A related rates problem (shouldn't be complicated)

[Null_]

Homework Statement

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?

Homework Equations

v= (1/3) \pi r^2 * h

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.

 

[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.

 

[HallsofIvy]

Homework Statement

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?

Homework Equations

v= (1/3) \pi r^2 * h

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)

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.

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.

 

[Null_]

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