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Help with polar coordinates.

  1. Feb 22, 2012 #1

    ElijahRockers

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    1. The problem statement, all variables and given/known data

    Let [itex]\hat r = <x_r , y_r>[/itex] and [itex]\hat\theta = <x_\theta , y_\theta>[/itex]

    Draw these vectors at points (x,y) = (1,0), (2,0), (3,0), (1,1), (0,1), (0,2).

    Here is the entire homework assignment so you can see what context it is in.

    Our teacher uses two distinct methods. He gives us questions from the book that we do online, and then his written assignments are generally questions he makes up that are seemingly meant to get us to think outside the box.

    This assignment has everyone pretty much stumped, and he's extended it another week.

    3. The attempt at a solution

    I assume by [itex]\hat r[/itex] he's talking about a unit vector, r. When he says [itex]x_r[/itex], to me, that means the partial of x with respect to r, which would be simply cos(theta).

    But when I start thinking about it, it just doesn't seem to make any sense to me.

    I am familiar with radial unit vectors and angular unit vectors (which he mentions in a later question) from my calculus based physics class, but I can't seem to make the connection from what he's trying to show us, to what I learned in physics.

    I'm trying to understand the process he's trying to have us use to get there.
     
  2. jcsd
  3. Feb 22, 2012 #2

    Dick

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    I don't think [itex]\hat r[/itex] and [itex]\hat \theta[/itex] are supposed to be unit vectors. If you look later on the assignment [itex]e_r[/itex] and [itex]e_{\theta}[/itex] are the unit vectors. They are just the vector of partial derivatives of x and y, like you thought.
     
  4. Feb 23, 2012 #3

    ElijahRockers

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    Ok. So [itex]\hat r = <cos\theta , sin\theta >[/itex]
    and [itex]\hat \theta = <-rsin\theta , rcos\theta >[/itex]

    The points are given to me in x,y form. These new functions are with respect to r and theta. I am confused on what he's asking me to plot. (1,0) is in cartesian coords. Is he asking me to convert these to polar and put them into the vectors?

    EDIT: I did that, and I suppose I can see where he's going with this. I am now trying to solve for i and j.
     
    Last edited: Feb 23, 2012
  5. Feb 23, 2012 #4

    Dick

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    I think so, yes.
     
  6. Feb 27, 2012 #5

    ElijahRockers

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    I'm stuck on number 6. The instructor explained it to me earlier today, and at the time I thought I understood it, but now that I'm trying to reproduce it, my mind is drawing a blank.

    How can I take the partial derivative of the function if it hasn't been defined?

    Using the chain rule to express df/dxwould be like [itex]\frac{\partial f}{\partial r}\frac{\partial r}{\partial x}[/itex] and [itex]\frac{\partial f}{\partial \theta}\frac{\partial \theta}{\partial x}[/itex]... but what does this mean?

    I'm afraid I don't really understand this. It all seems kind of vague and hand-wavy when he explains it.
     
  7. Feb 27, 2012 #6

    ElijahRockers

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    Also, he wants it expressed in terms of r and theta.... but df/dx is by definition goin to be in terms of x and y, right?

    I'm confused.
     
  8. Feb 27, 2012 #7

    Dick

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    df/dx doesn't have to be expressed in terms of x and y. It can be expressed in terms of r and θ. To do that you are going to have to express something like dr/dx in terms of r and θ. r=sqrt(x^2+y^2). What's dr/dx? Now express that in terms of r and θ.
     
  9. Feb 27, 2012 #8

    ElijahRockers

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    if r=sqrt(x^2+y^2) then dr/dx would be x/sqrt(x^2+y^2) which is cos(theta)...

    is that right? so (df/dr)(dr/dx) would be cos(theta)(df/dr)?
     
  10. Feb 27, 2012 #9

    Dick

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    Looks good so far. Now you need dθ/dx.
     
  11. Feb 27, 2012 #10

    ElijahRockers

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    theta = arctan(y/x). dtheta/dx would be... -y/(x^2+y^2) ? if so I'm not really sure how to put that in terms of r and/or theta. i know y would be the opposite side, x^2+y^2 is r^2... so we get df/dx = -sin(theta)/r (df/dtheta)

    then i just need to do the same for df/dy.

    i think i got it now, thanks :)
     
  12. Feb 28, 2012 #11

    ElijahRockers

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    Alright I'm almost done. For number 7 I substituted in my expressions and everything simplified pretty nicely, except im confused at the very end.

    After simplifying, terms cancelled out and sin^2+cos^2 turned into 1, and I was left with

    [itex]\nabla f = \frac{\partial f}{\partial r}e_{\hat r} + (\frac{sin^2\theta + cos^2\theta}{r})\frac{\partial f}{\partial \theta} e_{\hat \theta}[/itex]

    So I get the feeling that 'r' in the denominator isn't supposed to be there. It's there because of my calculations from [itex]\frac{\partial f}{\partial x}[/itex] and [itex]\frac{\partial f}{\partial y}[/itex]. More specifically the partials with respect to theta, since for x and y, they were -sin(theta)/r and cos(theta)/r respectively.

    I can use [itex]x=\cos \theta[/itex] and [itex]y=\sin \theta[/itex] to get [itex]r^2[/itex] on top, which would leave an 'r' behind, but I'm assuming the correct answer is

    [itex]\nabla f = \frac{\partial f}{\partial r}e_{\hat r} + \frac{\partial f}{\partial \theta} e_{\hat \theta}[/itex]

    without the r. Is this a wrong assumption? If not, what is supposed to happen to the r?
     
  13. Feb 28, 2012 #12

    Dick

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    You were right the first time. The r in the denominator should be there.
     
  14. Feb 28, 2012 #13

    ElijahRockers

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    ....Oh! Ok! This assignment is due tomorrow, so I finished just in time. Thanks so much for all your help!
     
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