# What's the value of this partial derivative?

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.

-
What is $z$? You did not define it... – Dennis Gulko Nov 19 '12 at 23:56
basically, z = f(x,y) – George Nov 20 '12 at 0:21
When writing on this site, you should use $\LaTeX$ formatting: use **$** before and after each equation, to get$z = f(x,y)$instead of z = f(x,y). See my edit for more info and meta.math.stackexchange.com/questions/107/… – Dennis Gulko Nov 20 '12 at 0:39 ## 1 Answer Assuming$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*} Usingx=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$\frac{\partial^2 z}{\partial y\partial x}=\frac{\partial^2 z}{\partial x\partial y}$- can you use simpler notations? – George Nov 20 '12 at 0:20 where do those two other terms come from? i usually get 4 terms and not 6. – George Nov 20 '12 at 0:22 Simpler? what is simpler than the derivative notation? Now you should open the parentheses, and plug the derivatives of$x$and$y$, which you can calculate. The terms come from the way we derive multiplication:$\frac{\partial}{\partial t}(fg)=\frac{\partial f}{\partial t}\cdot g+f\cdot\frac{\partial g}{\partial t}$. The second line comes from using the first formula I wrote for$w=\frac{\partial z}{\partial x}$and$w=\frac{\partial z}{\partial y}\$ – Dennis Gulko Nov 20 '12 at 0:25
i don't understand why sometimes we get 4 and sometimes we get 6 terms. – George Nov 20 '12 at 0:46
We always get 6 terms. Sometimes, you can write only 5 because of what I wrote in the end of my answer – Dennis Gulko Nov 20 '12 at 0:49