1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    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..
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook