Chapter 3. Applications on Differentiation 3.1 Extrema on ...
Transcript of Chapter 3. Applications on Differentiation 3.1 Extrema on ...
Lecture Note
3.1 Extrema on an Interval
3.2 Rolleβs 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 Newtonβs Method
3.9 Differentials
Chapter 3. Applications on Differentiation
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.
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
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
Example 2.
Which point on the graph of π¦ = 4 β π₯2 are closest to the point (0, 2)?
Lecture Note
Applied Minimum and Maximum Problems
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
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
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
Lecture Note
3.1 Extrema on an Interval
3.2 Rolleβs 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 Newtonβs Method
3.9 Differentials
Chapter 3. Applications on Differentiation
Lecture Note
3.8 Newtonβs Method
Objectives
β’ Approximate a zero of a function using Newtonβs Method.
Newtonβs Method is a technique for approximating the real zeros of a function. The following formula
π₯π+1 = π₯π βπ π₯ππβ² π₯π
is used for (π + 1)-th approximation to a zero of π(π₯).
The π₯π+1 formula is the π₯-intercept of the tangent line to the graph of π¦ = π(π₯)
at the point π₯π, π π₯π .
Lecture Note
Newtonβs Method
IdeaThe π₯-intercept of the tangent line approximates a zero of π.
Lecture Note
Newtonβs Method
Newtonβs Method for Approximation the Zeros of a FunctionLet π π = 0, where π is differentiable on an open interval containing π. Then to approximate π, use these steps.1. Make an initial guess π₯1.2. Use the following formula to obtain the next
approximation
π₯π+1 = π₯π βπ π₯ππβ² π₯π
3. Repeat the formula until |π₯π β π₯π+1| is within the desired accuracy.
This procedure is called an iteration.
Example 1. Calculate three iterations of Newtonβs Method to approximate a zero of π π₯ = π₯2 β 2. Use π₯1 = 1 as the initial guess.
Lecture Note
Newtonβs Method
Example 2. Use Newtonβs Method to approximate the zeros ofπ π₯ = 2π₯3 + π₯2 β π₯ + 1.
Continue the iterations until two successive approximations differ by less than 0.0001. Use π₯1 = β1.2 as the initial guess.
Lecture Note
Newtonβs Method
Convergence of Newtonβs Method
When the approximations approach a limit, the sequence of approximationsπ₯1, π₯2, π₯3, β― , π₯π, β―
is said to converge. Moreover, when the limit is π, it can be shown that π must be a zero of π.Newtonβs Method does not always yield a convergent sequence. One way it can fail to do so is when πβ² π₯π = 0 for some π. Because Newtonβs Method involves division by πβ²(π₯π), it is clear that the method will fail when the derivative is zero for any π₯π in the sequence. When you encounter this problem, you can usually overcome it by choosing a different value for π₯1. Other ways Newtonβs Method can fail is shown in the next examples.
Lecture Note
Newtonβs Method
Example 3. The function π π₯ = π₯1/3 is not differentiable at π₯ = 0. Show that Newtonβs Method fails to converge using π₯1 = 0.1.
Example 4. Show that Newtonβs Method for the function π π₯ = π₯3 β 2π₯ β 2fails to converge using π₯1 = 0.
Sufficient Condition for the Convergence of Newtonβs Method
π π₯ πβ²β² π₯
πβ² π₯ 2< 1 on an open interval containing the zero.
Lecture Note
Newtonβs Method
Lecture Note
3.1 Extrema on an Interval
3.2 Rolleβs 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 Newtonβs Method
3.9 Differentials
Chapter 3. Applications on Differentiation
Lecture Note
3.9 Differentials
Objectives
β’ Understand the concept of a tangent line approximation.
β’ Compare the value of the differential, ππ¦, with the actual change in π¦, Ξπ¦.
β’ Estimate a propagated error using a differential.
β’ Find the differential of a function using differentiation formulas.
Newtonβs 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.
Consider a function π that is differentiable at π. The equation for the tangent line at the point (π, π π ) is
π¦ β π π = πβ² π π₯ β π β π = π π + πβ²(π)(π β π)
and is called the tangent line approximation (or linear approximation) of π at π.
Because π is a constant, π¦ is a linear function of π₯. Moreover, by restricting the values of π₯ to those sufficiently close to π, the values of π¦ can be used as approximations (to any desired degree of accuracy) of the values of the function π. In other words, as π₯ approaches π, the limit of π¦ is π(π).
Lecture Note
Tangent Line Approximations
π¦ = π(π₯)
π π₯0
π π + πβ²(π)(π₯0 β π)
π(π₯0)
Example 1. Find the tangent line approximation of π π₯ = 1 + sin π₯ at the point (0, 1). Then use a table to compare the π¦-values of the linear function with those of π(π₯) on an open interval containing π₯ = 0. For example, try at π₯ =β 0.5, β0.1, β0.01, 0, 0.01, 0.1, and 0.5.
Lecture Note
Tangent Line Approximations
Tangent line
π π₯ = 1 + sin π₯
When the tangent line to the graph of π at the point (π, π π )π¦ = π π + πβ²(π)(π₯ β π)
is used as an approximation of the graph of π, the quantity π₯ β π is called the change in π₯, and is denoted by Ξπ₯. When Ξπ₯ is small, the change in π¦ (denoted by Ξπ¦ can be approximated as shown.
Ξπ¦ = π π + Ξπ₯ β π π β πβ² π Ξπ₯For such an approximation, the quantity Ξπ₯ is traditionally denoted by ππ₯ and is called the differential of π. The expression πβ² π₯ ππ₯ is denoted by ππ¦ and is called the differential of π.
Lecture Note
Differentials
π¦ = π(π₯)
π π + Ξπ₯
(π + Ξπ₯, π(π + Ξπ₯)
π, π π
Ξπ¦
π(π)
πβ² π Ξπ₯ π(π + Ξπ₯)
π¦ = π π + πβ²(π)(π₯ β π)
When the tangent line to the graph of π at the point (π, π π )π¦ = π π + πβ²(π)(π₯ β π)
is used as an approximation of the graph of π, the quantity π₯ β π is called the change in π₯, and is denoted by Ξπ₯. When Ξπ₯ is small, the change in π¦ (denoted by Ξπ¦ can be approximated as shown.
Ξπ¦ = π π + Ξπ₯ β π π β πβ² π Ξπ₯For such an approximation, the quantity Ξπ₯ is traditionally denoted by ππ₯ and is called the differential of π. The expression πβ² π₯ ππ₯ is denoted by ππ¦ and is called the differential of π.
Lecture Note
Differentials
Example 2. Let π¦ = π₯2. Find ππ¦ when π₯ = 1 and ππ₯ = 0.01. Compare this value with Ξπ¦ for π₯ = 1 and Ξπ₯ = 0.01.
Physicists and engineers tend to make liberal use of the approximation of Ξπ¦ and ππ¦. 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₯ + Ξπ₯ represent the exact value, then Ξπ₯ is the error in measurement. Finally, if the measured value π₯ is used to compute another value π(π₯), then the difference between π(π₯ + Ξπ₯) and π(π₯) is the propagated error.
π π₯ + Ξπ₯ β π π₯ = Ξπ¦
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 π 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 ππ and π. The ratio
ππ
π=4ππ2ππ
43ππ
3=3ππ
πβ3 Β±0.01
0.7β Β±0.0429
is called the relative error. The corresponding percent error is approximately 4.29%.
Each of the differentiation rules that you studied in Chapter 2 can be written in differential form. For example, let π’ and π£ be differentiable functions of π₯. By the definition of differentials, you have
ππ’ = π’β²ππ₯ and ππ£ = π£β²ππ₯.So, you can write the differential form of the Product Rule as shown below.
π π’π£ =π
ππ₯π’π£ ππ₯ = π’π£β² + π£π’β² ππ₯ = π’π£β²ππ₯ + π£π’β²ππ₯ = π’ ππ£ + π£ ππ’
Lecture Note
Calculating Differentials
Differential FormulasLet π’ and π£ be differentiable functions of π₯ and π be a constant.
Constant multiple: π ππ’ = π ππ’Sum or difference: π π’ Β± π£ = ππ’ Β± ππ£Product: π π’π£ = π’ ππ£ + π£ ππ’
Quotient: ππ’
π£=π£ ππ’ β π’ ππ£
π£2
Example 4 Find the differential of each function.
π. π¦ = π₯2 π. π¦ = π₯ π. π¦ = 2 sin π₯ π. π¦ = π₯ cos π₯ π. π¦ =1
π₯
Lecture Note
Calculating Differentials
Example 5-6 Find the differential of each function.
π. π¦ = π π₯ = sin 3π₯ π. π¦ = π π₯ = π₯2 + 1 1/2
Differentials can be used to approximate function values. To do this for the function given by π¦ = π(π₯), use the formula
π π₯ + Ξπ₯ β π π₯ + ππ¦ = π π₯ + πβ² π₯ ππ₯which is derived from the approximation
Ξπ¦ = π π₯ + Ξπ₯ β π π₯ β ππ¦The 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.