Help with polar coordinates.


by ElijahRockers
Tags: coordinates, polar
ElijahRockers
ElijahRockers is offline
#1
Feb22-12, 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.
Phys.Org News Partner Science news on Phys.org
NASA's space station Robonaut finally getting legs
Free the seed: OSSI nurtures growing plants without patent barriers
Going nuts? Turkey looks to pistachios to heat new eco-city
Dick
Dick is offline
#2
Feb22-12, 08:36 PM
Sci Advisor
HW Helper
Thanks
P: 25,171
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.
ElijahRockers
ElijahRockers is offline
#3
Feb23-12, 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.

Dick
Dick is offline
#4
Feb23-12, 03:27 PM
Sci Advisor
HW Helper
Thanks
P: 25,171

Help with polar coordinates.


Quote Quote by ElijahRockers View Post
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?
I think so, yes.
ElijahRockers
ElijahRockers is offline
#5
Feb27-12, 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 hand-wavy when he explains it.
ElijahRockers
ElijahRockers is offline
#6
Feb27-12, 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.
Dick
Dick is offline
#7
Feb27-12, 07:53 PM
Sci Advisor
HW Helper
Thanks
P: 25,171
Quote Quote by ElijahRockers View Post
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.
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.
ElijahRockers
ElijahRockers is offline
#8
Feb27-12, 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)?
Dick
Dick is offline
#9
Feb27-12, 08:11 PM
Sci Advisor
HW Helper
Thanks
P: 25,171
Quote Quote by ElijahRockers View Post
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)?
Looks good so far. Now you need dθ/dx.
ElijahRockers
ElijahRockers is offline
#10
Feb27-12, 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 :)
ElijahRockers
ElijahRockers is offline
#11
Feb28-12, 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?
Dick
Dick is offline
#12
Feb28-12, 03:18 PM
Sci Advisor
HW Helper
Thanks
P: 25,171
Quote Quote by ElijahRockers View Post
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?
You were right the first time. The r in the denominator should be there.
ElijahRockers
ElijahRockers is offline
#13
Feb28-12, 05:43 PM
P: 195
Quote Quote by Dick View Post
You were right the first time. The r in the denominator should be there.
....Oh! Ok! This assignment is due tomorrow, so I finished just in time. Thanks so much for all your help!


Register to reply

Related Discussions
Converting Polar coordinates to Cartesian coordinates Introductory Physics Homework 9
conversion of cartesian coordinates to polar coordinates Calculus & Beyond Homework 3
Changing rectangular coordinates to polar coordinates ? Calculus & Beyond Homework 6
xy coordinates to polar coordinates for double integral. hepl please! Calculus & Beyond Homework 3
From polar coordinates to heliocentric ecliptic coordinates Astrophysics 0