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I need to find $\frac{\partial^2x}{\partial t^2}$, where $x = r\sin t$ and $y = r\cos t$.
I get the following:

$\frac{\partial^2z}{\partial x^2} r^2\sin^2 t + 2\frac{\partial^2z}{\partial y\partial x} r^2\cos t\sin t + \frac{\partial^2z}{\partial y^2} r^2\cos^2t$

I heard that it's something else with 6 terms instead of 4, how come?

I don't understand. Please explain it to me in a way my feeble mind can understand.

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    When writing on this site, you should use $\LaTeX$ formatting: use **$** before and after each equation, to get $z = f(x,y)$ instead of$z$= f(x,y). See my edit for more info and http://meta.math.stackexchange.com/questions/107/faq-for-math-stackexchange/117#1172012-11-20

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Assuming $z=z(x,y)$, then $\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}$ Hence: $\begin{align*}\frac{\partial^2 z}{\partial t^2}&=\frac{\partial}{\partial t}\left(\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}\right)=\frac{\partial}{\partial t}\left(\frac{\partial z}{\partial x}\right)\frac{\partial x}{\partial t}+\frac{\partial z}{\partial x}\frac{\partial^2 x}{\partial t^2}+\frac{\partial}{\partial t}\left(\frac{\partial z}{\partial y}\right)\frac{\partial y}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial^2 y}{\partial t^2}\\ &= \left(\frac{\partial^2 z}{\partial x^2}\frac{\partial x}{\partial t}+\frac{\partial^2 z}{\partial y\partial x}\frac{\partial y}{\partial t}\right)\frac{\partial x}{\partial t}+\frac{\partial z}{\partial x}\frac{\partial^2 x}{\partial t^2}+\left(\frac{\partial^2 z}{\partial x\partial y}\frac{\partial x}{\partial t}+\frac{\partial^2 z}{\partial y^2}\frac{\partial y}{\partial t}\right)\frac{\partial y}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial^2 y}{\partial t^2} \end{align*}$ Using $x=r\sin t$, $y=r\cos t$, we have $\frac{\partial x}{\partial t}=r\cos t, \hspace{5pt} \frac{\partial y}{\partial t}=-r\sin t, \hspace{5pt} \frac{\partial^2 x}{\partial t^2}=-r\sin t, \hspace{5pt} \frac{\partial^2 y}{\partial t^2}=-r\cos t$ Substituting, we have: $\frac{\partial^2 z}{\partial x^2}r^2\cos^2t-\left(\frac{\partial^2 z}{\partial x\partial y}+\frac{\partial^2 z}{\partial y\partial x}\right)r^2\sin t\cos t-\frac{\partial z}{\partial x}r\sin t+\frac{\partial^2 z}{\partial y^2}r^2\sin^2 t-\frac{\partial z}{\partial y}r\cos t$ And if you know that $\frac{\partial^2 z}{\partial y\partial x}$ and $\frac{\partial^2 z}{\partial x\partial y}$ are continuous, then $\frac{\partial^2 z}{\partial y\partial x}=\frac{\partial^2 z}{\partial x\partial y}$

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    thanks the second edit made the difference i think2012-11-20