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Homework Help: Partial of a Sine where the PHASE is the variable?

  1. Oct 11, 2009 #1
    1. The problem statement, all variables and given/known data

    For example: the function is A0sin(w0*t - B*z)

    If I take the partial derivative with respect to z, how do you go about this? In my years at uni, I don't know if this has ever come up.

    2. Relevant equations

    3. The attempt at a solution

    My initial thoughts are that it is either 0, or to use the chain rule, although I'm having trouble thinking of how I would do that in this case. Any help is greatly appreciated, thanks!
  2. jcsd
  3. Oct 11, 2009 #2


    Staff: Mentor

    I'm assuming that your function is defined a f(t, z) = A0sin(w0*t - B*z)

    [tex]\frac{\partial f}{\partial t}~=~f_t(t,z) = A_0~cos(w_0t - Bz)*w_0[/tex]
    That last factor is the partial of (w0*t - B*z) with respect to t. The only difference between what I've done and what you want is that you want the partial of the same expression with respect to z. Both partials use the chain rule.
  4. Oct 11, 2009 #3
    Er... nevermind. I think you just neglect the first part since we're only concerned with the z variable, and treat it as a sin(-B*z), right?
  5. Oct 11, 2009 #4


    Staff: Mentor

    No, that's not how the chain rule works. If your function is truly a function of two variables, which it seems to be, you need to take the partial as I did, only you want the partial with respect to z, not the one with respect to t.
  6. Oct 11, 2009 #5
    Yes I understand that, but since I'm taking the partial with respect to z, the answer would be...

    -B*sin(w0*t - Bz)

    I didn't explain myself clearly, thats why your confused. I think that we agree however.
  7. Oct 11, 2009 #6


    Staff: Mentor

    Close. You forgot the A0.
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