chapter 3. applications on differentiation 3.1 extrema on

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Lecture Note 3.1 Extrema on an Interval 3.2 Rolle’s Theorem and the Mean Value Theorem 3.3 Increasing and Decreasing Functions and the First Derivative Test 3.4 Concavity and the Second Derivative Test 3.5 Limits at Infinity 3.6 A Summary of Curve Sketching 3.7 Optimization Problems 3.8 Newton’s Method 3.9 Differentials Chapter 3. Applications on Differentiation

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Page 1: Chapter 3. Applications on Differentiation 3.1 Extrema on

Lecture Note

3.1 Extrema on an Interval

3.2 Rolle’s Theorem and the Mean Value Theorem

3.3 Increasing and Decreasing Functions and the First Derivative Test

3.4 Concavity and the Second Derivative Test

3.5 Limits at Infinity

3.6 A Summary of Curve Sketching

3.7 Optimization Problems

3.8 Newton’s Method

3.9 Differentials

Chapter 3. Applications on Differentiation

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Lecture Note

3.7 Optimization Problems

Objectives

β€’ Solve applied minimum and maximum problems.

One of the most common applications of calculus involves the determination of minimum and maximum values. Consider how frequently you hear or read terms such as greatest profit, least cost, least time, greatest voltage, optimum size, least size, greatest strength, and greatest distance. In this section, we will see how derivatives can be used to solve such problems.

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Guidelines for Solving Applied Minimum and Maximum Problems.

1. Identify all given quantities and all quantities to be determined. If possible, make a sketch.

2. Write a primary equation for the quantity that to be maximized or minimized.

3. Reduce the primary equation to one having a single independent variable. This may involve the use of secondary equations relating the independent variables of the primary equation.

4. Determine the feasible domain of the primary equation. That is, determine the values for which the stated problem makes sense.

5. Determine the desired maximum or minimum value by the calculus techniques discussed in Sections 3.1 through 3.4.

Lecture Note

Applied Minimum and Maximum Problems

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Example 1.

A manufacturer wants to design an open box having a square base and a surface area of 108 square inches. What dimensions will produce a box with maximum volume?

Lecture Note

Applied Minimum and Maximum Problems

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Example 2.

Which point on the graph of 𝑦 = 4 βˆ’ π‘₯2 are closest to the point (0, 2)?

Lecture Note

Applied Minimum and Maximum Problems

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Example 3.

A rectangular page is to contain 24 square inches of print. The margins at the top and bottom of the page are to be 1Β½ inches, and the margins on the left and right are to be 1 inch. What should the dimensions of the page be so that the least amount of paper is used?

Lecture Note

Applied Minimum and Maximum Problems

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Example 4.

Two posts, one 12 feet high and the other 28 feet high, stand 30 feet apart. They are to be stayed by two wires, attached to a single stake, running from ground level to the top of each post. Where should the stake be placed to use the least amount of wire?

Lecture Note

Applied Minimum and Maximum Problems

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Example 5.

Four feet of wire is to be used to form a square and a circle. How much of the wire should be used for the square and how much should be used for the circle to enclose the maximum total area?

Lecture Note

Applied Minimum and Maximum Problems

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Lecture Note

3.1 Extrema on an Interval

3.2 Rolle’s Theorem and the Mean Value Theorem

3.3 Increasing and Decreasing Functions and the First Derivative Test

3.4 Concavity and the Second Derivative Test

3.5 Limits at Infinity

3.6 A Summary of Curve Sketching

3.7 Optimization Problems

3.8 Newton’s Method

3.9 Differentials

Chapter 3. Applications on Differentiation

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Lecture Note

3.8 Newton’s Method

Objectives

β€’ Approximate a zero of a function using Newton’s Method.

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Newton’s Method is a technique for approximating the real zeros of a function. The following formula

π‘₯𝑛+1 = π‘₯𝑛 βˆ’π‘“ π‘₯𝑛𝑓′ π‘₯𝑛

is used for (𝑛 + 1)-th approximation to a zero of 𝑓(π‘₯).

The π‘₯𝑛+1 formula is the π‘₯-intercept of the tangent line to the graph of 𝑦 = 𝑓(π‘₯)

at the point π‘₯𝑛, 𝑓 π‘₯𝑛 .

Lecture Note

Newton’s Method

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IdeaThe π‘₯-intercept of the tangent line approximates a zero of 𝑓.

Lecture Note

Newton’s Method

Newton’s Method for Approximation the Zeros of a FunctionLet 𝑓 𝑐 = 0, where 𝑓 is differentiable on an open interval containing 𝑐. Then to approximate 𝑐, use these steps.1. Make an initial guess π‘₯1.2. Use the following formula to obtain the next

approximation

π‘₯𝑛+1 = π‘₯𝑛 βˆ’π‘“ π‘₯𝑛𝑓′ π‘₯𝑛

3. Repeat the formula until |π‘₯𝑛 βˆ’ π‘₯𝑛+1| is within the desired accuracy.

This procedure is called an iteration.

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Example 1. Calculate three iterations of Newton’s Method to approximate a zero of 𝑓 π‘₯ = π‘₯2 βˆ’ 2. Use π‘₯1 = 1 as the initial guess.

Lecture Note

Newton’s Method

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Example 2. Use Newton’s Method to approximate the zeros of𝑓 π‘₯ = 2π‘₯3 + π‘₯2 βˆ’ π‘₯ + 1.

Continue the iterations until two successive approximations differ by less than 0.0001. Use π‘₯1 = βˆ’1.2 as the initial guess.

Lecture Note

Newton’s Method

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Convergence of Newton’s Method

When the approximations approach a limit, the sequence of approximationsπ‘₯1, π‘₯2, π‘₯3, β‹― , π‘₯𝑛, β‹―

is said to converge. Moreover, when the limit is 𝑐, it can be shown that 𝑐 must be a zero of 𝑓.Newton’s Method does not always yield a convergent sequence. One way it can fail to do so is when 𝑓′ π‘₯𝑛 = 0 for some 𝑛. Because Newton’s Method involves division by 𝑓′(π‘₯𝑛), it is clear that the method will fail when the derivative is zero for any π‘₯𝑛 in the sequence. When you encounter this problem, you can usually overcome it by choosing a different value for π‘₯1. Other ways Newton’s Method can fail is shown in the next examples.

Lecture Note

Newton’s Method

Example 3. The function 𝑓 π‘₯ = π‘₯1/3 is not differentiable at π‘₯ = 0. Show that Newton’s Method fails to converge using π‘₯1 = 0.1.

Example 4. Show that Newton’s Method for the function 𝑓 π‘₯ = π‘₯3 βˆ’ 2π‘₯ βˆ’ 2fails to converge using π‘₯1 = 0.

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Sufficient Condition for the Convergence of Newton’s Method

𝑓 π‘₯ 𝑓′′ π‘₯

𝑓′ π‘₯ 2< 1 on an open interval containing the zero.

Lecture Note

Newton’s Method

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Lecture Note

3.1 Extrema on an Interval

3.2 Rolle’s Theorem and the Mean Value Theorem

3.3 Increasing and Decreasing Functions and the First Derivative Test

3.4 Concavity and the Second Derivative Test

3.5 Limits at Infinity

3.6 A Summary of Curve Sketching

3.7 Optimization Problems

3.8 Newton’s Method

3.9 Differentials

Chapter 3. Applications on Differentiation

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Lecture Note

3.9 Differentials

Objectives

β€’ Understand the concept of a tangent line approximation.

β€’ Compare the value of the differential, 𝑑𝑦, with the actual change in 𝑦, Δ𝑦.

β€’ Estimate a propagated error using a differential.

β€’ Find the differential of a function using differentiation formulas.

Newton’s Method is an example of the use of a tangent line to approximatethe graph of a function. In this section, you will study other situations inwhich the graph of a function can be approximated by a straight line which isthe tangent line. We will study some related notations.

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Consider a function 𝑓 that is differentiable at 𝑐. The equation for the tangent line at the point (𝑐, 𝑓 𝑐 ) is

𝑦 βˆ’ 𝑓 𝑐 = 𝑓′ 𝑐 π‘₯ βˆ’ 𝑐 β‡’ π’š = 𝒇 𝒄 + 𝒇′(𝒄)(𝒙 βˆ’ 𝒄)

and is called the tangent line approximation (or linear approximation) of 𝑓 at 𝑐.

Because 𝑐 is a constant, 𝑦 is a linear function of π‘₯. Moreover, by restricting the values of π‘₯ to those sufficiently close to 𝑐, the values of 𝑦 can be used as approximations (to any desired degree of accuracy) of the values of the function 𝑓. In other words, as π‘₯ approaches 𝑐, the limit of 𝑦 is 𝑓(𝑐).

Lecture Note

Tangent Line Approximations

𝑦 = 𝑓(π‘₯)

𝑐 π‘₯0

𝑓 𝑐 + 𝑓′(𝑐)(π‘₯0 βˆ’ 𝑐)

𝑓(π‘₯0)

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Example 1. Find the tangent line approximation of 𝑓 π‘₯ = 1 + sin π‘₯ at the point (0, 1). Then use a table to compare the 𝑦-values of the linear function with those of 𝑓(π‘₯) on an open interval containing π‘₯ = 0. For example, try at π‘₯ =βˆ’ 0.5, βˆ’0.1, βˆ’0.01, 0, 0.01, 0.1, and 0.5.

Lecture Note

Tangent Line Approximations

Tangent line

𝑓 π‘₯ = 1 + sin π‘₯

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When the tangent line to the graph of 𝑓 at the point (𝑐, 𝑓 𝑐 )𝑦 = 𝑓 𝑐 + 𝑓′(𝑐)(π‘₯ βˆ’ 𝑐)

is used as an approximation of the graph of 𝑓, the quantity π‘₯ βˆ’ 𝑐 is called the change in π‘₯, and is denoted by Ξ”π‘₯. When Ξ”π‘₯ is small, the change in 𝑦 (denoted by Δ𝑦 can be approximated as shown.

Δ𝑦 = 𝑓 𝑐 + Ξ”π‘₯ βˆ’ 𝑓 𝑐 β‰ˆ 𝑓′ 𝑐 Ξ”π‘₯For such an approximation, the quantity Ξ”π‘₯ is traditionally denoted by 𝑑π‘₯ and is called the differential of 𝒙. The expression 𝑓′ π‘₯ 𝑑π‘₯ is denoted by 𝑑𝑦 and is called the differential of π’š.

Lecture Note

Differentials

𝑦 = 𝑓(π‘₯)

𝑐 𝑐 + Ξ”π‘₯

(𝑐 + Ξ”π‘₯, 𝑓(𝑐 + Ξ”π‘₯)

𝑐, 𝑓 𝑐

Δ𝑦

𝑓(𝑐)

𝑓′ 𝑐 Ξ”π‘₯ 𝑓(𝑐 + Ξ”π‘₯)

𝑦 = 𝑓 𝑐 + 𝑓′(𝑐)(π‘₯ βˆ’ 𝑐)

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When the tangent line to the graph of 𝑓 at the point (𝑐, 𝑓 𝑐 )𝑦 = 𝑓 𝑐 + 𝑓′(𝑐)(π‘₯ βˆ’ 𝑐)

is used as an approximation of the graph of 𝑓, the quantity π‘₯ βˆ’ 𝑐 is called the change in π‘₯, and is denoted by Ξ”π‘₯. When Ξ”π‘₯ is small, the change in 𝑦 (denoted by Δ𝑦 can be approximated as shown.

Δ𝑦 = 𝑓 𝑐 + Ξ”π‘₯ βˆ’ 𝑓 𝑐 β‰ˆ 𝑓′ 𝑐 Ξ”π‘₯For such an approximation, the quantity Ξ”π‘₯ is traditionally denoted by 𝑑π‘₯ and is called the differential of 𝒙. The expression 𝑓′ π‘₯ 𝑑π‘₯ is denoted by 𝑑𝑦 and is called the differential of π’š.

Lecture Note

Differentials

Example 2. Let 𝑦 = π‘₯2. Find 𝑑𝑦 when π‘₯ = 1 and 𝑑π‘₯ = 0.01. Compare this value with Δ𝑦 for π‘₯ = 1 and Ξ”π‘₯ = 0.01.

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Physicists and engineers tend to make liberal use of the approximation of Δ𝑦 and 𝑑𝑦. One way this occurs in practices is in the estimation of errors propagated by physical measuring devices. For example, if you let π‘₯ represent the measured value of a variable and let π‘₯ + Ξ”π‘₯ represent the exact value, then Ξ”π‘₯ is the error in measurement. Finally, if the measured value π‘₯ is used to compute another value 𝑓(π‘₯), then the difference between 𝑓(π‘₯ + Ξ”π‘₯) and 𝑓(π‘₯) is the propagated error.

𝑓 π‘₯ + Ξ”π‘₯ βˆ’ 𝑓 π‘₯ = Δ𝑦

Lecture Note

Error Propagation

Example 3. The measured radius of a ball bearing is 0.7 inch, as shown in the figure. The measurement is correct to within 0.01 inch. Estimate the propagated error in the volume 𝑉 of the ball bearing.

Would you say that the propagated error is large or small? The answer is best given in relative terms by comparing 𝑑𝑉 and 𝑉. The ratio

𝑑𝑉

𝑉=4πœ‹π‘Ÿ2π‘‘π‘Ÿ

43πœ‹π‘Ÿ

3=3π‘‘π‘Ÿ

π‘Ÿβ‰ˆ3 Β±0.01

0.7β‰ˆ Β±0.0429

is called the relative error. The corresponding percent error is approximately 4.29%.

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Each of the differentiation rules that you studied in Chapter 2 can be written in differential form. For example, let 𝑒 and 𝑣 be differentiable functions of π‘₯. By the definition of differentials, you have

𝑑𝑒 = 𝑒′𝑑π‘₯ and 𝑑𝑣 = 𝑣′𝑑π‘₯.So, you can write the differential form of the Product Rule as shown below.

𝑑 𝑒𝑣 =𝑑

𝑑π‘₯𝑒𝑣 𝑑π‘₯ = 𝑒𝑣′ + 𝑣𝑒′ 𝑑π‘₯ = 𝑒𝑣′𝑑π‘₯ + 𝑣𝑒′𝑑π‘₯ = 𝑒 𝑑𝑣 + 𝑣 𝑑𝑒

Lecture Note

Calculating Differentials

Differential FormulasLet 𝑒 and 𝑣 be differentiable functions of π‘₯ and 𝑐 be a constant.

Constant multiple: 𝑑 𝑐𝑒 = 𝑐 𝑑𝑒Sum or difference: 𝑑 𝑒 Β± 𝑣 = 𝑑𝑒 Β± 𝑑𝑣Product: 𝑑 𝑒𝑣 = 𝑒 𝑑𝑣 + 𝑣 𝑑𝑒

Quotient: 𝑑𝑒

𝑣=𝑣 𝑑𝑒 βˆ’ 𝑒 𝑑𝑣

𝑣2

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Example 4 Find the differential of each function.

𝐚. 𝑦 = π‘₯2 𝐛. 𝑦 = π‘₯ 𝐜. 𝑦 = 2 sin π‘₯ 𝐝. 𝑦 = π‘₯ cos π‘₯ 𝐞. 𝑦 =1

π‘₯

Lecture Note

Calculating Differentials

Example 5-6 Find the differential of each function.

𝐚. 𝑦 = 𝑓 π‘₯ = sin 3π‘₯ 𝐛. 𝑦 = 𝑓 π‘₯ = π‘₯2 + 1 1/2

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Differentials can be used to approximate function values. To do this for the function given by 𝑦 = 𝑓(π‘₯), use the formula

𝑓 π‘₯ + Ξ”π‘₯ β‰ˆ 𝑓 π‘₯ + 𝑑𝑦 = 𝑓 π‘₯ + 𝑓′ π‘₯ 𝑑π‘₯which is derived from the approximation

Δ𝑦 = 𝑓 π‘₯ + Ξ”π‘₯ βˆ’ 𝑓 π‘₯ β‰ˆ 𝑑𝑦The key to using this formula is to choose a value for π‘₯ that makes the calculations easier.

Lecture Note

Calculating Differentials

Example 7 Use differentials to approximate 16.5.