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Thread: Finding a partial derivative using differentials

  1. #1
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    Finding a partial derivative using differentials

    Hi,

    I'll just get straight into the question then say what I'm confused about at the bottom...

    x = rcos(t) and y = rsin(t), find dr/dx and dr/dy (partial derivatives)

    The way to solve it in my notes goes:

    dx = (dx/dr)dx + (dx/dt)dt = cos(t)dr - rsin(t)dt (sorry, don't know how to do partial d's)

    cos(t)dx = cos^2(t)dr -rsin(t)cos(t)dt

    dy = sin(t)dr + rcos(t)dt

    sin(t)dy = sin^2(t)dr + rsin(t)cos(t)dt

    so: cos(t)dx + sin(t)dy = dr
    = (dr/dx)dx + (dr/dy)dy

    -> (dr/dx) = cos(t) and (dr/dy) = sin(t)


    I get all that, but if we go back to x = rcos(t) and y = rsin(t), couldn't we just rearrange these to make r the subject, the take the partial derivatives of that? But this gets (dr/dx) = 1/cos(t) and (dr/dy) = 1/sin(t), so it obviously can't be right, but where is it incorrect? I'm probably missing something simple...

    Thanks
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  2. #2
    Senior Member x3bnm's Avatar
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    Re: Finding a partial derivative using differentials

    x = rcos(t) and y = rsin(t), find dr/dx and dr/dy (partial derivatives)...........I get all that, but if we go back to x = rcos(t) and y = rsin(t), couldn't we just rearrange these to make r the subject, the take the partial derivatives of that? But this gets (dr/dx) = 1/cos(t) and (dr/dy) = 1/sin(t), so it obviously can't be right, but where is it incorrect? I'm probably missing something simple...
    Partial derivative of what?

    If you meant to take the derivative with respect to $\displaystyle x$ and $\displaystyle y$ of the function $\displaystyle r$ composed of $\displaystyle x$ and $\displaystyle y$ Then:

    $\displaystyle x^2 + y^2 = r^2(\cos^2{(t)} + \sin^2{(t)}) = r^2......\text{[by pluggin' in the values of } x \text{ and } y \text{ and by trigonometry]}$

    Then do implicit differentiation of $\displaystyle r^2 = x^2 + y^2$ with respect to $\displaystyle x$ which is:

    $\displaystyle 2r\frac{d r}{d x} = 2x \,\,\,\therefore \frac{d r}{d x} = \frac{x}{r} = \cos{(t)}.....\text{[after rearrangement and substituting }\dfrac{x}{r}\text{ ]}$

    And then with respect to $\displaystyle y$:

    $\displaystyle 2r\frac{d r}{d y} = 2y \,\,\, \therefore \frac{d r}{d y} = \frac{y}{r} = \sin{(t)}......\text{[after rearrangement and substituting }\dfrac{y}{r}\text{ ]}$


    Or if you meant $\displaystyle \frac{dr}{dx}$ of $\displaystyle x = r \cos{(t)}$ and $\displaystyle \frac{dr}{dy}$ of $\displaystyle y = r \sin{(t)}$ , then the answer $\displaystyle \frac{dr}{dx} = \frac{1}{\cos{(t)}}$ and $\displaystyle \frac{dr}{dy} = \frac{1}{\sin{(t)}}$ is correct.

    Two different answers based on which function you choose to differentiate.
    Last edited by x3bnm; May 12th 2013 at 07:52 AM.
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