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!

Rate of change

  1. May 9, 2004 #1
    Sorry to bore you all with this rate of change question, I really hate this area of maths and struggle to get my hea round it. Here it is, I'd be greatful for any help you have to offer:

    Fluid flows out of a cylindrical tank with constant corss section. At the time t minutes (t is greater of equal to 0), the volume of fluid remaining in the tank is V m^3. The rate at which the fluid flows, in m^3min^-1, is proportional to the square root of V. Show that the depth h metre's of fluid in the tank satisfies the differential equation:

    dh/dt = -kh^(1/2) where k is a +ve constant


  2. jcsd
  3. May 9, 2004 #2


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    Dearly Missed

    dV/dt=-aV^(1/2), by assumption.

    V(t)=pi*R^(2)*h(t)=b*h(t) (where R is the radius)
    Rearranging a bit, and introducing k yields the desired result
  4. May 9, 2004 #3
    thanks v much, great help

  5. May 9, 2004 #4


    User Avatar
    Science Advisor

    In other words, the whole point of "rate of change problems" is to use a static equation (typically a geometry formula or something like that) that has no time variable, t, in it and then use the chain rule to differentiate both sides of the equation- giving a formula connecting the rates of change of the quantities.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook