Explanation Given information: The distance between the man and the helipad is 50 ft . The distance of helicopter from the helipad is y = 120 ft . The helicopter is rising at the speed of 44 ft/sec Calculation: Since the helicopter is rising at the speed of 44 ft/sec , Note that, the derivative of the function is rate of change of that function. Speed is changing with respect to time t , d y d t = 44 ft/sec Let the distance between the helicopter and the man is x ft . The diagram of the helicopter and man is shown below in Figure 1, Figure 1 From Figure 1, it is observed that the given triangle is right angle triangle. Now, apply the Pythagorean Theorem to the right angle triangle in the above Figure 1, x 2 = y 2 + 50 2 (1) Substitute 120 for y in the equation (1) to find the value of x . x 2 = ( 120 ) 2 + ( 50 ) 2 x 2 = 16900 x = 16900 = 130 Thus, the value of x is 130 feet

Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

10th Edition

ISBN: 9781305657861

Author: Soo T. Tan

Publisher: Cengage Learning

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Chapter 3.6, Problem 56E

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To find: The rate at which the distance between helicopter and the man is changing.

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Chapter 3.6, Problem 55E

Chapter 3.6, Problem 57E

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(b) Find the (instantaneous) rate of change of y at x = 5. In the previous part, we found the average rate of change for several intervals of decreasing size starting at x = 5. The instantaneous rate of change of fat x = 5 is the limit of the average rate of change over the interval [x, x + h] as h approaches 0. This is given by the derivative in the following limit. lim h→0 - f(x + h) − f(x) h The first step to find this limit is to compute f(x + h). Recall that this means replacing the input variable x with the expression x + h in the rule defining f. f(x + h) = (x + h)² - 5(x+ h) = 2xh+h2_ x² + 2xh + h² 5✔ - 5 )x - 5h Step 4 - The second step for finding the derivative of fat x is to find the difference f(x + h) − f(x). - f(x + h) f(x) = = (x² x² + 2xh + h² - ])- = 2x + h² - 5h ])x-5h) - (x² - 5x) = ]) (2x + h - 5) Macbook Pro

Evaluate the integral using integration by parts. Sx² cos (9x) dx

Let f be defined as follows. y = f(x) = x² - 5x (a) Find the average rate of change of y with respect to x in the following intervals. from x = 4 to x = 5 from x = 4 to x = 4.5 from x = 4 to x = 4.1 (b) Find the (instantaneous) rate of change of y at x = 4. Need Help? Read It Master It

Chapter 3 Solutions

Applied Calculus for the Managerial, Life, and Social Sciences (MindTap Course List)

Ch. 3.1 - State the following rules of differentiation in...Ch. 3.1 - Prob. 2CQ Ch. 3.1 - Prob. 3CQ Ch. 3.1 - Prob. 4CQ Ch. 3.1 - Prob. 1E Ch. 3.1 - Prob. 2E Ch. 3.1 - Prob. 3E Ch. 3.1 - Prob. 4E Ch. 3.1 - Prob. 5E Ch. 3.1 - Prob. 6E

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AUTO TRANSMISSIONS .AND FUEL ECONOMY In trying to...Ch. 3.5 - Prob. 32E Ch. 3.5 - Prob. 33E Ch. 3.5 - Prob. 34E Ch. 3.5 - Prob. 35E Ch. 3.5 - Prob. 36E Ch. 3.5 - Prob. 37E Ch. 3.5 - Prob. 38E Ch. 3.5 - Prob. 39E Ch. 3.5 - Prob. 40E Ch. 3.5 - Prob. 41E Ch. 3.5 - Prob. 42E Ch. 3.5 - Prob. 43E Ch. 3.5 - Prob. 44E Ch. 3.5 - Prob. 45E Ch. 3.5 - Prob. 46E Ch. 3.5 - Prob. 47E Ch. 3.5 - Prob. 48E Ch. 3.5 - Prob. 49E Ch. 3.5 - Prob. 50E Ch. 3.5 - Prob. 1TE Ch. 3.5 - Prob. 2TE Ch. 3.5 - Prob. 3TE Ch. 3.5 - Prob. 4TE Ch. 3.5 - In Exercises 1-8, find the value of the second...Ch. 3.5 - Prob. 6TE Ch. 3.5 - Prob. 7TE Ch. 3.5 - Prob. 8TE Ch. 3.6 - Prob. 1CQ Ch. 3.6 - Prob. 2CQ Ch. 3.6 - Prob. 3CQ Ch. 3.6 - Prob. 4CQ Ch. 3.6 - Prob. 1E Ch. 3.6 - Prob. 2E Ch. 3.6 - Prob. 3E Ch. 3.6 - Prob. 4E Ch. 3.6 - Prob. 5E Ch. 3.6 - Prob. 6E Ch. 3.6 - Prob. 7E Ch. 3.6 - Prob. 8E Ch. 3.6 - Prob. 9E Ch. 3.6 - Prob. 10E Ch. 3.6 - Prob. 11E Ch. 3.6 - Prob. 12E Ch. 3.6 - Prob. 13E Ch. 3.6 - Prob. 14E Ch. 3.6 - Prob. 15E Ch. 3.6 - 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Volume of a Cylinder The volume of a...Ch. 3.6 - Prob. 40E Ch. 3.6 - Prob. 41E Ch. 3.6 - Prob. 42E Ch. 3.6 - Prob. 43E Ch. 3.6 - Prob. 44E Ch. 3.6 - Prob. 45E Ch. 3.6 - Prob. 46E Ch. 3.6 - Prob. 47E Ch. 3.6 - PRODUCTION FUNCTION FOR A FURNITURE COMPANY The...Ch. 3.6 - Prob. 49E Ch. 3.6 - Prob. 50E Ch. 3.6 - Prob. 51E Ch. 3.6 - Prob. 52E Ch. 3.6 - Prob. 53E Ch. 3.6 - Prob. 54E Ch. 3.6 - Prob. 55E Ch. 3.6 - Prob. 56E Ch. 3.6 - Prob. 57E Ch. 3.6 - Prob. 58E Ch. 3.6 - Prob. 59E Ch. 3.6 - Prob. 60E Ch. 3.6 - Prob. 61E Ch. 3.6 - Prob. 62E Ch. 3.6 - Prob. 63E Ch. 3.6 - Prob. 64E Ch. 3.6 - Prob. 65E Ch. 3.6 - 66. 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The rate at which the distance between helicopter and the man is changing.

Solution Summary: The author explains how the distance between the helicopter and the man is increasing at the rate of 40.61feet/mathrmsec.

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To find: The rate at which the distance between helicopter and the man is changing.

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Chapter 3 Solutions