Applications of the Derivative Chapter 6 Applications of the Derivative

Section 6.1 Absolute Extrema

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Your Turn 1 Find the absolute extrema of the function on the interval [0,8]. Solution: First look for critical numbers in the interval (0,8). Continued

Your Turn 1 continued Notice that at x = 2/5 and 2/5 is in the interval (0,8) . The derivative is undefined at x = 0 but the function is defined there, so 0 is also a critical number. Evaluate the function at the critical numbers and the endpoints. x - Value Extrema Candidates Value of the Function 2/5 0.977 Absolute Maximum 8 -84 Absolute Minimum

Your Turn 2 Find the locations and values of the absolute extrema, if they exist, for the function Solution: In this example, the extreme value theorem does not apply since the domain is an open interval (−∞,∞), which has no endpoints. Begin as before by finding any critical numbers. x = 0 or x = − 4 or x = 1. Evaluate the function at the critical numbers. Continued

Your Turn 2 continued For an open interval, rather than evaluating the function at the endpoints, we evaluate the limit of the function when the endpoints are approached. Because the negative x4 term dominates the other terms as x becomes small, it has no absolute minimum. The absolute maximum 148, occurs at x = − 4. x - Value Extrema Candidates Value of the Function -4 148 Absolute Maximum 20 1 23

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Applications of Extrema Section 6.2 Applications of Extrema

Your Turn 1 Find two nonnegative numbers x and y for which x + 3y = 30, such that x2y is maximized. Solution: Step 1, reading and understanding the problem, is up to you. Step 2 does not apply in this example; there is nothing to draw. We proceed to Step 3, in which we decide what is to be maximized and assign a variable to that quantity. Here x2y, is to be maximized, so let M = x2y. According to Step 3, we must express M in terms of just one variable, which can be done using x + 3y = 30 the equation by solving for either x or y. Solving for y gives Continued

Your Turn 1 Continued x + 3y = 30 Substitute for y in the expression for M to get We are now ready for Step 4, when we find the domain of the function. Because of the nonnegativity requirement, x must be at least 0. Since y must also be at least 0, we require Continued

Your Turn 1 Continued Thus x is confined to the interval [0,30]. Moving on to Step 5, we find the critical points for M by finding and then solving

Your Turn 1 Continued x = 0 or x = 20 Finally, at Step 6, we find M for the critical numbers and endpoint of the domain. The other endpoint has already been included as a critical number. We see in the table that the maximum value of the function occurs when x = 20, y = 10/3 and the maximum value of x2y is 1333.3 Continued

Your Turn 1 continued x – value Extrema Candidates M 20 20 1333.3 Absolute Maximum 30

Your Turn 2 - Figure 7 A math professor participating in the sport of orienteering must get to a specific tree in the woods as fast as possible. He can get there by traveling east along the trail for 300 m and then north through the woods for 800 m. He can run 160 m per minute along the trail but only 40 m per minute through the woods. Running directly through the woods toward the tree minimizes the distance, but he will be going slowly the whole time. He could instead run 300 m along the trail before entering the woods, maximizing the total distance but minimizing the time in the woods. Perhaps the fastest route is a combination, as shown in Figure 7. Find the path that will get him to the tree in the minimum time.

Your Turn 2 Continued Solution : We are trying to minimize the total amount of time, which is the sum of the time on the trail and the time through the woods. We must express this time as a function of x. Since Time = Distance/ Speed, the total time is Notice in this equation that 0 ≤ x ≤ 300. Continued

Your Turn 2 Continued x T(x) 21.88 207 21.24 Absolute Minimum 300 21.88 207 21.24 Absolute Minimum 300 21.36

Your Turn 2 Continued We see from the table that the time is minimized when x = 207 m. That is, when the professor goes 93 m along the trail and then heads into the woods.

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Section 6.3 Further Business Applications: Economic Lot Size; Economic Order Quantity; Elasticity of Demand

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Implicit Differentiation Section 6.4 Implicit Differentiation

Your Turn 1 Solution: Differentiate with respect to x on both sides of the equation. Now differentiate each term on the left side of the equation. Think of xy as the product (x)(y) and use the product rule and the chain rule. Continued

Your Turn 1 Continued

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Your Turn 3 The graph of is called the devil’s curve. Find the equation of the tangent line at the point (1, 1). Solution: We can calculate by implicit differentiation. Continued

Your Turn 3 Continued To find the slope of the tangent line at the point (1, 1), let x = 1 and y = 1 The slope is The equation of the tangent line is then found by using the point-slope form of the equation of a line.

Section 6.5 Related Rates

Your Turn 1 Solution: We start by taking the derivative of the relationship, using the product and chain rules. Keep in mind that both x and y are functions of t. The result is Continued

Your Turn 1 Continued

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Your Turn 2 A 25-ft ladder is placed against a building. The base of the ladder is slipping away from the building at a rate of 3 ft per minute. Find the rate at which the top of the ladder is sliding down the building when the bottom of the ladder is 7 ft from the base of the building. Continued

Your Turn 2 Continued Solution: Starting with Step 1, let y be the height of the top of the ladder above the ground, and let x be the distance of the base of the ladder from the base of the building. We are trying to find dy/dt when x = 7. To perform Step 2, use the Pythagorean theorem to write Both x and y are functions of time t (in minutes) after the moment that the ladder starts slipping. According to Step 3, take the derivative of both sides of the equation with respect to time, getting Continued

Your Turn 2 Continued To complete Step 4, we need to find the values of x, y, and dx/dt. Once we find these, we can substitute them into Equation to find dy/dt. Since the base is sliding at the rate of 3 ft per minute, dx/dt = 3. Also, the base of the ladder is 7 ft from the base of the building, so x =7 . Use this to find y. Continued

Your Turn 2 Continued In summary, x = 7, y = 24, and dx/dt = 3.

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Differentials: Linear Approximation Section 6.6 Differentials: Linear Approximation

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Your Turn 1 Solution:

Your Turn 2 Approximate Solution: We will use the linear approximation for a small value of ∆x to form this estimate. We first choose a number x that is close to 99 for which we know its square root.

Your Turn 3 In a precision manufacturing process, ball bearings must be made with a radius of 1.25 mm, with a maximum error in the radius of ± 0.025 mm. Estimate the maximum error in the volume of the ball bearing. Solution: The formula for the volume of a sphere is If an error of ∆r is made in measuring the radius of the sphere, the maximum error in the volume is Replacing r with 1.25 and dr = ∆r with ± 0.025 gives