Polar Coordinates

  • #1
1. The question
The position of a particle is given by r(t) = acos(wt) i + bsin(wt) j. Assume a and b are both positive and a > b. The plane polar coordinates of a particle at a time t equal to 1/8 of the time period T will be given by _

Homework Equations


r(t) = acos(wt) i + bsin(wt) j.

The Attempt at a Solution


at t = T/8 the value of wt=π/4.
therefore in Cartesian coordinates the vector is r= a/√2 + b/√2.
so in polar coordinates this transforms to r = √(a2 + b2)/2 and tanθ = b/a. My answer does not match with the given one. Also if we were to write the polar equation of the curve how would that follow ?
 

Answers and Replies

  • #2
BvU
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Hi,
Strange. Don't see anything wrong. In what respect does your answer not match with the given one ?
 
  • #3
Hi,
Strange. Don't see anything wrong. In what respect does your answer not match with the given one ?
The angle does not match. tanθ = b/a√2 is given
Also this is one of the part of the question, and I still can't understand how to wite the polar equation for this curve
 
  • #4
BvU
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Still strange. Where would the ##\sqrt 2## come from ?
Your ##\tan\theta## matches the expression here (##\tan t=1##).
And the answer for the polar equation is right in the next paragraph :rolleyes:
 
  • #5
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The angle does not match. tanθ = b/a√2 is given
Also this is one of the part of the question, and I still can't understand how to wite the polar equation for this curve
You have $$r\cos \theta = a \cos \omega t$$and $$r\sin \theta = b \sin \omega t$$
Just solve for r and ##\theta##
 
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  • #6
Still strange. Where would the ##\sqrt 2## come from ?
Your ##\tan\theta## matches the expression here (##\tan t=1##).
And the answer for the polar equation is right in the next paragraph :rolleyes:
I calculated the polar equation to be r = √(a2cos2(wt) + b2sin2(wt)) r^
and then I differentiated it to get the velocity as v = ((b2 - a2)sin(wt)cos(wt)w)/√(a2cos2(wt) + b2sin2(wt)) r^ + √(a2cos2(wt) + b2sin2(wt))w θ^
 
  • #7
I calculated the polar equation to be r = √(a2cos2(wt) + b2sin2(wt)) r^
and then I differentiated it to get the velocity as v = ((b2 - a2)sin(wt)cos(wt)w)/√(a2cos2(wt) + b2sin2(wt)) r^ + √(a2cos2(wt) + b2sin2(wt))w θ^
Is this wrong ?
 
  • #8
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Is this wrong ?
It looks correct. For aesthetic purposes, I would go to the double angle formulas:
$$2\sin \omega t\cos \omega t=\sin2\omega t$$
$$\cos^2 \omega t=\frac{1+\cos 2\omega t}{2}$$
$$\sin^2 \omega t=\frac{1-\cos 2\omega t}{2}$$
 
  • #9
BvU
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Are we sure the exercise composer wants r(t) and v(t) instead of ##r(\theta)## when asking for the polar form ?
 
  • #10
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I calculated the polar equation to be r = √(a2cos2(wt) + b2sin2(wt)) r^
and then I differentiated it to get the velocity as v = ((b2 - a2)sin(wt)cos(wt)w)/√(a2cos2(wt) + b2sin2(wt)) r^ + √(a2cos2(wt) + b2sin2(wt))w θ^
Oops. I don't confirm your coefficient of ##\hat{\theta}##. I get ##\frac{\omega ab}{r}##.
 
  • #11
that is the problem I have been having, I can't figure out as to why I am not getting the correct angular component
 
  • #12
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that is the problem I have been having, I can't figure out as to why I am not getting the correct angular component
The angular component is ##rd\theta/dt##, not ##r\omega##. In this problem ##d\theta/dt## is not the same thing as what they call ##\omega##. You need to be able to show that ##\frac{d\theta}{dt}=\frac{\omega ab}{r^2}##.
 
  • #13
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$$\tan \theta = \frac{b}{a}\tan \omega t$$
$$\sec^2 \theta \frac{d\theta}{dt}=\frac{b\omega}{a}\sec^2\omega t$$$$r\cos \theta=a\cos \omega t$$

Combine 2nd and 3rd equations to eliminate ##\sec \theta##.
 

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