = 2π; press <Ctrl- Enter> to accept this interval for all the

level curves simultaneously.

9. Answer Worksheet Questions #2 –#4.

10. In all the examples that we’ve looked at so far, the level curves f (x, y) = c where pretty straight

forward to graph since they turned out to be nice “drawable” curves like circles or ellipses; once we

knew what the level curves looked like, putting them together to “build” the surface and finding the




relative extrema was easy. Often, however, finding the level curves z = c is not so easy. In this case, we

may want to look instead at the curves x = c or y = c on the surface. The curves y = c, for example,

give us the cross-section obtained by slicing the surface with the plane y = c.

11. Consider now the surface given by the equation z = −5cos

3πx10

sin

πy5

. Author

F (x, y) := −5cos

3πx

10

sin

πy

5

12. Derive actually has 3d graphing capabilities which we haven’t explored. Choose Plot Under <Return>

Plot. To improve the detail of the graph, choose Grid and change the number of grid lines for both x

and y to 30; then Plot again.

13. As you can see, the graph has three “mountains” above the xy plane; we want to find the coordinates

of the “peak” of one of these mountains, namely the one in the upper-left corner of the screen.

14. Notice that the curves z = c are not easy to graph; for example we would need to solve −5cos

3πx10

sin

πy5

=

c for y to get it into the form y = f (x) (Can you do that? Derive can, but it’s still pretty messy!). On the

other hand the curves traced by the planes y = c have equations of the form z = −5cos 3πx

10 sin πc5 =

C 1 cos(C 2x) for some constants C 1 and C 2, and are generally easy to graph (although we will let Derive

do this for us).

15. We will use window #3 (in the upper-right corner) to graph the level curves y = c (where we will think

of the horizontal axis as the x axis, and the vertical axis as the z axis). Window #4 (in the lower-right

corner) will be used to “build” the 3d surface. Reset these two window as before.

16. Suppose we want to look at the level curve traced out by the plane y = −4. On this plane, z =

F (x, y) = F (x, −4), so to graph this curve on the xz coordinate axes in window #3, Author

[x, F (x, −4)]



297

go to window #3, choose Options Color Auto Yes, and Plot from x = −5 to x = 5. To “place” this

curve on the plane y = −4 in window #4, Author

ISOMETRIC[x, −4, F (x, −4)]

go to window #4 and Plot from x = −5 to x = 5.

17. Repeat for y = −3, i.e. Author and Plot [x, F (x, −3)] in window #3 and ISOMETRIC[x, −3, F (x, −3)]

in window #4.

18. Repeat for y = −2, i.e. Author and Plot [x, F (x, −2)] in window #3 and ISOMETRIC[x, −2, F (x, −2)]

in window #4.


20. Go to window #4, choose Options Color Auto No, then choose Options Color Plot and change the

Next Plot color to your favorite color. Then “build” the graph of z = f (x, y) as follows: Author and

Simplify

VECTOR (ISOMETRIC[x,c,F (x, c)], c, −5, 5, .2)

go to window #4 and Plot from x = −5 to x = 5. You will obtain a plot of the level curves y = c from

c = −5 to c = 5 in increments of .2.

21. Let’s focus our attention on the part of the graph corresponding to −4 ≤ y ≤ −2, since the peak that

we are looking for should be there. Author

VECTOR([x, F (x, c)], c, −4, −2, .2)

go to window #3 and Plot from x = −5 to x = 5. Then Author and Simplify

VECTOR (ISOMETRIC[x, y, F (x, y)], y, −4, −2, .2)




go to window #4, change the Next Plot color to your second favorite color, and Plot from x = −5 to

x = 5.




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Functions of 2 Variables

NAME: DATE:


1. Let f (x, y) = x2 + y2, so that the level curves are x2 + y2 = c.

(a) Describe the level curves corresponding to c = 1 and c = 4, and c = 9.

(b) The easiest way in Derive to graph a circle centered at the origin with radius r is to use the

parametric equations x = r cos t, y = r sin t, 0 ≤ t ≤ 2π (i.e. Author [r cos t, r sin t], replacing r

with the desired radius, then Plot from t = 0 to t = 2π). Use Derive to obtain graphs in Window

#2 of the level curves corresponding to c = 1, c = 4, and c = 9. Then copy the graphs below.

6

-

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5

5

−5

−5

299



300 Worksheet # 25. FUNCTIONS OF 2 VARIABLES

(c) Are there any level curves corresponding to c < 0? Explain.

2. (a) On the axes below, sketch the surface z = x2 + y2 .

6z

HHHHHHHHHj

x y

(b) Use your sketch to identify the coordinates of all relative extreme points (i.e peaks and valleys)

of z = f (x, y); identify each point as a relative maximum or a relative minimum and indicate

whether it is an absolute maximum or minimum point.

Reset window #3 by Deleting All graphs, then replotting AXES from 0 to 20 <Ctrl-Enter>. Then go

to window #2, Delete All graphs, and choose Options Color Plot to change the axes color to 15.

3. Next consider f (x, y) = 1 − x2

4 − y2.

(a) Show that the level curves f (x, y) = c are ellipses (recall that the equation of an ellipse is x2

a2 + y2

b2 =



Name: 301

1, a,b > 0) and identify a and b (as functions of c).

(b) For what values of c are there no level curves?

(c) The easiest way to use Derive to graph the ellipse x2

a2 + y2

b2 = 1 is to use the parametric equations

x = a cos t, y = b sin t, 0 ≤ t ≤ 2π (i.e. Author [a cos t, b sin t] where you have replaced a and b by

the appropriate numbers, then Plot from t = 0 to t = 2π). In window #2, obtain graphs of the

level curves corresponding to c = 3/4, 0, −3, −8 (recall that formulas for a and b were calculated

in (a)). Copy the graphs below.

6

-

?

5

5

−5

−5

(d) Now plot these level curves at the appropriate level z = c in Window # 3 by Authoring

ISOMETRIC[a cos t, b sin t, c] where you’ve replaced a and b by the expressions involving c in

your answer to (a); then for each value of c above, Manage Substitute it in, go to Window #3,

and Plot from t = 0 to t = 2π. Finally, to obtain graphs of all the level curves f (x, y) = c from

c = −10 to c = 1 in increments of .5, say, Author and Simplify

VECTOR (ISOMETRIC[a cos t, b sin t, c], c, −10, 1, .5)




again replacing a and b by the appropriate expressions, go to Window #3 and Plot from t = 0 to

t = 2π <Ctrl-Enter>. On the axes below, sketch the surface z = 1 − x2

4 − y2.

6z

HHHHHHHH

Hj

x y

(e) Use your sketch to identify the coordinates of all relative extreme points of z = f (x, y) and identify

each point as a relative maximum or a relative minimum and indicate whether it is an absolute

maximum or minimum point.

Reset windows #2 and #3 as before.

4. Consider the function z = 21−x2−y2 .

(a) Show that the level curve 21−x2−y2 = c is a circle by solving for x2 + y2 . What is the radius of the

circle (as a function of c)?



Name: 303

(b) For what values of c are there no level curves?

(c) Graph the level curves for c = −2/3, −1/4, 2, 9/4, 8/3 below.

6

-

?

5

5

−5

−5

(d) What is the limiting value of the radii of the circles 21−x2−y2 = c as c → ∞?

To “build” the part of the surface z = 21−x2−y2 that lies above the xy plane, Author and Simplify

VECTOR (ISOMETRIC[r cos t, r sin t, c], c, 2, 7, .3)

where you’ve replaced r by the expression for the radius that you obtained in (a). then go to

window #3 and Plot from t = 0 to t = 2π <Ctrl-Enter>.

(e) What is the limiting value of the radii of the circles 21

−x2

−y2 = c as c

→ −∞?

To “build” the part of the surface z = 21−x2−y2 that lies below the xy plane, Author and Simplify

VECTOR (ISOMETRIC[r cos t, r sin t, c], c, −7, 0, .3)




where you’ve again replaced r appropriately. Then go to window #3 and Plot from t = 0 to

t = 2π <Ctrl-Enter>.

(f) What is the limiting value of the radii of the circles 21−x2−y2 = c as c → 0−?

To “build” the part of the surface z = 21−x2−y2 that lies below and near the xy plane, Author and

Simplify

VECTOR (ISOMETRIC[r cos t, r sin t, c], c, −1, 0, .1)

again replacing r appropriately, then go to window #3 and Plot from t = 0 to t = 2π <Ctrl-

Enter>.

(g) Sketch the surface z = 21−x2−y2 below:

6z

HHHHHHHHHj

x y



Name: 305

(h) Use your sketch to identify the coordinates of all relative extreme points of z = f (x, y) and identify

each point as a relative maximum or a relative minimum and indicate whether it is an absolute

maximum or minimum point.

5. (a) At what value of x does F (x, −4) = −5cos

3πx10

sin

−4π5

“peak” and what’s the value of

F (x, −4) for this value of x (i.e. what are the (x, z) coordinates of the relative maximum of

your first graph in window #3) ? Give a decimal approximation accurate to 5 places. [Hint: at

what point (x, y) should you evaluate F (x, y)?]

(b) At what value of x does F (x, −3) = −5cos

3πx10

sin

−3π5

peak? What’s the value of F (x, −3)

for this value of x?

(c) At what value of x does F (x, −2) = −5cos

3πx10

sin

−2π5

peak? What’s the value of F (x, −3)

for this value of x?

6. (a) At what value of x do all the curves y = c , −4 ≤ c ≤ −2, (which you plotted in window #3)

appear to peak?




(b) Then how can you evaluate the highest peak of z = F (x, c) for a given value of c, −4 ≤ c ≤ −2?

(c) So if I want the highest of all the peaks, what derivative should I take and set equal to 0? Do

this (use Calculus Differentiate and soLve). What c value (between −4 and −2) corresponds to

the level curve y = c that “peaks” the highest?

(d) What are the coordinates (x, y, z) of this peak? [Hint: to verify your answer, Author and Simplify

ISOMETRIC[x, y, z] (where you’ve replaced [x, y, z] with your point, then in window #4 Move

your cross to these coordinates to see if it coincides with the peak.]



Laboratory # 26

Directional Derivatives

Teaches:

• directional derivatives

• the gradient

1. Set up Derive as follows:

- Choose Window Split Horizontal <Enter>.

- Choose Window Split Vertical <Enter> to split window #1.

- Go to window #3 and choose Window Split Vertical, then type 26 <Enter>.

- Go to window #4 and choose Window Split Vertical <Enter>.

- Go to window #1 and choose Window Designate 3D-plot; choose Grids to change the number of

grid lines in both the x and y directions to 30.

- Window Designate window #2 as a 2D-plot window and set the x and y scales both to 6; Window

Designate window #4 as a 2D-plot window and set the x and y scales both to 6; Window Designate

window #5 as a 2D- plot window and set the x and y scales both to 2.

307



308 LABORATORY # 26. DIRECTIONAL DERIVATIVES

2. Author and Plot

F (x, y) :=

8

1 + x2 + y2

3. Derive’s 3d plotting window is useful for obtaining graphs of functions of 2 variables; however, it is

limited in that we cannot graph more than one function at a time.

4. Go to window #2 and choose Options Color Auto No; then choose Options Color Plot and change the

Next plot color to 15 and the axes color to 0; then go to the algebra window (window #3) and Transfer

Load the Utility file LABS.MTH. Then Author

AXES

and Plot in window #2 from 0 to 10 <Ctrl-Enter>.

Notes:

- We are going to think of window #2 as a 3D window, even though Derive thinks of it as a 2D

window; for this reason we have “hidden” the xy axes and drawn the xyz axes. On the other

hand, windows #4 and #5 are to be treated as ordinary 2D windows so we don’t want to hide

the xy axes in these windows. Unfortunately, each time you change the color options, it affects

all the graphic windows, so it may be necessary to change the color options from time to time to

keep the axes visible in windows #4 and #5 and invisible in window #2.

- In the following instructions, I’ve suggested colors for you to use for the different graphs. You

are not bound to use these particular colors, although you should choose a relatively light color

for the graph of F (x, y) and a relatively bright color for the cross sections so that they will show

up clearly. Since we have turned Color Auto off, the colors of subsequent graphs do not change

automatically, so use Options Color Plot to change the color of the next plot.

5. Choose Options Color Plot to change the Next plot color to 1, Author and Simplify

PLOT3D(F (x, y))



309

and Plot in window #2 from −5 to 5 <Ctrl-Enter>.

6. We are now going to use the graph in window #2 to study the function F (x, y) at the point (1, 0, 4)..

Author

CROSS SEC (F (x, y), 1, 0, 0)

go to window #2, change the Next plot color to 14, and Plot from 0 to 5 <Enter>. The graph that

you will obtain is a cross-section of the surface in the direction of the positive x axis. In other words, if

you started on the surface at the point (1 , 0, F (1, 0)) = (1, 0, 4), and walked in the direction parallel to

the positive x axis while staying on the surface, you would be walking along this curve. More generally,

use CROSS SEC (F (x, y), p , q , θ) to obtain a graph of the curve traced out by starting at the point

( p, q, F ( p, q )) and walking in the direction that makes an angle θ with the positive x axis, as illustrated

by the following picture:

-

6

s

ZZZZ~

( p, q, F ( p, q ))

( p, q, 0)θs

7. So now to see how the surface changes as you walk in the direction of the positive y axis, you would

let θ = π/2, so Author and Plot (in window #2, from 0 to 5)

CROSS SEC (F (x, y), 1, 0, π/2)




8. Author

CROSS SEC (F (x, y), 1, 0, 4π/3)

go to window #2 and Plot from 0 to 5 <Enter>.

9. As you can see, depending on which direction we take, we may be climbing or descending, and the

path might be flat or steep. We will use window #4 to obtain a better view of these cross-sectional

paths. To look at the first path (in the direction of the positive x axis) Author and Simplify

C 1(F (x, y), 1, 0, 0)

then go to window #4 and Plot; ignore the part of the graph that lies to the left of the y axis. Think

of the x axis as denoting time t and the y axis as denoting our height z = F (x, y) as we walk at a

steady rate in the direction of the positive x axis. At time t = 0, we start at height F (1, 0) = 4 and as

we begin to walk we descend down the surface.

10. How steep is our descent as we start out? Well, this can be determined by evaluating the slope of the

curve at time t = 0. The utility function TANGENT(y, t , t0) simplifies to the line tangent to expression

y(t) at t = t0. So Author and Simplify

TANGENT(#n,t, 0)

where you have replaced n by the number of the simplified expression C 1(F (x, y), 1, 0, 0). Plot this

line in window #4 to verify that it’s tangent to the cross-section at t = 0.


12. To plot the cross-section of the surface z = F (x, y) at z = 4 in window #2, Author

ISOMETRIC[cos t, sin t, 4]



311

go to window #2, and Plot from 0 to 2π. Then to plot the level curve in window #5 (i.e. projection

of this cross-section onto the xy plane), Author

[cos t, sin t]

go to window #5, and Plot from 0 to 2π.

13. Finally, I want you to plot the unit vector u that you found in (g) that points in the direction of steepest

climb. You will first graph it in window #2. Since we want the initial point to be (1, 0, 4), Author

V 3(a,b, 0, 1, 0, 4) where you have replaced a and b by the horizontal and vertical components of u that

you found in Worksheet Question #1. Then go to window #2, choose Options State Rectangular

Connected Small <Enter>, then Plot.

14. To plot the projection of this vector onto the xy plane in window #5, Author V 1(a,b, 1, 0) (where

you’ve replaced a and b the horizontal and vertical components of the unit vector u), then go to

window #5, choose Options State Rectangular Connected Small <Enter>, then Plot.


16. To plot the cross-section on the surface z = F (x, y) formed by the plane

z = 4/3 in window #2, Author

ISOMETRIC[√

5cos t,√

5sin t, 4/3]

go to window #2, and Plot from 0 to 2 π. Then to plot the projection of this cross-section onto the xy

plane in window #5, Author

[√

5cos t,√

5sin t]

go to window #5, and Plot from 0 to 2π.




17. Finally, I want you to plot the unit vector u that you found that points in the direction of steepest

climb. You will first graph it in window #2. Since we want the initial point to be (1, 2, 4/3), Author

V 3(a,b, 0, 1, 2, 4/3) where you have. Th replaced a and b by the horizontal and vertical components of

u that you found in Worksheet Question #4. Then Plot in window #2.

18. To plot the projection of this vector onto the xy plane in window #5, Author V 1(a,b, 1, 2) (where

you’ve replaced a and b the horizontal and vertical components of the unit vector u), then Plot in

window #5.


20. We would now like to introduce the gradient vector of a function f (x, y) at a point (a, b). Recall

that f x = ∂f ∂x is the partial derivative of f with respect to x; likewise f y = ∂f

∂y is the partial derivative

of f with respect to y.

Definition 26.1 The vector f x(a, b) i + f y(a, b) j is called the gradient vector of f at (a, b) and is

denoted by ∇f . It is also called “del f ” because of the upside-down delta ∇.

21. For example, if f (x, y) = x2 + y3, then f x(x, y) = 2x and f y(x, y) = 3y2, so ∇f (1, 2) = 2 · 1, 3(2)2 =

2 i + 12 j.




’

&

$

%

’

&

$

%


Directional Derivatives

NAME: DATE:


1. Let F (x, y) = 81+x2+y2 , ( p, q ) = (1, 0).

(a) Author, Simplify,andPlot (in window #4) C 1(F (x, y), p , q , θ) for each of the angles θ = 0, π/2, 4π/3,

to obtain frontal views of the cross-sections to the surface z = F (x, y) in each of these directions.

Copy the graphs below, (for t > 0 only), and label them with the appropriate θ:

-

6

5

5

t

(b) Fill in the following chart:

313



314 Worksheet # 26. DIRECTIONAL DERIVATIVES

θ C 1(F (x, y), 1, 0, θ) tangent line slope

simplified at t = 0 of tangent l ine

0 8t2+2t+2 4 − 4t −4

π/3

2π/3

π

4π/3

5π/3

(c) Which angle θ from the above table represents the direction of the steepest climb of the surface

and which represents the direction of steepest descent?

(d) Plot the 6 points (θ, slope of tangent line) from the above table and label them:

6

?

-

5

5

π 2π

(e) To obtain a complete graph of θ vs the slope of the surface in direction θ, we need to find the



Name: 315

slope for general θ; so Author

C 1(F (x, y), 1, 0, θ)

(θ is typed as alt-h).

i. Author and Simplify TANGENT(#n,t, 0) where you have replaced n by the number of the

last simplified expression.

ii. What is the slope of the above line (as a function of θ)?

(Check your formula by plugging in values of θ from the above table.)

(f) Now Delete All graphs from window #4 (choose Options Color Plot to change the axes color back

to 15) and Plot the slope function obtained above as a function of θ; this graph tells you, for each

θ, the instantaneous rate of change of the height z of the surface in that direction starting from

the point (1, 0, 4). In (d) you graphed six points from this graph; now complete the graph in (d)

by copying this graph onto it (it should go through the points that you have already graphed).

Which angle θ, 0 ≤ θ < 2π, represents the direction of the steepest climb of the surface and which

represents the direction of steepest descent?

(g) Find unit vectors u and v that point in the direction of steepest ascent and steepest descent,




respectively (i.e. calculate cos θ, sin θ for the values of θ that you’ve calculated).

2. Recall that the level curves of a function F (x, y) are the curves F (x, y) = c. Since F (1, 0) = 4, we are

interested now in the level curve F (x, y) = 4 (which necessarily goes through the point (1 , 0)). Show

that this level curve is a circle centered at the origin with radius 1.

3. Copy the graphs of the level curve F (x, y) = 4 and the vector which points in the direction of steepest

ascent below (from window #5):

-

6

?

2−2

2

−2

4. Delete All graphs from window #2; choose Options Color Plot to change the axes color back to 0 and

the Next Plot color back to 1. Also Delete All graphs from window #4, and change the axes color back

to 15. Then replot the original function (i.e. the simplified expression PLOT3D(F (x, y)), in window



Name: 317

#2. Change the Next plot color to 14.

(a) Plot the cross section of the surface in each of the directions 0, π/2, π and 3π/2 starting

at the point (1, 2, F (1, 2)) = (1, 2, 43

) (i.e. for each of these values of θ, Author and Plot

CROSS SEC(F (x, y), 1, 2, θ) from 0 to 5). Among these four directions, which appears to provide

the steepest climb?

(b) Plot these cross sectional curves in window #4, where as you recall the horizontal axis measures

time and the vertical axis the height z (i.e. for each of these values of θ , Author and Simplify

C 1(F (x, y), 1, 2, θ), then go to window #4 and Plot). Copy the graphs below (t ≥ 0 only), labeling

them with the appropriate θ.

-

6

5

5

t

(c) Find the exact initial instantaneous rate of ascent or descent along each of these four directions by

Authoring, Simplifying, and Plotting (in window #4) TANGENT(#n,t, 0) where n is the number

of the expression C 1(F (x, y), 1, 2, θ) (at each value θ) Simplified. Fill in the following table:




θ C 1(F (x, y), 1, 2, θ) tangent line slope

simplified at t = 0 of tangent l ine

0 8t2+2t+6

43 − 4t

9 −49

π/2

π

3π/2

(d) Find a general expression for the tangent line to the cross-section in any direction θ by Authoring

and Simplifying C 1(F (x, y), 1, 2, θ), then Authoring and Simplifying TANGENT(#n,t, 0) where

n is the number of the Simplified C 1 expression.

(e) Find a general expresssion for C (θ) = the slope of the above tangent line in any direction θ .

(f) Delete All graphs in window #4 and plot C (θ) as a function of θ, 0 ≤ θ < 2π. Again remember

that this graph gives you, for each θ, the instantaneous rate of change of the height z of the

surface at (1, 2, 4/3) in that direction. Copy the graph below; label the points corresponding to

θ = 0, π/2, π, 3π/2 whose y coordinates C (θ) you calculated in the table above:



Name: 319

-

6

?

1

−1

π 2π

(g) We are again interested in finding the directions θ of steepest ascent and steepest descent, i.e. the

relative min and max of the above graph. We can do this by taking the derivative with respect

to θ of C (θ) and setting it equal to 0. Do this (using Calculus Differentiate, Simplify, and soLve)

and approXimate the answer. What is this value of θ and does it correspond to the direction of

steepest ascent or descent? Author

CROSS SEC(F (x, y), 1, 2, θ)

for this value of θ and Plot in window #2 from 0 to 5.

(h) The other critical point of C (θ) can be found as follows: in your Algebra window set Options

Precision to Approximate, highlight the simplified derivative of C (θ) (whose other zero you want

to find) and choose soLve; you will need to give Derive an interval in which to start searching for

the root; the graph should help you come up with an appropriate interval. What is this value of




θ and does it correspond to the direction of steepest ascent or descent? Author

CROSS SEC(F (x, y), 1, 2, θ)

for this value of θ and Plot in window #2 from 0 to 5.

(i) Find unit vectors u and v that point in the direction of steepest ascent and steepest descent,

respectively –i.e. calculate cos θ, sin θ for the values of θ that you’ve calculated:

-

6

?

3θ

cos θ

sinθu

(j) Finally, let’s look at the level curve that contains the point (1, 2). Since F (1, 2) = 4/3, this is the

curve F (x, y) = 4/3. Show that this is a circle with radius√

5.

5. (a) Copy the graphs of the level curve F (x, y) = 4/3 and the vector that points in the direction of

steepest ascent below (from window #5):



Name: 321

-

6

?

2−2

2

−2

(b) What do you notice about the relationship of each vector that you’ve graphed to the corresponding

level curve? (e.g. do they appear tangent to the level curve, normal to the level curve, etc.)

6. Let F (x, y) = 81+x2+y2 . Use Derive to help you with the following calculations:

(a) Find F x(x, y) and F y(x, y).

(b) Find F x(1, 0) and F y(1, 0), F x(1, 2), and F y(1, 2).




(c) Find ∇F (1, 0) = F x(1, 0) i + F y(1, 0) j and ∇F (1, 2) = F x(1, 2) i + F y(1, 2) j.

(d) Graph ∇F (0, 1) in window #5 (use (0, 1) as initial point, so Author and Plot V 1(a,b, 0, 1) where

a and b are the components of ∇F (0, 1)). Also graph ∇F (1, 2) (using (1, 2) as initial point).

What relationship do you notice between these gradient vectors and the vectors (graphed earlier)

pointing in the direction of steepest ascent?

(e) Based on your observations, we might conjecture the following theorem (fill in the blanks):

Theorem 26.1 The gradient vector ∇F (a, b) points in the direction of

of F at (a,b),and is to the level curve at (a, b).



Laboratory # 27

Relative Extrema for Functions of

Two Variables

Teaches:

• criteria for finding relative maxima/minima for a function of 2 variables

1. When we studied relative maxima/minima for functions of one variable, we discovered that for a twice-

differentiable function y = f (x), a necessary condition for (a, f (a)) to be a relative extreme point was

that f (a) = 0, i.e. y = f (x) must have a horizontal tangent line at x = a:

(a, f (a))

(a, f (a)) (a, f (a))

s

ss

2. In the first picture, the graph of f is concave down at x = a (i.e. f (a) < 0) so we have a relative

323



324 LABORATORY # 27. RELATIVE EXTREMA FOR FUNCTIONS OF TWO VARIABLES

maximum. In the second picture, f is concave up at x = a (i.e. f (a) > 0) so we have a relative

minimum. It’s also possible that x = a is not a relative extremum, as in the third picture where

f (a) = 0 and we have an inflection point.

3. Transfer Load the Utility file LABS.MTH. Choose Plot Beside Enter to open a graphics window.

Change both x and y scales to 5, then choose Window Split Horizontal Enter to open another window.

4. Author

F (x, y) := −5cos

3πx

10

cos

πy

10

then choose Plot Under Enter to open a 3D-plot window. Change Grids to 30 in both directions, then

Plot to obtain a graph of z = F (x, y).

5. Now we will set up window #3 (upper right-hand corner) as we usually do to study 3-dimensional

graphs. Go to window #3, then:

• Choose Options Color Plot and change the axes and cross colors both to 0.


• Author AXES, then go to window #3 and Plot from 0 to 20 <Ctrl-Enter>.

• Choose Options Color Plot to change the next plot color to 1.

6. Now Author, Simplify, and Plot

PLOT3D(F (x, y))

from −5 to 5 <Ctrl-Enter>.


8. We wish to explore the graph of z = F (x, y) at the points (±10/3, 0). Author and Simplify

CROSS SEC(F (x, y), 10/3, 0, 0)



325

then go to window #3, choose Options Color Plot to change the next plot color to 14, and Plot

from −5 to 5.You will obtain a graph of the cross-section of the graph of z = F (x, y) at the point

(10/3, 0, F (10/3, 0)) in the direction of the positive x axis.

9. In order to obtain a frontal view of this space curve, Author and Simplify

C 1(F (x, y), 10/3, 0, 0)

then plot in window #4. In this graph, t = 0 (where we think of the horizontal axis as the t axis)

corresponds to the point (10/3, 0) on the xy plane of window #3, and the vertical axis (we will call it

the z axis) measures the height z = F (x, y), so that the z intercept of the graph in window #4 occurs

at z = F (10/3, 0).


11. Now Author and Simplify

CROSS SEC(F (x, y), 10/3, 0, π/2)

then Plot in window #3 from −5 to 5 to obtain the graph of the cross-section at (10 /3, 0) in the

direction of the positive y axis. Again, to obtain a frontal view of this, Author and Simplify

C 1(F (x, y), 10/3, 0, π/2)

then Plot in window #4.


Definition 27.1 A point ( p, q ) for which F x( p, q ) = F y( p, q ) = 0 is called a critical point of z =

F (x, y).




13. Author

F (x, y) :=

3

2 cos yesin x

(recall the e is entered as alt-e) then Plot in Window #2. Then Delete All graphs from windows #3

and #4 and reset the axes as necessary. Author and Simplify

PLOT3D(F (x, y))

then go to window #3, change the Next Plot color to 1, and Plot from −5 to 5 <Ctrl-Enter>.


15. Author and Simplify

CROSS SEC(F (x, y), π/2, 0, 0)

then go to window #3, choose Options Color Plot to change the next plot color to 14, and Plot from

−5 to 5 to obtain a plot of the cross-section of z = F (x, y) at (π/2, 0) in the direction of the x axis.

Then Author and Simplify

C 1(F (x, y), π/2, 0, 0)

and plot in window #4 to obtain a frontal view of this cross-section. (You may first want to delete all

other graphs from window #4 and set the axes color to 15).


CROSS SEC(F (x, y), π/2, 0, π/2)

then Plot in window #3 from −5 to 5 to obtain the graph of the cross-section at (π/2, 0) in the direction

of the y axis. Again, to obtain a frontal view of this, Author and Simplify

C 1(F (x, y), π/2, 0, π/2)



327




CROSS SEC(F (x, y), −π/2, 0, 0)

then go to window #3 and Plot from −5 to 5 to obtain the cross-sectional graph at (−π/2, 0) in the

direction of the x axis. Then Author and Simplify

C 1(F (x, y), −π/2, 0, 0)

and plot in window #4 to obtain a frontal view of this cross-section (after deleting all other graphs).

19. Then Author and Simplify

CROSS SEC(F (x, y), −π/2, 0, π/2)

and Plot in window #3 from −

5 to 5 to obtain the graph of the cross-section at (−

π/2, 0) in the

direction of the y axis. Again, to obtain a frontal view of this, Author and Simplify

C 1(F (x, y), −π/2, 0, π/2)



21. For our final example, Author

F (x, y) := x3y − y3 x + 1

2 x2 + 12 y2 + 2xy

50

then Plot in Window #2.





23. Delete All graphs from windows #3 and #4 and reset the axes as necessary. Author and Simplify

PLOT3D(F (x, y))

then go to window #3, change the Next Plot color to 1, and Plot from −5 to 5 <Ctrl-Enter>.

24. Author and Simplify

CROSS SEC(F (x, y), 0, 0, 0)

then go to window #3, choose Options Color Plot to change the next plot color to 14, and Plot from

−5 to 5 to obtain a plot of the cross-section of z = F (x, y) at (0, 0) in the direction of the x axis. Then

Author and Simplify

C 1(F (x, y), 0, 0, 0)

and plot in window #4 to obtain a frontal view of this cross-section.


CROSS SEC(F (x, y), 0, 0, π/2)

then Plot in window #3 from −5 to 5 to obtain the graph of the cross-section at (0, 0) in the direction

of the y axis. Again, to obtain a frontal view of this, Author and Simplify

C 1(F (x, y), 0, 0, π/2)


26. Finally, let’s look at some other cross-sections: Author and Simplify



then Plot both in window #3 from −5 to 5.



329

27. Also, Author and Simplify

C 1(F (x, y), 0, 0, π/6)

C 1(F (x, y), 0, 0, π/3)

and Plot both in window #4.


29. We saw in the last example that determining that the concavity at a critical point of a function

z = F (x, y) in the x and y directions are both positive (or both negative) does not ensure that we have

a relative minimum (or a relative maximum). The function z = F (x, y) will have a relative minimum

(relative maximum) at ( p, q ) only if every cross-sectional curve at ( p, q ) is concave up (concave down)

at ( p, q ). We know that, in order to examine the concavity in the x and y directions we need only look

at the sign of F xx( p, q ) and F yy( p, q ). It turns out that the concavity in any direction θ is given by the

sign of

G(θ) = a cos2 θ + b sin2 θ + c sin2θ

where a = F xx( p, q ), b = F yy( p, q ), and c = F xy( p, q ).


31. Since G(θ) gives the concavity of F (x, y) at ( p, q ) for any direction θ, we would like to determine for

what values of a, b, and c will G(θ) be positive (negative) for all θ (thus insuring that ( p, q ) is a relative

minimum (relative maximum)).

32. Close windows #2,#3, and #4. Then Author

G(θ) := a cos2 θ + b sin2 θ + c sin2θ




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&

$

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Relative Extrema for Functions of Two Variables

NAME: DATE:


1. Let F (x, y) = −5cos

3πx10

cos

πy10

.

Using Derive’s Calculus Differentiate command to help you, calculate the following:

(a) F x(x, y) = ; F x(±10/3, 0) =

(b) F y(x, y) = ; F y(

±10/3, 0) =

(c) F xx(x, y) = ;F xx(±10/3, 0) =

(d) F yy(x, y) = ;F yy(±10/3, 0) =

2. Does the behavior of the cross-section of the graph in window #4 at t = 0 (i.e. at (x, y) = (10/3, 0))

agree with your answer to Worksheet Questions #1a and #1c? Explain.

331



332 Worksheet # 27. RELATIVE EXTREMA FOR FUNCTIONS OF TWO VARIABLES

3. Does the behavior of the cross-section of the graph in the direction of the y axis agree with your answer

to Worksheet Questions #1b and #1d? Explain.

4. Generalize your observations by completing the following theorem:

Theorem 27.1 Let z = F (x, y) be a continuous function for which all second-order derivatives ex-

ist. A necessary condition for z = F (x, y) to have a relative maximum at (x, y) = ( p, q ) is that

F x( p, q ) = F y( p, q ) = and F xx( p, q ) and F yy( p, q ) both be .A nec-

essary condition for z = F (x, y) to have a relative minimum at (x, y) = ( p, q ) is that F x( p, q ) =

F y( p, q ) = and F xx( p, q ) and F yy( p, q ) both be .

5. Let F (x, y) = 32 cos yesin x.

Using Derive to help you, calculate the following:

(a) F x(x, y) = ; F x(±π/2, 0) =

(b) F y(x, y) = ; F y(±π/2, 0) =

(c) F xx(x, y) = ; F xx(π/2, 0) = ;

F xx(−π/2, 0) =

(d) F yy(x, y) = ; F yy(π/2, 0) = ;

F yy(−π/2, 0) =



Name: 333

6. Does the behavior of the cross-sections in window #4 at t = 0 (i.e. at (x, y) = (π/2, 0)) agree with

your answer to Worksheet Question #5? Explain.

7. Does the behavior of the cross-sections in window #4 at t = 0 (i.e. at (x, y) = (−π/2, 0)) agree with

your answer to Worksheet Question #5? Explain.

8. Explain how the graphs of the two cross-sections at (−π/2, 0) make it clear that (−π/2, 0, F (−π/2, 0))

is not a relative extreme point.

9. Let F (x, y) = x3y−y3x+ 1

2 x2+ 1

2y2+2xy

50 .

Using Derive to help you, calculate the following:

(a) F x(x, y) = ; F x(0, 0) =

(b) F y(x, y) = ; F y(0, 0) =

(c) F xx(x, y) = ; F xx(0, 0) =



334 Worksheet # 27. RELATIVE EXTREMA FOR FUNCTIONS OF TWO VARIABLES

(d) F yy(x, y) = ; F yy(0, 0) =

(e) According to what you learned in this lab, do you think that z = F (x, y) could have a relative

maximum or minimum at (0, 0)? Explain.

10. (a) Does F (x, y) = x3y

−y3x+ 1

2

x2+ 1

2

y2+2xy

50 have a relative extreme point at (0, 0)? Explain.

(b) Is the converse of Theorem 27.1 true? Explain.

11. Verify that G(0) gives the concavity of F (x, y) at ( p, q ) in the x direction, and G(π/2) gives the

concavity of F (x, y) in the y direction.

calculus laboratory(with derive)

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