rate of change of the area of the rebgion

fifaking7

1. The problem statement, all variables and given/known data

A board 5 feet long slides down a wall. at the instant the bottom end is 4 feet from the wall, the other end is moving down the wall the rate of 2 feet per second.at that moment how fast is the area of the region between the board, the ground and the wall changing?

2. Relevant equations
a=1/2bh


3. The attempt at a solution
x^2 + 4^2= 5^2
x=3
da/dt= 1/2b(db/dt)(dh/dt)

I think i started it up wrong and I don't know where to go next..

SammyS

What is the shape of the region below the board and above the ground at the moment under consideration?

fifaking7

just a normal triangle drawn on a line like a typical triangle leaning against a wall problem.

SammyS

Neither end of the board is touching the ground, so it's not a triangle.

aralbrec

Have another look at that derivative. It looks like the product rule hasn't been applied properly. Also one of those b/h is x. You'll have to stay consistent with your variable names.

fifaking7

that is how it looks

SammyS

You're right. I misread the problem.

When you said x=3, that should be h=3 .

In general, how are b and h related, considering that they're legs of a right triangle?

fifaking7

i said b= 4ft and h=3 ft

SammyS

That's what b & h are at the instant that h = 4 ft, but how are they related in general?

(Use the Pythagorean theorem.)

Ray Vickson

The crucial point that you seem to be missing is that you need to figure out what is happening as the board slides down the wall, so h moves from more than 3 ft to less than 3 ft (and, at the same time, b moves from less than 4 ft to more than 4 ft). When that is happening, the area of the triangle is changing, and that is what you are supposed to be reckoning. So, you need to let b and h be variables, not fixed numbers.

RGV

SammyS

Of course, that's a typo !!!

It should have said:

That's what b & h are at the instant that b = 4 ft, but how are they related in general?