0

$$\frac{\partial }{\partial \theta}\frac{\partial z }{\partial x}$$

Can any kind soul help me on this? I've tried numerous calculators and resources but I can't seem to find it.

  1. Let $z=f(x,y)$ be a function in two variables so that $x=r\cos\theta$ and $y=r\sin\theta$.$\tag6$

    [Assume that all the partial derivatives of the function exist and are continuous.]

    (i) Show that $$\frac{\partial z}{\partial\theta}=-\frac{\partial z}{\partial x}r\sin\theta+\frac{\partial z}{\partial y}r\cos\theta$$

    (ii) Show that $$\frac{\partial^2z}{\partial\theta^2}=\frac{\partial^2z}{\partial x^2}r^2\sin^2\theta-2\frac{\partial z^2}{\partial x\partial y}r^2\cos\theta\sin\theta+\frac{\partial^2 z}{\partial y^2}r^2\cos^2\theta-\frac{\partial z}{\partial x}r\cos\theta-\frac{\partial z}{\partial y}r\sin\theta$$

Continuing from 7i, I know I have to use product rule to get the second derivative, but I run into trouble of differentiating (dz/dx) wrt theta, hence this question.

Community
  • 1
ConfusedGuest
  • 1

2 Answers2

1

I assume you can exchange partial derivatives, so that:

$\frac{\partial}{\partial \theta} \frac{\partial z}{\partial x} = \frac{\partial}{\partial x}\frac{\partial z}{\partial \theta}$

Then you can use your answer form part (i) to plug into $\frac{\partial z}{\partial \theta}$, and take the partial of that with respect to $x$.

Machine Caliber
  • 714
  • Thanks so much for your reply! I get what you mean and did the calculations, and I managed to get the final answer except for the mixed term.. instead of

    $-2\frac{\partial z^2}{\partial x \partial y}r^2cos(O)sin(O))$ i get

    $-2\frac{\partial^2 z}{\partial x \partial y}r^2cos(O)sin(O))$

    is there something im doing wrong..?

     – ConfusedGuest Mar 29 '15 at 06:19
  • Your answer is correct, and the solution key is wrong. $\frac{\partial z^2}{\partial x\partial y}$ makes no mathematical sense, and it should be what you have. – Machine Caliber Mar 29 '15 at 15:28
0

By the chain rule, the operation becomes: $\dfrac{\partial ~~}{\partial\theta}=\dfrac{\partial x}{\partial\theta}{\cdot}\dfrac{\partial ~~}{\partial x}+\dfrac{\partial y}{\partial\theta}{\cdot}\dfrac{\partial ~~}{\partial y}$

So applying this operation to $\dfrac{\partial z}{\partial x}$, we have

$$\dfrac{\partial~~}{\partial\theta}\left[\dfrac{\partial z}{\partial x}\right]=\dfrac{\partial x}{\partial\theta}{\cdot}\dfrac{\partial^2 z}{\partial x~^2}+\dfrac{\partial y}{\partial\theta}{\cdot}\dfrac{\partial^2 z}{\partial y~\partial x}$$

So we use chain rule, product rule, chain rule, and some algebra:

$$\begin{align}\dfrac{\partial^2 z}{\partial\theta~^2} &=\dfrac{\partial~~}{\partial\theta}\left[\dfrac{\partial x}{\partial\theta}{\cdot}\dfrac{\partial z}{\partial x}+\dfrac{\partial y}{\partial\theta}{\cdot}\dfrac{\partial z}{\partial y}\right]\\&=\dfrac{\partial^2 x}{\partial\theta~^2}{\cdot}\dfrac{\partial z}{\partial x}+\dfrac{\partial x}{\partial \theta}{\cdot}\dfrac{\partial~~}{\partial\theta}\left[\dfrac{\partial z}{\partial x}\right]+\dfrac{\partial y}{\partial \theta}{\cdot}\dfrac{\partial~~}{\partial\theta}\left[\dfrac{\partial z}{\partial y}\right]+ \dfrac{\partial^2 y}{\partial\theta~^2}{\cdot}\dfrac{\partial z}{\partial y}\\&=\dfrac{\partial^2 x}{\partial\theta~^2}{\cdot}\dfrac{\partial z}{\partial x}+\dfrac{\partial x}{\partial \theta}{\cdot}\left(\dfrac{\partial x}{\partial\theta}{\cdot}\dfrac{\partial^2 z}{\partial x~^2}+\dfrac{\partial y}{\partial\theta}{\cdot}\dfrac{\partial^2 z}{\partial y~\partial x}\right)+\dfrac{\partial y}{\partial \theta}{\cdot}\left(\dfrac{\partial x}{\partial\theta}{\cdot}\dfrac{\partial^2 z}{\partial x~\partial y}+\dfrac{\partial y}{\partial\theta}{\cdot}\dfrac{\partial^2 z}{\partial y~^2}\right)+ \dfrac{\partial^2 y}{\partial\theta~^2}{\cdot}\dfrac{\partial z}{\partial y}\\&=\dfrac{\partial^2 x}{\partial\theta~^2}{\cdot}\dfrac{\partial z}{\partial x}+{\left(\dfrac{\partial x}{\partial \theta}\right)}^2{\cdot}\dfrac{\partial^2 z}{\partial x~^2}+2{\cdot}\dfrac{\partial x}{\partial \theta}{\cdot}\dfrac{\partial y}{\partial\theta}{\cdot}\dfrac{\partial^2 z}{\partial y~\partial x}+{\left(\dfrac{\partial y}{\partial \theta}\right)}^2{\cdot}\dfrac{\partial^2 z}{\partial y~^2}+ \dfrac{\partial^2 y}{\partial\theta~^2}{\cdot}\dfrac{\partial z}{\partial y}\end{align}$$

Graham Kemp
  • 129,094