# Finding y prime (derivative)

1. Sep 28, 2004

### buffgilville

What is the y prime of:
A 15 foot board rests against a vertical wall. If the bottom of the board slides away from the wall at the rate of 3 feet per second, how fast is the area of the triangle formed by the board, the wall and the ground changing at the instant the bottom of the board is 9 feet from the wall?

I got this far:
y= square root of (225 - x^2)

I'm stuck after this step.

Can the answer for this be [(-9 yprime /2) + 18] square feet per second.

Last edited: Sep 28, 2004
2. Sep 28, 2004

### vsage

Hint x in terms of t. The displacement x away from the wall is equal to what in terms of t? It will be easy to then sub in this for x.

Area of a triangle is 1/2 * base * height
Base is x height is sqrt(225-x^2) right?

so A = 1/2 * x * sqrt(225-x^2)

However you need to plug in t to this I just didn't want to give away that much work.

Differentiate with respect to t in the new equation you find using the product rule (gets a little messy). Since 3t = 9, t = 3 goes into this new derivative and bam you got your answer.

Last edited by a moderator: Sep 28, 2004
3. Sep 28, 2004

### cepheid

Staff Emeritus
I'm not sure what you mean by the question "what is the y prime of....?". If anything, the question seems to be asking for A'(t), where A is the area of the right triangle formed by the ladder, the wall, and the ground.

Draw a diagram, label the horizontal distance between the bottom of the ladder and wall x, and the corresponding vertical distance y

Given:

$\frac{dx}{dt}$ = 3 ft/s

This is a related rates problem. A is a function of x, so given the rate at which x changes with time, you should be able to find the rate at which A changes with time using the functional relationship:

$$A = \frac{1}{2}xy = \frac{1}{2}x\sqrt{225 - x^2}$$

You identified this relationship correctly!

$$\frac{dA}{dt} = \frac{dA}{dx}\frac{dx}{dt}$$
$$= \frac{1}{2}\frac{d}{dx}(x\sqrt{225 - x^2})(3 \text{ft/s})$$