WeBwork: 20 x WeBWork: Spring20 x Chegg Study I Guided Sc x Secure h ps://webwor

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WeBwork: 20 x WeBWork: Spring20 x Chegg Study I Guided Sc x Secure h ps://webwork uncc edu /webwork2/Spring 2017-Math1241-Commo n/Sec2.7/9/? effective User cyarrisoBuser cyarriso&key; aMu917puuPbrCFNn2QZntK2LtjLLRBft webwork spring2017-math1241-common sec2.7 9 MAIN MENU Courses Homework Sets Sec2.7: Problem 9 Sec2.7 Problem 9 Previous Problem List Next User Settings Grades 1 point) Book Problem 18 Problems A boat is pulled into a dock by means of a rope attached to a pulley on the dock. The rope is attached to the front of the boat, which is 6 feet below the level of the pulley. (There is a diagram of this situation with problem 18 on p132 of the text.) Problem 1 ft/min Problem 2 Problem 3 Problem 4 Preview My Answers Submt Answers Problem 5 Problem 6 You have attempted this problem 0 times. Problem 7 You have unlimited attempts remaining. Problem 8 Problem 9 Problem Email instructor Problem 11 Problem 12 Problem 13 Problem 14 Problem 15 P 51 AM 3/8/2017

Explanation / Answer

Let x be the horizontal distance of the boat from dock, and h be the length of rope out.

Then:
x^2 = h^2 - 6^2
When h = 90;
x = sqrt(90^2 - 6^2)
= 89.79ft.

Differentiating:
2x dx/dt = 2h dh/dt
x dx/dt = h dh/dt

Substituting dh/dt = -16, h = 90, x = 89.79;
89.79 dx/dt = - 16 * 90
dx/dt = -16 * 90 / 89.79
= - 16.03 ft/min

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