A related rates problem (shouldn't be complicated)

1. Apr 26, 2010

Null_

1. The problem statement, all variables and given/known data
An inverted conical container has a diameter of 42 in and a depth of 15 in. If water is flowing out of the vertex at a rate of 35 $$\pi$$ in^3 / sec, how fast is the depth of the water dropping when the height is 5 in?

2. Relevant equations
v= (1/3) $$\pi$$ r^2 * h

3. The attempt at a solution
What I know:
D=42 in
R=21 in
h=15 in
(dV/dT)=35 $$\pi$$ in^3/min
(dH/dT)=? When h=5.

(r1/h1)=(r2/h2) ... (21/15)= (r2/5) ...r2=7

plugging in 7 for r, I get...
v= (1/3) $$\pi$$ (49) * h

Taking the derivative in terms of T, I get...
(dV/dT)=(49 $$\pi$$ /3) (dH/dT)
35 $$\pi$$ = (49 $$\pi$$ / 3) (dH/dT)

I get (dH/dT) to = 15/7 in/sec, but the back of my book says 5/7 in/sec. Is it just supposed to be the ratio of height to radius? Thanks in advance.

2. Apr 27, 2010

Null_

Did I go wrong in plugging 7 in for r before I took the derivative? But if I took the derivative I'd have (dr/dt) in the equation, which I wouldn't be able to solve for.

3. Apr 27, 2010

HallsofIvy

Staff Emeritus
Here is your error- you are treating r as if it were a constant. Since (r/h)= (21/7), r= (21/7)h and $V= (1/3)\pi r^2 h= (1/3)\pi(21/7)^2 h^3$. differentiate both sides of that with respect to h.

4. Apr 28, 2010

Null_

Thank you very much. I felt iffy when I used r as a constant..
:)