Understanding Divergence: Solving the Mystery of Vector Functions | Jim L.

[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.

 

[Astronuc]

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.

 

[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.

 

[Astronuc]

The way the problem is written

F (of x,y,z)= i (f of x) + j (f of y)+k f (-2z).

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}

 

[HallsofIvy]

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

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.

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.

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.

 

[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.