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A related rates problem (shouldn't be complicated)

  1. Apr 26, 2010 #1
    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.
  2. jcsd
  3. Apr 27, 2010 #2
    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.
  4. Apr 27, 2010 #3


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    Science Advisor

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

  5. Apr 28, 2010 #4
    Thank you very much. I felt iffy when I used r as a constant..
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