How fast is the boat approaching the dock when the amount of rope out is 20ft.?

Question

a rope is attached to a boat at water level, and a woman on a dock is pulling on the rope at the rate of 50ft./min. if her hands are 16ft. above the water level, how fast is the boat approaching the dock when the amount of rope out is 20ft.?

(this question is about calculus)

(topic about: Related Rates)

(pls. help me to answer this question)

(pls. show your solutions)

(thanks)

Answer 1

I agree, you don't need calculus for that one, just trigonometry!

Why make anything more difficult than it need be?

Answer 2

i don't think it's necessarily a calculus problem since you are imparting a constant velocity on the rope.  but as the boat approaches the dock, the vertical component (16' above the waterline) is going to slow down the horizontal velocity.  the numbers work out perfectly for a 3-4-5 right triangle.  the horizontal component when the hypotenuse is 20' should be 12' or 6/10ths of the 50fpm velocity of the rope.  that's 30 fpm at that particular second when the line is 20' long.

at this point you'd better start thinking about braking, though.

Answer 3

Ok, here's how you do it.

First of all you have to draw a triangle representing the lengths. The length of the rope, the length between the dock and the boat and the height the hands of the woman are. Then you will the that it's a rectangle triangle and that the length of the rope will be the hypotenuse.

let's say that x will be the distance between the dock and the boat, y will be the distance between the water level and the woman's hands and z will be the length of the rope. Then with pitagore:

x^2 = z^2 - y^2

you see from the statement of the problem that the hands of the woman are not changing their height, so you can say that y will be constant. Then you derive the equation for x and z

2xdx = 2zdz

Y disappears because it's a constant. Then you multiply your equation for dt/dt and you get

2xdx/dt = 2zdz/dt

you know that dx/dt is defined as velocity and is the same with dz/dt so:

2xVx = 2zVz

solving for Vx

Vx = 2zVz/(2x) = zVz/(x)

and now you just substitute for the data given by the problem

z = 20ft

Vz = 50ft/min

x = sqrt (20ft^2-16ft^2) = 12ft

so Vx will be

Vx = (20ft)(50ft/min)/(12ft) = 83.3ft/min

then the boat is approaching the dock at 83.3 ft/min

hope this helps.