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Math Help - Partial Differentiation Q

  1. #1
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    Question Partial Differentiation Q

    Ok here goes....

    <br />
z=x^{2}+y^{2}

    x=rcos\vartheta

    y=r sin\vartheta

    Find:

    \frac{\partial z}{\partial x}_{y},<br />
\frac{\partial z}{\partial \vartheta}_{x}, <br />
  \frac{\partial z}{\partial r}_{y}, <br />
  \frac{\partial z}{\partial r}_{ \vartheta}, <br />

    ___________________________

    \frac{\partial z}{\partial x}_{y} = 2x
    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....
    ___________________________

    <br />
\frac{\partial z}{\partial \vartheta}_{x}=

    I've been at this one for hours, i think i can use the following, but i'm getting nowhere.

     dz= \frac{\partial z}{\partial x}_{y}dx + \frac{\partial z}{\partial y}_{x}dy


     dx= \frac{\partial x}{\partial r}_{\vartheta}dr +  \frac{\partial x}{\partial \vartheta}_{r}d\vartheta


    dy= \frac{\partial y}{\partial r}_{\vartheta}dr + \frac{\partial y}{\partial \vartheta}_{r}d\vartheta

    so....

     dz= \frac{\partial z}{\partial x}_{y} \left[ \frac{\partial x}{\partial r}_{\vartheta}dr +  \frac{\partial x}{\partial \vartheta}_{r}d\vartheta\right]+  \frac{\partial z}{\partial y}_{x}\left[ \frac{\partial y}{\partial r}_{\vartheta}dr + \frac{\partial y}{\partial \vartheta}_{r}d\vartheta\right]

    can i divide through by  \partial \vartheta_{x} and then work it all out to get
    <br />
 \frac{\partial z}{\partial \vartheta}_{x} ?

    please help, i don't have a clue what i'm doing!!!
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  2. #2
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    Quote Originally Posted by UbikPkd View Post
    Ok here goes....

    <br />
z=x^{2}+y^{2}

    x=rcos\vartheta

    y=r sin\vartheta
    I am not sure what you want to find.

    Let z=f(x,y) where x=r\cos \theta \mbox{ and }y=r\sin \theta and f(x,y) = x^2+y^2.

    Then,
    \frac{\partial z}{\partial r} = \frac{\partial f}{\partial x}\cdot \frac{\partial x}{\partial r} + \frac{\partial f}{\partial y}\cdot \frac{\partial y}{\partial r}
    Then,
    \frac{\partial z}{\partial r} = (2x)(\sin \theta) + (2y)(\cos \theta) = 2r\sin \theta \cos \theta + 2r\sin \theta \cos \theta = 2r\sin 2\theta

    The other one is similar.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    I am not sure what you want to find.

    Let z=f(x,y) where x=r\cos \theta \mbox{ and }y=r\sin \theta and f(x,y) = x^2+y^2.

    Then,
    \frac{\partial z}{\partial r} = \frac{\partial f}{\partial x}\cdot \frac{\partial x}{\partial r} + \frac{\partial f}{\partial y}\cdot \frac{\partial y}{\partial r}
    Then,
    \frac{\partial z}{\partial r} = (2x)(\sin \theta) + (2y)(\cos \theta) = 2r\sin \theta \cos \theta + 2r\sin \theta \cos \theta = 2r\sin 2\theta

    The other one is similar.
    \frac{\partial z}{\partial r} = (2x)(\cos (\theta)) + (2y)(\sin (\theta)) = 2r\cos^2(\theta)+2r\sin^2(\theta)=2r

    RonL
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  4. #4
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    thanks both of you, i understand captain blacks correction, theta is constant in both of those partial diffs.

    quickly doing
    <br />
\frac{\partial z}{\partial \vartheta}_{x}
    i get
    <br />
 \frac{\partial z}{\partial \vartheta}_{x}=0

    which makes sense to me. thanks for you help
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  5. #5
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    Quote Originally Posted by UbikPkd View Post
    <br />
z=x^{2}+y^{2}

    x=rcos\vartheta

    y=r sin\vartheta

    Find:

    \frac{\partial z}{\partial x}_{y},<br />
\frac{\partial z}{\partial \vartheta}_{x}, <br />
  \frac{\partial z}{\partial r}_{y}, <br />
  \frac{\partial z}{\partial r}_{ \vartheta}, <br />
    This is a very strange question, and I'd like to be sure that I understand the notation before I try to answer it.

    I'm guessing that those little subscripts indicate a quantity that is to be held constant. If so, then \frac{\partial z}{\partial x}_{y}, for example, makes good sense: z is a function of x and y, and when you take the partial derivative of z with respect to x you would naturally interpret this to mean that y is being held constant. But the next derivative, \frac{\partial z}{\partial \vartheta}_{x}, is another matter altogether. The variables r and ϑ go together, and when you form a partial derivative with respect to ϑ you would normally expect that r, not x, is the variable being held constant. To keep x constant when differentiating with respect to ϑ is unusual, not to say perverse. It can be done, but it needs to be done with care. Am I right in thinking that this is what this problem is really about, or am I misinterpreting it?
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  6. #6
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    Quote Originally Posted by Opalg View Post
    This is a very strange question, and I'd like to be sure that I understand the notation before I try to answer it.

    I'm guessing that those little subscripts indicate a quantity that is to be held constant. If so, then \frac{\partial z}{\partial x}_{y}, for example, makes good sense: z is a function of x and y, and when you take the partial derivative of z with respect to x you would naturally interpret this to mean that y is being held constant. But the next derivative, \frac{\partial z}{\partial \vartheta}_{x}, is another matter altogether. The variables r and ϑ go together, and when you form a partial derivative with respect to ϑ you would normally expect that r, not x, is the variable being held constant. To keep x constant when differentiating with respect to ϑ is unusual, not to say perverse. It can be done, but it needs to be done with care. Am I right in thinking that this is what this problem is really about, or am I misinterpreting it?
    yer i have copied the question exactly and correctly so it's probably just unusual but not necessarily wrong.

    For the last one I get:

    <br />
  \frac{\partial z}{\partial r}_{ \vartheta}=2r<br />

    yer and the subscripts are the constants

    hmm, your guess is better than mine, though i think it's right, the question isn't likely to be wrong at least
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  7. #7
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    Quote Originally Posted by UbikPkd View Post
    yer i have copied the question exactly and correctly so it's probably just unusual but not necessarily wrong.

    For the last one I get:

    <br />
  \frac{\partial z}{\partial r}_{ \vartheta}=2r<br />
That one's definitely right.

    yer and the subscripts are the constants

    hmm, your guess is better than mine, though i think it's right, the question isn't likely to be wrong at least
    Okay, for \frac{\partial z}{\partial \vartheta}_x, I would interpret this as meaning that r and ϑ are variables, and x is constant. (But r and ϑ are not independent variables, because they are constrained by the fact that x has to be held constant.)

    Differentiate z=x^2+y^2 with respect to ϑ, using the chain rule and treating x as a constant: \frac{\partial z}{\partial \vartheta} = 2y\frac{\partial y}{\partial \vartheta}. Now we have to find \frac{\partial y}{\partial \vartheta}.

    Differentiate y=r\sin\vartheta with respect to ϑ, using the product rule and the chain rule: \frac{\partial y}{\partial \vartheta} = r\cos\vartheta + \frac{\partial r}{\partial \vartheta}\sin\vartheta.

    Finally, we have to find \frac{\partial r}{\partial \vartheta}. For this, differentiate the equation x=r\cos\vartheta with respect to ϑ (remembering that x is constant): 0 = -r\sin\vartheta + \frac{\partial r}{\partial \vartheta}\cos\vartheta.

    Solve that last equation for \frac{\partial r}{\partial \vartheta}, substitute it into the equation for \frac{\partial y}{\partial \vartheta}, and then substitute that into the equation for \frac{\partial z}{\partial \vartheta}. When you put it all together, you should find that \frac{\partial z}{\partial \vartheta} = 2y\Bigl(r\cos\vartheta + \frac{r\sin^2\vartheta}{\cos\vartheta}\Bigr). Since r and ϑ are the variables, it's probably best to express this in terms of them, namely (after a bit of simplification) \frac{\partial z}{\partial \vartheta} = 2r^2\tan\vartheta.

    (Note: all the above partial derivatives should really have have that little subscript _x tacked on to them, just to emphasise that x is being held constant throughout.)

    You could find the other derivative \frac{\partial z}{\partial r}_y by exactly the same sort of method, this time keeping y fixed throughout. In practice, however, it would be much simpler to notice that z=x^2+y^2=r^2, so \frac{\partial z}{\partial r}=2r, regardless of what else is being kept constant.
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  8. #8
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    Opalg, thanks, that's really helpful, i'm sorry i was being a bit of an idiot, that is indeed the way i believe they want me to do it, sorry for your trouble!
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