# Partial differentiation and changing variables

1. Nov 29, 2003

### fudge

Maths Question: I am having a lot of problems with this question, can any undergrad physicists or mathematicians help me?

(note: p before a differntial= partial derivative) .

Spherical polar coordinates (r, (thetha), (phi)) are defined in terms of Cartesian coorindates (x,y,z) by:

x=rsin(theta)cos(phi)
y=rsin(theta)sin(phi)
z=rcos(theta)

given that f is a function of r only, independent of theta and phi, show that

p(df)/p(dx) = (x/r).(df/dr)

p(d^2f)/p(dx^2) = (1/r).(df/dr) + (x^2/r).d[(1/r).(df/dr)]/dr

and hence deduce that:

p(d^2f)/p(dx^2) + p(d^2f)/p(dy^2) + p(d^2f)/p(dz^2) =
(1/r^2).d[r^2.(df/dr)]/dr

a) is straigthforward, any thoughts on how to appraoch b) ???

thanks

2. Nov 29, 2003

### jamesrc

Consider the fact that x2 + y2 + z2 = r2

What you've already shown for x also applies to y and z:

$\frac{\partial f}{\partial x} = \frac{x}{r} \cdot \frac{df}{dr}$

$\frac{\partial f}{\partial y} = \frac{y}{r} \cdot \frac{df}{dr}$

$\frac{\partial f}{\partial z} = \frac{z}{r} \cdot \frac{df}{dr}$

$\frac{\partial^2 f}{\partial x^2} = \frac{1}{r} \cdot \frac{df}{dr} + \frac{x^2}{r}\,\frac{d}{dr}\left(\frac{1}{r}\,\frac{df}{dr}\right)$

$\frac{\partial^2 f}{\partial y^2} = \frac{1}{r} \cdot \frac{df}{dr} + \frac{y^2}{r}\,\frac{d}{dr}\left(\frac{1}{r}\,\frac{df}{dr}\right)$

$\frac{\partial^2 f}{\partial z^2} = \frac{1}{r} \cdot \frac{df}{dr} + \frac{z^2}{r}\,\frac{d}{dr}\left(\frac{1}{r}\,\frac{df}{dr}\right)$

Add them together and use the first equation to get:

$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} = r\frac{d}{dr}\left(\frac{1}{r}\,\frac{df}{dr}\right) + \frac{3}{r}\frac{df}{dr}$

which can be simplified to:

$\frac{\partial^2 f}{\partial x^2} + \frac{\partial^2 f}{\partial y^2} + \frac{\partial^2 f}{\partial z^2} = \frac{d^2 f}{dr^2} + \frac{2}{r}\frac{df}{dr} = \frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{df}{dr}\right)$

3. Nov 29, 2003

### fudge

thanks

Thanks so much, you've made it very clear!

4. Dec 31, 2003

### sam2

Hi,

I would much appreciate some help in deriving the second expression in the original question below:

p(d^2f)/p(dx^2) = (1/r).(df/dr) + (x^2/r).d[(1/r).(df/dr)]/dr

Thanks for the help!

Regards,

Sam

5. Dec 31, 2003

### jamesrc

Use the product rule to find the derivative of this equation:

$\frac{\partial f}{\partial x} = \frac{x}{r} \cdot \frac{df}{dr}$

6. Dec 31, 2003

### sam2

Thanks for the reply,

I am trying this but am getting very confsed...

first of all:

I have a term: dr/dx .dx/dr. But I dont see why this is equal to one. If i accept this I get the result, but we know:

x = r.cos(phi).sin(theta)

and r^2 = x^2 + y^2 + z^2

so we can verify that the 2 partial derivatives are not eqal.

secondly, f is a function of r ony.... but df/dr (partial) is also a function of p and theta...

Therefore, you have a lot more terms than you would expect, because you have to take the derivative wrt phi and theta and then these wrt to x.

I dont know if I have completely misunderstood partial derivatives, but this is gettig messy!

Regards,

Sam

7. Dec 31, 2003

### jamesrc

You are given that f = f(r) (function of r only).

Using the chain rule:
$$\frac{\partial f}{\partial x} = \frac{df}{dr}\cdot\frac{\partial r}{\partial x}$$

(notice that it's df/dr (not partials) because f is a function of r only)

Now we need to find $$\frac{\partial r}{\partial x}$$.

To do this, you can use the fact that x2 + y2 + z2 = r2.

Take the derivative of both sides with respect to x to get:

$$2x = 2r\frac{\partial r}{\partial x}$$

rearranging:

$$\frac{\partial r}{\partial x} = \frac{x}{r}$$

plug that in to the results from the chain rule:

$$\frac{\partial f}{\partial x} = \frac{x}{r}\cdot\frac{df}{dr}$$

Okay so far?

Now on to the second part using the product rule:

$$\frac{\partial^2 f}{\partial x^2} = x\frac{\partial}{\partial x}\left(\frac{1}{r}\cdot\frac{df}{dr} \right) + \frac{1}{r}\cdot\frac{df}{dr}$$

from before: $$\partial x = \frac{r}{x}\partial r$$

substitute that in to get:

$$\frac{\partial^2 f}{\partial x^2} = x\frac{x}{r}\frac{\partial}{\partial r}\left(\frac{1}{r}\cdot\frac{df}{dr} \right) + \frac{1}{r}\cdot\frac{df}{dr}$$

which becomes:

$\frac{\partial^2 f}{\partial x^2} = \frac{1}{r} \cdot \frac{df}{dr} + \frac{x^2}{r}\,\frac{d}{dr}\left(\frac{1}{r}\,\frac{df}{dr}\right)$

Hope that helps.

8. Jan 1, 2004

### sam2

Much Appreciated.

Sam

9. Oct 3, 2004

### wai

the function f(x,y) is transformed to F(r,a) by change of variables x=rcos a and y=rsin a. Show that
1. p(d^2F)/p(dr^2)
2. p(d^2F)/p(da^2)+r[p(dF/dr)]