Finding the Rate of Change of Water Level in a Conical Tank at a Specific Depth

[Cefari]

Homework Statement

A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.

The Attempt at a Solution

First I derived the volume equation of a cone ( v = hπr^2 ) with respect to time which yielded:

dv/dt = π/3(2rh(dr/dt) + r^2 (dh/dt))

Next, I substituted in all known variables:

10 = π/3((2)(10)(8)(dr/dt) + (10^2)(dh/dt))
10 = π3(160(dr/dt) + 100(dh/dt))

I'm suppose to solve for dh/dt but I can't do so until I figure out dr/dt, I'm wondering if it involves replacing r with terms of h out of the volume equation?

 

[Mark44]

You need a relationship between r and h, the radius of the water in the tank at time t, and the height of water in the tank at time t. You can get these from the dimensions of the conical tank by using similar triangles.

After you have an equation involving r and h, differentiate to get an equation involving dr/dt and dh/dt.

 

[Cefari]

I don't quite see how to do that, although it does look similar to something my teacher may of gone over in class.

I'm thinking the similar triangles use radius and height as their legs, so the larger triangle I guess would be width of 5 and a height of 12 and the similar triangle would be a height of 8 with a width of 10/3. This is as far as I've gotten.

 

[Mark44]

I don't quite see how to do that, although it does look similar to something my teacher may of gone over in class.

I'm thinking the similar triangles use radius and height as their legs, so the larger triangle I guess would be width of 5 and a height of 12 and the similar triangle would be a height of 8 with a width of 10/3. This is as far as I've gotten.

The larger triangle has legs of 5 and 12 as you said, but the smaller triangle has legs that correspond to the radius of the water level in the tank at a given time, and the depth of water in the tank at the same time.

 

[Cefari]

Alright so here's what I did, I don't have an answer key so not sure if it's 100% right.

1) I rewrote r in terms of h as r/h = 10/12 so r = 5h/6
2) I substituted that in for r in the original and simplified getting a new equation of (25h^3 pi)/108.
3) I derived with respect to t and got dv/dt = (25h^2 pi / 36) dh/dt.
4) Plugging in h and dv/dt I get 9/(40 pi) = dh/dt

Any concerns?

 

[Mark44]

Alright so here's what I did, I don't have an answer key so not sure if it's 100% right.

1) I rewrote r in terms of h as r/h = 10/12 so r = 5h/6

The radius of the tank is not 10 ft.

2) I substituted that in for r in the original and simplified getting a new equation of (25h^3 pi)/108.

An equation has an = in it.

3) I derived with respect to t and got dv/dt = (25h^2 pi / 36) dh/dt.

You differentiated with respect to t.

4) Plugging in h and dv/dt I get 9/(40 pi) = dh/dt

Any concerns?

Only one major concern, the first, but that one affects your results all the way through. Your final answer should include units, however. What are the units for dh/dt?