Second partial derivative?

**billym** · October 2nd 2009, 08:38 AM

I have a bunch of questions like this but am a little mixed up with what the method is. Can somebody get me started on this?

$<br /> f(x,y) , x = e^rcos(\theta), y=e^rsin(\theta)$

$f^2_r+f^2_\theta=<br />$

**Amer** · October 2nd 2009, 09:40 AM

Originally Posted by billym

I have a bunch of questions like this but am a little mixed up with what the method is. Can somebody get me started on this?

$<br /> f(x,y) , x = e^rcos(\theta), y=e^rsin(\theta)$

$f^2r+f^2\theta=<br />$
(not sure about the latex for that - its second derivative of f with respect to r and theta - I think)

use \partial ok nvm

$\frac{\partial ^2 f}{\partial r^2} = \frac{\partial }{\partial r}(\frac{\partial f}{\partial r})$

$\frac{\partial f}{\partial r} = \frac{\partial f}{\partial y}\cdot \frac{\partial y}{\partial r}+\frac{\partial f}{\partial x}\cdot \frac{\partial x}{\partial r}$

but you did not give f(x,y) = ?? what is equal

**billym** · October 2nd 2009, 09:54 AM

if you look at the top its

$x=e^rcos(\theta), y=e^rsin(\theta)$

What is the difference between $f^2_r$ and $frr$ ?

**Amer** · October 2nd 2009, 10:19 AM

Originally Posted by billym

if you look at the top its

$x=e^rcos(\theta), y=e^rsin(\theta)$

What is the difference between $f^2_r$ and $frr$ ?

I do not know but I think the question should give you f(x,y) equal what for example

$f(x,y)=x+y$ or $f(x,y) = xy$ or $f(x,y) = \frac{x}{2} + \frac{y}{x}$ etc

since you want to find the partial derivatives of f so you need the rule of f and the rule of x,y you give me x,y rules but I need f(x,y) = ?? to solve the question

**billym** · October 2nd 2009, 10:36 AM

Well that's what is causing me problems.

Here is the question in all it's glory:

If $f=f(x,y), x=e^rcos(\theta)$ and $y=e^rsin(\theta)$ , prove the following identidy:

$f^2_r+f^2_\theta=e^{2r}(f^2_x+f^2_y)$

Wouldn't this just work out to 0 = 0 ?

**Amer** · October 2nd 2009, 10:51 AM

Originally Posted by billym

Well that's what is causing me problems.

Here is the question in all it's glory:

If $f=f(x,y), x=e^rcos(\theta)$ and $y=e^rsin(\theta)$ , prove the following identidy:

$f^2_r+f^2_\theta=e^{2r}(f^2_x+f^2_y)$

Wouldn't this just work out to 0 = 0 ?

ok I get it

$f_r ^2 + f_\theta ^2 = e^{2r}(f_x ^2 + f_y ^2 )$ ok take the left side

I will write d instead of partial to make is easier for me ok
$f_r = \frac{df}{dx} \cdot \frac{dx}{dr} + \frac{df}{dy} \cdot \frac{dy}{dr}$

$f_r = f_x (x) + f_y (y)$ since dx/dr =x and dy/dr =y

$f_r ^2 = f_x ^2 x^2 + f_x f_y yx + f_y ^2 y^2$

$f_r ^2 = f_x ^2 (e^{2r} \cos ^2 \theta ) + f_x f_y (e^{2r}\sin \theta \cos \theta ) + f_y ^2 (e^{2r} \sin ^2 \theta )$

this is the first thing now

$\frac{df}{d\theta} = f_x dx/d\theta + f_y dy/d\theta$

$x_\theta = -y , y_\theta = x$

$\frac{df}{d\theta} = f_x (-y) + f_y (x)$

simplify it and find the sum you will get the solution

$\sin ^2 x + \cos ^2 x =1$ use this

**Amer** · October 2nd 2009, 11:25 AM

$x= e^r \cos \theta$

$y=e^r \sin \theta$

$x_\theta = \frac{\partial x}{\partial \theta } = e^r (-\sin \theta) = -e^r\sin \theta = -y$

$y_\theta = \frac{\partial y}{\partial \theta } = e^r \cos \theta = x$

**billym** · October 2nd 2009, 12:10 PM

er... could you start me off on this one?

$f_{rr}+f_{\theta\theta}=e^{2r}(f_{xx}+f_{yy})$

**Amer** · October 2nd 2009, 12:38 PM

Originally Posted by billym

er... could you start me off on this one?

$f_{rr}+f_{\theta\theta}=e^{2r}(f_{xx}+f_{yy})$

$f_r = f_x \cdot x_r + f_y \cdot y_r$

$f_{rr} = (f_{xx}\cdot x_r)x_r + f_x\cdot x_{rr} + (f_{yy} \cdot y_r)y_r + f_y \cdot y_{rr}$ now

$x_{rr} = x_r=x , y_{rr} = y_r=y$

$f_{rr} = f_{xx}\cdot x^2 + f_x \cdot x + f_{yy}\cdot y^2 + f_y \cdot y$ .................(1)

other one I will symbol theta with t ok

$f_{t} = f_x \cdot x_t + f_y \cdot y_t$

$f_{tt}= (f_{xx} x_t)x_t + f_x \cdot x_{tt} + (f_{yy} \cdot y_t)y_t + f_y \cdot y_{tt}$

$x_t = -y , y_t = x$ and $x_{tt} = -x , y_{tt} = -y$

$f_{tt} = f_{xx} \cdot y^2 + f_x \cdot -x + f_{yy} \cdot x^2 + f_y \cdot -y$ .........(2)

sub x value and y value in 1,2 and fin sum and simplify

$f_{rr}+f_{tt} = f_{xx}\cdot x^2 + f_x \cdot x + f_{yy}\cdot y^2 + f_y \cdot y + f_{xx} \cdot y^2 + f_x \cdot -x + f_{yy} \cdot x^2 + f_y \cdot -y$

you can continue from here just do what I say

best wishes

**Amer** · October 3rd 2009, 07:21 AM

I made a mistake

$f_r = f_x (x_r) + f_y(y_r)$

$f_{rr} = \frac{\partial f_x}{\partial r} (x_r ) + f_x (x_{rr}) + \frac{\partial f_y}{\partial r}(y_r) + f_y(y_{rr})$

$\frac{\partial f_x}{\partial r} = f_{xx} (x_r) + f_{xy} (y_r) \ne f_{xx} (x_r)$

same thing for all second partial for f with respect to r and theta

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