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 boat 10 feet below the pulley. If the rope is pulled through the pulley at a rate of 20 ft/min, at what rate will the boat be approaching the dock when 125 ft of rope is out?

101.6 ft/min

Step-by-step explanation:

let x be the horizontal distance to the dock. And the rope is attached to the boat 10 feet below the pulley

Using pythagorean theorem

x^2=R^2-10^2

where R is rope length to the pulley.

Differentiates with respect to time t

2x dx/dt=2RdR/dt

Xdx/dt = RdR/dt (1)

If the boat be approaching the dock when 125 ft of rope is out

This is R = 125ft

Using pythagorean theorem again

X^2 = R^2 - 10^2

X^2 = 125^2 - 10^2

X^2 = 15525

X = 124.6 ft

The rate the boat be approaching the dock = dx/dt

While

the rope is pulled through the pulley at a rate of 20 ft/min = dR/dt

Solve for dx/dt when R is 125, x= 124.6 and dR/dt= 20ft/min

Substitute all in equation 1

124.6 dx/dt = 125 × 20

dx/dt = 2500/124.6

dx/dt = 101.6 ft/min

won't that be plagrisim thought?

step-by-step explanation: