Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Related Rates

  1. Feb 11, 2008 #1
    Water is leaking out of a cylindrical tank , with circular base radius 0.8 m at the rate of 0.2 m3/min. The water level is falling at ...? ..m/min. What is the question mark? (Hint: the radius does not change)

    V=(pi)r^2h, dV/dt = 0.2, r = 0.8 ?

    Can anyone give me a hand in getting this one started?
  2. jcsd
  3. Feb 11, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    Differentiate both sides of V= (pi)r^2 h, remembering that "the radius does not change". Since h is the "water level", "the rate at which the water level is falling" is dh/dt.
  4. Feb 11, 2008 #3
    So, dV/dt = (pi)r^2*dh/dt
    0.2 = (pi)(o.8)^2*dh/dt
    dh/dt =0.0995 ?

    Hypothetically, how would you solve for volume if the rates of change of the height and radii are given, as well as the radius and height when the derivative would seem to cancel the h in the equation?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Related Rates
  1. Related Rates (Replies: 2)

  2. The concept of a rate (Replies: 4)