coursera calculus i fake midterm 2 (march 2013)

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Coursera Calculus I Fake Midterm 2 (March 2013) Jabari Zakiya Question 1 Evaluate or state that the limit does not exist. Because use L' Hopital's Rule: = Question 2 Evaluate or state that the limit does not exist. Because do: = = = = = = = = Question 3 Which of the following is the derivative of arcsin(tan(x))? a) b) c) d) none of the above Question 4 Three changing quantities a, b, and c have the property that a 4 + b 4 + c 4 = 4, even as they change. At this particular moment, a is increasing and b is decreasing. What can be said about c? a) The quantity c is decreasing b) The quantity c is staying the same c) The quantity c is increasing d) There is not enough information to determine the answer . 4a 3 da/dt + 4b 3 db/dt + 4c 3 dc/dt = 0 where = not enough information to characterize c

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Page 1: Coursera Calculus I Fake Midterm 2 (March 2013)

Coursera Calculus I Fake Midterm 2 (March 2013)

Jabari Zakiya

Question 1

Evaluate or state that the limit does not exist.

Because use L' Hopital's Rule: =

Question 2

Evaluate or state that the limit does not exist.

Because do: = =

= =

= = = =

Question 3

Which of the following is the derivative of arcsin(tan(x))?

a) b) c) d) none of the above

Question 4

Three changing quantities a, b, and c have the property that a4 + b4 + c4 = 4, even as they change.

At this particular moment, a is increasing and b is decreasing. What can be said about c?

a) The quantity c is decreasing b) The quantity c is staying the same

c) The quantity c is increasing d) There is not enough information to determine the answer.

4a3 da/dt + 4b3 db/dt + 4c3 dc/dt = 0 where

= not enough information to characterize c

Page 2: Coursera Calculus I Fake Midterm 2 (March 2013)

Question 5

A giant hailstone gathers even more ice as it hurdles towards me, in such a way that

• it remains in the shape of a perfect sphere, of radius length r, and • the rate of change in its volume is proportional to its surface area.

At the beginning of our story, r is half a centimeter, but after one minute has elapsed r is one cm.

How much longer (in minutes) must I wait until r is 2 centimeters? a) 4 b) 2 c) 6 d) The answer cannot be determined from information given

We need to find an expression of r in terms of t, i.e. r(t).

volume of sphere is: and thus

at t = 0, r =

at t = 1, r = 1

therefore: = =

equals some value k so:

k = and so

and thus: r(t) = t + C

now use initial conditions to solve for k and C:

for t = 0, r = = (0) + C gives C =

for t = 1, r = 1 1 = (1) + gives k =

therefore: r(t) = t +

now find t when r is 2: 2 = t +

t = = 3 minutes

Since r was 1 after 1 minute, and 2 after 3 minutes, thus we had to wait 2 more minutes to get there.

Page 3: Coursera Calculus I Fake Midterm 2 (March 2013)

Question 6

Model a space-faring civilization's territory as a perfect circle of increasing radius. The area that this space-faring civilization controls grows at a rate proportional to its frontier. It took this civilization one thousand years to control 100 square light-years, and then another nine thousand years to control ten thousand square light-years. How long will it take this civilization to control the whole galaxy, which isa disk of area 1010 square light-years?a) 100 million years b) 500 million years c) > 100 billion years d) 1 million years e) 10 million years

We need to find an expression of r in terms of t, i.e. r(t), which controls the circle's rate of growth.

area of circle is: and thus

t = 1000, A = 100 100 = gives r =

t = 10000, A = 10000 10000 = gives r =

t = ? A = 1010 1010 = gives r =

t = 1000, r =

t = 10000, r =

therefore: =

equals some value k so:

k = and so

thus, using time shifting: r(t) = (t – 1000) + C

now use initial conditions to solve for k and C:

t = 1000, r = = (1000 – 1000) + C gives C =

t = 10000, r = = (10000 – 1000) + gives k =

therefore: r(t) = (t – 1000) +

now find t when r is : = (t – 1000) +

t =

t = 107 (or 10 million) years

Page 4: Coursera Calculus I Fake Midterm 2 (March 2013)

Question 7

I have spilled 2 cm3 of truffle oil on my kitchen floor. This delicious oil slick can be modeled as a cylinder with a very small height; the height is decreasing at a rate of cm/sec and the radius is currently 5 cm.

How quickly is the radius of the oil slick increasing, in cm/sec? Your answer should be a fraction of whole numbers.

First find cylinder height h of volume of 2 cm3 cylinder with 5 cm radius:

V = , h = cm

Because V is constant, is 0, and because height is decreasing is negative, therefore:

and thus:

0 =

cm/sec

So the radius is increasing, while the cylinder is getting flatter.

Page 5: Coursera Calculus I Fake Midterm 2 (March 2013)

Question 8

I want to take a one meter long length of wire, and build a "house" shaped figure,like the one shown to the right.

The bottom of my house is to be a square, and the roof is to be an isoceles triangle.I must use all of the wire.

Which value is closest to the area of the largest such figure I can build, using all ofthe wire?

This problem is somewhat harder than the corresponding problem in the real exam. a) 520 cm2 b) 625 cm2 c) 300 cm2 d) 430 cm 2

Let x be length of square sides and y the length of the two roof (triangle) sides, then

1 (meter) = 4x + 2y; y = (1 – 4x)/2 and from triangle ;

Total Area (TA) = Area of Square + Area of Isoceles Triangle = +

TA = + = + = +

TA = +

now differentiate TA with respect to x,

= 2x + (1/4) (1 – 8x + 15x2)1/2 + (x/8)(1 – 8x + 15x2)-1/2 (30x – 8) = 0

2x + (1/4) (1 – 8x + 15x2)1/2 + x(30x – 8)/8(1 – 8x + 15x2)1/2 = 0

16x(1 – 8x + 15x2)1/2 + 2(1 – 8x + 15x2) + x(30x – 8) = 0

16x(1 – 8x + 15x2)1/2 + 2 – 16x + 30x2 + 30x2 – 8x = 0

16x(1 – 8x + 15x2)1/2 + 60x2 – 24x + 2 = 0

To solve for x (find a root of the expressions) use Newton's Method.

Let y(x) = 16x(1 – 8x + 15x2)1/2 + 60x2 – 24x + 2 = 0

then y'(x) = 120x – 24 + 16(1 – 8x + 15x2)1/2 + 8x(1 – 8x + 15x2)-1/2 (30x – 8)

and thus: xn+1 = xn – y(xn)/y'(xn)

Ruby code to perform Newton's Method.

def y(x); 16*x*(1 ­ 8*x + 15*x*x)**(0.5) + 60*x*x ­ 24*x + 2 end 

def yy(x); 120*x­24+16*(1­8*x+15*x*x)**(0.5)+8*x*(1­8*x+15*x*x)**(­0.5)*(30*x­8)end

x=0.17; 5.times{ x ­= y(x)/yy(x)}; x => 0.19215377322680077                     

with this value for x, we get TA = + = 0.0431 meters2 = 431 cm2

[Can also check graphically by plotting y(x) (say with http://www.geogebra.org, et al) and see that y(x) = 0 close to x = 0.19]

Page 6: Coursera Calculus I Fake Midterm 2 (March 2013)

Question 9

To build a beautiful trough with a semicircular cross section in celebration of п day on March 14, I tooka п-by-п piece of metal, bent it in semicircle, and welded two semicircles of radius 1 meter to each of the open sides. To keep up the celebration of day, I am pouring water into my trough at a rate of п cubicmeters per second, and at this moment there is cubic meters of water in my trough.

How quickly is the height of the water changing, in meters/sec?

This problem is quite a bit more challenging than the corresponding problem on the real exam; it might help to think about how to use trigonometry on this problem. a) 2 m/sec b) 3 m/sec c) 5 m/sec d) 1 m/sec e) п m/sec f) 4 m/sec

Fig. 1

The volume of the full trough is half the volume of a cylinder:

A line parallel to the top of the semicircle constitutes the current volume of water in the trough.

Let h be the height of the amount of water from the bottom of the trough to the waterline and let H be the remaining height from the waterline to the top of the trough. Thus r = h + H and H = r – h.

A line from the midpoint at the top of the trough to the point on the semicircle where the waterline meets creates a right triangle of height H, and an angle from top of the trough to the waterline of .

The area above the waterline is the area of the two right triangle of height H and the area of the two arcsections subtended by angle .

Area of Triangle = Area of Arcsection =

The area of the cross-section of water in the trough at any time is, thus, the total area of the semicircle minus twice the area of the right triangle of height H and twice the area of the arcsection of angle ,

A = – – = – –

and because r = 1 meter; A = – bH –

where b is the base of the right triangle from the top of the waterline to the edge of the circle in Fig. 1.

[When empty, H = r, but b = 0, so bH = 0 and = , so A = 0. When full, and H are 0, so A = = ]

Now write b, H and is term of h, using fact the r = 1 meter.

h + H = r; h + H = 1; H = 1 – h

and therefore

and also so

Page 7: Coursera Calculus I Fake Midterm 2 (March 2013)

So now the volume of the water in the trough at any height is: V = A * length = A* gives

1) V = ( – bH – )

2) V = ( – (1 – h) – arcsin(1 – h))

Using the initial volume condition V = equate this to expression 1)

= ( – bH – )

= – bH –

3) + = bH +

[Here is where mathematical insight and inspiration comes into play to make this problem simple.]

Notice in 3) that the term is the value of an angle (in radians), therefore equate: =

Therefore must equal bH. Remember, we are trying to find h (the height at this given volume of water). There are two ways to do this, the hard way and the easy way! Let's do the easy way first.

Since we know = , then from Fig. 1, b = cos = cos( ) = 1/2

4) = bH = H/2 = (1 – h)/2

= 1 – h

h = 1 – = 1 – 0.87 = 0.13

Now let's find h the hard(er) way. Using and H = (1 – h) we get 4) to be:

5) = bH = (1 – h)

3/16 = (1 – h)2 (2h – h2)

3/16 = 2h – 5h2 + 4h3 – h4

Newton's Method can now be used to find h from y(h) (a 4th degree polynomial with 4 roots).

Let y(h) = 2h – 5h2 + 4h3 – h4 – 3/16 = 0

and then y'(h) = 2 – 10h + 12h2 – 4h3

and to find h iterate: hn+1 = hn – y(hn)/y'(hn)

Below is Ruby code to perform Newton's Method with h0 that give correct root value of h.

def y(h); 2*h ­ 5*h*h + 4*h**3 ­ h**4 ­ 3.0/16 end                                 

def yy(h); 2 ­ 10*h + 12*h*h ­ 4*h**3 end

h = 0.0;  10.times { h = h ­ y(h)/yy(h) }; h  => 0.13397459621556132 

h = 0.25; 10.times { h = h ­ y(h)/yy(h) }; h  => 0.13397459621556135

h = 1 – (3**0.5)/2                            => 0.1339745962155614

[Other h0 values < 2 give all 4 roots of y(h): h1,2,3,4 = 1± 0.5 (0.5,1.5) and 1± (0.13,1.87) but h must be < r=1, so h = 1.5 or 1.87 are invalid and h = 0.5 would reduce the value of from .]

The really interesting question may be how would you find h if you didn't see (have the insight) to equate = and then = bH ? (This is frequently done with complex numbers, where with anexpression like (3 – x) + (5 + y)i = 7 + 4i you equate the real and imaginary parts of both sides to solve for x and y). The answer is YES!, we can find h using Newton's Method.

Page 8: Coursera Calculus I Fake Midterm 2 (March 2013)

We can “brute force” our way to finding h from 3) using Newton's Method.

+ = bH +

+ = (1 – h) + arcsin(1 – h)

Let y(h) = – h + arcsin(1 – h) – ( + ) = 0

and then y'(h) = – – +

y'(h) = = = =

and to find h iterate: hn+1 = hn – y(hn)/y'(hn)

The Ruby code to do this is below:

include Math

def y(h); (2*h ­h*h)**0.5 ­ h*(2*h ­h*h)**0.5 + asin(1­h) ­ (PI/3 +(3**0.5)/4) end

def yy(h); ­2*(2*h ­h*h)**0.5 end 

h =0.0;   5.times {h= h ­ y(h)/yy(h)}; h    => (NaN+NaN*i) 

h =0.1;   5.times {h= h ­ y(h)/yy(h)}; h    => 0.13397459621556132

h =0.25;  5.times {h= h ­ y(h)/yy(h)}; h    => 0.13397459621556143

Thus we see, we can find the height h for any initial volume of water by “brute force” using Newton's Method. In fact, this becomes the generic process for solving for h for any initial volume amount without the need of applying any mathematical insight to the form or value of the volume expression.

Now that we have determined h we are ready to find dh/dt for the given dV/dt, using 2) V(h).

V = [ – (1 – h) – arcsin(1 – h)]

V = – [(1 – h) + arcsin(1 – h)]

dV/dt = d h{ – [(1 – h) + arcsin(1 – h)] } /dt

= 0 – d h[ – h + arcsin(1 – h)] /dt

-1 = d h[ – h + arcsin(1 – h)] /dt

-1 = [ – – + ]

-1 = [ – – – ]

-1 =

-1 = = =

= = = 0.99999999 = 1 meter/sec

Page 9: Coursera Calculus I Fake Midterm 2 (March 2013)

Question 10

Consider the ellipse plotted by 3x2 + 2y2 = 4. Find a polynomial with a root at the x-coordinate of the point on the ellipse that is closest to the point (2,2).

This problem is quite a bit harder than the corresponding problem onthe real exam.

a) 3x4 + 24x3 + 116x2 – 31x – 64 b) 3x4 + 24x3 + 116x2 – 33x – 64c) 3x4 + 24x3 + 116x2 – 30x – 64 d) 3x 4 + 24x 3 + 116x 2 – 32x – 64

To find the x-coordinate that minimizes h, the distance from the point(2,2) to the ellipse, write h in terms of x, then differentiate it and set tozero.

3x2 + 2y2 = 4 and

h2 = (2 – y)2 + (2 – x)2 = 4 – 4y + y2 + 4 – 4x + x2 = 8 – 4(x +y) + y2 + x2

h2 =

h2 = 10 – 4x – 4 –

2h = -4 – 2 (-3x) – x

= (-4 + 6x – x)/2h = 0

-4 + 6x – x = 0

6x = x + 4

36x2 = (x + 4)2

72x2 = (x + 4)2

72x2 = (x + 4)2 (4 – 3x2) 72x2 – (x2 + 8x + 16)(4 – 3x2) = 0 72x2 – (4x2 + 32x + 64 – 3x4 – 24x3 – 48x2 ) = 0 72x2 – 4x2 – 32x – 64 + 3x4 + 24x3 + 48x2 = 0 3x4 + 24x3 + 116x2 – 32x – 64 = 0

Newton's Method can now be used to find x from y(x) (a 4th degree polynomial with 4 roots).

Let y(x) = 3x4 + 24x3 + 116x2 – 32x – 64 = 0

and then y'(x) = 12x3 + 72x2 + 232x – 32

and to find x iterate: xn+1 = xn – y(xn)/y'(xn) with x0 = 0.5

The best root is x = 0.80888 which makes y = 1.00924 and then h = 1.54931 is shortest distance.

Page 10: Coursera Calculus I Fake Midterm 2 (March 2013)

Question 11

You have a 10 cm by 15 cm sheet of cardboard that you wish to fold into a box with an open top. You plan to cut x centimeter squares out of each of the four corners, and fold up the sides. Find the value of x that maximizes the volume.

a) cm b) cm c) cm d) cm

V = (10 – 2x)(15 – 2x)x = (150 – 50x + 4x2)x = 4x3 – 50x2 + 150x

= 12x2 – 100x + 150 = 0

6x2 – 50x + 75 = 0 and x = or gives

x = =

x = – = – = – cm, because twice ( + ) is > 10 cm.

Question 12A certain acting troupe puts on plays inspired by calculus. The troupe agrees to put on a show in your home town, provided that at least one person shows up. The troupe charges $10 if only one person is in the audience, $9.75 per person if there are two audience members, $9.50 per person if there are three members, and so on.

At most how much money can the troupe expect to bring in?

a) $75.50 b) $105.06 c) $20.50 d) $21.00 e) $105.00 f) $21.00

You can create a chart to find the revenue per number of people, as below:

number of people 1 2 3 4 5 ….........$ price per person 10 9.75 9.50 9.25 9.00 ….........total revenue 10 19.50 28.50 37.00 45.00 ….........

We can translate this chart into a curve. Let p be the number of people attending a show.

Total Revenue (TR) = p(10 – (p – 1)(0.25)) = p(10 – 0.25p + 0.25) = 10.25p – 0.25p2

= 10.25 – 0.5p = 0, so p = 10.25/0.5 = 20.5 to maximum TR

Since there can only be whole numbers of people 20 or 21 people will maximize revenue:

Maximum TR is: TR(20) = TR(21) = 10.25(20) – 0.25(20)2 = 10.25(21) – 0.25(21)2 = $105.00

Page 11: Coursera Calculus I Fake Midterm 2 (March 2013)

Question 13

Find the x coordinate of the point on the curve that is closest to the origin.

This question is quite a bit harder than the corresponding question on the real exam; to make it a bit easier, remember that minimizing distance to (0,0) is the same as minimizing the value of x2 + y2, whichis a bit easier to deal with than .

a) 0 b) 2 c) 3 d) -1

Want to find the x value that minimizes distance h from origin to curve y.

= 0

Set numerator to zero and solve for x using Newton's Method: xn+1 = xn – y(xn)/y'(xn)

Let y(x) = = 0

and then y'(x) = 1 +

Ruby code to perform Newton's Method with x0 = 0:

def y(x); x + (3.3*x*x ­ 48*x + 36043.0/435)*(6.6*x ­48) end

def yy(x); 1 + (6.6*x ­48)**2 + (3.3*x*x ­ 48*x + 36043.0/435)*6.6 end

x=0; 5.times{ x = x ­ y(x)/yy(x)}; x  => 1.9999999999997298 

x=0; 6.times{ x = x ­ y(x)/yy(x)}; x  => 2.0

When x = 2.0 then y = 0.05747 and h = 2.00083 is shortest distance.

Page 12: Coursera Calculus I Fake Midterm 2 (March 2013)

Question 14

Which of the following could be a slope field for = cos x + sin y?

a b

c d

The answer is b.

Page 13: Coursera Calculus I Fake Midterm 2 (March 2013)

Question 15

Which of the following could be a slope field for ?

a b

c d

The answer is d.

Page 14: Coursera Calculus I Fake Midterm 2 (March 2013)

Question 16

For which of the following functions f is it the case that f '(x) = 2x + 2f(x)x?

In other words, which function satisfies the differential equation ?

a) f(x) = ex b) f(x) = sin(x) c) f(x) = e2x d) f(x) =

The answer is d): y = , y' = =

Question 17

Suppose f is a function for which f(2) = 0,

and which satisfies the equation

f ' (x) = f(x)2 – 4

To get a sense of the function f, you maywish

to consider the slope field ,

since you can recover f by starting at (2,0)and following the lines.

Place a checkmark next to the truestatements below.

a) f is increasing at x = 4

b)

c) f ( x ) 2 is < 4 for all x

d) f is decreasing at x = 2

e)

The answer is: c), d), and e) are true.

Page 15: Coursera Calculus I Fake Midterm 2 (March 2013)

Question 18

Consider the interval I = [8,14]. Break I into three subintervals of equal length, namely the three subintervals [8,10], [10,12], [12,14]. Suppose that f(8) = 2, f '(8) = 3, f '(10) = 4, f '(12) = -5. What is the approximate value of f(14) by applying Euler's method?

for h =2: yn+1 = yn + h*f '(yn)

f(10) = f(8) + 2*f '(8) = 2 + 2(3) = 2 + 6 = 8

f(12) = f(10) + 2*f '(10) = 8 + 2(4) = 8 + 8 = 16

f(14) = f(12) + 2*f '(12) = 16 + 2(-5) = 16 – 10 = 6

Question 19

The function f satisfies the differential equation , meaning f ' = –x2 and also suppose

f(0) = 1. Will applying Euler's method over-approximate or under-approximate the actual value f(2)?

a) over-approximate b) under-approximate

try h = 0.5: yn+1 = yn + h*f '(yn) with x0 = 0 and f(0) = 1.

f(0.5) = f(0) + (0.5)*f '(0) = 1 + (0.5)(-(0)2) = 1 – 0 = 1

f(1.0) = f(0.5) + (0.5)*f '(0.5) = 1 + (0.5)(-(0.5)2) = 1 – (0.125) = 0.875

f(1.5) = f(1.0) + (0.5)*f '(1.0) = 0.875 + (0.5)(-(1.0)2) = 0.875 – 0.5 = 0.375

f(2.0) = f(1.5) + (0.5)*f '(1.5) = 0.375 + (0.5)(-(1.5)2)= 0.375 – 1.25 = –0.75

Because f '(x) = –x2 then f (x)= –x3/3 + C and f(0) = 1 = –(0)3/3 + C makes C = 1

Then f(2) = –(2)3/3 + 1 = 1 – 8/3 = –1.6666 is < –0.75 so Euler's Method over-approximates f(2).

Question 20

Suppose f satisfies the differential equation and f(1) = 3. In other words,

f '(x) = f(x) + f(x)2 and f(1) = 3.

Use Euler's method with a step size of 0.01 to approximate f(1.02).

a) b) c) d)

for h = 0.01: yn+1 = yn + h*f '(yn), with x0 = 1 and f(1) = 3.

f(1.01) = f(1.00) + (0.01)*f '(1.00) = 3 + (0.01)(3+32) = 3.12

f(1.02) = f(1.01) + (0.01)*f '(1.01) = 3.12 + (0.01)(3.12+(3.12)2) = 3.248544

therefore, f(1.02) = 3.248544 =

Page 16: Coursera Calculus I Fake Midterm 2 (March 2013)

Question 21

Let f(x) = x5 – x – 2 . Then f has a root somewhere between x = 0 and x = 2.

Using x0 = 2 as your first value for Newton's method, compute the third approximation x2.

a) b) c)

Newton's Method gives: xn+1 = xn – f(xn)/f '(xn) = xn – with x0 = 2

Ruby code to perform this is given below:

x = 2.0; 2.times{x = x ­ (x**5 ­ x ­ 2)/(5*x**4 ­1)}; x => 1.4094484436882508

Therefore, x2 = 1.4094484436882508 =

Question 22

How many iterations may Newton's method require to approximate a root of a polynomial to within 1/100 ?

a) 100 iterations b) 17 iterations c) Newton's method may never come within of a root .

d) Newton's method is guaranteed to find a root to within 1/100 as long as you repeat until two successive approximations are within, say, 1/1000.

Question 23

Which of the following is an antiderivative of x3cos(x) + 3x2sin(x) with regard to x ?

a) sin2(x) + 2 b) sin(x3) + cos(3) c) x2sin(x) + e2 d) x 3 sin(x) + Π

Question 24Is x3/3 an antiderivative of (x sin(x))2 + (x cos(x))2 with respect to x? a) No b) Yes

(x sin(x))2 + (x cos(x))2 = x2(sin2x + cos2x) = x2(1) = x2 which is the derivative of x3/3.

Question 25

Suppose F(x) is an antiderivative of f(x) = log log x.Is F increasing or decreasing near x = 10? a) Decreasing b) Increasing

Since F(x) is an antiderivative of f(x) = log log x then f(x) is the derivative of F(x). Therefore, the sign of f(x) indicates whether F(x) is increasing (positive) or decreasing (negative). At x = 10 the derivative f(10) = log log (10) = 0.834032445247956 is positive so F(x) is increasing.