2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e Slide...

© 2010 Pearson Education Inc. Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e – Slide 1 of 115

Chapter 1

The Derivative


The Slope of a Straight Line The Slope of a Curve at a Point The Derivative Limits and the Derivative Some Rules for Differentiation More About Derivatives The Derivative as a Rate of Change

Chapter Outline


Definition ExampleEquations of Nonvertical Lines: A nonvertical line L has an equation of the form

The number m is called the slope of L and the point (0, b) is called the y-intercept. The equation above is called the slope-intercept equation of L.

For this line, m = 3 and b = -4.

Nonvertical Lines

.bmxy 43 xy


Lines – Positive SlopeEXAMPLEEXAMPLE

The following are graphs of equations of lines that have positive slopes.


Lines – Negative SlopeEXAMPLEEXAMPLE

The following are graphs of equations of lines that have negative slopes.


Interpretation of a GraphEXAMPLEEXAMPLE

SOLUTIONSOLUTION

A salesperson’s weekly pay depends on the volume of sales. If she sells x units of goods, then her pay is y = 5x + 60 dollars. Give an interpretation of the slope and the y-intercept of this straight line.

First, let’s graph the line to help us understand the exercise.

-40

60

160

260

360

460

-20 30 80 130

# of sales

pay

(80, 460)


Interpretation of a Graph

The slope is 5, or 5/1. Since the numerator of this fraction represents the amount of change in her pay relative to the amount of change in her sales, the denominator, for every 1 sale that she makes, her pay increases by 5 dollars.

The y-intercept is 60 and occurs on the graph at the point (0, 60). This point suggests that when she has executed 0 sales, her pay is 60 dollars. This $60 could be referred to as her base pay.

CONTINUECONTINUEDD


Properties of the Slope of a Nonvertical Line


Properties of the Slope of a Line


Finding Slope and y-intercept of a LineEXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find the slope and y-intercept of the line

First, we write the equation in slope-intercept form.

.3

1

xy

31

xy This is the given equation.

31

3

xy Divide both terms of the numerator of the right side by 3.

31

31

xy Rewrite . 31 as

3xx

Since the number being multiplied by x is 1/3, 1/3 is the slope of the line. Since the other 1/3 is the number being added to the term containing x, 1/3, or (0, 1/3), is the y-intercept.

Incidentally, it was a complete coincidence that the slope and y-intercept were the same number. This does not normally occur.


Sketching Graphs of LinesEXAMPLEEXAMPLE

SOLUTIONSOLUTION

Sketch the graph of the line passing through (-1, 1) with slope ½.

We use Slope Property 1. We begin at the given point (-1, 1) and from there, move up one unit and to the right two units to find another point on the line.

-5

-3

-1

1

3

5

-5 -3 -1 1 3 5

(-1, 1)

-5

-3

-1

1

3

5

-5 -3 -1 1 3 5

(-1, 1)


Sketching Graphs of Lines

Now we connect the two points that have already been determined, since two points determine a straight line.

-5

-3

-1

1

3

5

-5 -3 -1 1 3 5

CONTINUECONTINUEDD


Making Equations of LinesEXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find an equation of the line that passes through the points (-1/2, 0) and (1, 2).

To find an equation of the line that passes through those two points, we need a point (we already have two) and a slope. We do not yet have the slope so we must find it. Using the two points we will determine the slope by using Slope Property 2.

34

322

232

21102

m

We now use Slope Property 3 to find an equation of the line. To use this property we need the slope and a point. We can use either of the two points that were initially provided. We’ll use the second (the first would work just as well).


Making Equations of Lines

11 xxmyy

1342 xy

34342 xy

3234 xy

CONTINUECONTINUEDD

This is the equation from Property 3.

(x1, y1) = (1, 2) and m = 4/3.

Distribute.

Add 2 to both sides of the equation.

NOTE: Technically, we could have stopped when the equation looked like since it is an equation that represents the line and is equivalent to our final equation.

1342 xy


Making Equations of LinesEXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find an equation of the line that passes through the point (2, 0) and is perpendicular to the line y = 2x.

To find an equation of the line, we need a point (we already have one) and a slope. We do not yet have the slope so we must find it. We know that the line we desire is perpendicular to the line y = 2x. Using Slope Property 5, we know that the product of the slope of the line desired and the slope of the line y = 2x is -1. We recognize that the line y = 2x is in slope-intercept form and therefore the slope of the line is 2. We can now find the slope of the line that we desire. Let the slope of the new line be m.

This is Property 5.(slope of a line)(slope of a new line) = -1

The slope of one line is 2 and the slope of the desired line is denoted by m.

2m = -1

Divide.m = -0.5

Now we can find the equation of the desired line using Property 3.


Making Equations of Lines

11 xxmyy

25.00 xy

15.0 xy

CONTINUECONTINUEDD

This is the equation from Property 3.

(x1, y1) = (2, 0) and m = -0.5.

Distribute.


Slope as a Rate of ChangeEXAMPLEEXAMPLE

SOLUTIONSOLUTION

Compute the rate of change of the function over the given intervals.

We first get y by itself in the given equation.

This is the given equation.

5.0 ,0, 1 ,0272 xy

272 xy

Subtract 2x from both sides.272 xy

Since this is clearly a linear function (since it’s now in slope-intercept form) it has constant slope, namely -2. Therefore, by definition, it also has a constant rate of change, -2. Therefore, no matter what interval is considered for this function, the rate of change will be -2. Therefore the answer, for both intervals, is -2.


Tangent Lines

Definition ExampleTangent Line to a Circle at a Point P: The straight line that touches the circle at just the one point P


Slope of a Curve & Tangent Lines

Definition ExampleThe Slope of a Curve at a Point P: The slope of the tangent line to the curve at P

(Enlargements)


Slope of a GraphEXAMPLEEXAMPLE

SOLUTIONSOLUTION

Estimate the slope of the curve at the designated point P.

The slope of a graph at a point is by definition the slope of the tangent line at that point. The figure above shows that the tangent line at P rises one unit for each unit change in x. Thus the slope of the tangent line at P is

.111

in changein change

xy


Slope of a Curve: Rate of Change


Equation & Slope of a Tangent LineEXAMPLEEXAMPLE

Find the slope of the tangent line to the graph of y = x2 at the point (-0.4, 0.16) and then write the corresponding equation of the tangent line.

SOLUTIONSOLUTIONThe slope of the graph of y = x2 at the point (x, y) is 2x. The x-coordinate of (-0.4, 0.16) is -0.4, so the slope of y = x2 at this point is 2(-0.4) = -0.8.

We shall write the equation of the tangent line in point-slope form. The point is (-0.4, 0.16) and the slope (which we just found) is -0.8. Hence the equation is:

. 4.08.016.0 xy


The Derivative

Definition ExampleDerivative: The slope formula for a function y = f (x), denoted:

Given the function f (x) = x3, the derivative is

. xfy . 3 2xxf


Differentiation

Definition ExampleDifferentiation: The process of computing a derivative.

No example will be given at this time since we do not yet know how to compute derivatives. But don’t worry, you’ll soon be able to do basic differentiation in your sleep.


Differentiation Examples

These examples can be summarized by the following rule.


Differentiation ExamplesEXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find the derivative of

This is the given equation.

.17 x

xf

7

1x

xf

Rewrite the denominator as an exponent. 71

1x

xf

Rewrite with a negative exponent. 71xxf

What we’ve done so far has been done for the sole purpose of rewriting the function in the form of f (x) = xr.

Use the Power Rule where r = -1/7 and then simplify.

787771171

71

71

71 xxxxf


Differentiation ExamplesEXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find the slope of the curve y = x5 at x = -2.

This is the given function. 5xxf

We must first find the derivative of the given function.

Use the Power Rule. 45xxf

Since the derivative function yields information about the slope of the original function, we can now use to determine the slope of the original function at x = -2. xf

Replace x with -2. 4252 f

Evaluate. 801652 f

Therefore, the slope of the original function at x = -2 is 80 (or 80/1).


Equation of the Tangent Line to the Graph of y = f (x) at the point (a, f (a))


Equation of the Tangent LineEXAMPLEEXAMPLE

SOLUTIONSOLUTION

Find the equation of the tangent line to the graph of f (x) = 3x at x = 4.

This is the given function. xxf 3

We must first find the derivative of the given function.

Differentiate. 3 xf

Notice that in this case the derivative function is a constant function, 3. Therefore, at x = 4, or any other value, the value of the derivative will be 3. So now we use the Equation of the Tangent Line that we just saw.

444 xffy This is the Equation of the Tangent Line.

4312 xy f (4) = 12 and .3 xf

xy 3 Simplify.


Leibniz Notation for Derivatives

Ultimately, this notation is a better and more effective notation for working with derivatives.


Calculating Derivatives Via the Difference Quotient

The Difference Quotient is .

hxfhxf



EXAMPLEEXAMPLE

SOLUTIONSOLUTION

Apply the three-step method to compute the derivative of the following function:

This is the difference quotient.

hxfhxf

STEP 1: We calculate the difference quotient and simplify as much as possible.

. 2 where,1 2 xxff

Evaluate f (x + h) and f (x). h

xhx 22 22

Simplify. h

xhxhx 222 222

Simplify.h

xhxhx 222 222

Simplify.h

hxh 22



STEP 2: As h approaches zero, the expression -2x – h approaches -2x, and hence the difference quotient approaches -2x.

Factor.

hhhx

2

CONTINUECONTINUEDD

Cancel and simplify.hx 2

STEP 3: Since the difference quotient approaches the derivative of f (x) = -x2 + 2 as h approaches zero, we conclude that . 2x-xf


Limit Definition of the Derivative


Limit Calculation of the DerivativeEXAMPLEEXAMPLE

SOLUTIONSOLUTION

Using limits, apply the three-step method to compute the derivative of the following function:

This is the difference quotient.

hxfhxf

. 25.02 xxxf

Evaluate f (x + h) and f (x). h

xxhxhx 25.025.0 22

Simplify. h

xxhxhxhx 25.025.02 222

Simplify.h

xxhxhxhx 25.025.05.02 222

Simplify.h

hhxh 5.02 2

STEP 1:


Limit Calculation of the Derivative

Factor.

hhhx 5.02

CONTINUECONTINUEDD

Cancel and simplify.5.02 hx

STEP 2: As h approaches 0, the expression -2x – h + 0.5 approaches -2x + 0.5.

STEP 3: Since the difference quotient approaches , we conclude that xf . 5.02 xxf


Limit Theorems


Using Limits to Calculate a Derivative


Using Limits to Calculate a DerivativeEXAMPLEEXAMPLE

SOLUTIONSOLUTION

Use limits to compute the derivative for the function .52

1

x

xf

We must calculate .33lim0 h

fhfh

3f

h

hh

fhfhh

5321

5321

lim33lim00

hh

h

561

5261

lim0

hh

h

111

2111

lim0

hhh

hh

211211

111

2111

1111

lim0

hh

hh

h

22121211

2212111

lim0

hhh

h

2212121111

lim0


Using Limits to Calculate a DerivativeCONTINUECONTINUE

DD

hh

h

h

221212

lim0

1

221212

lim0 h

hh

h

hhh

h

122121

2lim0

hhh

h

122121

2lim0

hh 221212lim

0

Now that replacing h with 0 will not cause the denominator to be equal to 0, we use Limit Theorem VIII.

0221212

1212


More Work With Derivatives and LimitsEXAMPLEEXAMPLE

SOLUTIONSOLUTION

Match the limit with a derivative. Then find the limit by computing the derivative.

The idea here is to identify the given limit as a derivative given by for a

specific choice of f and x. Toward this end, let us rewrite the limit as follows.

1

111lim

0 hhh

h

xfhxfh

0

lim

hh

hh hh

11

1

lim11

11lim00

Now go back to . Take f (x) = 1/x and evaluate according to the limit

definition of the derivative:

h

xfhxfh

0

lim 1f

.1

11

lim10 h

hfh


More Work With Derivatives and Limits

On the right side we have the desired limit; while on the left side can be computed using the power rule (where r = -1):

.1111

11lim0

f

hhh

Hence,

1f

,2 xxf

CONTINUECONTINUEDD

11 f


Limits as x Increases Without BoundEXAMPLEEXAMPLE

SOLUTIONSOLUTION

Calculate the following limit.

Both 10x + 100 and x2 – 30 increase without bound as x does. To determine the limit of their quotient, we employ an algebraic trick. Divide both numerator and denominator by x2 (since the highest power of x in either the numerator or the denominator is 2) to obtain

3010010lim 2

x

xx

.301

10010

lim3010010lim

2

2

2

x

xxxx

xx

As x increases without bound, 10/x approaches 0, 100/x2 approaches 0, and 30/x2 approaches 0. Therefore, as x increases without bound, 10/x + 100/x2 approaches 0 + 0 = 0 and 1 - 30/x2 approaches 1 – 0 = 1. Therefore,


Limits as x Increases Without Bound

.010

0100

301

10010

lim3010010lim

2

2

2

x

xxxx

xx

CONTINUECONTINUEDD


Rules of Differentiation


Differentiation

SOLUTIONSOLUTION

Differentiate

This is the given function.

.1

13

xx

xf

EXAMPLEEXAMPLE

1

13

xx

xf

We begin to differentiate.

11

3 xxdxdxf

Rewrite the rational expression with a negative exponent. 13 1

xx

dxdxf

Use the General Power Rule taking x3 + x + 1 to be g(x).

111 323 xx

dxdxxxf

Use the Sum Rule.

111 323

dxdx

dxdx

dxdxxxf


Differentiation

Differentiate. 01311 223 xxxxf

CONTINUECONTINUEDD

Simplify. 1311 223 xxxxf

Simplify. 131

11 223

xxx

xf

Simplify. 23

2

113

xxxxf


Differentiation

SOLUTIONSOLUTION

Differentiate

This is the given function.

.1

45xx

xf

EXAMPLEEXAMPLE


Rewrite the rational expression with a negative exponent.

1145 xx

dxdxf

Use the General Power Rule taking to be g(x).

1145 xx

dxdxf Use the Constant Multiple Rule.

xx

xf

1

45

xxdx

dxf1

45

xx 1 xx

dxdxxxf

11145

2


Differentiation

Use the Sum Rule and rewrite as x1/2.

21211145 x

dxdx

dxd

dxdxxxf

CONTINUECONTINUEDD

x

Differentiate.

212

21101145 xxxxf

Simplify.

212

2111145 xxxxf

Simplify.

xxx

xf 15.011

145 2

Simplify. 21

5.0145

xx

xxf


The Derivative as a Rate of Change


The Derivative as a Rate of Change

SOLUTIONSOLUTION

Let S(x) represent the total sales (in thousands of dollars) for month x in the year 2005 at a certain department store. Represent each statement below by an equation involving S or .

EXAMPLEEXAMPLE

S

(a) The sales at the end of January reached $120,560 and were rising at the rate of $1500 per month.

(b) At the end of March, the sales for this month dropped to $80,000 and were falling by about $200 a day (Use 1 month = 30 days).

(a) Since the sales at the end of January (the first month, so x = 1) reached $120,560 and S(x) represents the amount of sales for a given month, we have: S(1) = 120,560. Further, since the rate of change of sales (rate of change means we will use the derivative of S(x)) for the month of January is a positive $1500 per month, we have: .15001 S

(b) At the end of March (the third month, so x = 3), the sales dropped to $80,000. Therefore, sales for the month of March was $80,000. That is: S(3) = 80,000. Additionally, since sales were dropping by $200 per day during March, this means that the rate of change of the function S(x) was (30 days) x (-200 dollars) = -6000 dollars per month. Therefore, we have: .60003 S


Differentiating Various Independent Variables

SOLUTIONSOLUTION

Find the first derivative.

EXAMPLEEXAMPLE

3221 ttT

We first note that the independent variable is t and the dependent variable is T. This is significant inasmuch as they are considered to be two totally different variables, just as x and y are different from each other. We now proceed to differentiate the function.

3221 ttT This is the given function.

3221 ttdtd

dtdT


3221 tdtdt

dtd

dtdT

Use the Sum Rule.

21 321212 ttdtdt

dtdT

Use the General Power Rule.


Differentiating Various Independent Variables

21 32212 ttdtdT

Finish differentiating.

CONTINUECONTINUEDD

23214 ttdtdT

Simplify.

2384 ttdtdT

Simplify.


Second Derivatives

SOLUTIONSOLUTION

Find the first and second derivatives.

EXAMPLEEXAMPLE

513 PPf

This is the given function. 513 PPf

This is the first derivative. 41315 PPf

This is the second derivative. 34 131801315 PPdPdPf

dPdPf


Second Derivatives Evaluated at a Point

SOLUTIONSOLUTION

Compute the following.

EXAMPLEEXAMPLE

2

242

2

43

x

xxdxd

Compute the first derivative. xxxxdxd

dxdy 81243 324

Compute the second derivative. 836812 232

2

xxx

dxd

dxdy

dxd

dxyd

Evaluate the second derivative at x = 2. 15284368236836 2

2

2

22

2

x

x

xdx

yd


Average Rate of Change


Average Rate of Change

SOLUTIONSOLUTION

Suppose that f (x) = -6/x. What is the average rate of change of f (x) over the interval 1 to 1.2?

EXAMPLEEXAMPLE

The average rate of change over the interval is2.11 x

.52.0

12.0

6512.1

16

2.16


Instantaneous Rate of Change


Instantaneous Rate of Change

SOLUTIONSOLUTION

Suppose that f (x) = -6/x. What is the (instantaneous) rate of change of f (x) when x = 1?

EXAMPLEEXAMPLE

2

6x

xf

The rate of change of f (x) at x = 1 is equal to . We have 1f

.616

161 2 f

That is, the rate of change is 6 units per unit change in x.


Average Velocity

Definition ExampleAverage Velocity: Given a position function s(t), the average velocity from time t = a to t = a + h is

Suppose a car is 3 miles from its starting point after 5 minutes and 7 miles from its starting point after an additional 6 minutes (after a total of 11 minutes). The average velocity of the car between the two given locations is

miles per minute where a = 5 and h = 6.

.elapsed time traveleddistance

hashas

ahaashas

3

264

637

565565

ss


Position, Velocity & Acceleration

s(t) is the position function, v(t) is the velocity function, and a(t) is the acceleration function.



SOLUTIONSOLUTION

A toy rocket fired straight up into the air has height s(t) = 160t – 16t2 feet after t seconds.

EXAMPLEEXAMPLE

216160 ttts

(a) What is the rocket’s initial velocity (when t = 0)?

(b) What is the velocity after 2 seconds?

(c) What is the acceleration when t = 3?

(d) At what time will the rocket hit the ground?

(e) At what velocity will the rocket be traveling just as it smashes into the ground?

(a) To determine what the rocket’s initial velocity is, we must first have a velocity function and then evaluate t = 0 in that function.

This is the given position function.

ttstv 32160 Differentiate to get the velocity function.



Therefore, the initial velocity of the rocket is 160 feet per second.

16001600321600 v Now replace t with 0 and evaluate.

CONTINUECONTINUEDD

(b) To determine the velocity after 2 seconds we evaluate v(2).

ttv 32160 This is the velocity function.

96641602321602 v Replace t with 2 and evaluate.

Therefore, the velocity of the rocket after 2 seconds is 96 feet per second.

(c) To determine the acceleration when t = 3, we must first find the acceleration function.


32 tvta Differentiate to get the acceleration function.

Since the acceleration function is a constant function, the acceleration of the rocket is a constant -32 ft/s2. Therefore, the acceleration when t = 3 is -32ft/s2.


Position, Velocity & AccelerationCONTINUECONTINUE

DD

Therefore, the rocket will be 0 feet above the ground at times t = 0 and t = 10. t = 0 corresponds to when the rocket first began its flight, so that would not be the solution. So, the rocket hit the ground after 10 seconds, when t – 10 = 0.

(d) To determine at what time the rocket will hit the ground we must determine what we know about the position, velocity, or acceleration of the rocket when the rocket hits the ground. We know that at the time of impact, the position of the rocket will be 0 feet above the ground. Therefore, we will use the position function and replace s(t) with 0.

216160 ttts This is the given position function.2161600 tt Replace s(t) with 0.

2100 tt Factor 16 out of both terms on the right and divide both sides by 16.

tt 100 Factor.

100 tt Solve for t.


Position, Velocity & AccelerationCONTINUECONTINUE

DD

Therefore, when the rocket hits the ground, it will be have a velocity of -160 ft/s. That is, it will be traveling 160 ft/s in the downward direction.

(e) To determine at what velocity the rocket will be traveling just as it smashes into the ground, we must use the velocity function. The question is, what do we use for t? From part (d), we know that the rocket will hit the ground at t = 10 seconds. Therefore, we will find v(10).


160320160103216010 v Replace t with 10 and evaluate.

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2010 Pearson Education Inc.Goldstein/Schneider/Lay/Asmar, CALCULUS AND ITS APPLICATIONS, 12e Slide...

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