# Homework Help: Rate of change of a right triangle

1. Sep 11, 2010

### vande060

Imagine a boat on the water being putted
towards the shore by a winch mounted a height
h above the water. The winch reels in cable
(shortening the hypotenuse) at constant rate
u in m/s . When the boat is a distance d from
the shore, find its speed and acceleration.
This is an example of motion at non-uniform
acceleration.

im letting hypotenuse here = x, so x^2 = d^2 + h^2

Im thinking i want to solve for d' to get the velocity at of d at distance d. okay im going to give it a shot, but bear with me, i have little confidence in this answer and i understand that reading these next steps will probably be a little frightening for those who know how to solve this.
-------------------------------

x^2 = d^2 + h^2

d^2 = x^2 - h^2

2dd' = 2xx' - 2hh'

d' = [2xx' - 2hh']/(2d) <--- this is what i was thinking to explain the speed of d at distance d . my intuition was the to allow x' be u to explain the constant rate of the shortening of the hypotenuse, but im not sure about it. i want to differentiate again to find d'' to explain acceleration, but i need to establish the correct way to do this first step before moving on.

can anyone give me a pointer or two.

Last edited: Sep 11, 2010
2. Sep 11, 2010

### rl.bhat

Whether the height h is variable or constant?

3. Sep 11, 2010

### vande060

4. Sep 12, 2010

### rl.bhat

5. Sep 12, 2010

### vande060

okay, i understand how you got 2, you multiplied both sides by d then differentiated right?

okay so now i just simplify..

(x'*x' - d'*d')/d = d"

x' here represent the constant rate, u m/s, that the hypotenuse is shortening.

Last edited: Sep 12, 2010
6. Sep 12, 2010

### rl.bhat

Yes.

7. Sep 12, 2010

### vande060

thank you greatly