What Is the Potential Function U for a Given Gradient?

[fresh_42]

My mistake. f(y,z) = y + g(z).
Although, what about the z?
Yesterday, I had f(y,z) = $y^4 z^2 + y^6 / 3 + y^2 z^4 + g(z)$.

That's been a different $f$. In both cases $f$ is just a term within a calculation. 2 calculations, different terms within.
Imagine a polynomial, like $a_{390567} z^{28768} y^{9808} + ... + a_{57657} z^2 + a_{764} y + 5$. The numbers are just to prevent you from calculating with it! Now, if you differentiate this w.r.t. $y$ and receive $0$ and then w.r.t $z$ and receive $0$ again? How does it look like?

 

[Amrator]

That's been a different $f$. In both cases $f$ is just a term within a calculation. 2 calculations, different terms within.
Imagine a polynomial, like $a_{390567} z^{28768} y^{9808} + ... + a_{57657} z^2 + a_{764} y + 5$. The numbers are just to prevent you from calculating with it! Now, if you differentiate this w.r.t. $y$ and receive $0$ and then w.r.t $z$ and receive $0$ again? How does it look like?

Oh, then it's C.

 

[fresh_42]

ευρηκα!

 

[Amrator]

Therefore $U = (x^2 + y^2 + z^2)^3 / 3 + C = r^3 / 3 + C$

Question: How would I write "integral of f(y,z) w.r.t. y and z = C" symbolically? What would the notation look like?

 

[fresh_42]

Therefore $U = (x^2 + y^2 + z^2)^3 / 3 + C = r^3 / 3 + C$

Question: How would I write "integral of f(y,z) w.r.t. y and z = C" symbolically? What would the notation look like?

Firstly, if I remember the definition of $r$ right, then $U(x,y,z) = U(r)= \frac{1}{3}r^6 +C$.
Secondly, $f$ itself is constant, not its integral. So I would write $f(y,z) = C$. It is the direct consequence of the equations $\frac{\partial }{\partial y} f(y,z) = 0$ and $\frac{\partial }{\partial z} f(y,z) = 0$ holding both.

Thirdly, if you look up what Chestermiller said $\frac{dU}{dr} = 2 r^5$ you could have gotten the result immediately.
Nevertheless, it's been a good exercise. And to be honest, I've done the long way first as well.

 

[Amrator]

Firstly, if I remember the definition of $r$ right, then $U(x,y,z) = U(r)= \frac{1}{3}r^6 +C$.
Secondly, $f$ itself is constant, not its integral. So I would write $f(y,z) = C$. It is the direct consequence of the equations $\frac{\partial }{\partial y} f(y,z) = 0$ and $\frac{\partial }{\partial z} f(y,z) = 0$ holding both.

Thirdly, if you look up what Chestermiller said $\frac{dU}{dr} = 2 r^5$ you could have gotten the result immediately.
Nevertheless, it's been a good exercise. And to be honest, I've done the long way first as well.

Sorry, I meant to say the integral of PD of f.

By what Chestermiller said, you mean spherical coordinates? I'll go learn gradients using spherical coordinates right now then.

 

[fresh_42]

Sorry, I meant to say the integral of PD of f.

I don't know whether there is an elegant short notation. Basically it is the same as we did before:

$\frac{\partial}{\partial y}f(y,z) = 0 ⇒ f(y,z) = f(z) + C' ⇒ 0 = \frac{\partial}{\partial z}f(y,z) = \frac{\partial}{\partial z} (f(z) + C') = \frac{\partial}{\partial z} f(z) ⇒ f(y,z) = f(z) = f = constant = C$

 

[Amrator]

Alright, thanks for the help everyone.

 

[fresh_42]

By what Chestermiller said, you mean spherical coordinates? I'll go learn gradients using spherical coordinates right now then.

In this case you should definitely look up ehild's first post (#32). The one with the figure. (S)he explained it very well. Mainly how $r^4 \vec r$ becomes $r^5$.

 

[SammyS]

${\partial U} / {\partial y} = 2 y (x^2 + y^2 + z^2)^2 + {\partial f(y,z)} / {\partial y}$
Setting that equal to the $\hat j$ component will give you ${\partial f(y,z)} / {\partial y} = 0$.

Yes. That's correct !

This means ƒ(y,z) is not actually a function of y.

$\partial f(y,z) /\partial y = 0$
$\partial f(y,z) /\partial z = 0$
Yeah, that doesn't seem right to me.

Also, it appears that you found a similar result regarding ƒ(y,z) not depending on z.

What does that leave you with?

ƒ(y,z) is just an ordinary constant, call it C.