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Calc 3 partial derivative review for PDE's class

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data

    I am suppose to use polar coordinate data to find derivatives, ie

    x = r cos(theta)
    y = r sin(theta)

    r^2 = x^2 + y^2

    2. Relevant equations

    show dtheta/dy = cos(theta)/r
    show dtheta/dx = -sin(theta)/r

    in other words since i dont have the math script
    find the equation for theta and take the derivatives
    These are partial derivatives by the way (as you can tell by inspection)

    3. The attempt at a solution

    d theta / dy = cos(theta)/r

    I separate and integrate

    dtheta/cos(theta) = dy/sqrt(x^2+y^2)

    ln(cos(theta)) = ln(sqrt(x^2+y^2))/(0.5(x^2+y^2)^-.5*2y) => per chain rule

    ln(cos(theta))=ln(r)*r/ r*sin(theta)

    ln(cos(theta)) = ln(r)/ sin(theta)

    now what?

    any help is very much appreciated. Thanks for looking. have a good one.
  2. jcsd
  3. Sep 17, 2009 #2


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    Homework Helper

    how about trying implictly differentiating both sides of all your equations w.r.t x & y, then a little subtitution to get the form you want?

    you'll need to consider
    r = r(x,y)
    theta = theta(x,y)
  4. Sep 17, 2009 #3
    do you mind refreshing me on implicit differentiation?
  5. Sep 17, 2009 #4


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    Staff Emeritus
    Science Advisor
    Gold Member

    Implicit differentiation is where you differentiate every term with respect to one of them without solving one term as a function of another.

    For example

    x2 + y2 = 1

    I want to find dy/dx. Rather than calculate y as a function of x, I just differentiate both sides with respect to x. I get a dy/dx because of the chain rule

    2x + 2y dy/dx = 0

    dy/dx = -x/y

    Similarly, you could differentiate your equations implicitly with respect to e.g. y in an attempt to find dtheta/dy
  6. Sep 20, 2009 #5
    Thanks, this worked and was actually quite a clever way to solve the problem.
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