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Help with polar coordinates. 
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#1
Feb2212, 06:56 PM

P: 195

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
Feb2212, 08:36 PM

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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.



#3
Feb2312, 03:20 PM

P: 195

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. 


#4
Feb2312, 03:27 PM

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Help with polar coordinates.



#5
Feb2712, 07:36 PM

P: 195

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 handwavy when he explains it. 


#6
Feb2712, 07:47 PM

P: 195

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. 


#7
Feb2712, 07:53 PM

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#8
Feb2712, 08:07 PM

P: 195

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)? 


#9
Feb2712, 08:11 PM

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#10
Feb2712, 08:34 PM

P: 195

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 :) 


#11
Feb2812, 01:59 PM

P: 195

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? 


#12
Feb2812, 03:18 PM

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#13
Feb2812, 05:43 PM

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