Calculating Water Flow Rate in Sloped Reservoir | Simple Calculus Question

[Darkiekurdo]

Homework Statement

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.

Homework Equations

Not stated

The Attempt at a Solution

I don't know how to begin, I'm sorry.

 

[Dick]

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

 

[Darkiekurdo]

What shape is the reservoir?

 

[Dick]

What shape is the reservoir?

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

 

[Darkiekurdo]

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

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

 

[Dick]

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

I read the first sentence of the problem you posted.

 

[d_leet]

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

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.

 

[Darkiekurdo]

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.

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

 

[Darkiekurdo]

Anyway, the volume of a truncated pyramid is

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

right?

 

[Dick]

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

 

[Darkiekurdo]

What should I do next?

 

[Dick]

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

 

[Darkiekurdo]

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

 

[Dick]

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

 

[Darkiekurdo]

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

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

 

[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".

 

[Darkiekurdo]

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

Thus,

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

 

[Dick]

Better and better. What next?

 

[Darkiekurdo]

I'm stuck. :(

 

[Darkiekurdo]

Maybe we should take time in consideration?

 

[Dick]

Maybe we should take time in consideration?

Sure. You must have done some other rate problems, right? Review them.

 

[Darkiekurdo]

I'm just beginning with these so I'm not very familiar with those. I'll try though:

*something*time = $$p^2h + 2ph^2 + \frac{4}{3}h^3$$

 

[Darkiekurdo]

*something* = c. :tongue2:

So:

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

 

[Dick]

Think derivatives. dV/dt is the rate water is coming in. dh/dt is the rate that it's rising.

 

[Darkiekurdo]

So we need to find dh/dt?

 

[Dick]

That's what they are asking for, right?

 

[Darkiekurdo]

Sure. That derivative is

$$\frac{c}{(p + 2h)^2}$$

Then it's just substituting the required values.

 

[Dick]

No. How did you get that? Differentiate your expression for V with respect to t, remember h=h(t) is a function of time.

 

[Darkiekurdo]

t = dt/dh. So c*dt/dh = p² + 4ph + 4h² = (p + 2h)². Thus we get

dt/dh = $$\frac{c}{(p + 2h)^2}$$

 

[Dick]

t = dt/dh. So c*dt/dh = p² + 4ph + 4h² = (p + 2h)². Thus we get

dt/dh = $$\frac{c}{(p + 2h)^2}$$

Actually, that's right. Apologies. But it's dh/dt, not dt/dh.

 

[Darkiekurdo]

Actually, that's right. Apologies. But it's dh/dt, not dt/dh.

You're right.

Thanks a lot though! I know sometimes I look a little bit stupid but I hope that's the age. :rofl: