# Question on Divergence

1. Sep 23, 2007

### Jim L

Been working my way thru H.M. Schey-been out of college for 50 yrs.

This problem has me stumped. F (of x,y,z)= i (f of x) + j (f of y)+k f (-2z).

F is a vector function, and i,j,k are unit vectors for x,y,z axis.

The problem is to find Div F., and then show it is 0 for the point c,c, -c/2.

Cannot for the life of me see how one can get a unique solution that would allow one to insert the point c,c, -c/2, which are for x,y,z.

What am I missing, please and thanks. Jim L.

Last edited: Sep 23, 2007
2. Sep 23, 2007

### Staff: Mentor

Assuming it's the same f is each case,

one ends up with f'(c) + f'(c) - 2f'(c) = 0, if in last term df(-2z)/dz = -2 f'(-2z), and f'(-c) would have to equal f'(c).

Does the problem state c, c, c/2 or c, c, -c/2 or is f' an even function.

Last edited: Sep 23, 2007
3. Sep 23, 2007

### Jim L

Sorry, the value IS -c/2. The way I see it:

f of x can be anything, such as x^2 +3y.

f of y can be anything , such as 4y-2x

f of(-2z) can be anything , such as 4z+2y

Also, cannot see how df(-2z)dz= -2 f'(-2z), assuming f' is the partial of -2z wrt z. Or is df mean partial der? Either way, does not seem correct.

The values c,c,-c/2 are to be inserted in the resulting expression for Div F, as I see it.

Obviously I am still missing something. Thanks, Jim.

Last edited: Sep 23, 2007
4. Sep 24, 2007

### Staff: Mentor

The way the problem is written

I take to mean F (x,y,z) = i f(x) + j f(y) + k f(-2z).

True that $$\nabla\,\cdot{F}\,=\,\frac{\partial{f(x)}}{\partial{x}}+\frac{\partial{f(y)}}{\partial{y}}+\frac{\partial{f(-2z)}}{\partial{z}}$$,

but since f(x), f(y), f(-2z) are the same function in one variable the partial derivative is just the standard derivative in one variable, i.e.

$$\frac{\partial{f(x)}}{\partial{x}} = \frac{d f(x)}{dx}$$

Last edited: Sep 24, 2007
5. Sep 24, 2007

### HallsofIvy

Staff Emeritus
No! You do NOT use the same symbol, f, to mean different things in the same formula. It also does not make sense to write "f(x)" and then give a formula that involves both x and y.

The original problem was :F(x,y,z)= i (f(x)) + j (f(y))+ kf (-2z) where f is some differentiable function of a single variable. The divergence would be, using the chain rule, div F= f'(x) + f'(y)-2f'(-2z). evaluating that at (c, c, -c/2) would give div F= f'(c)i+ f'(c)- 2f'(-2(-c/2)= 2f'(c)- 2f'(c)= 0.

6. Sep 24, 2007

### Jim L

I've got it. I was equating the authors F sub x, which can contain x,y,z and is the multiplier of the unit vector i, with f of x.
Still cannot see how the partial der. of -2Z wrt z can be anything but -2. How the chain rule applies and generates a multiplier of 2evades me. But it has been a long time. Sorry and many thanks. jim.

Last edited: Sep 24, 2007