- #1
Null_
- 231
- 0
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 [tex]\pi[/tex] in^3 / sec, how fast is the depth of the water dropping when the height is 5 in?
Homework Equations
v= (1/3) [tex]\pi[/tex] r^2 * h
The Attempt at a Solution
What I know:
D=42 in
R=21 in
h=15 in
(dV/dT)=35 [tex]\pi[/tex] 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) [tex]\pi[/tex] (49) * h
Taking the derivative in terms of T, I get...
(dV/dT)=(49 [tex]\pi[/tex] /3) (dH/dT)
35 [tex]\pi[/tex] = (49 [tex]\pi[/tex] / 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.