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Partial Differentiation of an expression.

  1. Oct 11, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the partial of z with respect to x keeping r constant.

    2. Relevant equations

    z=x2+y2

    x= rcos(t)
    y= rsin(t)

    3. The attempt at a solution


    = r^2(cos(t))^2 + r^2(sin(t))^2

    use product rule on "x" and hold r and y constant
    = [0(cos(t))^2 + r^2(2cos(t))(-sin(t)))] + 0

    simplify terms

    = -2r^2(cos(t)sin(t))

    I am confused because to hold r constant, does that mean i hold both r and y constant?

    thanks
     
    Last edited: Oct 11, 2008
  2. jcsd
  3. Oct 11, 2008 #2

    Mark44

    Staff: Mentor

    You're given z as a function of x and y, and asked for [tex]\frac{\partial z}{\partial x}[/tex]. The bit about holding r constant seems to be a red herring. To help you out a little, [tex]\frac{\partial z}{\partial y}[/tex] = 2y.

    BTW, I corrected the definitions of x and y per your later submission.
     
  4. Oct 11, 2008 #3
    In this case we are holding x constant, but what is happening to r? How can i hold r constant if it is connected to theta? Do I hold theta constant as well?
     
  5. Oct 12, 2008 #4

    Mark44

    Staff: Mentor

    It doesn't matter what happens with r and t (your original post gives t, not theta). You have three functions: one with z as a function of x and y, one with x as a function of r and t, and one with y as a function of r and t.

    The question you asked was, what is [tex]\frac{\partial z}{\partial x}[/tex]. As a hint, I showed you what [tex]\frac{\partial z}{\partial y}[/tex] was.

    You didn't ask, but there are other partial derivatives that could be gotten, namely [tex]\frac{\partial z}{\partial r}[/tex] and [tex]\frac{\partial z}{\partial t}[/tex]. As it turns out, both of these partials are zero. For each of these partials, you do have take r and t into consideration, which you do by using a form of the chain rule for partial derivatives.

    For the functions you have in this problem,
    [tex]\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} * \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}*\frac{\partial y}{\partial t}
    [/tex]

    The chain rule form for [tex]\frac{\partial z}{\partial r}[/tex] is similar, but involves [tex]\frac{\partial x}{\partial r}[/tex] and [tex]\frac{\partial y}{\partial r}[/tex].


    Mark
     
  6. Oct 12, 2008 #5

    Mark44

    Staff: Mentor

    It doesn't matter what happens with r and t (your original post gives t, not theta). You have three functions: one with z as a function of x and y, one with x as a function of r and t, and one with y as a function of r and t.

    The question you asked was, what is [tex]\frac{\partial z}{\partial x}[/tex]. As a hint, I showed you what [tex]\frac{\partial z}{\partial y}[/tex] was.

    You didn't ask, but there are other partial derivatives that could be gotten, namely [tex]\frac{\partial z}{\partial r}[/tex] and [tex]\frac{\partial z}{\partial t}[/tex]. As it turns out, both of these partials are zero. For each of these partials, you do have to take r and t into consideration, which you do by using a form of the chain rule for partial derivatives.

    For the functions you have in this problem,
    [tex]\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} * \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}*\frac{\partial y}{\partial t}
    [/tex]

    The chain rule form for [tex]\frac{\partial z}{\partial r}[/tex] is similar, but involves partials of x and y with respect to r, instead of with respect to t.


    Mark
     
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