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## Homework Statement

Change of coordinates from rectangular (x,y) to polar (r,θ). Not sure what's wrong with my working..

- Thread starter unscientific
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- #1

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Change of coordinates from rectangular (x,y) to polar (r,θ). Not sure what's wrong with my working..

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pasmith

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Your error is that you have differentiated [itex]y = x \tan\theta[/itex] incorrectly. You should get

[tex]\frac{\partial y}{\partial \theta} = x\sec^2\theta + \frac{\partial x}{\partial \theta}\tan\theta[/tex]

because [itex]x[/itex] is also a function of [itex]\theta[/itex].

Also, to work out [itex]\partial x/\partial r[/itex], you would need to differentiate

[tex]r^2 = x^2 + y^2[/tex]

with respect to [itex]r[/itex]*with [itex]\theta[/itex] held constant*, which again gives

[tex]2r = 2x\frac{\partial x}{\partial r} + 2y\frac{\partial y}{\partial r}[/tex]

because y is also a function of [itex]r[/itex].

If you're trying to find derivatives with respect to r and [itex]\theta[/itex], it's best to start from [itex]x = r\cos\theta[/itex], [itex]y = r\sin\theta[/itex].

[tex]\frac{\partial y}{\partial \theta} = x\sec^2\theta + \frac{\partial x}{\partial \theta}\tan\theta[/tex]

because [itex]x[/itex] is also a function of [itex]\theta[/itex].

Also, to work out [itex]\partial x/\partial r[/itex], you would need to differentiate

[tex]r^2 = x^2 + y^2[/tex]

with respect to [itex]r[/itex]

[tex]2r = 2x\frac{\partial x}{\partial r} + 2y\frac{\partial y}{\partial r}[/tex]

because y is also a function of [itex]r[/itex].

If you're trying to find derivatives with respect to r and [itex]\theta[/itex], it's best to start from [itex]x = r\cos\theta[/itex], [itex]y = r\sin\theta[/itex].

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