Partial differentiation and changing variables

In summary, we are looking at how to transform a function from Cartesian coordinates (x,y) to spherical polar coordinates (r,a) and want to find the partial derivatives of F with respect to r and a.
  • #1
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
 
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  • #2
Consider the fact that x2 + y2 + z2 = r2

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

[itex] \frac{\partial f}{\partial x} = \frac{x}{r} \cdot \frac{df}{dr} [/itex]

[itex] \frac{\partial f}{\partial y} = \frac{y}{r} \cdot \frac{df}{dr} [/itex]

[itex] \frac{\partial f}{\partial z} = \frac{z}{r} \cdot \frac{df}{dr} [/itex]

[itex] \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) [/itex]

[itex] \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) [/itex]

[itex] \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) [/itex]

Add them together and use the first equation to get:

[itex] \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} [/itex]

which can be simplified to:

[itex] \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) [/itex]
 
  • #3
thanks

Thanks so much, you've made it very clear!
 
  • #4
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
Use the product rule to find the derivative of this equation:

[itex] \frac{\partial f}{\partial x} = \frac{x}{r} \cdot \frac{df}{dr} [/itex]
 
  • #6
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 don't 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 don't know if I have completely misunderstood partial derivatives, but this is gettig messy!

Regards,

Sam
 
  • #7
You are given that f = f(r) (function of r only).

Using the chain rule:
[tex] \frac{\partial f}{\partial x} = \frac{df}{dr}\cdot\frac{\partial r}{\partial x} [/tex]

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

Now we need to find [tex] \frac{\partial r}{\partial x} [/tex].

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:

[tex] 2x = 2r\frac{\partial r}{\partial x} [/tex]

rearranging:

[tex] \frac{\partial r}{\partial x} = \frac{x}{r} [/tex]

plug that into the results from the chain rule:

[tex] \frac{\partial f}{\partial x} = \frac{x}{r}\cdot\frac{df}{dr} [/tex]

Okay so far?

Now on to the second part using the product rule:

[tex] \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} [/tex]

from before: [tex] \partial x = \frac{r}{x}\partial r [/tex]

substitute that into get:

[tex] \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} [/tex]

which becomes:

[itex] \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) [/itex]

Hope that helps.
 
  • #8
Much Appreciated.

Sam
 
  • #9
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)]
 

1. What is partial differentiation?

Partial differentiation is a mathematical concept used to describe how a function changes when only one of its independent variables is changed, while holding the other variables constant.

2. Why is partial differentiation useful?

Partial differentiation allows us to analyze the rate of change of a function with respect to a specific variable, which is useful in many fields such as physics, economics, and engineering.

3. How do you find partial derivatives?

To find a partial derivative, you treat all other variables as constants and use the standard rules of differentiation to differentiate with respect to the variable of interest. For example, if the function is f(x,y) and you want to find the partial derivative with respect to x, you would differentiate with respect to x while treating y as a constant.

4. What is the chain rule in partial differentiation?

The chain rule in partial differentiation is used when the function has multiple variables that are functions of other variables. It allows us to find the derivative of the composite function by taking the derivatives of each individual function and multiplying them together.

5. How does changing variables affect partial derivatives?

When changing variables, the partial derivatives can be affected. This is because the new variables may not be independent, and therefore, the partial derivatives may change. However, the overall function remains the same regardless of the choice of variables.

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