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Homework Help: Help with basic integration?

  1. Sep 13, 2009 #1
    This is the integral that I am having trouble with
    [tex] \frac{dh}{dt} = -0.064*\sqrt{h} [/tex]

    I hope you will kindly nudge me to the point where I have made a mistake
    I did this problem by hand (and can scan and present if anyone is interested.)
    Anyway, I used Maple to check my work and found that I must have made a mistake during integration because I am missing a term.

    Here is my work:

    [tex] \int \frac{1}{\sqrt{h}} dh = -0.064 \int dt [/tex]

    Using integration by substitution:

    [tex] u = \sqrt{h} [/tex]

    [tex] du = \frac{1}{\sqrt{h}} dh [/tex]

    [tex] \int \frac{1}{\sqrt{h}} dh \rightarrow 2\int \frac{u}{u} du \rightarrow 2*\sqrt{h} [/tex]

    After integration by substitution and the simple integration on the RHS:
    [tex]2*\sqrt{h} = -0.064*t+C_{1}[/tex]

    Putting the equation in General form:

    The problem is occurring before this point (I believe)
    [tex]h = -0.001024*t^{2}+C_{1}[/tex]

    This was an initial value problem with h(0) = 10;
    But, the problem occurs before this point (as I stated above).
    Because, the general solution of the IVP in Maple includes a single term in t, where the solution I found below - does not.

    Using h(0)=10;
    [tex]h = -0.001024*t^{2} + 10;[/tex]

    The answer I found with Maple (for my own reference, I suppose
    [tex]h(t) = \frac{16}{15625}*t^{2}-\frac{8}{125}*t*\sqrt{10}+10[/tex]
  2. jcsd
  3. Sep 13, 2009 #2
    After you get √h by itself in the middle equation, square both sides entirely.
  4. Sep 13, 2009 #3
    [tex] h=(\frac{16}{125})^2(t+C)^2 [/tex]
  5. Sep 13, 2009 #4
    Thanks all.

    Got it.
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