What Is the Potential Function U for a Given Gradient?

[Amrator]

Homework Statement

$\nabla U = 2 r^4 \vec r$ Find U.

Homework Equations

$\vec r = x \hat i + y \hat j + z \hat j$
$r = \sqrt (x^2 + y^2 + z^2)$

The Attempt at a Solution

$\nabla U = 2 (x^2 + y^2 + z^2)^2 (x \hat i + y \hat j + z \hat j)$

I multiplied everything out,

$\nabla U = (2 x^4 + 4 x^2 y^2 + 4 x^2 z^2 + 4 y^2 z^2 + 2 y^4 + 2 z^4)x\hat i + (2 x^4 + 4 x^2 y^2 + 4 x^2 z^2 + 4 y^2 z^2 + 2 y^4 + 2 z^4)y\hat j + (2 x^4 + 4 x^2 y^2 + 4 x^2 z^2 + 4 y^2 z^2 + 2 y^4 + 2 z^4)z\hat k$

Since $\nabla U = \frac{\partial U}{\partial x} \hat i + \frac{\partial U}{\partial y} \hat j + \frac{\partial U}{\partial z} \hat k$, we really only need one of the partial derivatives.

$\frac{\partial U}{\partial x} = (2 x^5 + 4 x^3 y^2 + 4 x^3 z^2 + 4 x y^2 z^2 + 2 x y^4 + 2 x z^4)$

Then I took the integral of only the x terms,

$\frac{\partial U}{\partial x} = \frac{\partial}{\partial x} (\frac {x^6}{3} + x^4 y^2 + x^4 z^2 + 2 x^2 y^2 z^2 + x^2 y^4 + x^2 z^4)$

Therefore,

$U = \frac {x^6}{3} + x^4 y^2 + x^4 z^2 + 2 x^2 y^2 z^2 + x^2 y^4 + x^2 z^4$

Am I correct so far? If so, where do I go from here?

 

[SteamKing]

Homework Statement

$\nabla U = 2 r^2 \vec r$ Find U.

Homework Equations

$\vec r = x \hat i + y \hat j + z \hat j$
$r = \sqrt (x^2 + y^2 + z^2)$

The Attempt at a Solution

$\nabla U = 2 (x^2 + y^2 + z^2)^2 (x \hat i + y \hat j + z \hat j)$
I multiplied everything out.
$\frac{\partial U}{\partial x} = (2 x^5 + 4 x^3 y^2 + 4 x^3 z^2 + 4 x y^2 z^2 + 2 x y^4 + 2 x z^4)\hat i$
Then I took the integral of only the x terms,
$\frac{\partial U}{\partial x} = \frac{\partial}{\partial x} (\frac {x^6}{3} + x^4 y^2 + x^4 z^2 + 2 x^2 y^2 z^2 + x^2 y^4 + x^2 z^4)\hat i$
Therefore,
$U = \frac {x^6}{3} + x^4 y^2 + x^4 z^2 + 2 x^2 y^2 z^2 + x^2 y^4 + x^2 z^4$
Am I correct so far? If so, where do I go from here?

If $r = \sqrt{x^2 + y^2 + z^2}$, then what is r²?

 

[Amrator]

If $r = \sqrt{x^2 + y^2 + z^2}$, then what is r²?

Sorry about that, I screwed up the question. I'll edit right now. It should be r^4.

 

[SammyS]

Homework Statement

$\nabla U = 2 r^4 \vec r$ Find U.
...

The Attempt at a Solution

[/B]
...
Then I took the integral of only the x terms,

$\frac{\partial U}{\partial x} = \frac{\partial}{\partial x} (\frac {x^6}{3} + x^4 y^2 + x^4 z^2 + 2 x^2 y^2 z^2 + x^2 y^4 + x^2 z^4)$

Therefore,

$\displaystyle U = \frac {x^6}{3} + x^4 y^2 + x^4 z^2 + 2 x^2 y^2 z^2 + x^2 y^4 + x^2 z^4$

Am I correct so far? If so, where do I go from here?

You need a "constant" of integration. Of, course it's only constant w.r.t. $x$ . It's potentially a function of $y$ and $z$ .

 

[Amrator]

You need a "constant" of integration. Of, course it's only constant w.r.t. $x$ . It's potentially a function of $y$ and $z$ .

So $\frac{\partial U}{\partial x} = \frac{\partial}{\partial x} (\frac {x^6}{3} + x^4 y^2 + x^4 z^2 + 2 x^2 y^2 z^2 + x^2 y^4 + x^2 z^4 + constant)$?

 

[SammyS]

So $\frac{\partial U}{\partial x} = \frac{\partial}{\partial x} (\frac {x^6}{3} + x^4 y^2 + x^4 z^2 + 2 x^2 y^2 z^2 + x^2 y^4 + x^2 z^4 + constant)$?

I would put it more like:

$\displaystyle U = \frac {x^6}{3} + x^4 y^2 + x^4 z^2 + 2 x^2 y^2 z^2 + x^2 y^4 + x^2 z^4 + f(y,z)$​

Now, take the partial derivative w.r.t. y or z.

 

[Amrator]

I would put it more like:

$\displaystyle U = \frac {x^6}{3} + x^4 y^2 + x^4 z^2 + 2 x^2 y^2 z^2 + x^2 y^4 + x^2 z^4 + f(y,z)$​

Now, take the partial derivative w.r.t. y or z.

But f(y,z) is a function, not a constant? And why do I need to take the PD w.r.t. y or z? Just trying to understand.

 

[SammyS]

But f(y,z) is a function, not a constant? And why do I need to take the PD w.r.t y or z? Just trying to understand.

$\displaystyle \ \frac{\partial}{\partial x}f(y\,,z)=0\,,\$ right ?

So ƒ(y, z) plays the same role here as the constant of integration plays for a single variable integral .

 

[Amrator]

$\displaystyle \ \frac{\partial}{\partial x}f(y\,,z)=0\,,\$ right ?

So ƒ(y, z) plays the same role here as the constant of integration plays for a single variable integral .

Yes.

So taking the PD w.r.t. y yields,

$U = 2 x^4 y + 4 x^2 y z^2 + 4 x^2 y^3 + f(y,z)$

 

[fresh_42]

Yes.

So taking the PD w.r.t. y yields,

$U = 2 x^4 y + 4 x^2 y z^2 + 4 x^2 y^3 + f(y,z)$

Yes (but), if I made no mistake by scrolling up and down. However, can the constant here be a function of y?

 

[vela]

Yes.

So taking the PD w.r.t. y yields,

$U = 2 x^4 y + 4 x^2 y z^2 + 4 x^2 y^3 + f(y,z)$

Why didn't you take the partial derivative of f(y,z)? And that's supposed to be $\partial U/\partial y$ on the LHS, right?

 

[Ray Vickson]

But f(y,z) is a function, not a constant? And why do I need to take the PD w.r.t. y or z? Just trying to understand.

Because you have not yet matched $\partial U/ \partial y$ and $\partial U/ \partial z$ to the given forms that you started with. Notice that if you did not have the $f(y,z)$ term, the $y$ and $z$ partials would come out wrong. Try it and see! Don't agonize about it; just do it.

 

[Amrator]

${\partial U}/{\partial y} = 2 x^4 y + 4 x^2 y z^2 + 4 x^2 y^3$

${\partial U}/{\partial z} = 4 x^2 y^2 z$

 

[Chestermiller]

Are you not allowed to solve this in spherical coordinates? dU/dr = 2 r⁵?

 

[Amrator]

Are you not allowed to solve this in spherical coordinates? dU/dr = 2 r⁵?

I don't see why not. The book doesn't say I can't. Although, I am trying to get used to partial derivatives and the del operator.

 

[Chestermiller]

I don't see why not. The book doesn't say I can't.

So...

 

[fresh_42]

Are you not allowed to solve this in spherical coordinates? dU/dr = 2 r⁵?

Thanks a lot. I just did it by hand and at the end I came out with: "There must be a way to see it immediately!" Thank you.

 

[vela]

${\partial U}/{\partial y} = 2 x^4 y + 4 x^2 y z^2 + 4 x^2 y^3$

${\partial U}/{\partial z} = 4 x^2 y^2 z$

You're missing a bunch of terms in ${\partial U}/{\partial z}$.

$\nabla U = (2 x^4 + 4 x^2 y^2 + 4 x^2 z^2 + 4 y^2 z^2 + 2 y^4 + 2 z^4)x\hat i + (2 x^4 + 4 x^2 y^2 + 4 x^2 z^2 + 4 y^2 z^2 + 2 y^4 + 2 z^4)y\hat j + (2 x^4 + 4 x^2 y^2 + 4 x^2 z^2 + 4 y^2 z^2 + 2 y^4 + 2 z^4)z\hat k$

$\nabla U = \frac{\partial U}{\partial x} \hat i + \frac{\partial U}{\partial y} \hat j + \frac{\partial U}{\partial z} \hat k$

Note that the result you got for ${\partial U}/{\partial y}$ by neglecting f(y,z) only has the first three terms of $\hat{j}$ component of $\nabla U$. The missing terms only depend on y and z. That's why the f(y,z) has to be there.

 

[Amrator]

You're missing a bunch of terms in ${\partial U}/{\partial z}$.Note that the result you got for ${\partial U}/{\partial y}$ by neglecting f(y,z) only has the first three terms of $\hat{j}$ component of $\nabla U$. The missing terms only depend on y and z. That's why the f(y,z) has to be there.

I see, and f(y,z) contains only y and z terms, correct?
So $U = x^6 / 3 + x^4 y^2 + x^4 z^2 + 2 x^2 y^2 z^2 + x^2 y^4 + x^2 z^4 + f(y,z)$
$f(y,z) = \int {4 y^3 z^2 + 2 y^5 +2 y z^4} dy$
Something doesn't seem right with that above equation though. I have not learned multiple integrals yet. That's after this section.

 

[SammyS]

${\partial U}/{\partial y} = 2 x^4 y + 4 x^2 y z^2 + 4 x^2 y^3$

${\partial U}/{\partial z} = 4 x^2 y^2 z$

What you have for $\displaystyle {\partial U}/{\partial y}\$ is missing $\displaystyle \frac{\partial }{\partial y}f(y,\,z)\$.

 

[Amrator]

What you have for $\displaystyle {\partial U}/{\partial y}\$ is missing $\displaystyle \frac{\partial }{\partial y}f(y,\,z)\$.

But I don't know what f(y, z) is. Don't I have to find it first?

Perhaps I should start over? I'm getting very confused.

 

[fresh_42]

But I don't know what f(y, z) is. Don't I have to find it first?

Perhaps I should start over? I'm getting very confused.

You can get $f(y,z)$ by comparison of $δU / δx$ with the $î$ coordinate you computed in the beginning.

 

[Amrator]

You can get $f(y,z)$ by comparison of $δU / δx$ with the $î$ coordinate you computed in the beginning.

What? What is $δU / δx$?

 

[fresh_42]

What? What is $δU / δx$?

Sorry I took what I got. I haven't found the correct delta. You have $U = ... + f(y,z)$ on one hand, which allows you to differentiate along $y$. This result can be compared to your $j$-coordinate in the beginning. That gives you the partial differentiation of $f(y,z)$ along $y$ which you could integrate to something like $f(y,z) = ... + g(z)$ and so on.

 

[Amrator]

Sorry I took what I got. I haven't found the correct delta. You have $U = ... + f(y,z)$ on one hand, which allows you to differentiate along $y$. This result can be compared to your $j$-coordinate in the beginning. That gives you the partial differentiation of $f(y,z)$ along $y$ which you could integrate to something like $f(y,z) = ... + g(z)$ and so on.

So you want me to take PD of U = ... + f(y, z) w.r.t. y, and set it equal to the $\hat j$ component?
Isn't that what I did in one of my last posts on the previous page?

 

[fresh_42]

So you want me to take PD of U = ... + f(y, z) w.r.t. y, and set it equal to the $\hat j$ component?

Yes.

 

[fresh_42]

In order to keep confusion in limits here is what you already got: $\frac{\partial}{\partial y} U(x,y,z) = 2 x^4 y + 4 x^2 y z^2 + 4 x^2 y^3 + \frac{\partial}{\partial y} f(y,z)$.

 

[vela]

I see, and f(y,z) contains only y and z terms, correct?
So $U = x^6 / 3 + x^4 y^2 + x^4 z^2 + 2 x^2 y^2 z^2 + x^2 y^4 + x^2 z^4 + f(y,z)$
$f(y,z) = \int {4 y^3 z^2 + 2 y^5 +2 y z^4} dy$
Something doesn't seem right with that above equation though. I have not learned multiple integrals yet. That's after this section.

Which equation seems wrong? You're not doing a multiple integral. You're integrating with respect to y, treating z as a constant. You're just reversing the partial differentiation. You'll end up with another "constant" of integration, which is a function of z.

 

[Amrator]

Which equation seems wrong? You're not doing a multiple integral. You're integrating with respect to y, treating z as a constant. You're just reversing the partial differentiation. You'll end up with another "constant" of integration, which is a function of z.

So $f(y,z) = y^4 z^2 + y^6 / 3 + y^2 z^4 + g(z)$?
And then do the same thing with z?

 

[fresh_42]

So $f(y,z) = y^4 z^2 + y^6 / 3 + y^2 z^4 + g(z)$?
And then do the same thing with z?

Both correct.
And once you got the $g(z)$ you can all put together to write down $U(x,y,z)$. (Don't forget the constant which is this time really constant.)

 

[SammyS]

So $f(y,z) = y^4 z^2 + y^6 / 3 + y^2 z^4 + g(z)$?
And then do the same thing with z?

At this point, you may be able guess what g(z) is.

- and to quote fresh 42, "Don't forget the constant ..."

 

[ehild]

Homework Statement

$\nabla U = 2 r^4 \vec r$ Find U.

The method you follow will result in the potential function, but there is a simpler way when the gradient is given in terms of the position vector $\vec r$: $\nabla U = 2 ((\vec r)^2)^2 \vec r$
You did not include among the "Relevant equations" how the gradient of a function is defined.
So the total differential of a function U is $dU= \nabla (U) \cdot \vec{dr}$. ($\nabla (U)$ is a vector.)
You get the change of U from point O (the origin) to point P if you integrate both sides between O and P.
$\Delta U = U(P)-U(O)=\int_O^P{\nabla U}\cdot \vec{dr}$. In this problem $\nabla U = 2 r^4 \vec r$,
so you have the integral $\int_O^P{2 r^4 \vec r \cdot \vec {dr}}$.
You can choose the integration path arbitrary, the integral will be the same. See figure: You can reach P from O along several paths, why not along the straight line connecting them. In this case, $\vec r$ and $\vec {dr}$ have the same direction, so $\vec r \cdot \vec {dr}=r dr$, and the integral becomes $\int_0^{r(P)}{2 r^4 r dr}$.

 

[ehild]

Homework Statement

$\nabla U = 2 r^4 \vec r$ Find U.

The Attempt at a Solution

$\nabla U = 2 (x^2 + y^2 + z^2)^2 (x \hat i + y \hat j + z \hat j)$

I multiplied everything out,

$\nabla U = (2 x^4 + 4 x^2 y^2 + 4 x^2 z^2 + 4 y^2 z^2 + 2 y^4 + 2 z^4)x\hat i + (2 x^4 + 4 x^2 y^2 + 4 x^2 z^2 + 4 y^2 z^2 + 2 y^4 + 2 z^4)y\hat j + (2 x^4 + 4 x^2 y^2 + 4 x^2 z^2 + 4 y^2 z^2 + 2 y^4 + 2 z^4)z\hat k$

.

It was not necessary to expand the square.
You know that
$\frac{\partial U}{\partial x} = 2 x(x^2 + y^2 + z^2 )^2$.
Integral it with respect to x. Notice that you can do u-substitution with u= x²+y²+z². What do you get? Include the integration constant, which is a function of y and z: (f(yz).
So your integral with respect to x becomes $\frac{(x^2+y^2+z^2)^3 }{3}+f(yz)$
Take the partial derivative of the above expression with respect y. It must be equal to $2 y(x^2 + y^2 + z^2 )^2$. So what is the partial derivative of f(x,y) with respect to y?

 

[Amrator]

.

It was not necessary to expand the square.
You know that
$\frac{\partial U}{\partial x} = 2 x(x^2 + y^2 + z^2 )^2$.
Integral it with respect to x. Notice that you can do u-substitution with u= x²+y²+z². What do you get? Include the integration constant, which is a function of y and z: (f(yz).
So your integral with respect to x becomes $\frac{(x^2+y^2+z^2)^3 }{3}+f(yz)$
Take the partial derivative of the above expression with respect y. It must be equal to $2 y(x^2 + y^2 + z^2 )^2$. So what is the partial derivative of f(x,y) with respect to y?

f(x, y)? Don't you mean g(z)?

 

[ehild]

f(x, y)? Don't you mean g(z)?

A stupid error! I meant f(y,z) and then g(z).

 

[Amrator]

.

It was not necessary to expand the square.
You know that
$\frac{\partial U}{\partial x} = 2 x(x^2 + y^2 + z^2 )^2$.
Integral it with respect to x. Notice that you can do u-substitution with u= x²+y²+z². What do you get? Include the integration constant, which is a function of y and z: (f(yz).
So your integral with respect to x becomes $\frac{(x^2+y^2+z^2)^3 }{3}+f(yz)$
Take the partial derivative of the above expression with respect y. It must be equal to $2 y(x^2 + y^2 + z^2 )^2$. So what is the partial derivative of f(x,y) with respect to y?

I don't understand. How do you find f(y, z) using this method?

 

[SammyS]

I don't understand. How do you find f(y, z) using this method?

What is your result for $\displaystyle \ \frac{\partial f(y,z)}{\partial y} \$ if you use ehild's method ?

 

[Amrator]

What is your result for $\displaystyle \ \frac{\partial f(y,z)}{\partial y} \$ if you use ehild's method ?

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

 

[Ray Vickson]

I don't understand. How do you find f(y, z) using this method?

You find $f(y,z)$ by first finding $\partial f(y,z) /\partial y$ and $\partial f(y,z) /\partial z$ . So, go ahead and do that: find these partial derivatives, using the method that has been explained to you many times already.

 

[Amrator]

You find $f(y,z)$ by first finding $\partial f(y,z) /\partial y$ and $\partial f(y,z) /\partial z$ . So, go ahead and do that: find these partial derivatives, using the method that has been explained to you many times already.

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

 

[Amrator]

$\nabla U = 2 x (x^2 + y^2 + z^2)^2 \hat i + 2 y (x^2 + y^2 + z^2)^2 \hat j +2 z (x^2 + y^2 + z^2)^2 \hat k$
$U = \int \partial U /\partial x dx = (x^2 + y^2 + z^2)^3 / 3 + f(y,z)$
$\partial U /\partial y = 2 y (x^2 + y^2 + z^2)^2 + \partial f(y,z) /\partial y$
$2 y (x^2 + y^2 + z^2)^2 + \partial f(y,z) /\partial y = 2 y (x^2 + y^2 + z^2)^2$
$\partial f(y,z) /\partial y = 0$

 

[Ray Vickson]

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

Why not? It IS what you get!

So, assuming you believe your own work, what do those two formulas above tell you about $f(y,z)$?

 

[Amrator]

Why not? It IS what you get!

So, assuming you believe your own work, what do those two formulas above tell you about $f(y,z)$?

f(y, z) = g(z)?

 

[vela]

f(y, z) = g(z)?

Are you just guessing?

 

[Amrator]

Are you just guessing?

Well the integral of 0 is a constant of integration.

 

[fresh_42]

Well the integral of 0 is a constant of integration.

Yes. And what does it mean to be constant w.r.t. y and z?

 

[Amrator]

Yes. And what does it mean to be constant w.r.t. y and z?

Oh right, f(y,z) = x since y and z have constant slopes.

 

[fresh_42]

Oh right, f(y,z) = x since y and z have constant slopes.

But f(y,z) isn't a function of x. How can this be?

 

[Amrator]

But f(y,z) isn't a function of x. How can this be?

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

 

[fresh_42]

You already have had it. Just add up what you've said:
$f(y,z)$ does not have $x$ as variable and it's dependency on $y$ is constant, as you have said. But the partial derivative w.r.t. $z$ is also zero. And you have said that means $f$ does not change if $z$ does. So it has to be?