# Homogeneous of degree n

1. Mar 27, 2005

### Callisto

Hi

" A function f is called homogeneous of degree n if it satisfies the equation

f(tx,ty,tz)=t^n*f(x,y,z) for all t, where n is a positive integer and f has continuous second order partial derivatives".

I dont have equation editor so let curly d=D

I need help to show that

x(Df/Dx)+y(Df/Dy)+z(Df/Dz) = nf(x,y,z)

The hint that is given is to use the chain rule to differentiate f(tx,ty,tz) with respect to t.

I am at a total loss, can somebody offer help as to how i show this.
Thanks

Callisto

2. Mar 27, 2005

### dextercioby

Don't u know how to use chain rule for partial derivatives...?

Compute

$$\frac{\partial f}{\partial x}$$

,where

$$f=f(r,r^{2})$$,where r=r(x,y,z).

Daniel.

3. Mar 27, 2005

### Callisto

Sorry DEXTECIOBY

your reply is of no use to me

"In the beginning was the symmetry" ???

Werner Heisenberg.
Why quote what you cant prove?

Callisto

4. Mar 27, 2005

### dextercioby

That's my signature :rofl: I asked you a very good question...Do you have any idea what it means to use the chain rule for partial derivatives...?

Daniel.

5. Mar 27, 2005

### Callisto

The short answer is no

Chain rule for partial derivatives is knew for me, hence my seeking help for this problem. Your reply was vague and of no assitance. Thanks anyway.

In the beginning was symmetry?

Callisto

6. Mar 27, 2005

### dextercioby

That's what Werner Heisenberg thought.Advice:learn the theory before trying to solve problems...

Daniel.

7. Apr 7, 2005

### estalniath

You'll have to let A=f(tx,ty,tz)=t^kf(x,y,z)
Then find dA/dt= df/d(tx)*dx/dt+df/d(ty)*dy/dt+df/d(tz)*dz/dt
=df/d(tx)*x+df/d(ty)*y+df/d(tz)*z
For the right hand side, we'll get, k*t^(k-1)f(x,y,z)

Then put t=1, and we'll get the equation xdf/dx+ydf/dx+zdf/dz=Kf(x,y,z)

Though I'm not so sure what the rational behind using the substitution t=1 is in solving this question =/ I guess its only for simplicity since the equation works for all t and t=1 is a good way to simplify both sides of the equation ^^;;

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