Parametrization of Witch of Agnesi

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Homework Statement



upload_2015-2-15_17-11-41.png


The question is completely described in the photo.

Homework Equations


Trigonometric translation properties

The Attempt at a Solution


The problem is in two dimensions, so I'm ignoring the z coordinates. For a circle centered at (0,a), the position vector of P is ##(a## ##sin(θ),a-a## ##cos(θ))## (by taking into consideration what theta is in this problem) since ##a## ##cos(\frac{3π}{2}+θ)=a## ##sin(θ)## and ##a+a## ##sin(\frac{3π}{2}+θ)=a-a## ##cos(θ)## .
Therefore, the y coordinates of R should be ##a-a## ##cos(θ)## . R's x coordinate equals to the x coordinate of Q, which is given by ##x=\frac{2a}{m}## , where ##m=\frac{1-cos(θ)}{sin(θ)}## .
So my answer is ##γ(θ) = (\frac{2asin(θ)}{1-cos(θ)},a-cos(θ))##. However, the correct answer is ##γ(θ) = (2a## ##cot(θ), a(1-cos(2θ))##. Where did I go wrong?
 

Answers and Replies

  • #2
Orodruin
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What you are calling ##\theta##, your solution manual calls ##2\theta##. Make this substitution and use some trigonometric identities and you will get the same. This is of course only a different parametrisation. You are also missing an a multiplying the cos in your y coordinate.
 
  • #3
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Oops, I made a typo at the end there. And yes, if it is as you say, then my answer would be correct, although why my book refers to ##\theta## as ##2\theta## is beyond me. Thank you.
 

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