What are the steps for drawing vectors in polar coordinates?

In summary, the conversation was about a homework assignment involving vectors and partial derivatives. The person was confused about the concept of unit vectors and how to plot them at various points. They also struggled with solving for the partial derivatives and expressing them in terms of r and θ. The conversation ends with a clarification on the correct answer for a specific problem involving partial derivatives.
  • #1
ElijahRockers
Gold Member
270
10

Homework Statement



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 http://www.math.tamu.edu/~vargo/courses/251/HW5.pdf 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.

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.
 
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  • #2
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
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:
  • #4
ElijahRockers said:
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.
 
  • #5
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.
 
  • #6
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
ElijahRockers said:
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.
 
  • #8
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
ElijahRockers said:
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.
 
  • #10
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
Alright I'm almost done. For number 7 I substituted in my expressions and everything simplified pretty nicely, except I am confused at the very end.

After simplifying, terms canceled 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
ElijahRockers said:
Alright I'm almost done. For number 7 I substituted in my expressions and everything simplified pretty nicely, except I am confused at the very end.

After simplifying, terms canceled 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.
 
  • #13
Dick said:
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!
 

1. What are polar coordinates?

Polar coordinates are a system of describing the position of a point in a two-dimensional space using a distance from a fixed point and an angle from a fixed direction.

2. How are polar coordinates different from cartesian coordinates?

Polar coordinates use a different set of variables (distance and angle) to describe a point, while cartesian coordinates use x and y coordinates. Polar coordinates are often used for describing circular or curved objects, while cartesian coordinates are used for straight lines and rectangular shapes.

3. How do you convert polar coordinates to cartesian coordinates?

To convert polar coordinates to cartesian coordinates, you can use the following formulas: x = r * cos(theta) and y = r * sin(theta). R represents the distance from the origin and theta represents the angle from the positive x-axis.

4. How do you plot a point using polar coordinates?

To plot a point using polar coordinates, first identify the distance from the origin and the angle from the positive x-axis. Then, draw a line from the origin at the given angle, and measure the distance to plot the point.

5. What are some real-world applications of polar coordinates?

Polar coordinates are used in various fields, such as physics, engineering, and navigation. They are particularly useful for describing circular motion, such as the movement of planets or satellites, and for mapping points on a globe.

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