(adsbygoogle = window.adsbygoogle || []).push({}); 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 [tex]\pi[/tex] in^3 / sec, how fast is the depth of the water dropping when the height is 5 in?

2. Relevant equations

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

3. 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.

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# Homework Help: A related rates problem (shouldn't be complicated)

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