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.