# Partial Differentiation Q

• Dec 18th 2007, 07:01 AM
UbikPkd
Partial Differentiation Q
Ok here goes....

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

$x=rcos\vartheta$

$y=r sin\vartheta$

Find:

$\frac{\partial z}{\partial x}_{y},
\frac{\partial z}{\partial \vartheta}_{x},
\frac{\partial z}{\partial r}_{y},
\frac{\partial z}{\partial r}_{ \vartheta},
$

___________________________

$\frac{\partial z}{\partial x}_{y} = 2x$
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....
___________________________

$
\frac{\partial z}{\partial \vartheta}_{x}=$

I've been at this one for hours, i think i can use the following, but i'm getting nowhere.

$dz= \frac{\partial z}{\partial x}_{y}dx + \frac{\partial z}{\partial y}_{x}dy$

$dx= \frac{\partial x}{\partial r}_{\vartheta}dr + \frac{\partial x}{\partial \vartheta}_{r}d\vartheta$

$dy= \frac{\partial y}{\partial r}_{\vartheta}dr + \frac{\partial y}{\partial \vartheta}_{r}d\vartheta$

so....

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

can i divide through by $\partial \vartheta_{x}$ and then work it all out to get
$
\frac{\partial z}{\partial \vartheta}_{x}$
?

• Dec 18th 2007, 07:08 AM
ThePerfectHacker
Quote:

Originally Posted by UbikPkd
Ok here goes....

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

$x=rcos\vartheta$

$y=r sin\vartheta$

I am not sure what you want to find.

Let $z=f(x,y)$ where $x=r\cos \theta \mbox{ and }y=r\sin \theta$ and $f(x,y) = x^2+y^2$.

Then,
$\frac{\partial z}{\partial r} = \frac{\partial f}{\partial x}\cdot \frac{\partial x}{\partial r} + \frac{\partial f}{\partial y}\cdot \frac{\partial y}{\partial r}$
Then,
$\frac{\partial z}{\partial r} = (2x)(\sin \theta) + (2y)(\cos \theta) = 2r\sin \theta \cos \theta + 2r\sin \theta \cos \theta = 2r\sin 2\theta$

The other one is similar.
• Dec 18th 2007, 07:16 AM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
I am not sure what you want to find.

Let $z=f(x,y)$ where $x=r\cos \theta \mbox{ and }y=r\sin \theta$ and $f(x,y) = x^2+y^2$.

Then,
$\frac{\partial z}{\partial r} = \frac{\partial f}{\partial x}\cdot \frac{\partial x}{\partial r} + \frac{\partial f}{\partial y}\cdot \frac{\partial y}{\partial r}$
Then,
$\frac{\partial z}{\partial r} = (2x)(\sin \theta) + (2y)(\cos \theta) = 2r\sin \theta \cos \theta + 2r\sin \theta \cos \theta = 2r\sin 2\theta$

The other one is similar.

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

RonL
• Dec 18th 2007, 07:34 AM
UbikPkd
thanks both of you, i understand captain blacks correction, theta is constant in both of those partial diffs.

quickly doing
$
\frac{\partial z}{\partial \vartheta}_{x}$

i get
$
\frac{\partial z}{\partial \vartheta}_{x}=0$

which makes sense to me. thanks for you help :)
• Dec 18th 2007, 08:47 AM
Opalg
Quote:

Originally Posted by UbikPkd
$
z=x^{2}+y^{2}$

$x=rcos\vartheta$

$y=r sin\vartheta$

Find:

$\frac{\partial z}{\partial x}_{y},
\frac{\partial z}{\partial \vartheta}_{x},
\frac{\partial z}{\partial r}_{y},
\frac{\partial z}{\partial r}_{ \vartheta},
$

This is a very strange question, and I'd like to be sure that I understand the notation before I try to answer it.

I'm guessing that those little subscripts indicate a quantity that is to be held constant. If so, then $\frac{\partial z}{\partial x}_{y}$, for example, makes good sense: z is a function of x and y, and when you take the partial derivative of z with respect to x you would naturally interpret this to mean that y is being held constant. But the next derivative, $\frac{\partial z}{\partial \vartheta}_{x}$, is another matter altogether. The variables r and ϑ go together, and when you form a partial derivative with respect to ϑ you would normally expect that r, not x, is the variable being held constant. To keep x constant when differentiating with respect to ϑ is unusual, not to say perverse. It can be done, but it needs to be done with care. Am I right in thinking that this is what this problem is really about, or am I misinterpreting it?
• Dec 18th 2007, 09:18 AM
UbikPkd
Quote:

Originally Posted by Opalg
This is a very strange question, and I'd like to be sure that I understand the notation before I try to answer it.

I'm guessing that those little subscripts indicate a quantity that is to be held constant. If so, then $\frac{\partial z}{\partial x}_{y}$, for example, makes good sense: z is a function of x and y, and when you take the partial derivative of z with respect to x you would naturally interpret this to mean that y is being held constant. But the next derivative, $\frac{\partial z}{\partial \vartheta}_{x}$, is another matter altogether. The variables r and ϑ go together, and when you form a partial derivative with respect to ϑ you would normally expect that r, not x, is the variable being held constant. To keep x constant when differentiating with respect to ϑ is unusual, not to say perverse. It can be done, but it needs to be done with care. Am I right in thinking that this is what this problem is really about, or am I misinterpreting it?

yer i have copied the question exactly and correctly so it's probably just unusual but not necessarily wrong.

For the last one I get:

$
\frac{\partial z}{\partial r}_{ \vartheta}=2r
$

yer and the subscripts are the constants

hmm, your guess is better than mine, though i think it's right, the question isn't likely to be wrong at least
• Dec 18th 2007, 11:12 AM
Opalg
Quote:

Originally Posted by UbikPkd
yer i have copied the question exactly and correctly so it's probably just unusual but not necessarily wrong.

For the last one I get:

$
\frac{\partial z}{\partial r}_{ \vartheta}=2r
$
That one's definitely right.

yer and the subscripts are the constants

hmm, your guess is better than mine, though i think it's right, the question isn't likely to be wrong at least

Okay, for $\frac{\partial z}{\partial \vartheta}_x$, I would interpret this as meaning that r and ϑ are variables, and x is constant. (But r and ϑ are not independent variables, because they are constrained by the fact that x has to be held constant.)

Differentiate $z=x^2+y^2$ with respect to ϑ, using the chain rule and treating x as a constant: $\frac{\partial z}{\partial \vartheta} = 2y\frac{\partial y}{\partial \vartheta}$. Now we have to find $\frac{\partial y}{\partial \vartheta}$.

Differentiate $y=r\sin\vartheta$ with respect to ϑ, using the product rule and the chain rule: $\frac{\partial y}{\partial \vartheta} = r\cos\vartheta + \frac{\partial r}{\partial \vartheta}\sin\vartheta$.

Finally, we have to find $\frac{\partial r}{\partial \vartheta}$. For this, differentiate the equation $x=r\cos\vartheta$ with respect to ϑ (remembering that x is constant): $0 = -r\sin\vartheta + \frac{\partial r}{\partial \vartheta}\cos\vartheta$.

Solve that last equation for $\frac{\partial r}{\partial \vartheta}$, substitute it into the equation for $\frac{\partial y}{\partial \vartheta}$, and then substitute that into the equation for $\frac{\partial z}{\partial \vartheta}$. When you put it all together, you should find that $\frac{\partial z}{\partial \vartheta} = 2y\Bigl(r\cos\vartheta + \frac{r\sin^2\vartheta}{\cos\vartheta}\Bigr)$. Since r and ϑ are the variables, it's probably best to express this in terms of them, namely (after a bit of simplification) $\frac{\partial z}{\partial \vartheta} = 2r^2\tan\vartheta$.

(Note: all the above partial derivatives should really have have that little subscript _x tacked on to them, just to emphasise that x is being held constant throughout.)

You could find the other derivative $\frac{\partial z}{\partial r}_y$ by exactly the same sort of method, this time keeping y fixed throughout. In practice, however, it would be much simpler to notice that $z=x^2+y^2=r^2$, so $\frac{\partial z}{\partial r}=2r$, regardless of what else is being kept constant.
• Dec 19th 2007, 03:19 AM
UbikPkd
Opalg, thanks, that's really helpful, i'm sorry i was being a bit of an idiot, that is indeed the way i believe they want me to do it, sorry for your trouble!