# Apostol's Calculus Vol. II Question on Gradients

## Homework Statement

If $$\nabla f(x,y,z)$$ is always parallel to $$x \hat i + y \hat j + z \hat k$$, show that f must assume equal values at the points (0,0,a) and (0,0,-a).

## The Attempt at a Solution

I tried a number of things - inspecting the values arrived at when computing the cross product of the gradient and the position vector, writing $$r(t)=(0,0,t)$$ and then integrating $$\nabla f(r(t))\cdot r'(t)$$ from -a to a, nothing is getting me anywhere. I also found this old thread: https://www.physicsforums.com/showthread.php?t=165790 but I don't see what Mathgician is getting at. In particular, f(0,0,a)-f(0,0,-a) is not necessarily equal to -2ak as the poster mentions toward the bottom.

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Dick
Homework Helper
Look at the path integral of grad(f) along a circle connecting (0,0,a) and (0,0,-a).

LCKurtz
Homework Helper
Gold Member
If ##f_x = \lambda x,\ f_y=\lambda y,\ f_z=\lambda z## can't you derive what ##f(x,y,z)## must be by taking anti-partial derivatives?

Look at the path integral of grad(f) along a circle connecting (0,0,a) and (0,0,-a).
Unfortunately we have not covered path integrals yet. Thanks though!

If ##f_x = \lambda x,\ f_y=\lambda y,\ f_z=\lambda z## can't you derive what ##f(x,y,z)## must be by taking anti-partial derivatives?
I have no reason to think that ##\nabla f(x,y,z)=\lambda(x,y,z)## - although I could solve this if I could assume that. The conditions, as far as I can tell, just say that for some scalar field ##h## we have that ##\nabla f(x,y,z) = [h(x,y,z)](x,y,z)##. In particular, I tried this approach: Let ##r(t)=(0,0,t)##. Then let ##g(t)=f[r(t)]##, so ##g'(t)=\nabla f [r(t)] \cdot r'(t)##. Now consider $$\int_{-a}^a \nabla f[r(t)]\cdot r'(t) dt$$. Now since the gradient is parallel to the position we have ##\nabla f (0,0,t) = (0,0,h(t) t)##, and since ##r'(t)=(0,0,1)## our integral is just $$\int_{-a}^a h(t) t dt$$. If ##h(t)## is even, the result follows. It may still be possible if ##h(t)## is not even, however note that ##h(t)## cannot be odd (which is an interesting result as well... I think anyway).

I doubt, however, that this was the intended direction for the proof to take... and since I can't prove that there's any other reason to think that ##h(t)## should be even, it's not conclusive. On the other hand, if someone comes up with a way to have ##\nabla f(x,y,z)= [h(x,y,z)](x,y,z)## where the function ##h## is odd with respect to ##z##, that would indicate that I have over-generalized the intended problem, and it probably is only intended to be ##\nabla f(x,y,z) = \lambda (x,y,z)##.

Last edited:
LCKurtz
Homework Helper
Gold Member
I have no reason to think that ##\nabla f(x,y,z)=\lambda(x,y,z)## - although I could solve this if I could assume that.
Oh, OK. I misunderstood the problem. I may have time to look at it some more tomorrow.

Oh, OK. I misunderstood the problem. I may have time to look at it some more tomorrow.
Thanks!

Dick
I have no reason to think that ##\nabla f(x,y,z)=\lambda(x,y,z)## - although I could solve this if I could assume that. The conditions, as far as I can tell, just say that for some scalar field ##h## we have that ##\nabla f(x,y,z) = [h(x,y,z)](x,y,z)##. In particular, I tried this approach: Let ##r(t)=(0,0,t)##. Then let ##g(t)=f[r(t)]##, so ##g'(t)=\nabla f [r(t)] \cdot r'(t)##. Now consider $$\int_{-a}^a \nabla f[r(t)]\cdot r'(t) dt$$. Now since the gradient is parallel to the position we have ##\nabla f (0,0,t) = (0,0,h(t) t)##, and since ##r'(t)=(0,0,1)## our integral is just $$\int_{-a}^a h(t) t dt$$. If ##h(t)## is even, the result follows. It may still be possible if ##h(t)## is not even, however note that ##h(t)## cannot be odd (which is an interesting result as well... I think anyway).