Polar equation problem

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Homework Statement





Using this polar equation:

[tex]r = \theta + sin(2\theta)[/tex]

Find the angle [tex]\theta[/tex] that relates to the point on the curve when x = -2


I'm not sure where to start...but my guess is to convert the equation to another form...any help is appreciated.
 

Answers and Replies

  • #2
hunt_mat
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Use [tex]x=r\cos\theta[/tex], you know that theta has to be in the range to make cos theta negative, so I might write:

[tex]
\frac{-2}{\cos\theta}=\theta +\sin (2\theta )
[/tex]

and look for a possible numerical solution.
 
  • #3
HallsofIvy
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Don't convert the equation, convert the condition. [itex]x= r cos(\theta)[/itex] so the condition that x= -2 becomes [itex]rcos(\theta)= -2[/itex]. Since on this curve, [itex]r= \theta+ sin(2\theta)[/itex] you want to solve [itex](\theta+ sin(2\theta))cos(\theta)= -2[/itex].
 
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  • #4
SammyS
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x=r·cos(θ), so if x=2, r=2/cos(θ)

That leaves you to solve: 2/cos(θ) = θ + sin(2θ) for θ.

Looks like a numerical solution.
 
  • #5
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x=r·cos(θ), so if x=2, r=2/cos(θ)

That leaves you to solve: 2/cos(θ) = θ + sin(2θ) for θ.

Looks like a numerical solution.

yes, this is what I initially had...[tex] \theta + sin2\theta = \frac{-2}{cos\theta}[/tex]

but got stuck on finding [tex] \theta[/tex]
 
  • #6
hunt_mat
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Use Newton Raphson method. if you have an equation to solve [tex]f(\theta )=0[/tex] then to iterate to a solution use the process:

[tex]
\theta_{n+1}=\theta_{n}-\frac{f(\theta_{n})}{f'(\theta_{n})}
[/tex]

You have to have an initial stating guess mind you, and you have to work in radians. I would start with an initial guess of [tex]\pi /2[/tex]
 
  • #7
hunt_mat
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Having done the calculation myself (just out of interest), the iteration process out should use is:

[tex]
\theta_{n+1}=\theta_{n}-\frac{\theta_{n}\cos\theta_{n} +\sin 2\theta_{n}\cos\theta_{n}+2}{\cos\theta_{n} (1+2\cos 2\theta_{n} )-\sin\theta_{n} (\theta_{n} +\sin 2\theta_{n} ))}
[/tex]

After about 3 iterations it settled down to the value of [tex]\theta =2.7861[/tex] radians.
 

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