math 31 lessons chapter 1: limits 1. linear functions and tangents

MATH 31 LESSONS

Chapter 1: Limits

1. Linear Functions and Tangents

Section 1.1: Linear Functions and

The Tangent Problem

Read Textbook pp. 5 - 9

A. Linear Functions

Recall

y = m x + b

where

m = slope

b = y-intercept

x

y

b

x

y

(x1, y1)

(x2, y2)

x = x2 - x1

y = y2 - y1

12

12

Δ

Δslope

xx

yy

x

y

Note:

is called the rate of change of y with respect to x

i.e. how quickly y changes relative to changes in

x

x

y

Δ

Δ

is called the rate of change of y with respect to x

i.e. how quickly y changes relative to changes in

x

If y is positive, then y is increasing

If y is negative, then y is decreasing

x

y

Δ

Δ

e.g.

How can this be interpreted?

x

y

x = 2

y = 52

5

Δ

Δ

x

y

When x increases by 2 units, y increases by 5 units.

x

y

x = 2

y = 52

5

Δ

Δ

x

y

Clearly,

the greater the slope

the greater the rate of change of y (with respect to x)

the “faster” y changes

Low rate of changeHigh rate of change

Note:

x

y

Δ

Δ

Ex. 1 Find the equation of the linear function that passes

through A(2, 1) and B(-4, -9).

Answer in general form.

Try this example on your own first.Then, check out the solution.

A(2, 1) and B(-4, -9)

Find slope

x1 y1 x2 y2

24

19

12

12

xx

yym

6

10

3

5

A(2, 1)

Use point-slope formula:

x1 y1

11 xxmyy

23

51 xy

3

5m

11 xxmyy

23

51 xy

2513 xy Multiply both sides by 3 to remove the denominator

2513 xy

10533 xy

0735 yx

Ax + By + C = 0

A > 0

A, B, and C are integers

Ex. 2 For the linear function 32x + 4y - 19 = 0,

if x decreases by 2, how does y change?

Try this example on your own first.Then, check out the solution.

Find slope

019432 yx

019432 yx

19324 xy

4

198 xy

Put in slope-intercept form: y = mx + b

The slope is -8

019432 yx

19324 xy

4

198 xy

Find the change in y (y) :

8m8Δ

Δm

x

y

2x

8m8Δ

Δm

x

y

xy 8

2x

28

16

So, y increases by 16

8Δ

Δm

x

y

xy 8

28

16

B. The Tangent Problem

What is a tangent line?

The question is more challenging than it appears,

and as we will see, only calculus can truly answer it.

For circles, the tangent line is readily defined.

A tangent is a straight line

that touches the circle

only once. Tangent, t

Not tangent

A tangent line is a straight line that touches the

circle only once

However, the definition above is not adequate for

general curves.

To show why, consider the next two illustrations.

Consider the following parabola:

y = f (x)

Both l and t touch the curve only once.

However, only t is a tangent.

l

t

This line t is a tangent.

t

For an even more interesting example, consider

the cubic function below:

y = f (x)

The line t is a tangent line, even though it crosses

the curve twice!

t

So, the statement ...

“a tangent line is a straight line that touches a curve

only once”

is clearly inadequate.

C. An Approach to the Tangent Problem

Much of this unit is devoted to solving the

tangent problem.

To illustrate the approach we will take, consider

the following problem.

Question: Find the slope of the tangent line to the

parabola y = x2 at the point P(2, 4).

y = x2

y

x2

4

t

P(2, 4)

Problem:

In order to find the slope

of the tangent line, we need

to know two points on

the tangent line.

However, we only know one.

y = x2

y

x2

4

t

P(2, 4)

Solution:

We introduce a second point

on the curve y = x2,

somewhere close to P.

We will call this point Q,

and it will have the

coordinates (x, x2).

y = x2

x 2

4

t

P(2, 4)

Q(x, x2)

x2

The line l that connects

P and Q is called a

secant line.

It can be thought of as

an approximation of

the tangent line.

y = x2

x 2

4

t

P(2, 4)

Q(x, x2)

x2

l

The slope of the secant

line l is given byy = x2

x 2

4

t

P(2, 4)

Q(x, x2)

x2

l

12

12

xx

yyml

The slope of the secant

line l is given byy = x2

x 2

4

t

P(2, 4)

Q(x, x2)

x2

l

12

12

xx

yyml

2

42

x

x

The slope of the secant

line l is given byy = x2

x 2

4

t

P(2, 4)

Q(x, x2)

x2

l

12

12

xx

yyml

2

42

x

x

2

22

x

xx

The slope of the secant

line l is given byy = x2

x 2

4

t

P(2, 4)

Q(x, x2)

x2

l

12

12

xx

yyml

2

42

x

x

2,2 xx

2

22

x

xx

How can we use the secant

slope to find the tangent

slope?

The main idea is to

start moving Q closer

and closer to the point P.

y = x2

x 2

4

t

P(2, 4)

Q(x, x2)

x2

l

Notice that as Q

approaches P ...

x approaches 2

l approaches t

mPQ approaches mt

y = x2

x 2

4 P(2, 4)

Q(x, x2)x2

t l

In fact, if Q gets really close

to P, the secant line l is almost

identical to the tangent line t

It follows that the slope

of the secant line would

be almost identical

to the slope of the tangent.

y = x2

x2

4 P(2, 4)Q(x, x2)

x2

t l

In calculus, we attempt to

bring Q right to P, so that

the line l actually becomes

the tangent line t.

In this way, the slope is

exactly the same.

But how do we do this?

We use the idea of limits.

y = x2

2

4 P(2, 4)Q

t l

We will now find the

pattern of slopes as Q

approaches P

(i.e. as x approaches 2

“from the left”)y = x2

x 2

4

t

P(2, 4)

Q(x, x2)

x2

l

x mt = x + 2

1

1.5

1.9

1.99

1.999

Slope of the secant as x approaches 2 from the left:

x mt = x + 2

1

1.5

1.9

1.99

1.999

3

3.5

3.9

3.99

3.999

Slope of the secant as x approaches 2 from the left:

x mt = x + 2

1

1.5

1.9

1.99

1.999

3

3.5

3.9

3.99

3.999

2 4

As x approaches 2 from the left, the slopes appear to be approaching 4

We should also find the slope of the secants as

x approaches 2 from the right as well:

x mt = x + 2

3

2.5

2.1

2.01

2.001

So, as x gets extremely close to 2,

the slope of the secant lines get extremely close to 4.

In fact, in calculus we state that if x gets infinitely close to 2

(e.g. x = 1.9 ), the slopes would become so close to 4

(e.g. slope = 3.9 ) as to become actually equal to 4.

So, as x gets extremely close to 2,

the slope of the secant lines get extremely close to 4.

In fact, in calculus we state that if x gets infinitely close to 2

(e.g. x = 1.9 ), the slopes would become so close to 4

(e.g. slope = 3.9 ) as to become actually equal to 4.

This value of 4 would represent the true slope of the tangent.

We say that the slope of the tangent is the limit

of the secant slopes, because it represents the limit (endpoint)

of slopes as x gets infinitely close to 2.

Notation:

If the tangent slope is the limit of the secant slopes,

then we write:

tPQPQ

mm

lim

We say, “The limit as Q approaches P of the slopes of secant line PQ is the slope of the tangent.”

If the tangent slope is the limit of the secant slopes,

then we write:

or

tPQPQ

mm

lim

42

4lim

2

2

x

xx

Note:

We could also find the

equation of the tangent,

since we now know the

slope and one point.

y = x2

2

4 P(2, 4)

t

mt = 4

Equation of tangent: P(2, 4) mt = 4

11 xxmyy

244 xy

844 xy

44 xy or 044 yx

Ex. 3 For the function y = x2 - 5x, determine the equation

of the tangent line at the point P(2, -6).

Be certain to:

- include a sketch

- find the formula for the secant slope

- see what the slopes approach as x -2

from both sides to get the tangent slope

Try this example on your own first.Then, check out the solution.

Sketch: y = x2 - 5xy

x2

-6

t

P(2, -6)

We introduce

a second point

Q (x, x2 - 5x)

on the curve.

The point PQ is the

secant line.

As Q approaches P,

the secant approaches the tangent.

y = x2 - 5xy

x2

-6

t

Q(x, x2 - 5x)

P(2, -6)

Slope of secant:

y = x2 - 5xy

x2

-6

t

P(2, -6)

Q(x, x2 - 5x)

12

12

xx

yyml

2

652

x

xx

Slope of secant:

12

12

xx

yyml

2

652

x

xx

2

652

x

xx 2

23

x

xx

Slope of secant:

12

12

xx

yyml

2

652

x

xx

2

652

x

xx 2

23

x

xx

2,3 xx

Slope of the secant as x approaches 2 from the left:

x mt = x - 3

1

1.5

1.9

1.99

1.999

Slope of the secant as x approaches 2 from the left:

x mt = x - 3

1

1.5

1.9

1.99

1.999

-2

-1.5

-1.1

-1.01

-1.001

2 -1

As x approaches 2 from the left, the slopes appear to be approaching -1

Slope of the secant as x approaches 2 from the right:

x mt = x - 3

3

2.5

2.1

2.01

2.001

Slope of the secant as x approaches 2 from the left:

x mt = x - 3

3

2.5

2.1

2.01

2.001

0

-0.5

-0.9

-0.99

-0.999

2 -1

As x approaches 2 from the right, the slopes appear to be approaching -1 as well

The tangent slope is the limit of the secant slopes.

So, it follows that

PQPQ

t mm

lim

2

65lim

2

2

x

xxx

1

Equation of tangent: P(2, -6) mt = -1

11 xxmyy

216 xy

Equation of tangent: P(2, -6) mt = -1

11 xxmyy

216 xy

26 xy

4 xy or 04 yx