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Separable Differential Equation

  1. Feb 11, 2010 #1
    1. The problem statement, all variables and given/known data
    The volume V of water in a particular container is related to the depth h of the water by the equation dV/dh below. If V = 0 when h = 0, find V when h = 4.


    2. Relevant equations
    dV/dh = 16 sqrt(4-(h-2)2)

    V(0) = 0
    V(4) = ?


    3. The attempt at a solution

    I have not gotten the ODE in a form that I can integrate using separation of variables. I have gotten to the point dV/dh = 16 sqrt(-h2+4h) but this doesn't help much.
     
    Last edited: Feb 11, 2010
  2. jcsd
  3. Feb 11, 2010 #2

    jks

    User Avatar

    Unless you miswrote the equation, you don't need to use separation of variables in this problem, since the right side is an equation of 1 variable. But you can do it like this:

    dV/dh = √(4-(h-2)²)

    move the dh over
    ∫dV = ∫√(4-(h-2)²) dh

    V = (integrate that expression with respect to h) + C
     
  4. Feb 11, 2010 #3
    The problem was from my textbook under the "separable equations" section so I listed it as that.

    Integrating the expression √(4-(h-2)²) is what I was having trouble with. I tried u substitution, integration by parts, and other methods for awhile. Am I missing something obvious to integrate the expression?
     
  5. Feb 11, 2010 #4

    Mark44

    Staff: Mentor

    I would try a trig substitution. To make things a little simpler, I would make an ordinary substitution u = h - 2, and then the trig substitution.

    I usually draw a right triangle, with t being one of the acute angles, hypotenuse 2 and opposite (to angle t) side u. So the substitution here is sin t = u/2.

    Don't forget that you'll need to undo two substitutions when you finally evaluate your antiderivative.
     
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