Question

a boat at a A Rope passing through a capstan son dock is attachted offshore. The rope is pulled in Leonstant rate of 3ft/s and the capstan 5ft vertically above the water, HOW FAST IS THE BOAT TRAVELING WHEN IT IS 7 ft FROM THE Dock? (Round Final Final Answer to a decimal places) 15

Answer 1

Solution A Capstan) z 5ft C (Boat) B dz dt Let the position of Capstan be 'A', position of boat be 'C' and 'B' be the point in front of the boat at the dock. Thus, we can draw the right-angled triangle as shown in the figure. The length of AC will be the length of the rope let it be z and BC will be the horizontal distance from the front of the boat let it be x. The capstan is attached at a point 5ft higher than the front of the boat so, the length of the side AB will be 5ft. The rope is being pulled through the capstan at a rate of 3ft/s. i.e. = 3ft/s We have to know how fast is the boat approaching the dock which means how fast the length BC (which is denoted by x) is changing. i.e. we have to find the value of ** when the boat is 7ft from the dock. Which means we have to find the value of when x = 7ft. We can relate the sides of the triangle using Pythagoras theorem i.e. z2 = x2 +52 So, when boat is 7ft away from the dock i.e. x = 7ft z = 152 + 72 = 174ft dt dx dt So, the value of z will be x = v74ft when x=7ft.

We have the equation z2 = x² + 52 In order to find the rate of change of x with repsect to time t we have to differentiate both sides with respect to t. So, we will get dz dx 2z = 2x dt dt We have to find the value of dx when x = 7ft and the value of z at that time dt will be z = 774.The rate at which the rope is pulling 2 = 3ft/s Substituting all the known values we get dx z dz V74 3774 X 3 =- 3.69ft/s dt x dt 7 So, the boat is approaching the dock at a rate of 3.69ft/s 7