1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Partial Differentiation Q

  1. Dec 18, 2007 #1
    Ok here goes....

    [tex]z=x^{2}+y^{2} [/tex]


    [tex]y=r sin\vartheta[/tex]


    [tex]\[ \frac{\partial z}{\partial x}_{y},
    \[ \frac{\partial z}{\partial

    \[ \frac{\partial z}{\partial r}_{y},
    \[ \frac{\partial z}{\partial r}_{



    [tex]\[ \frac{\partial z}{\partial x}_{y}

    = 2x [/tex]

    this seems right to me (though i'm not

    sure if i'm supposed to use a chain rule),

    it's the next ones i'm not sure about....

    \[ \frac{\partial z}{\partial

    \vartheta}_{x}= [/tex]

    I've been at this one for hours, i think i

    can use the following, but i'm getting


    [tex] dz= \[ \frac{\partial z}{\partial

    x}_{y}dx + \[ \frac{\partial z}{\partial


    [tex] dx= \[ \frac{\partial x}{\partial

    r}_{\vartheta}dr + \[ \frac{\partial

    x}{\partial \vartheta}_{r}d\vartheta[/tex]

    [tex] dy= \[ \frac{\partial y}{\partial

    r}_{\vartheta}dr + \[ \frac{\partial

    y}{\partial \vartheta}_{r}d\vartheta[/tex]


    [tex] dz= \[ \frac{\partial z}{\partial

    x}_{y} \left[\[ \frac{\partial x}{\partial

    r}_{\vartheta}dr + \[ \frac{\partial


    \vartheta}_{r}d\vartheta\right]+ \[

    \frac{\partial z}{\partial y}_{x}\left[\[

    \frac{\partial y}{\partial

    r}_{\vartheta}dr + \[ \frac{\partial



    can i divide through by [tex] \partial

    \vartheta_{x}[/tex] and then work it all

    out to get
    \[ \frac{\partial z}{\partial

    \vartheta}_{x} [/tex] ?

    please help, i don't have a clue what i'm

    doing!!! :confused:
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 18, 2007 #2


    User Avatar
    Science Advisor
    Homework Helper

    The direct way to do this is to simply write z as a function only of x and theta. This is pretty easy, since y/x=tan(theta). Try it that way before you go back to struggling with the chain rule.
  4. Dec 18, 2007 #3
    okay thanks for the tip, i worked it out using chain rules in the end before i got to read it, the answers i got were:

    [tex]\[ \frac{\partial z}{\partial x}_{y}=2x[/tex]

    [tex]\[ \frac{\partial z}{\partial \vartheta}_{x}=0[/tex]

    [tex]\[ \frac{\partial z}{\partial r}_{y}=2r[/tex]

    [tex]\[ \frac{\partial z}{\partial r}_{\vartheta}=2r


    I think these are right, thanks for your help!
  5. Dec 18, 2007 #4


    User Avatar
    Science Advisor
    Homework Helper

    Well, e.g. for the second one, y/x=tan(theta), so z=x^2+x^2*tan^2(theta). It doesn't look to me like the derivative wrt theta at constant x is 0.
  6. Dec 19, 2007 #5
    sorry! arghh i've made such a mess of this, yer the last two are right, as [tex]z=r^2[/tex] so it doesn't matter what's being held constant

    [tex]\[ \frac{\partial z}{\partial \vartheta}_{x}=2r^{2}tan\vartheta[/tex]

    not 0, sorry, here's my full working:

    [tex]\[ \frac{\partial z}{\partial \vartheta}_{x}=2y\[ \frac{\partial y}{\partial \vartheta}_{x}[/tex]



    [tex]\[ \frac{\partial y}{\partial \vartheta}_{x}=rcos\vartheta+\[ \frac{\partial r}{\partial \vartheta}_{x}sin\vartheta[/tex]



    [tex]0=-rsin\vartheta+\[ \frac{\partial r}{\partial \vartheta}_{x}cos\vartheta[/tex]

    [tex]\[ \frac{\partial r}{\partial \vartheta}_{x}=r\[ \frac{sinv}{cos\vartheta}=rtan\vartheta[/tex]


    [tex]\[ \frac{\partial z}{\partial \vartheta}_{x}=2y(rcos\vartheta+rsin\vartheta tan\vartheta)[/tex]

    [tex]\[ \frac{\partial z}{\partial \vartheta}_{x}=2r^{2}\frac{sin\vartheta}{cos\vartheta}=2r^{2}tan\vartheta[/tex]

    I did it that way, as I thought differentiating via the [tex] tan^{2}\vartheta[/tex] way would probably go wrong for me somehow! I think the above is OK, seems like that's the way they want me to do it as well. Thanks for your help!
  7. Dec 19, 2007 #6


    User Avatar
    Science Advisor
    Homework Helper

    Doing it the other way gives you 2x^2*sin(theta)/cos^3(theta), which if you put x=r*cos(theta) gives you the same thing you got. So I think your answer is right.
  8. Dec 19, 2007 #7
    cool thanks!
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Partial Differentiation Q
  1. Partial Differentiation (Replies: 11)

  2. Partial differentiation (Replies: 10)