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Implicit differentiation help three variables!

  1. Oct 24, 2012 #1
    1. The problem statement, all variables and given/known data

    I have a question. How in general would one differentiate a composite function like F(x,y,z)=2x^2-yz+xz^2 where x=2sint , y=t^2-t+1 , and z = 3e^-1 ? I want to find the value of dF/dt evaluated at t=0 and I don't know how. Can someone please walk me through this?


    2. Relevant equations
    Mod note: Fixed the equation below to match the OP's change above.
    F(x,y,z)=2x^2-yz+xz^2

    dF/dx=4x-z^2 , dF/dy= -z , dF/dz = 2xz-y , dz/dt=0 , dx/dt=2cost, dy/dt=2t-1

    dF/dx dx/dt + dF/dy dy/dt + dF/dz dz/dt= dF/dt

    3. The attempt at a solution

    I tried a couple of things, including chain rules and jacobians. I know that dF/dt should equal dF/dx dx/dt + dF/dy dy/dt + dF/dz dz/dt but for some reason this doesn't work, or I am doing something wrong. I start out by differentiating to get dF/dx=4x-z^2 , dF/dy= -z , dF/dz = 2xz-y , dz/dt=0 , dx/dt=2cost, dy/dt=2t-1 but this doesn't match the answer, which my book says is 24.

    How do they get this, and where is my error? Thanks.
     
    Last edited by a moderator: Oct 24, 2012
  2. jcsd
  3. Oct 24, 2012 #2

    Mark44

    Staff: Mentor

    Do you have a typo in your definition for F(x, y, z)? As you wrote it, it could be simplified to 3x2 - yz.
    These three should be partial derivatives, and a couple of them are incorrect, at least based on what you wrote.

    $$ \frac{\partial F}{\partial x} = 6x $$

     
  4. Oct 24, 2012 #3
    Sorry, yes there was a typo, I fixed it. Are the partials still wrong? Thanks for responding so quickly!
     
  5. Oct 24, 2012 #4
    I want to rephrase the question. Since everyone else I have talked to thinks there was an error in the book, does everyone here agree?
     
  6. Oct 24, 2012 #5

    Mark44

    Staff: Mentor

    Here's your work from post #1.
    Your partial with respect to x has a sign error. The other partials are correct, and your three derivatives are correct.
     
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