# Partial derivatives, word problem help

• Dec 13th 2010, 01:34 PM
Tweety
Partial derivatives, word problem help
Heat is being conducted radially through a cylindrical pipe. The temperature at a radius r is T(r). In Cartesian co-ordinates, $\displaystyle r = \sqrt{(x^{2}+ y^{2}})$

show that $\displaystyle \displaystyle \frac{\partial T}{\partial x} = \frac{x}{r} \frac{dT}{dr}$
• Dec 13th 2010, 01:45 PM
$\displaystyle T(r)=[\math]T(\sqrt{x^2+y^2))$

You have to differentiate r with restpect to x, this is just $\displaystyle \frac{\partial r}{\partial x}=\frac{2x}{2\sqrt{x^2+y^2}}=\frac{x}{r}$.

So $\displaystyle \frac{\partial T}{\partial x}=\frac{\partial r}{\partial x}\frac{dT}{dr}=\frac{x}{r}\frac{dT}{dr}$
• Dec 13th 2010, 01:53 PM
Tweety
Quote:

$\displaystyle T(r)=[\math]T(\sqrt{x^2+y^2))$

You have to differentiate r with restpect to x, this is just $\displaystyle \frac{\partial r}{\partial x}=\frac{2x}{2\sqrt{x^2+y^2}}=\frac{x}{r}$.

So $\displaystyle \frac{\partial T}{\partial x}=\frac{\partial r}{\partial x}\frac{dT}{dr}=\frac{x}{r}\frac{dT}{dr}$

Thank you so much, but I dont understand what the first expression is and how you got it?

your saying $\displaystyle T(r) = T(\sqrt{ x^{2} + y^{2}})$ ?

How comes?

Also for $\displaystyle \frac{\partial r}{\partial x}$ , you took the partial derivative of r with respect to x, from the equation $\displaystyle r = \sqrt{(x^{2} + y^{2})$ ?

so you get this expression ; $\displaystyle \frac{2x}{2\sqrt{x^2+y^2}}$
but how does that equal the expression; $\displaystyle \frac{x}{r}$ ?

thank you.
• Dec 13th 2010, 01:55 PM
Tweety
edit; I actually understand how you got the first expression, thanks, could please explain the rest?

Thank you
• Dec 13th 2010, 02:08 PM
Quote:

Thank you so much, but I dont understand what the first expression is and how you got it?

your saying $\displaystyle T(r) = T(\sqrt{ x^{2} + y^{2}})$ ?

How comes?
All I'm doing is substituting the equation $\displaystyle r=\sqrt{x^2+y^2}$ into the expression$\displaystyle T(r)$ so you can see that T is a composite function of x and y, and therefore the partial of T with respect to x can be calculated using a chain rule.

Quote:

Also for $\displaystyle \frac{\partial r}{\partial x}$ , you took the partial derivative of r with respect to x, from the equation $\displaystyle r = \sqrt{(x^{2} + y^{2})$ ?

so you get this expression ; $\displaystyle \frac{2x}{2\sqrt{x^2+y^2}}$
but how does that equal the expression; $\displaystyle \frac{x}{r}$ ?

thank you.
$\displaystyle r=\sqrt{x^2+y^2}$

The derivative of a function like $\displaystyle \sqrt{u}$ is $\displaystyle \frac{u'}{2\sqrt{u}}$.

You have to apply the chain rule to find the partial of r with respect to x.

$\displaystyle \frac{\partial r}{\partial x}=\frac{1}{2\sqrt{x^2+y^2}}\frac{\partial}{\parti al x}(x^2+y^2)=\frac{2x}{2\sqrt{x^2+y^2}}$

Remember that $\displaystyle r=\sqrt{x^2+y^2}$ so just substitute that into the partial derivative for r with respect to x, and you get $\displaystyle \frac{x}{r}$
• Dec 13th 2010, 02:33 PM
Tweety
Thank you , I understand your method now.
• Dec 13th 2010, 02:53 PM
Tweety
sorry just one question, how does $\displaystyle \frac{dT}{dr}$ come into the equation ? It seems like its just been put there?

Thank you