Question

I need some help with multivariable calculus. Stuck, and would like some clarification. Question in thread.

Answer 1

It is given that \(g(r,\theta)=f(r\cos \theta, r\sin \theta)\) i need to express \[\frac{\partial ^2 g}{\partial \theta^2}\] in terms of the partial derivatives of f.

So let \(x(r,\theta) = r\cos \theta \) and \(y(r,\theta) = r\sin \theta \), thus we have \(f(x(r,\theta),y(r,\theta))\)

\[\begin{split} \frac{\partial g}{\partial \theta} &= \frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta} +\frac{\partial f}{\partial y}\frac{\partial y}{\partial \theta} \\ &= \frac{\partial f}{\partial x}\cdot -r\sin\theta+\frac{\partial f}{\partial y}\cdot r\cos\theta\\\\ \end{split} \]

Therefore,

\[\frac{\partial^2 g }{\partial \theta^2} =\frac{d}{d \theta} \left[ \frac{\partial f}{\partial x}\cdot -r\sin\theta +\frac{\partial f}{\partial y}\cdot r\cos\theta\right] \]

This is what i have done so far, but i am not sure how to carry on. I know i would need to use product rule, but i am not sure what \[\frac{d}{d \theta} \left[ \frac{\partial f}{\partial x}\right] \] is. Now Wio did try explaining that

\[\begin{split}
\frac{d}{d\theta }\left[\frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta}\right] &=\frac{d}{d\theta}\left[\frac{\partial f}{\partial\theta}\right]\frac{\partial x}{\partial \theta}+\frac{\partial f}{\partial x}\frac{d}{d\theta}\left[\frac{\partial x}{\partial \theta}\right] \\
&= \left(\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial \theta}\right)\frac{\partial x}{\partial\theta}+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial \theta^2} \\
&=\frac{\partial^2 f}{\partial x^2}\left(\frac{\partial x}{\partial \theta}\right)^2+\frac{\partial f}{\partial x}\frac{\partial^2 x}{\partial \theta^2}
\end{split} \]

But I do need a bit of clarification about how the term on the RHS
\[\frac{d}{d\theta}\left[\frac{\partial f}{\partial\theta}\right]\] came from \[\frac{d}{d\theta }\left[\frac{\partial f}{\partial x}\frac{\partial x}{\partial \theta}\right] \].

Answer 2

Seems like the df/d(theta) right after the equals sign should be a df/dx

Answer 3

oh ok, so if thats the case that result still holds, and thats how i should finish off the question.

Answer 4

Yup looks good

Answer 5

one quick question why is
\[=\frac{d}{d\theta}\left[\frac{\partial f}{\partial x}\right]= \left(\frac{\partial^2 f}{\partial x^2}\frac{\partial x}{\partial \theta}\right)\]

Answer 6

@saifoo.khan do you think you can help me?

Answer 7

I'm really sorry. I can't. :S

Answer 8

have you looked through the whole post? do you know anyone who would be able to explain this to me?

Answer 9

Many helpers aren't online currently.

@jim_thompson5910 might know this.

Answer 10

ok, well i will leave it open for a while

Answer 11

@dpaInc you think you can help me?

Answer 12

@phi are you able to explain this?

Answer 13

@ash2326, @UnkleRhaukus anyone able to help?

Answer 14

Sure. Your question is hard. Everyone can't do it (including me). :/

Answer 15

ok well i need to go sleep, but going to leave the question open so feel free to answer it anyone. I think the thing i need help with is, from \[\frac{\partial^2 g }{\partial \theta^2} =\frac{d}{d \theta} \left[ \frac{\partial f}{\partial x}\cdot -r\sin\theta +\frac{\partial f}{\partial y}\cdot r\cos\theta\right]\] what \[\frac{d}{d\theta}\left[\frac{\partial f}{\partial x}\right]\] is, so i can carry out the product rule. Please, if you dont mind, tag other helpers in as i will not be able to bump the question

Answer 16

@hero, @zepdrix , @robtobey ,@Shane_B

Answer 17

@Wio

Answer 18

Have you looked at
http://tutorial.math.lamar.edu/Classes/CalcIII/ChainRule.aspx
Example 5 near the very bottom (but I would read the whole thing)

It looks quite messy.

Answer 19

@phi thank you for pointing me to that link, its funny as i use that site to help me with other things but missed that part.