Simple calculus question

1. Nov 1, 2007

Darkiekurdo

1. The problem statement, all variables and given/known data
A reservoir of square cross-section has sides sloping at an angle of 45 degrees with the vertical. The side of the bottom is p feet in length, and water flows in the reservoir at the rate of c cubic feet per minute. Find an expression for the rate at which the surface of the water is rising at the instant its depth is h feet. Calculate this rate when p = 17, h = 4 and c = 35.

2. Relevant equations
Not stated

3. The attempt at a solution
I don't know how to begin, I'm sorry.

2. Nov 1, 2007

Dick

Start by trying to find an expression for the volume V as a function of h and p.

3. Nov 1, 2007

Darkiekurdo

What shape is the reservoir?

4. Nov 1, 2007

Dick

Did you read the problem? It's a truncated pyramid with square cross section.

5. Nov 1, 2007

Darkiekurdo

Hmm. I didn't know that. How did you come up with that?

6. Nov 1, 2007

Dick

I read the first sentence of the problem you posted.

7. Nov 1, 2007

d_leet

Did you even read the problem? You asked the question and it seems like you haven't even read it. The first sentence describes the shape of the resevoir.

8. Nov 1, 2007

Darkiekurdo

Of course I read it but I didn't know what that was. Sorry.

9. Nov 1, 2007

Darkiekurdo

Anyway, the volume of a truncated pyramid is

$$V = \frac{H}{3}(A + a + \sqrt{Aa})$$

right?

Last edited: Nov 1, 2007
10. Nov 1, 2007

Dick

Right, if the A and a are the areas of the top and bottom.

11. Nov 1, 2007

Darkiekurdo

What should I do next?

12. Nov 1, 2007

Dick

Apply that formula to your problem. If the reservoir is filled to height h, what are the areas of the top and bottom?

13. Nov 1, 2007

Darkiekurdo

So the area of A = p2 and a = (p+2h)2.

14. Nov 1, 2007

Dick

Now you are cooking. What's the formula then for the volume V in terms of p and h?

15. Nov 1, 2007

Darkiekurdo

Then I'll just substitute those in the formula for the volume:

$$\frac{H}{3}[p^2 + p(p + 2h) + (p + 2h)^2]$$

16. Nov 1, 2007

Dick

H=h, right? Now to make life easier later, I would expand that out. Now you have V as a function of p and h. What next? Please don't say "I don't know".

17. Nov 1, 2007

Darkiekurdo

Yeah, I should have written it as 1/3h instead of H/3 actually.

Thus,

$$p^2h + 2ph^2 + \frac{4}{3}h^3$$

18. Nov 1, 2007

Dick

Better and better. What next?

19. Nov 1, 2007

Darkiekurdo

I'm stuck. :(

20. Nov 1, 2007

Darkiekurdo

Maybe we should take time in consideration?