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?
$\displaystyle
f(x,y) , x = e^rcos(\theta), y=e^rsin(\theta)$
$\displaystyle f^2_r+f^2_\theta=
$
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?
$\displaystyle
f(x,y) , x = e^rcos(\theta), y=e^rsin(\theta)$
$\displaystyle f^2_r+f^2_\theta=
$
use \partial ok nvm
$\displaystyle \frac{\partial ^2 f}{\partial r^2} = \frac{\partial }{\partial r}(\frac{\partial f}{\partial r}) $
$\displaystyle \frac{\partial f}{\partial r} = \frac{\partial f}{\partial y}\cdot \frac{\partial y}{\partial r}+\frac{\partial f}{\partial x}\cdot \frac{\partial x}{\partial r}$
but you did not give f(x,y) = ?? what is equal
I do not know but I think the question should give you f(x,y) equal what for example
$\displaystyle f(x,y)=x+y $ or $\displaystyle f(x,y) = xy $ or $\displaystyle f(x,y) = \frac{x}{2} + \frac{y}{x}$ 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
Well that's what is causing me problems.
Here is the question in all it's glory:
If $\displaystyle f=f(x,y), x=e^rcos(\theta)$ and $\displaystyle y=e^rsin(\theta)$, prove the following identidy:
$\displaystyle f^2_r+f^2_\theta=e^{2r}(f^2_x+f^2_y)$
Wouldn't this just work out to 0 = 0 ?
ok I get it
$\displaystyle f_r ^2 + f_\theta ^2 = e^{2r}(f_x ^2 + f_y ^2 )$ ok take the left side
I will write d instead of partial to make is easier for me ok
$\displaystyle f_r = \frac{df}{dx} \cdot \frac{dx}{dr} + \frac{df}{dy} \cdot \frac{dy}{dr} $
$\displaystyle f_r = f_x (x) + f_y (y) $ since dx/dr =x and dy/dr =y
$\displaystyle f_r ^2 = f_x ^2 x^2 + f_x f_y yx + f_y ^2 y^2 $
$\displaystyle f_r ^2 = f_x ^2 (e^{2r} \cos ^2 \theta ) + f_x f_y (e^{2r}\sin \theta \cos \theta ) + f_y ^2 (e^{2r} \sin ^2 \theta )$
this is the first thing now
$\displaystyle \frac{df}{d\theta} = f_x dx/d\theta + f_y dy/d\theta $
$\displaystyle x_\theta = -y , y_\theta = x $
$\displaystyle \frac{df}{d\theta} = f_x (-y) + f_y (x) $
simplify it and find the sum you will get the solution
$\displaystyle \sin ^2 x + \cos ^2 x =1 $ use this
$\displaystyle x= e^r \cos \theta $
$\displaystyle y=e^r \sin \theta $
$\displaystyle x_\theta = \frac{\partial x}{\partial \theta } = e^r (-\sin \theta) = -e^r\sin \theta = -y $
$\displaystyle y_\theta = \frac{\partial y}{\partial \theta } = e^r \cos \theta = x $
$\displaystyle f_r = f_x \cdot x_r + f_y \cdot y_r $
$\displaystyle 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
$\displaystyle x_{rr} = x_r=x , y_{rr} = y_r=y $
$\displaystyle 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
$\displaystyle f_{t} = f_x \cdot x_t + f_y \cdot y_t $
$\displaystyle 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} $
$\displaystyle x_t = -y , y_t = x $ and $\displaystyle x_{tt} = -x , y_{tt} = -y $
$\displaystyle 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
$\displaystyle f_{rr}+f_{tt} = f_{xx}\cdot x^2 + f_x \cdot x + f_{yy}\cdot y^2 + f_y \cdot y + f_{xx} \cdot y^2 + f_x \cdot -x + f_{yy} \cdot x^2 + f_y \cdot -y$
you can continue from here just do what I say best wishes
I made a mistake
$\displaystyle f_r = f_x (x_r) + f_y(y_r) $
$\displaystyle 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}) $
$\displaystyle \frac{\partial f_x}{\partial r} = f_{xx} (x_r) + f_{xy} (y_r) \ne f_{xx} (x_r)$
same thing for all second partial for f with respect to r and theta