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Calculus Volume 1

4.2 Linear Approximations and Differentials

Calculus Volume 14.2 Linear Approximations and Differentials
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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

  • 4.2.1. Describe the linear approximation to a function at a point.
  • 4.2.2. Write the linearization of a given function.
  • 4.2.3. Draw a graph that illustrates the use of differentials to approximate the change in a quantity.
  • 4.2.4. Calculate the relative error and percentage error in using a differential approximation.

We have just seen how derivatives allow us to compare related quantities that are changing over time. In this section, we examine another application of derivatives: the ability to approximate functions locally by linear functions. Linear functions are the easiest functions with which to work, so they provide a useful tool for approximating function values. In addition, the ideas presented in this section are generalized later in the text when we study how to approximate functions by higher-degree polynomials Introduction to Power Series and Functions.

Linear Approximation of a Function at a Point

Consider a function ff that is differentiable at a point x=a.x=a. Recall that the tangent line to the graph of ff at aa is given by the equation

y=f(a)+f(a)(xa).y=f(a)+f(a)(xa).

For example, consider the function f(x)=1xf(x)=1x at a=2.a=2. Since ff is differentiable at x=2x=2 and f(x)=1x2,f(x)=1x2, we see that f(2)=14.f(2)=14. Therefore, the tangent line to the graph of ff at a=2a=2 is given by the equation

y=1214(x2).y=1214(x2).

Figure 4.7(a) shows a graph of f(x)=1xf(x)=1x along with the tangent line to ff at x=2.x=2. Note that for xx near 2, the graph of the tangent line is close to the graph of f.f. As a result, we can use the equation of the tangent line to approximate f(x)f(x) for xx near 2. For example, if x=2.1,x=2.1, the yy value of the corresponding point on the tangent line is

y=1214(2.12)=0.475.y=1214(2.12)=0.475.

The actual value of f(2.1)f(2.1) is given by

f(2.1)=12.10.47619.f(2.1)=12.10.47619.

Therefore, the tangent line gives us a fairly good approximation of f(2.1)f(2.1) (Figure 4.7(b)). However, note that for values of xx far from 2, the equation of the tangent line does not give us a good approximation. For example, if x=10,x=10, the yy-value of the corresponding point on the tangent line is

y=1214(102)=122=−1.5,y=1214(102)=122=−1.5,

whereas the value of the function at x=10x=10 is f(10)=0.1.f(10)=0.1.

This figure has two parts a and b. In figure a, the line f(x) = 1/x is shown with its tangent line at x = 2. In figure b, the area near the tangent point is blown up to show how good of an approximation the tangent is near x = 2.
Figure 4.7 (a) The tangent line to f(x)=1/xf(x)=1/x at x=2x=2 provides a good approximation to ff for xx near 2. (b) At x=2.1,x=2.1, the value of yy on the tangent line to f(x)=1/xf(x)=1/x is 0.475. The actual value of f(2.1)f(2.1) is 1/2.1,1/2.1, which is approximately 0.47619.

In general, for a differentiable function f,f, the equation of the tangent line to ff at x=ax=a can be used to approximate f(x)f(x) for xx near a.a. Therefore, we can write

f(x)f(a)+f(a)(xa)forxneara.f(x)f(a)+f(a)(xa)forxneara.

We call the linear function

L(x)=f(a)+f(a)(xa)L(x)=f(a)+f(a)(xa)
4.1

the linear approximation, or tangent line approximation, of ff at x=a.x=a. This function LL is also known as the linearization of ff at x=a.x=a.

To show how useful the linear approximation can be, we look at how to find the linear approximation for f(x)=xf(x)=x at x=9.x=9.

Example 4.5

Linear Approximation of xx

Find the linear approximation of f(x)=xf(x)=x at x=9x=9 and use the approximation to estimate 9.1.9.1.

Solution

Since we are looking for the linear approximation at x=9,x=9, using Equation 4.1 we know the linear approximation is given by

L(x)=f(9)+f(9)(x9).L(x)=f(9)+f(9)(x9).

We need to find f(9)f(9) and f(9).f(9).

f(x)=xf(9)=9=3f(x)=12xf(9)=129=16f(x)=xf(9)=9=3f(x)=12xf(9)=129=16

Therefore, the linear approximation is given by Figure 4.8.

L(x)=3+16(x9)L(x)=3+16(x9)

Using the linear approximation, we can estimate 9.19.1 by writing

9.1=f(9.1)L(9.1)=3+16(9.19)3.0167.9.1=f(9.1)L(9.1)=3+16(9.19)3.0167.
The function f(x) = the square root of x is shown with its tangent at (9, 3). The tangent appears to be a very good approximation from x = 6 to x = 12.
Figure 4.8 The local linear approximation to f(x)=xf(x)=x at x=9x=9 provides an approximation to ff for xx near 9.

Analysis

Using a calculator, the value of 9.19.1 to four decimal places is 3.0166. The value given by the linear approximation, 3.0167, is very close to the value obtained with a calculator, so it appears that using this linear approximation is a good way to estimate x,x, at least for xx near 9.9. At the same time, it may seem odd to use a linear approximation when we can just push a few buttons on a calculator to evaluate 9.1.9.1. However, how does the calculator evaluate 9.1?9.1? The calculator uses an approximation! In fact, calculators and computers use approximations all the time to evaluate mathematical expressions; they just use higher-degree approximations.

Checkpoint 4.5

Find the local linear approximation to f(x)=x3f(x)=x3 at x=8.x=8. Use it to approximate 8.138.13 to five decimal places.

Example 4.6

Linear Approximation of sinxsinx

Find the linear approximation of f(x)=sinxf(x)=sinx at x=π3x=π3 and use it to approximate sin(62°).sin(62°).

Solution

First we note that since π3π3 rad is equivalent to 60°,60°, using the linear approximation at x=π/3x=π/3 seems reasonable. The linear approximation is given by

L(x)=f(π3)+f(π3)(xπ3).L(x)=f(π3)+f(π3)(xπ3).

We see that

f(x)=sinxf(π3)=sin(π3)=32f(x)=cosxf(π3)=cos(π3)=12f(x)=sinxf(π3)=sin(π3)=32f(x)=cosxf(π3)=cos(π3)=12

Therefore, the linear approximation of ff at x=π/3x=π/3 is given by Figure 4.9.

L(x)=32+12(xπ3)L(x)=32+12(xπ3)

To estimate sin(62°)sin(62°) using L,L, we must first convert 62°62° to radians. We have 62°=62π18062°=62π180 radians, so the estimate for sin(62°)sin(62°) is given by

sin(62°)=f(62π180)L(62π180)=32+12(62π180π3)=32+12(2π180)=32+π1800.88348.sin(62°)=f(62π180)L(62π180)=32+12(62π180π3)=32+12(2π180)=32+π1800.88348.
The function f(x) = sin x is shown with its tangent at (π/3, square root of 3 / 2). The tangent appears to be a very good approximation for x near π / 3.
Figure 4.9 The linear approximation to f(x)=sinxf(x)=sinx at x=π/3x=π/3 provides an approximation to sinxsinx for xx near π/3.π/3.
Checkpoint 4.6

Find the linear approximation for f(x)=cosxf(x)=cosx at x=π2.x=π2.

Linear approximations may be used in estimating roots and powers. In the next example, we find the linear approximation for f(x)=(1+x)nf(x)=(1+x)n at x=0,x=0, which can be used to estimate roots and powers for real numbers near 1. The same idea can be extended to a function of the form f(x)=(m+x)nf(x)=(m+x)n to estimate roots and powers near a different number m.m.

Example 4.7

Approximating Roots and Powers

Find the linear approximation of f(x)=(1+x)nf(x)=(1+x)n at x=0.x=0. Use this approximation to estimate (1.01)3.(1.01)3.

Solution

The linear approximation at x=0x=0 is given by

L(x)=f(0)+f(0)(x0).L(x)=f(0)+f(0)(x0).

Because

f(x)=(1+x)nf(0)=1f(x)=n(1+x)n1f(0)=n,f(x)=(1+x)nf(0)=1f(x)=n(1+x)n1f(0)=n,

the linear approximation is given by Figure 4.10(a).

L(x)=1+n(x0)=1+nxL(x)=1+n(x0)=1+nx

We can approximate (1.01)3(1.01)3 by evaluating L(0.01)L(0.01) when n=3.n=3. We conclude that

(1.01)3=f(1.01)L(1.01)=1+3(0.01)=1.03.(1.01)3=f(1.01)L(1.01)=1+3(0.01)=1.03.
This figure has two parts a and b. In figure a, the line f(x) = (1 + x)3 is shown with its tangent line at (0, 1). In figure b, the area near the tangent point is blown up to show how good of an approximation the tangent is near (0, 1).
Figure 4.10 (a) The linear approximation of f(x)f(x) at x=0x=0 is L(x).L(x). (b) The actual value of 1.0131.013 is 1.030301. The linear approximation of f(x)f(x) at x=0x=0 estimates 1.0131.013 to be 1.03.
Checkpoint 4.7

Find the linear approximation of f(x)=(1+x)4f(x)=(1+x)4 at x=0x=0 without using the result from the preceding example.

Differentials

We have seen that linear approximations can be used to estimate function values. They can also be used to estimate the amount a function value changes as a result of a small change in the input. To discuss this more formally, we define a related concept: differentials. Differentials provide us with a way of estimating the amount a function changes as a result of a small change in input values.

When we first looked at derivatives, we used the Leibniz notation dy/dxdy/dx to represent the derivative of yy with respect to x.x. Although we used the expressions dy and dx in this notation, they did not have meaning on their own. Here we see a meaning to the expressions dy and dx. Suppose y=f(x)y=f(x) is a differentiable function. Let dx be an independent variable that can be assigned any nonzero real number, and define the dependent variable dydy by

dy=f(x)dx.dy=f(x)dx.
4.2

It is important to notice that dydy is a function of both xx and dx.dx. The expressions dy and dx are called differentials. We can divide both sides of Equation 4.2 by dx,dx, which yields

dydx=f(x).dydx=f(x).
4.3

This is the familiar expression we have used to denote a derivative. Equation 4.2 is known as the differential form of Equation 4.3.

Example 4.8

Computing differentials

For each of the following functions, find dy and evaluate when x=3x=3 and dx=0.1.dx=0.1.

  1. y=x2+2xy=x2+2x
  2. y=cosxy=cosx

Solution

The key step is calculating the derivative. When we have that, we can obtain dy directly.

  1. Since f(x)=x2+2x,f(x)=x2+2x, we know f(x)=2x+2,f(x)=2x+2, and therefore
    dy=(2x+2)dx.dy=(2x+2)dx.

    When x=3x=3 and dx=0.1,dx=0.1,
    dy=(2·3+2)(0.1)=0.8.dy=(2·3+2)(0.1)=0.8.
  2. Since f(x)=cosx,f(x)=cosx, f(x)=sin(x).f(x)=sin(x). This gives us
    dy=sinxdx.dy=sinxdx.

    When x=3x=3 and dx=0.1,dx=0.1,
    dy=sin(3)(0.1)=−0.1sin(3).dy=sin(3)(0.1)=−0.1sin(3).
Checkpoint 4.8

For y=ex2,y=ex2, find dy.dy.

We now connect differentials to linear approximations. Differentials can be used to estimate the change in the value of a function resulting from a small change in input values. Consider a function ff that is differentiable at point a.a. Suppose the input xx changes by a small amount. We are interested in how much the output yy changes. If xx changes from aa to a+dx,a+dx, then the change in xx is dxdx (also denoted Δx),Δx), and the change in yy is given by

Δy=f(a+dx)f(a).Δy=f(a+dx)f(a).

Instead of calculating the exact change in y,y, however, it is often easier to approximate the change in yy by using a linear approximation. For xx near a,a, f(x)f(x) can be approximated by the linear approximation

L(x)=f(a)+f(a)(xa).L(x)=f(a)+f(a)(xa).

Therefore, if dxdx is small,

f(a+dx)L(a+dx)=f(a)+f(a)(a+dxa).f(a+dx)L(a+dx)=f(a)+f(a)(a+dxa).

That is,

f(a+dx)f(a)L(a+dx)f(a)=f(a)dx.f(a+dx)f(a)L(a+dx)f(a)=f(a)dx.

In other words, the actual change in the function ff if xx increases from aa to a+dxa+dx is approximately the difference between L(a+dx)L(a+dx) and f(a),f(a), where L(x)L(x) is the linear approximation of ff at a.a. By definition of L(x),L(x), this difference is equal to f(a)dx.f(a)dx. In summary,

Δy=f(a+dx)f(a)L(a+dx)f(a)=f(a)dx=dy.Δy=f(a+dx)f(a)L(a+dx)f(a)=f(a)dx=dy.

Therefore, we can use the differential dy=f(a)dxdy=f(a)dx to approximate the change in yy if xx increases from x=ax=a to x=a+dx.x=a+dx. We can see this in the following graph.

A function y = f(x) is shown along with its tangent line at (a, f(a)). The tangent line is denoted L(x). The x axis is marked with a and a + dx, with a dashed line showing the distance between a and a + dx as dx. The points (a + dx, f(a + dx)) and (a + dx, L(a + dx)) are marked on the curves for y = f(x) and y = L(x), respectively. The distance between f(a) and L(a + dx) is marked as dy = f’(a) dx, and the distance between f(a) and f(a + dx) is marked as Δy = f(a + dx) – f(a).
Figure 4.11 The differential dy=f(a)dxdy=f(a)dx is used to approximate the actual change in yy if xx increases from aa to a+dx.a+dx.

We now take a look at how to use differentials to approximate the change in the value of the function that results from a small change in the value of the input. Note the calculation with differentials is much simpler than calculating actual values of functions and the result is very close to what we would obtain with the more exact calculation.

Example 4.9

Approximating Change with Differentials

Let y=x2+2x.y=x2+2x. Compute ΔyΔy and dy at x=3x=3 if dx=0.1.dx=0.1.

Solution

The actual change in yy if xx changes from x=3x=3 to x=3.1x=3.1 is given by

Δy=f(3.1)f(3)=[(3.1)2+2(3.1)][32+2(3)]=0.81.Δy=f(3.1)f(3)=[(3.1)2+2(3.1)][32+2(3)]=0.81.

The approximate change in yy is given by dy=f(3)dx.dy=f(3)dx. Since f(x)=2x+2,f(x)=2x+2, we have

dy=f(3)dx=(2(3)+2)(0.1)=0.8.dy=f(3)dx=(2(3)+2)(0.1)=0.8.
Checkpoint 4.9

For y=x2+2x,y=x2+2x, find ΔyΔy and dydy at x=3x=3 if dx=0.2.dx=0.2.

Calculating the Amount of Error

Any type of measurement is prone to a certain amount of error. In many applications, certain quantities are calculated based on measurements. For example, the area of a circle is calculated by measuring the radius of the circle. An error in the measurement of the radius leads to an error in the computed value of the area. Here we examine this type of error and study how differentials can be used to estimate the error.

Consider a function ff with an input that is a measured quantity. Suppose the exact value of the measured quantity is a,a, but the measured value is a+dx.a+dx. We say the measurement error is dx (or Δx).Δx). As a result, an error occurs in the calculated quantity f(x).f(x). This type of error is known as a propagated error and is given by

Δy=f(a+dx)f(a).Δy=f(a+dx)f(a).

Since all measurements are prone to some degree of error, we do not know the exact value of a measured quantity, so we cannot calculate the propagated error exactly. However, given an estimate of the accuracy of a measurement, we can use differentials to approximate the propagated error Δy.Δy. Specifically, if ff is a differentiable function at a,a, the propagated error is

Δydy=f(a)dx.Δydy=f(a)dx.

Unfortunately, we do not know the exact value a.a. However, we can use the measured value a+dx,a+dx, and estimate

Δydyf(a+dx)dx.Δydyf(a+dx)dx.

In the next example, we look at how differentials can be used to estimate the error in calculating the volume of a box if we assume the measurement of the side length is made with a certain amount of accuracy.

Example 4.10

Volume of a Cube

Suppose the side length of a cube is measured to be 5 cm with an accuracy of 0.1 cm.

  1. Use differentials to estimate the error in the computed volume of the cube.
  2. Compute the volume of the cube if the side length is (i) 4.9 cm and (ii) 5.1 cm to compare the estimated error with the actual potential error.

Solution

  1. The measurement of the side length is accurate to within ±0.1±0.1 cm. Therefore,
    −0.1dx0.1.−0.1dx0.1.

    The volume of a cube is given by V=x3,V=x3, which leads to
    dV=3x2dx.dV=3x2dx.

    Using the measured side length of 5 cm, we can estimate that
    −3(5)2(0.1)dV3(5)2(0.1).−3(5)2(0.1)dV3(5)2(0.1).

    Therefore,
    −7.5dV7.5.−7.5dV7.5.
  2. If the side length is actually 4.9 cm, then the volume of the cube is
    V(4.9)=(4.9)3=117.649cm3.V(4.9)=(4.9)3=117.649cm3.

    If the side length is actually 5.1 cm, then the volume of the cube is
    V(5.1)=(5.1)3=132.651cm3.V(5.1)=(5.1)3=132.651cm3.

    Therefore, the actual volume of the cube is between 117.649 and 132.651. Since the side length is measured to be 5 cm, the computed volume is V(5)=53=125.V(5)=53=125. Therefore, the error in the computed volume is
    117.649125ΔV132.651125.117.649125ΔV132.651125.

    That is,
    −7.351ΔV7.651.−7.351ΔV7.651.

    We see the estimated error dVdV is relatively close to the actual potential error in the computed volume.
Checkpoint 4.10

Estimate the error in the computed volume of a cube if the side length is measured to be 6 cm with an accuracy of 0.2 cm.

The measurement error dx (x)(x) and the propagated error ΔyΔy are absolute errors. We are typically interested in the size of an error relative to the size of the quantity being measured or calculated. Given an absolute error ΔqΔq for a particular quantity, we define the relative error as Δqq,Δqq, where qq is the actual value of the quantity. The percentage error is the relative error expressed as a percentage. For example, if we measure the height of a ladder to be 63 in. when the actual height is 62 in., the absolute error is 1 in. but the relative error is 162=0.016,162=0.016, or 1.6%.1.6%. By comparison, if we measure the width of a piece of cardboard to be 8.25 in. when the actual width is 8 in., our absolute error is 1414 in., whereas the relative error is 0.258=132,0.258=132, or 3.1%.3.1%. Therefore, the percentage error in the measurement of the cardboard is larger, even though 0.25 in. is less than 1 in.

Example 4.11

Relative and Percentage Error

An astronaut using a camera measures the radius of Earth as 4000 mi with an error of ±80±80 mi. Let’s use differentials to estimate the relative and percentage error of using this radius measurement to calculate the volume of Earth, assuming the planet is a perfect sphere.

Solution

If the measurement of the radius is accurate to within ±80,±80, we have

−80dr80.−80dr80.

Since the volume of a sphere is given by V=(43)πr3,V=(43)πr3, we have

dV=4πr2dr.dV=4πr2dr.

Using the measured radius of 4000 mi, we can estimate

−4π(4000)2(80)dV4π(4000)2(80).−4π(4000)2(80)dV4π(4000)2(80).

To estimate the relative error, consider dVV.dVV. Since we do not know the exact value of the volume V,V, use the measured radius r=4000mir=4000mi to estimate V.V. We obtain V(43)π(4000)3.V(43)π(4000)3. Therefore the relative error satisfies

−4π(4000)2(80)4π(4000)3/3dVV4π(4000)2(80)4π(4000)3/3,−4π(4000)2(80)4π(4000)3/3dVV4π(4000)2(80)4π(4000)3/3,

which simplifies to

−0.06dVV0.06.−0.06dVV0.06.

The relative error is 0.06 and the percentage error is 6%.6%.

Checkpoint 4.11

Determine the percentage error if the radius of Earth is measured to be 3950 mi with an error of ±100±100 mi.

Section 4.2 Exercises

46.

What is the linear approximation for any generic linear function y=mx+b?y=mx+b?

47.

Determine the necessary conditions such that the linear approximation function is constant. Use a graph to prove your result.

48.

Explain why the linear approximation becomes less accurate as you increase the distance between xx and a.a. Use a graph to prove your argument.

49.

When is the linear approximation exact?

For the following exercises, find the linear approximation L(x)L(x) to y=f(x)y=f(x) near x=ax=a for the function.

50.

f(x)=x+x4,a=0f(x)=x+x4,a=0

51.

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

52.

f(x)=tanx,a=π4f(x)=tanx,a=π4

53.

f(x)=sinx,a=π2f(x)=sinx,a=π2

54.

f(x)=xsinx,a=2πf(x)=xsinx,a=2π

55.

f(x)=sin2x,a=0f(x)=sin2x,a=0

For the following exercises, compute the values given within 0.01 by deciding on the appropriate f(x)f(x) and a,a, and evaluating L(x)=f(a)+f(a)(xa).L(x)=f(a)+f(a)(xa). Check your answer using a calculator.

56.

[T] (2.001)6(2.001)6

57.

[T] sin(0.02)sin(0.02)

58.

[T] cos(0.03)cos(0.03)

59.

[T] (15.99)1/4(15.99)1/4

60.

[T] 10.9810.98

61.

[T] sin(3.14)sin(3.14)

For the following exercises, determine the appropriate f(x)f(x) and a,a, and evaluate L(x)=f(a)+f(a)(xa).L(x)=f(a)+f(a)(xa). Calculate the numerical error in the linear approximations that follow.

62.

[T] (1.01)3(1.01)3

63.

[T] cos(0.01)cos(0.01)

64.

[T] (sin(0.01))2(sin(0.01))2

65.

[T] (1.01)−3(1.01)−3

66.

[T] (1+110)10(1+110)10

67.

[T] 8.998.99

For the following exercises, find the differential of the function.

68.

y=3x4+x22x+1y=3x4+x22x+1

69.

y=xcosxy=xcosx

70.

y=1+xy=1+x

71.

y=x2+2x1y=x2+2x1

For the following exercises, find the differential and evaluate for the given xx and dx.dx.

72.

y=3x2x+6,y=3x2x+6, x=2,x=2, dx=0.1dx=0.1

73.

y=1x+1,y=1x+1, x=1,x=1, dx=0.25dx=0.25

74.

y=tanx,y=tanx, x=0,x=0, dx=π10dx=π10

75.

y=3x2+2x+1,y=3x2+2x+1, x=0,x=0, dx=0.1dx=0.1

76.

y=sin(2x)x,y=sin(2x)x, x=π,x=π, dx=0.25dx=0.25

77.

y=x3+2x+1x,y=x3+2x+1x, x=1,x=1, dx=0.05dx=0.05

For the following exercises, find the change in volume dVdV or in surface area dA.dA.

78.

dVdV if the sides of a cube change from 10 to 10.1.

79.

dAdA if the sides of a cube change from xx to x+dx.x+dx.

80.

dAdA if the radius of a sphere changes from rr by dr.dr.

81.

dVdV if the radius of a sphere changes from rr by dr.dr.

82.

dVdV if a circular cylinder with r=2r=2 changes height from 3 cm to 3.05cm.3.05cm.

83.

dVdV if a circular cylinder of height 3 changes from r=2r=2 to r=1.9cm.r=1.9cm.

For the following exercises, use differentials to estimate the maximum and relative error when computing the surface area or volume.

84.

A spherical golf ball is measured to have a radius of 5mm,5mm, with a possible measurement error of 0.1mm.0.1mm. What is the possible change in volume?

85.

A pool has a rectangular base of 10 ft by 20 ft and a depth of 6 ft. What is the change in volume if you only fill it up to 5.5 ft?

86.

An ice cream cone has height 4 in. and radius 1 in. If the cone is 0.1 in. thick, what is the difference between the volume of the cone, including the shell, and the volume of the ice cream you can fit inside the shell?

For the following exercises, confirm the approximations by using the linear approximation at x=0.x=0.

87.

1x112x1x112x

88.

11x2111x21

89.

c2+x2cc2+x2c

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