How do I solve for partial differentiation in multivariable calculus?

[UbikPkd]

Ok here goes...

z=x^{2}+y^{2}

x=rcos\vartheta

y=r sin\vartheta

Find:

\[ \frac{\partial z}{\partial x}_{y}, <br /> \[ \frac{\partial z}{\partial <br /> <br /> \vartheta}_{x}, <br /> \[ \frac{\partial z}{\partial r}_{y}, <br /> \[ \frac{\partial z}{\partial r}_{ <br /> <br /> \vartheta}, <br />

___________________________

\[ \frac{\partial z}{\partial x}_{y} <br /> <br /> = 2x

this seems right to me (though I'm not

sure if I'm supposed to use a chain rule),

it's the next ones I'm not sure about...
___________________________

<br /> \[ \frac{\partial z}{\partial <br /> <br /> \vartheta}_{x}=

I've been at this one for hours, i think i

can use the following, but I'm getting

nowhere.

dz= \[ \frac{\partial z}{\partial <br /> <br /> x}_{y}dx + \[ \frac{\partial z}{\partial <br /> <br /> y}_{x}dy


dx= \[ \frac{\partial x}{\partial <br /> <br /> r}_{\vartheta}dr + \[ \frac{\partial <br /> <br /> x}{\partial \vartheta}_{r}d\vartheta


dy= \[ \frac{\partial y}{\partial <br /> <br /> r}_{\vartheta}dr + \[ \frac{\partial <br /> <br /> y}{\partial \vartheta}_{r}d\vartheta

so...

dz= \[ \frac{\partial z}{\partial <br /> <br /> x}_{y} \left[\[ \frac{\partial x}{\partial <br /> <br /> r}_{\vartheta}dr + \[ \frac{\partial <br /> <br /> x}{\partial <br /> <br /> \vartheta}_{r}d\vartheta\right]+ \[ <br /> <br /> \frac{\partial z}{\partial y}_{x}\left[\[ <br /> <br /> \frac{\partial y}{\partial <br /> <br /> r}_{\vartheta}dr + \[ \frac{\partial <br /> <br /> y}{\partial <br /> <br /> \vartheta}_{r}d\vartheta\right]

can i divide through by \partial <br /> <br /> \vartheta_{x} and then work it all

out to get
<br /> \[ \frac{\partial z}{\partial <br /> <br /> \vartheta}_{x} ?

please help, i don't have a clue what I'm

doing!

 

[Dick]

The direct way to do this is to simply write z as a function only of x and theta. This is pretty easy, since y/x=tan(theta). Try it that way before you go back to struggling with the chain rule.

 

[UbikPkd]

okay thanks for the tip, i worked it out using chain rules in the end before i got to read it, the answers i got were:

\[ \frac{\partial z}{\partial x}_{y}=2x

\[ \frac{\partial z}{\partial \vartheta}_{x}=0

\[ \frac{\partial z}{\partial r}_{y}=2r

\[ \frac{\partial z}{\partial r}_{\vartheta}=2r<br /> <br />

I think these are right, thanks for your help!

 

[Dick]

Well, e.g. for the second one, y/x=tan(theta), so z=x^2+x^2*tan^2(theta). It doesn't look to me like the derivative wrt theta at constant x is 0.

 

[UbikPkd]

Well, e.g. for the second one, y/x=tan(theta), so z=x^2+x^2*tan^2(theta). It doesn't look to me like the derivative wrt theta at constant x is 0.

sorry! arghh I've made such a mess of this, yer the last two are right, as z=r^2 so it doesn't matter what's being held constant


\[ \frac{\partial z}{\partial \vartheta}_{x}=2r^{2}tan\vartheta

not 0, sorry, here's my full working:

\[ \frac{\partial z}{\partial \vartheta}_{x}=2y\[ \frac{\partial y}{\partial \vartheta}_{x}

_________

y=rsin\vartheta

\[ \frac{\partial y}{\partial \vartheta}_{x}=rcos\vartheta+\[ \frac{\partial r}{\partial \vartheta}_{x}sin\vartheta

________

x=rcos\vartheta

0=-rsin\vartheta+\[ \frac{\partial r}{\partial \vartheta}_{x}cos\vartheta

\[ \frac{\partial r}{\partial \vartheta}_{x}=r\[ \frac{sinv}{cos\vartheta}=rtan\vartheta

________

\[ \frac{\partial z}{\partial \vartheta}_{x}=2y(rcos\vartheta+rsin\vartheta tan\vartheta)

\[ \frac{\partial z}{\partial \vartheta}_{x}=2r^{2}\frac{sin\vartheta}{cos\vartheta}=2r^{2}tan\vartheta

I did it that way, as I thought differentiating via the tan^{2}\vartheta way would probably go wrong for me somehow! I think the above is OK, seems like that's the way they want me to do it as well. Thanks for your help!

 

[Dick]

Doing it the other way gives you 2x^2*sin(theta)/cos^3(theta), which if you put x=r*cos(theta) gives you the same thing you got. So I think your answer is right.

 

[UbikPkd]

Doing it the other way gives you 2x^2*sin(theta)/cos^3(theta), which if you put x=r*cos(theta) gives you the same thing you got. So I think your answer is right.

cool thanks!