Apostol's Calculus Vol. II Question on Gradients

[process91]

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.

 

[Dick]

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

 

[LCKurtz]

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?

 

[process91]

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!

 

[process91]

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 $\nabla f(x,y,z) = \lambda (x,y,z)$.

 

[LCKurtz]

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.

 

[process91]

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).

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 $\nabla f(x,y,z) = \lambda (x,y,z)$.

That IS a path integral. Now try changing the path to say r(t)=(0,a*sin(t),a*cos(t)) for t in [0,pi].

 

[process91]

That IS a path integral. Now try changing the path to say r(t)=(0,a*sin(t),a*cos(t)) for t in [0,pi].

Awesome - the book doesn't cover path integrals explicitly for two more chapters, good to know they aren't much different than what I already have covered. Thanks!