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Calculus Volume 1

3.1 Defining the Derivative

Calculus Volume 13.1 Defining the Derivative
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  1. Preface
  2. 1 Functions and Graphs
    1. Introduction
    2. 1.1 Review of Functions
    3. 1.2 Basic Classes of Functions
    4. 1.3 Trigonometric Functions
    5. 1.4 Inverse Functions
    6. 1.5 Exponential and Logarithmic Functions
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  3. 2 Limits
    1. Introduction
    2. 2.1 A Preview of Calculus
    3. 2.2 The Limit of a Function
    4. 2.3 The Limit Laws
    5. 2.4 Continuity
    6. 2.5 The Precise Definition of a Limit
    7. Key Terms
    8. Key Equations
    9. Key Concepts
    10. Chapter Review Exercises
  4. 3 Derivatives
    1. Introduction
    2. 3.1 Defining the Derivative
    3. 3.2 The Derivative as a Function
    4. 3.3 Differentiation Rules
    5. 3.4 Derivatives as Rates of Change
    6. 3.5 Derivatives of Trigonometric Functions
    7. 3.6 The Chain Rule
    8. 3.7 Derivatives of Inverse Functions
    9. 3.8 Implicit Differentiation
    10. 3.9 Derivatives of Exponential and Logarithmic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  5. 4 Applications of Derivatives
    1. Introduction
    2. 4.1 Related Rates
    3. 4.2 Linear Approximations and Differentials
    4. 4.3 Maxima and Minima
    5. 4.4 The Mean Value Theorem
    6. 4.5 Derivatives and the Shape of a Graph
    7. 4.6 Limits at Infinity and Asymptotes
    8. 4.7 Applied Optimization Problems
    9. 4.8 L’Hôpital’s Rule
    10. 4.9 Newton’s Method
    11. 4.10 Antiderivatives
    12. Key Terms
    13. Key Equations
    14. Key Concepts
    15. Chapter Review Exercises
  6. 5 Integration
    1. Introduction
    2. 5.1 Approximating Areas
    3. 5.2 The Definite Integral
    4. 5.3 The Fundamental Theorem of Calculus
    5. 5.4 Integration Formulas and the Net Change Theorem
    6. 5.5 Substitution
    7. 5.6 Integrals Involving Exponential and Logarithmic Functions
    8. 5.7 Integrals Resulting in Inverse Trigonometric Functions
    9. Key Terms
    10. Key Equations
    11. Key Concepts
    12. Chapter Review Exercises
  7. 6 Applications of Integration
    1. Introduction
    2. 6.1 Areas between Curves
    3. 6.2 Determining Volumes by Slicing
    4. 6.3 Volumes of Revolution: Cylindrical Shells
    5. 6.4 Arc Length of a Curve and Surface Area
    6. 6.5 Physical Applications
    7. 6.6 Moments and Centers of Mass
    8. 6.7 Integrals, Exponential Functions, and Logarithms
    9. 6.8 Exponential Growth and Decay
    10. 6.9 Calculus of the Hyperbolic Functions
    11. Key Terms
    12. Key Equations
    13. Key Concepts
    14. Chapter Review Exercises
  8. A | Table of Integrals
  9. B | Table of Derivatives
  10. C | Review of Pre-Calculus
  11. Answer Key
    1. Chapter 1
    2. Chapter 2
    3. Chapter 3
    4. Chapter 4
    5. Chapter 5
    6. Chapter 6
  12. Index

Learning Objectives

  • 3.1.1. Recognize the meaning of the tangent to a curve at a point.
  • 3.1.2. Calculate the slope of a tangent line.
  • 3.1.3. Identify the derivative as the limit of a difference quotient.
  • 3.1.4. Calculate the derivative of a given function at a point.
  • 3.1.5. Describe the velocity as a rate of change.
  • 3.1.6. Explain the difference between average velocity and instantaneous velocity.
  • 3.1.7. Estimate the derivative from a table of values.

Now that we have both a conceptual understanding of a limit and the practical ability to compute limits, we have established the foundation for our study of calculus, the branch of mathematics in which we compute derivatives and integrals. Most mathematicians and historians agree that calculus was developed independently by the Englishman Isaac Newton (1643–1727)(1643–1727) and the German Gottfried Leibniz (1646–1716),(1646–1716), whose images appear in Figure 3.2. When we credit Newton and Leibniz with developing calculus, we are really referring to the fact that Newton and Leibniz were the first to understand the relationship between the derivative and the integral. Both mathematicians benefited from the work of predecessors, such as Barrow, Fermat, and Cavalieri. The initial relationship between the two mathematicians appears to have been amicable; however, in later years a bitter controversy erupted over whose work took precedence. Although it seems likely that Newton did, indeed, arrive at the ideas behind calculus first, we are indebted to Leibniz for the notation that we commonly use today.

Photos of Newton and Leibniz.
Figure 3.2 Newton and Leibniz are credited with developing calculus independently.

Tangent Lines

We begin our study of calculus by revisiting the notion of secant lines and tangent lines. Recall that we used the slope of a secant line to a function at a point (a,f(a))(a,f(a)) to estimate the rate of change, or the rate at which one variable changes in relation to another variable. We can obtain the slope of the secant by choosing a value of xx near aa and drawing a line through the points (a,f(a))(a,f(a)) and (x,f(x)),(x,f(x)), as shown in Figure 3.3. The slope of this line is given by an equation in the form of a difference quotient:

msec=f(x)f(a)xa.msec=f(x)f(a)xa.

We can also calculate the slope of a secant line to a function at a value a by using this equation and replacing xx with a+h,a+h, where hh is a value close to 0. We can then calculate the slope of the line through the points (a,f(a))(a,f(a)) and (a+h,f(a+h)).(a+h,f(a+h)). In this case, we find the secant line has a slope given by the following difference quotient with increment h:h:

msec=f(a+h)f(a)a+ha=f(a+h)f(a)h.msec=f(a+h)f(a)a+ha=f(a+h)f(a)h.

Definition

Let ff be a function defined on an interval II containing a.a. If xaxa is in I,I, then

Q=f(x)f(a)xaQ=f(x)f(a)xa
3.1

is a difference quotient.

Also, if h0h0 is chosen so that a+ha+h is in I,I, then

Q=f(a+h)f(a)hQ=f(a+h)f(a)h
3.2

is a difference quotient with increment h.h.

Media

View the development of the derivative with this applet.

These two expressions for calculating the slope of a secant line are illustrated in Figure 3.3. We will see that each of these two methods for finding the slope of a secant line is of value. Depending on the setting, we can choose one or the other. The primary consideration in our choice usually depends on ease of calculation.

This figure consists of two graphs labeled a and b. Figure a shows the Cartesian coordinate plane with 0, a, and x marked on the x-axis. There is a curve labeled y = f(x) with points marked (a, f(a)) and (x, f(x)). There is also a straight line that crosses these two points (a, f(a)) and (x, f(x)). At the bottom of the graph, the equation msec = (f(x) - f(a))/(x - a) is given. Figure b shows a similar graph, but this time a + h is marked on the x-axis instead of x. Consequently, the curve labeled y = f(x) passes through (a, f(a)) and (a + h, f(a + h)) as does the straight line. At the bottom of the graph, the equation msec = (f(a + h) - f(a))/h is given.
Figure 3.3 We can calculate the slope of a secant line in either of two ways.

In Figure 3.4(a) we see that, as the values of xx approach a,a, the slopes of the secant lines provide better estimates of the rate of change of the function at a.a. Furthermore, the secant lines themselves approach the tangent line to the function at a,a, which represents the limit of the secant lines. Similarly, Figure 3.4(b) shows that as the values of hh get closer to 0,0, the secant lines also approach the tangent line. The slope of the tangent line at aa is the rate of change of the function at a,a, as shown in Figure 3.4(c).

This figure consists of three graphs labeled a, b, and c. Figure a shows the Cartesian coordinate plane with 0, a, x2, and x1 marked in order on the x-axis. There is a curve labeled y = f(x) with points marked (a, f(a)), (x2, f(x2)), and (x1, f(x1)). There are three straight lines: the first crosses (a, f(a)) and (x1, f(x1)); the second crosses (a, f(a)) and (x2, f(x2)); and the third only touches (a, f(a)), making it the tangent. At the bottom of the graph, the equation mtan = limx → a (f(x) - f(a))/(x - a) is given. Figure b shows a similar graph, but this time a + h2 and a + h1 are marked on the x-axis instead of x2 and x1. Consequently, the curve labeled y = f(x) passes through (a, f(a)), (a + h2, f(a + h2)), and (a + h1, f(a + h1)) and the straight lines similarly cross the graph as in Figure a. At the bottom of the graph, the equation mtan = limh → 0 (f(a + h) - f(a))/h is given. Figure c shows only the curve labeled y = f(x) and its tangent at point (a, f(a)).
Figure 3.4 The secant lines approach the tangent line (shown in green) as the second point approaches the first.

Media

You can use this site to explore graphs to see if they have a tangent line at a point.

In Figure 3.5 we show the graph of f(x)=xf(x)=x and its tangent line at (1,1)(1,1) in a series of tighter intervals about x=1.x=1. As the intervals become narrower, the graph of the function and its tangent line appear to coincide, making the values on the tangent line a good approximation to the values of the function for choices of xx close to 1.1. In fact, the graph of f(x)f(x) itself appears to be locally linear in the immediate vicinity of x=1.x=1.

This figure consists of four graphs labeled a, b, c, and d. Figure a shows the graphs of the square root of x and the equation y = (x + 1)/2 with the x-axis going from 0 to 4 and the y-axis going from 0 to 2.5. The graphs of these two functions look very close near 1; there is a box around where these graphs look close. Figure b shows a close up of these same two functions in the area of the box from Figure a, specifically x going from 0 to 2 and y going from 0 to 1.4. Figure c is the same graph as Figure b, but this one has a box from 0 to 1.1 in the x coordinate and 0.8 and 1 on the y coordinate. There is an arrow indicating that this is blown up in Figure d. Figure d shows a very close picture of the box from Figure c, and the two functions appear to be touching for almost the entire length of the graph.
Figure 3.5 For values of xx close to 1,1, the graph of f(x)=xf(x)=x and its tangent line appear to coincide.

Formally we may define the tangent line to the graph of a function as follows.

Definition

Let f(x)f(x) be a function defined in an open interval containing a.a. The tangent line to f(x)f(x) at aa is the line passing through the point (a,f(a))(a,f(a)) having slope

mtan=limxaf(x)f(a)xamtan=limxaf(x)f(a)xa
3.3

provided this limit exists.

Equivalently, we may define the tangent line to f(x)f(x) at aa to be the line passing through the point (a,f(a))(a,f(a)) having slope

mtan=limh0f(a+h)f(a)hmtan=limh0f(a+h)f(a)h
3.4

provided this limit exists.

Just as we have used two different expressions to define the slope of a secant line, we use two different forms to define the slope of the tangent line. In this text we use both forms of the definition. As before, the choice of definition will depend on the setting. Now that we have formally defined a tangent line to a function at a point, we can use this definition to find equations of tangent lines.

Example 3.1

Finding a Tangent Line

Find the equation of the line tangent to the graph of f(x)=x2f(x)=x2 at x=3.x=3.

Solution

First find the slope of the tangent line. In this example, use Equation 3.3.

mtan=limx3f(x)f(3)x3Apply the definition.=limx3x29x3Substitutef(x)=x2andf(3)=9.=limx3(x3)(x+3)x3=limx3(x+3)=6Factor the numerator to evaluate the limit.mtan=limx3f(x)f(3)x3Apply the definition.=limx3x29x3Substitutef(x)=x2andf(3)=9.=limx3(x3)(x+3)x3=limx3(x+3)=6Factor the numerator to evaluate the limit.

Next, find a point on the tangent line. Since the line is tangent to the graph of f(x)f(x) at x=3,x=3, it passes through the point (3,f(3)).(3,f(3)). We have f(3)=9,f(3)=9, so the tangent line passes through the point (3,9).(3,9).

Using the point-slope equation of the line with the slope m=6m=6 and the point (3,9),(3,9), we obtain the line y9=6(x3).y9=6(x3). Simplifying, we have y=6x9.y=6x9. The graph of f(x)=x2f(x)=x2 and its tangent line at 33 are shown in Figure 3.6.

This figure consists of the graphs of f(x) = x squared and y = 6x - 9. The graphs of these functions appear to touch at x = 3.
Figure 3.6 The tangent line to f(x)f(x) at x=3.x=3.

Example 3.2

The Slope of a Tangent Line Revisited

Use Equation 3.4 to find the slope of the line tangent to the graph of f(x)=x2f(x)=x2 at x=3.x=3.

Solution

The steps are very similar to Example 3.1. See Equation 3.4 for the definition.

mtan=limh0f(3+h)f(3)hApply the definition.=limh0(3+h)29hSubstitutef(3+h)=(3+h)2andf(3)=9.=limh09+6h+h29hExpand and simplify to evaluate the limit.=limh0h(6+h)h=limh0(6+h)=6mtan=limh0f(3+h)f(3)hApply the definition.=limh0(3+h)29hSubstitutef(3+h)=(3+h)2andf(3)=9.=limh09+6h+h29hExpand and simplify to evaluate the limit.=limh0h(6+h)h=limh0(6+h)=6

We obtained the same value for the slope of the tangent line by using the other definition, demonstrating that the formulas can be interchanged.

Example 3.3

Finding the Equation of a Tangent Line

Find the equation of the line tangent to the graph of f(x)=1/xf(x)=1/x at x=2.x=2.

Solution

We can use Equation 3.3, but as we have seen, the results are the same if we use Equation 3.4.

mtan=limx2f(x)f(2)x2Apply the definition.=limx21x12x2Substitutef(x)=1xandf(2)=12.=limx21x12x2·2x2xMultiply numerator and denominator by2xtosimplify fractions.=limx2(2x)(x2)(2x)Simplify.=limx2−12xSimplify using2xx2=−1,forx2.=14Evaluate the limit.mtan=limx2f(x)f(2)x2Apply the definition.=limx21x12x2Substitutef(x)=1xandf(2)=12.=limx21x12x2·2x2xMultiply numerator and denominator by2xtosimplify fractions.=limx2(2x)(x2)(2x)Simplify.=limx2−12xSimplify using2xx2=−1,forx2.=14Evaluate the limit.

We now know that the slope of the tangent line is 14.14. To find the equation of the tangent line, we also need a point on the line. We know that f(2)=12.f(2)=12. Since the tangent line passes through the point (2,12)(2,12) we can use the point-slope equation of a line to find the equation of the tangent line. Thus the tangent line has the equation y=14x+1.y=14x+1. The graphs of f(x)=1xf(x)=1x and y=14x+1y=14x+1 are shown in Figure 3.7.

This figure consists of the graphs of f(x) = 1/x and y = -x/4 + 1. The part of the graph f(x) = 1/x in the first quadrant appears to touch the other function’s graph at x = 2.
Figure 3.7 The line is tangent to f(x)f(x) at x=2.x=2.
Checkpoint 3.1

Find the slope of the line tangent to the graph of f(x)=xf(x)=x at x=4.x=4.

The Derivative of a Function at a Point

The type of limit we compute in order to find the slope of the line tangent to a function at a point occurs in many applications across many disciplines. These applications include velocity and acceleration in physics, marginal profit functions in business, and growth rates in biology. This limit occurs so frequently that we give this value a special name: the derivative. The process of finding a derivative is called differentiation.

Definition

Let f(x)f(x) be a function defined in an open interval containing a.a. The derivative of the function f(x)f(x) at a,a, denoted by f(a),f(a), is defined by

f(a)=limxaf(x)f(a)xaf(a)=limxaf(x)f(a)xa
3.5

provided this limit exists.

Alternatively, we may also define the derivative of f(x)f(x) at aa as

f(a)=limh0f(a+h)f(a)h.f(a)=limh0f(a+h)f(a)h.
3.6

Example 3.4

Estimating a Derivative

For f(x)=x2,f(x)=x2, use a table to estimate f(3)f(3) using Equation 3.5.

Solution

Create a table using values of xx just below 33 and just above 3.3.

xx x29x3x29x3
2.92.9 5.95.9
2.992.99 5.995.99
2.9992.999 5.9995.999
3.0013.001 6.0016.001
3.013.01 6.016.01
3.13.1 6.16.1

After examining the table, we see that a good estimate is f(3)=6.f(3)=6.

Checkpoint 3.2

For f(x)=x2,f(x)=x2, use a table to estimate f(3)f(3) using Equation 3.6.

Example 3.5

Finding a Derivative

For f(x)=3x24x+1,f(x)=3x24x+1, find f(2)f(2) by using Equation 3.5.

Solution

Substitute the given function and value directly into the equation.

f(x)=limx2f(x)f(2)x2Apply the definition.=limx2(3x24x+1)5x2Substitutef(x)=3x24x+1andf(2)=5.=limx2(x2)(3x+2)x2Simplify and factor the numerator.=limx2(3x+2)Cancel the common factor.=8Evaluate the limit.f(x)=limx2f(x)f(2)x2Apply the definition.=limx2(3x24x+1)5x2Substitutef(x)=3x24x+1andf(2)=5.=limx2(x2)(3x+2)x2Simplify and factor the numerator.=limx2(3x+2)Cancel the common factor.=8Evaluate the limit.

Example 3.6

Revisiting the Derivative

For f(x)=3x24x+1,f(x)=3x24x+1, find f(2)f(2) by using Equation 3.6.

Solution

Using this equation, we can substitute two values of the function into the equation, and we should get the same value as in Example 3.5.

f(2)=limh0f(2+h)f(2)hApply the definition.=limh0(3(2+h)24(2+h)+1)5hSubstitutef(2)=5andf(2+h)=3(2+h)24(2+h)+1.=limh03h2+8hhSimplify the numerator.=limh0h(3h+8)hFactor the numerator.=limh0(3h+8)Cancel the common factor.=8Evaluate the limit.f(2)=limh0f(2+h)f(2)hApply the definition.=limh0(3(2+h)24(2+h)+1)5hSubstitutef(2)=5andf(2+h)=3(2+h)24(2+h)+1.=limh03h2+8hhSimplify the numerator.=limh0h(3h+8)hFactor the numerator.=limh0(3h+8)Cancel the common factor.=8Evaluate the limit.

The results are the same whether we use Equation 3.5 or Equation 3.6.

Checkpoint 3.3

For f(x)=x2+3x+2,f(x)=x2+3x+2, find f(1).f(1).

Velocities and Rates of Change

Now that we can evaluate a derivative, we can use it in velocity applications. Recall that if s(t)s(t) is the position of an object moving along a coordinate axis, the average velocity of the object over a time interval [a,t][a,t] if t>at>a or [t,a][t,a] if t<at<a is given by the difference quotient

vave=s(t)s(a)ta.vave=s(t)s(a)ta.
3.7

As the values of tt approach a,a, the values of vavevave approach the value we call the instantaneous velocity at a.a. That is, instantaneous velocity at a,a, denoted v(a),v(a), is given by

v(a)=s(a)=limtas(t)s(a)ta.v(a)=s(a)=limtas(t)s(a)ta.
3.8

To better understand the relationship between average velocity and instantaneous velocity, see Figure 3.8. In this figure, the slope of the tangent line (shown in red) is the instantaneous velocity of the object at time t=at=a whose position at time tt is given by the function s(t).s(t). The slope of the secant line (shown in green) is the average velocity of the object over the time interval [a,t].[a,t].

This figure consists of the Cartesian coordinate plane with 0, a, and t1 marked on the t-axis. The function y = s(t) is graphed in the first quadrant along with two lines marked tangent and secant. The tangent line touches y = s(t) at only one point, (a, s(a)). The secant line touches y = s(t) at two points: (a, s(a)) and (t1, s(t1)).
Figure 3.8 The slope of the secant line is the average velocity over the interval [a,t].[a,t]. The slope of the tangent line is the instantaneous velocity.

We can use Equation 3.5 to calculate the instantaneous velocity, or we can estimate the velocity of a moving object by using a table of values. We can then confirm the estimate by using Equation 3.7.

Example 3.7

Estimating Velocity

A lead weight on a spring is oscillating up and down. Its position at time tt with respect to a fixed horizontal line is given by s(t)=sints(t)=sint (Figure 3.9). Use a table of values to estimate v(0).v(0). Check the estimate by using Equation 3.5.

A picture of a spring hanging down with a weight at the end. There is a horizontal dashed line marked 0 a little bit above the weight.
Figure 3.9 A lead weight suspended from a spring in vertical oscillatory motion.

Solution

We can estimate the instantaneous velocity at t=0t=0 by computing a table of average velocities using values of tt approaching 0,0, as shown in Table 3.1.

tt sintsin0t0=sinttsintsin0t0=sintt
−0.1−0.1 0.9983341660.998334166
−0.01−0.01 0.99998333330.9999833333
−0.001−0.001 0.9999998330.999999833
0.0010.001 0.9999998330.999999833
0.010.01 0.99998333330.9999833333
0.10.1 0.9983341660.998334166
Table 3.1 Average velocities using values of t approaching 0

From the table we see that the average velocity over the time interval [−0.1,0][−0.1,0] is 0.998334166,0.998334166, the average velocity over the time interval [−0.01,0][−0.01,0] is 0.9999833333,0.9999833333, and so forth. Using this table of values, it appears that a good estimate is v(0)=1.v(0)=1.

By using Equation 3.5, we can see that

v(0)=s(0)=limt0sintsin0t0=limt0sintt=1.v(0)=s(0)=limt0sintsin0t0=limt0sintt=1.

Thus, in fact, v(0)=1.v(0)=1.

Checkpoint 3.4

A rock is dropped from a height of 6464 feet. Its height above ground at time tt seconds later is given by s(t)=−16t2+64,0t2.s(t)=−16t2+64,0t2. Find its instantaneous velocity 11 second after it is dropped, using Equation 3.5.

As we have seen throughout this section, the slope of a tangent line to a function and instantaneous velocity are related concepts. Each is calculated by computing a derivative and each measures the instantaneous rate of change of a function, or the rate of change of a function at any point along the function.

Definition

The instantaneous rate of change of a function f(x)f(x) at a value aa is its derivative f(a).f(a).

Example 3.8

Chapter Opener: Estimating Rate of Change of Velocity

The same sports car speeding along a winding road from the beginning of the chapter.
Figure 3.10 (credit: modification of work by Codex41, Flickr)

Reaching a top speed of 270.49270.49 mph, the Hennessey Venom GT is one of the fastest cars in the world. In tests it went from 00 to 6060 mph in 3.053.05 seconds, from 0to1000to100 mph in 5.885.88 seconds, from 0to2000to200 mph in 14.5114.51 seconds, and from 0to229.90to229.9 mph in 19.9619.96 seconds. Use this data to draw a conclusion about the rate of change of velocity (that is, its acceleration) as it approaches 229.9229.9 mph. Does the rate at which the car is accelerating appear to be increasing, decreasing, or constant?

Solution

First observe that 6060 mph = 8888 ft/s, 100100 mph 146.67146.67 ft/s, 200200 mph 293.33293.33 ft/s, and 229.9229.9 mph 337.19337.19 ft/s. We can summarize the information in a table.

tt v(t)v(t)
00 00
3.053.05 8888
5.885.88 147.67147.67
14.5114.51 293.33293.33
19.9619.96 337.19337.19
Table 3.2 v(t)v(t) at different values of t

Now compute the average acceleration of the car in feet per second on intervals of the form [t,19.96][t,19.96] as tt approaches 19.96,19.96, as shown in the following table.

tt v(t)v(19.96)t19.96=v(t)337.19t19.96v(t)v(19.96)t19.96=v(t)337.19t19.96
0.00.0 16.8916.89
3.053.05 14.7414.74
5.885.88 13.4613.46
14.5114.51 8.058.05
Table 3.3 Average acceleration

The rate at which the car is accelerating is decreasing as its velocity approaches 229.9229.9 mph (337.19(337.19 ft/s).

Example 3.9

Rate of Change of Temperature

A homeowner sets the thermostat so that the temperature in the house begins to drop from 70°F70°F at 99 p.m., reaches a low of 60°60° during the night, and rises back to 70°70° by 77 a.m. the next morning. Suppose that the temperature in the house is given by T(t)=0.4t24t+70T(t)=0.4t24t+70 for 0t10,0t10, where tt is the number of hours past 99 p.m. Find the instantaneous rate of change of the temperature at midnight.

Solution

Since midnight is 33 hours past 99 p.m., we want to compute T(3).T(3). Refer to Equation 3.5.

T(3)=limt3T(t)T(3)t3Apply the definition.=limt30.4t24t+7061.6t3SubstituteT(t)=0.4t24t+70andT(3)=61.6.=limt30.4t24t+8.4t3Simplify.=limt30.4(t3)(t7)t3=limt30.4(t3)(t7)t3=limt30.4(t7)Cancel.=−1.6Evaluate the limit.T(3)=limt3T(t)T(3)t3Apply the definition.=limt30.4t24t+7061.6t3SubstituteT(t)=0.4t24t+70andT(3)=61.6.=limt30.4t24t+8.4t3Simplify.=limt30.4(t3)(t7)t3=limt30.4(t3)(t7)t3=limt30.4(t7)Cancel.=−1.6Evaluate the limit.

The instantaneous rate of change of the temperature at midnight is −1.6°F−1.6°F per hour.

Example 3.10

Rate of Change of Profit

A toy company can sell xx electronic gaming systems at a price of p=−0.01x+400p=−0.01x+400 dollars per gaming system. The cost of manufacturing xx systems is given by C(x)=100x+10,000C(x)=100x+10,000 dollars. Find the rate of change of profit when 10,00010,000 games are produced. Should the toy company increase or decrease production?

Solution

The profit P(x)P(x) earned by producing xx gaming systems is R(x)C(x),R(x)C(x), where R(x)R(x) is the revenue obtained from the sale of xx games. Since the company can sell xx games at p=−0.01x+400p=−0.01x+400 per game,

R(x)=xp=x(−0.01x+400)=−0.01x2+400x.R(x)=xp=x(−0.01x+400)=−0.01x2+400x.

Consequently,

P(x)=−0.01x2+300x10,000.P(x)=−0.01x2+300x10,000.

Therefore, evaluating the rate of change of profit gives

P(10000)=limx10000P(x)P(10000)x10000=limx10000−0.01x2+300x100001990000x10000=limx10000−0.01x2+300x2000000x10000=100.P(10000)=limx10000P(x)P(10000)x10000=limx10000−0.01x2+300x100001990000x10000=limx10000−0.01x2+300x2000000x10000=100.

Since the rate of change of profit P(10,000)>0P(10,000)>0 and P(10,000)>0,P(10,000)>0, the company should increase production.

Checkpoint 3.5

A coffee shop determines that the daily profit on scones obtained by charging ss dollars per scone is P(s)=−20s2+150s10.P(s)=−20s2+150s10. The coffee shop currently charges $3.25$3.25 per scone. Find P(3.25),P(3.25), the rate of change of profit when the price is $3.25$3.25 and decide whether or not the coffee shop should consider raising or lowering its prices on scones.

Section 3.1 Exercises

For the following exercises, use Equation 3.3 to find the slope of the secant line between the values x1x1 and x2x2 for each function y=f(x).y=f(x).

1.

f(x)=4x+7;x1=2,x2=5f(x)=4x+7;x1=2,x2=5

2.

f(x)=8x3;x1=−1,x2=3f(x)=8x3;x1=−1,x2=3

3.

f(x)=x2+2x+1;x1=3,x2=3.5f(x)=x2+2x+1;x1=3,x2=3.5

4.

f(x)=x2+x+2;x1=0.5,x2=1.5f(x)=x2+x+2;x1=0.5,x2=1.5

5.

f(x)=43x1;x1=1,x2=3f(x)=43x1;x1=1,x2=3

6.

f(x)=x72x+1;x1=0,x2=2f(x)=x72x+1;x1=0,x2=2

7.

f(x)=x;x1=1,x2=16f(x)=x;x1=1,x2=16

8.

f(x)=x9;x1=10,x2=13f(x)=x9;x1=10,x2=13

9.

f(x)=x1/3+1;x1=0,x2=8f(x)=x1/3+1;x1=0,x2=8

10.

f(x)=6x2/3+2x1/3;x1=1,x2=27f(x)=6x2/3+2x1/3;x1=1,x2=27

For the following functions,

  1. use Equation 3.4 to find the slope of the tangent line mtan=f(a),mtan=f(a), and
  2. find the equation of the tangent line to ff at x=a.x=a.
11.

f(x)=34x,a=2f(x)=34x,a=2

12.

f(x)=x5+6,a=−1f(x)=x5+6,a=−1

13.

f(x)=x2+x,a=1f(x)=x2+x,a=1

14.

f(x)=1xx2,a=0f(x)=1xx2,a=0

15.

f(x)=7x,a=3f(x)=7x,a=3

16.

f(x)=x+8,a=1f(x)=x+8,a=1

17.

f(x)=23x2,a=−2f(x)=23x2,a=−2

18.

f(x)=−3x1,a=4f(x)=−3x1,a=4

19.

f(x)=2x+3,a=−4f(x)=2x+3,a=−4

20.

f(x)=3x2,a=3f(x)=3x2,a=3

For the following functions y=f(x),y=f(x), find f(a)f(a) using Equation 3.3.

21.

f(x)=5x+4,a=−1f(x)=5x+4,a=−1

22.

f(x)=−7x+1,a=3f(x)=−7x+1,a=3

23.

f(x)=x2+9x,a=2f(x)=x2+9x,a=2

24.

f(x)=3x2x+2,a=1f(x)=3x2x+2,a=1

25.

f(x)=x,a=4f(x)=x,a=4

26.

f(x)=x2,a=6f(x)=x2,a=6

27.

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

28.

f(x)=1x3,a=−1f(x)=1x3,a=−1

29.

f(x)=1x3,a=1f(x)=1x3,a=1

30.

f(x)=1x,a=4f(x)=1x,a=4

For the following exercises, given the function y=f(x),y=f(x),

  1. find the slope of the secant line PQPQ for each point Q(x,f(x))Q(x,f(x)) with xx value given in the table.
  2. Use the answers from a. to estimate the value of the slope of the tangent line at P.P.
  3. Use the answer from b. to find the equation of the tangent line to ff at point P.P.
31.

[T] f(x)=x2+3x+4,P(1,8)f(x)=x2+3x+4,P(1,8) (Round to 66 decimal places.)

x Slope mPQmPQ x Slope mPQmPQ
1.1 (i) 0.9 (vii)
1.01 (ii) 0.99 (viii)
1.001 (iii) 0.999 (ix)
1.0001 (iv) 0.9999 (x)
1.00001 (v) 0.99999 (xi)
1.000001 (vi) 0.999999 (xii)
32.

[T] f(x)=x+1x21,P(0,−1)f(x)=x+1x21,P(0,−1)

x Slope mPQmPQ x Slope mPQmPQ
0.1 (i) −0.1−0.1 (vii)
0.01 (ii) −0.01−0.01 (viii)
0.001 (iii) −0.001−0.001 (ix)
0.0001 (iv) −0.0001−0.0001 (x)
0.00001 (v) −0.00001−0.00001 (xi)
0.000001 (vi) −0.000001−0.000001 (xii)
33.

[T] f(x)=10e0.5x,P(0,10)f(x)=10e0.5x,P(0,10) (Round to 44 decimal places.)

x Slope mPQmPQ
−0.1−0.1 (i)
−0.01−0.01 (ii)
−0.001−0.001 (iii)
−0.0001−0.0001 (iv)
−0.00001−0.00001 (v)
−0.000001 (vi)
34.

[T] f(x)=tan(x),P(π,0)f(x)=tan(x),P(π,0)

x Slope mPQmPQ
3.1 (i)
3.14 (ii)
3.141 (iii)
3.1415 (iv)
3.14159 (v)
3.141592 (vi)

[T] For the following position functions y=s(t),y=s(t), an object is moving along a straight line, where tt is in seconds and ss is in meters. Find

  1. the simplified expression for the average velocity from t=2t=2 to t=2+h;t=2+h;
  2. the average velocity between t=2t=2 and t=2+h,t=2+h, where (i)h=0.1,(i)h=0.1, (ii)h=0.01,(ii)h=0.01, (iii)h=0.001,(iii)h=0.001, and (iv)h=0.0001;(iv)h=0.0001; and
  3. use the answer from a. to estimate the instantaneous velocity at t=2t=2 second.
35.

s(t)=13t+5s(t)=13t+5

36.

s(t)=t22ts(t)=t22t

37.

s(t)=2t3+3s(t)=2t3+3

38.

s(t)=16t24ts(t)=16t24t

39.

Use the following graph to evaluate a. f(1)f(1) and b. f(6).f(6).

This graph shows two connected line segments: one going from (1, 0) to (4, 6) and the other going from (4, 6) to (8, 8).
40.

Use the following graph to evaluate a. f(−3)f(−3) and b. f(1.5).f(1.5).

This graph shows two connected line segments: one going from (−4, 3) to (1, 3) and the other going from (1, 3) to (1.5, 4).

For the following exercises, use the limit definition of derivative to show that the derivative does not exist at x=ax=a for each of the given functions.

41.

f(x)=x1/3,x=0f(x)=x1/3,x=0

42.

f(x)=x2/3,x=0f(x)=x2/3,x=0

43.

f(x)={1,x<1x,x1,x=1f(x)={1,x<1x,x1,x=1

44.

f(x)=|x|x,x=0f(x)=|x|x,x=0

45.

[T] The position in feet of a race car along a straight track after tt seconds is modeled by the function s(t)=8t2116t3.s(t)=8t2116t3.

  1. Find the average velocity of the vehicle over the following time intervals to four decimal places:
    1. [4, 4.1]
    2. [4, 4.01]
    3. [4, 4.001]
    4. [4, 4.0001]
  2. Use a. to draw a conclusion about the instantaneous velocity of the vehicle at t=4t=4 seconds.
46.

[T] The distance in feet that a ball rolls down an incline is modeled by the function s(t)=14t2,s(t)=14t2, where t is seconds after the ball begins rolling.

  1. Find the average velocity of the ball over the following time intervals:
    1. [5, 5.1]
    2. [5, 5.01]
    3. [5, 5.001]
    4. [5, 5.0001]
  2. Use the answers from a. to draw a conclusion about the instantaneous velocity of the ball at t=5t=5 seconds.
47.

Two vehicles start out traveling side by side along a straight road. Their position functions, shown in the following graph, are given by s=f(t)s=f(t) and s=g(t),s=g(t), where ss is measured in feet and tt is measured in seconds.

Two functions s = g(t) and s = f(t) are graphed. The first function s = g(t) starts at (0, 0) and arcs upward through roughly (2, 1) to (4, 4). The second function s = f(t) is a straight line passing through (0, 0) and (4, 4).
  1. Which vehicle has traveled farther at t=2t=2 seconds?
  2. What is the approximate velocity of each vehicle at t=3t=3 seconds?
  3. Which vehicle is traveling faster at t=4t=4 seconds?
  4. What is true about the positions of the vehicles at t=4t=4 seconds?
48.

[T] The total cost C(x),C(x), in hundreds of dollars, to produce xx jars of mayonnaise is given by C(x)=0.000003x3+4x+300.C(x)=0.000003x3+4x+300.

  1. Calculate the average cost per jar over the following intervals:
    1. [100, 100.1]
    2. [100, 100.01]
    3. [100, 100.001]
    4. [100, 100.0001]
  2. Use the answers from a. to estimate the average cost to produce 100100 jars of mayonnaise.
49.

[T] For the function f(x)=x32x211x+12,f(x)=x32x211x+12, do the following.

  1. Use a graphing calculator to graph f in an appropriate viewing window.
  2. Use the ZOOM feature on the calculator to approximate the two values of x=ax=a for which mtan=f(a)=0.mtan=f(a)=0.
50.

[T] For the function f(x)=x1+x2,f(x)=x1+x2, do the following.

  1. Use a graphing calculator to graph ff in an appropriate viewing window.
  2. Use the ZOOM feature on the calculator to approximate the values of x=ax=a for which mtan=f(a)=0.mtan=f(a)=0.
51.

Suppose that N(x)N(x) computes the number of gallons of gas used by a vehicle traveling xx miles. Suppose the vehicle gets 3030 mpg.

  1. Find a mathematical expression for N(x).N(x).
  2. What is N(100)?N(100)? Explain the physical meaning.
  3. What is N(100)?N(100)? Explain the physical meaning.
52.

[T] For the function f(x)=x45x2+4,f(x)=x45x2+4, do the following.

  1. Use a graphing calculator to graph ff in an appropriate viewing window.
  2. Use the nDerivnDeriv function, which numerically finds the derivative, on a graphing calculator to estimate f(−2),f(−0.5),f(1.7),f(−2),f(−0.5),f(1.7), and f(2.718).f(2.718).
53.

[T] For the function f(x)=x2x2+1,f(x)=x2x2+1, do the following.

  1. Use a graphing calculator to graph ff in an appropriate viewing window.
  2. Use the nDerivnDeriv function on a graphing calculator to find f(−4),f(−2),f(2),f(−4),f(−2),f(2), and f(4).f(4).
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