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Vector Calculus - not sure where to start here

by FrogPad
Tags: calculus, start, vector
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FrogPad
#1
Mar5-06, 07:31 PM
P: 837
I'm not sure what the question is asking. Any help on getting started with this would be cool.

Q: Show that [itex] u = r\cos \psi +\frac{1}{2}r^{-2}\cos \psi [/itex] satisfies [itex] \nabla^2 u [/itex] and also [itex] \frac{\partial u}{\partial r}=0 [/itex] on the unit sphere. Find the velocity field [itex] \vec v = \nabla u [/itex] for flow past the sphere.

Now we have not done problems like this yet. We have only just done background work that will be leading up to these problems by definining the laplacian\div\grad\... in cylindrical\spherical coordinates.

The first part is straight forward, show that [itex] u [/itex] satisfies [itex] \nabla^2 u [/itex]. I did that. Now the next two parts are where I am confused.

[itex] \frac{\partial u}{\partial r} = 0 [/itex] on the unit sphere? I have no idea what to do here. Here are my thoughts on it:
* It is asking for me to show that the partial derivative of [itex] u [/itex] with respect to [itex] r [/itex] equals 0, ON the unit sphere.

* But what does ON the unit sphere mean? That if we are in a space that is within the unit sphere the rate of change with respect to r is 0? Well how do I express that mathematically.

And the last part [itex] \vec v = \nabla u [/itex] for flow past the sphere. I have no idea. Any help... please.
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quasar987
#2
Mar5-06, 07:52 PM
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What is the equation of the unit sphere in spherical coordinates?
FrogPad
#3
Mar5-06, 08:36 PM
P: 837
In cylindrical coodrinates it would be:
[tex] x^2+y^2+z^2 = 1 [/tex]

Cartesian -> Spherical
[tex] x = r \sin \psi \cos \phi [/tex]
[tex] x = r \sin \psi \sin \phi [/tex]
[tex] z = r \cos \psi [/tex]

Unit Sphere
[tex] r^2 = 1 [/tex]
[tex] r = \pm 1 [/tex]

Right?

FrogPad
#4
Mar5-06, 08:39 PM
P: 837
Vector Calculus - not sure where to start here

Hmmm... so for [itex] u [/itex] to be on the unit sphere [itex] r = \pm 1 [/itex] which reduces [itex] u [/itex] to a function dependent on [itex] \psi [/itex]. So differentiating it with respect to [itex] r [/itex] will yield 0 ?
FrogPad
#5
Mar5-06, 08:42 PM
P: 837
If that is actually the answer (it seems like it makes sense, maybe not though) then that just leaves me with the last part.

I believe [itex] \vec v = \nabla u [/itex] is basically asking me to find a vector that when I take the gradient it is equal to [itex] u [/itex]. But the fact that it says for the flow past the sphere... I don't understand that at all.
quasar987
#6
Mar5-06, 08:49 PM
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A point in spherical coordinates is of the form [itex](r,\psi, \phi)[/itex]. Your calculations indicate that the constraint [itex]x^2+y^2+z^2=1[/itex] in rectagular coordinates takes the form [itex]r^2=1 \Leftrightarrow r=1[/itex] in spherical (r=-1 is to ban because the polar coordinate r is nonnegative by definition). So the other two coordinates, [itex]\psi[/itex] and [itex]\phi[/itex], are free to take any value.

So in spherical coordinates, the unit sphere is:

[tex]\mathbb{S}^3 = \{(r,\psi, \phi) : \mbox{what?}\}[/tex]
FrogPad
#7
Mar5-06, 09:06 PM
P: 837
Ahh, good to know on the [itex] r \neq -1 [/itex] that makes sense. Now as far as the unit sphere, I thought that [itex] r=1 [/itex] IS the equation for the unit sphere, where there are limits on [itex] \phi,\,\psi,\,r [/itex]

[tex]\mathbb{S}^3 = \{(r,\psi, \phi) : r=1, \,\, 0 \leq \psi \leq \pi, \,\, 0 \leq \phi \leq 2\pi \}[/tex]
quasar987
#8
Mar5-06, 09:13 PM
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Actually it is that psi that ranges from 0 to 2pi and phi from 0 to pi. But you get the idea.

Now, can you guess what "show that [itex]\partial u / \partial r =0[/itex] on the unit sphere" means?
FrogPad
#9
Mar5-06, 10:04 PM
P: 837
So I am left with:

[tex] u = r\cos \psi +\frac{1}{2}r^{-2}\cos \psi [/tex]

and the constraint:
[tex] r = 1 [/tex] and the limits on psi and phi.

[tex] u(r=1) = \cos \psi +\frac{1}{2} \cos \psi = u_1[/tex]

[tex] \frac{\partial u_1}{\partial r} = 0 [/tex]

Is that the right type of thought?
quasar987
#10
Mar5-06, 10:10 PM
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No. Of course if you set r=constant and then differentiate wrt r you'll get 0, since you don't have any r left in your equation!

What you want to do is calculate [itex]\partial u / \partial r[/itex] first, and THEN evaluate it at r=1, and show that for any psi and phi, the result is 0.
FrogPad
#11
Mar5-06, 10:23 PM
P: 837
:) hahah... yeah, that actually makes sense. Silly me. I was just trying to get to an answer (could you tell?). So yeah... when you differentiate:

[tex] \frac{du}{dr} = \cos \psi - \frac{\cos \psi}{r^3} [/tex]

So now ON the unit sphere means the values of r, psi, and phi are:

r= 1 (to actually be on the sphere)
psi = 0 to 2pi
phi = 0 to pi

So ON the unit sphere du/dr is equal to 0.

---
That's awesome man. It actually makes sense. Do you have time for the last part of the question?
quasar987
#12
Mar5-06, 10:34 PM
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Actually, I'm not sure what the first part of the question means...

Quote Quote by FrogPad
Q: Show that [itex] u = r\cos \psi +\frac{1}{2}r^{-2}\cos \psi [/itex] satisfies [itex] \nabla^2 u [/itex] and also [itex] \frac{\partial u}{\partial r}=0 [/itex] on the unit sphere.
If it is actually worded in this way in your textbook, I would take it that you must also show that [itex]\nabla^2 u = 0[/itex] on the unit sphere, because the statement "u satifies [itex]\nabla^2 u[/itex]" alone does not make sense.
quasar987
#13
Mar5-06, 10:41 PM
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The part that says

Quote Quote by FrogPad
Find the velocity field [itex] \vec v = \nabla u [/itex] for flow past the sphere.
means: calculate [itex] \vec v = \nabla u [/itex]. What is its value for r>1, psi in [0, 2pi) and phi in [0,pi)?
FrogPad
#14
Mar5-06, 10:45 PM
P: 837
:( grr... sorry about that. It says:

Show that [itex] u = r \cos \psi + \frac{1}{2}r^{-2}\cos \psi [/itex] satisfies [itex] \nabla^2 = 0 [/itex]

I did not type in the [itex] = 0 [/itex] part.


Anyways, that was just a straightforward calculation. I got that part. The part that I'm not sure what is being asked is:

Find the velocity field [itex] \vec v = \nabla u [/itex] for flow past a sphere.
FrogPad
#15
Mar5-06, 10:48 PM
P: 837
Oh hells yeah. That makes sense. So I just find a vector that is still satisfies with those conditions. Now if [itex] r >1 [/itex] that is past the sphere. :)

You rock man. That helps me so much.
quasar987
#16
Mar5-06, 10:50 PM
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To be on the safe side, let's say it means, what is [itex]\nabla u [/itex] everywhere but on the sphere.
FrogPad
#17
Mar5-06, 11:10 PM
P: 837
My books definition of the gradient in spherical coordinates:

[tex] \nabla f = \frac{\partial f}{\partial r} \hat r + \frac{1}{r \sin \psi} \frac{\partial f}{\partial \theta} \hat \theta + \frac{1}{r} \frac{\partial f}{\partial \psi} \hat \psi [/tex]

[tex] \frac{\partial f}{\partial r}= \cos \psi - \frac{\cos \psi}{r^3} [/tex]

[tex] \frac{\partial f}{\partial \theta}=0 [/tex]

[tex] \frac{\partial f}{\partial \psi} = \frac{1}{r} \left( \frac{-\sin \psi (2r^3+1)}{2r^2} \right) [/tex]


[tex] \vec v = \nabla u [/tex]

[tex] \vec v = \left[ \begin{array}{c}\cos \psi - \frac{\cos \psi}{r^3} \\ 0 \\ -\sin \psi (1+\frac{1}{2r^3}) \end{array} \right] = \left(\cos \psi - \frac{\cos \psi}{r^3} \right)\hat r -\sin \psi \left(1+\frac{1}{2r^3}\right) \hat \psi [/tex]
WHERE:
[tex] r>1,\,\, \psi = 0..\pi[/tex]

So if [itex] r\geq 1 [/itex] then if [itex] r=1 [/itex] the [itex] \hat r [/itex] component would be 0.
quasar987
#18
Mar5-06, 11:21 PM
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lol, I like the "IF YOU ARE READING THIS... I'm STILL EDITING IT. I made a mistake." warning sign :)


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