Solving Polar Equation r = θ + sin(2θ) for x = -2: Homework Problem

[rjs123]

Homework Statement

Using this polar equation:

$$r = \theta + sin(2\theta)$$

Find the angle $$\theta$$ 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.

 

[hunt_mat]

Use $$x=r\cos\theta$$, you know that theta has to be in the range to make cos theta negative, so I might write:

$$\frac{-2}{\cos\theta}=\theta +\sin (2\theta )$$

and look for a possible numerical solution.

 

[HallsofIvy]

Don't convert the equation, convert the condition. $x= r cos(\theta)$ so the condition that x= -2 becomes $rcos(\theta)= -2$. Since on this curve, $r= \theta+ sin(2\theta)$ you want to solve $(\theta+ sin(2\theta))cos(\theta)= -2$.

 

[SammyS]

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.

 

[rjs123]

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...$$\theta + sin2\theta = \frac{-2}{cos\theta}$$

but got stuck on finding $$\theta$$

 

[hunt_mat]

Use Newton Raphson method. if you have an equation to solve $$f(\theta )=0$$ then to iterate to a solution use the process:

$$\theta_{n+1}=\theta_{n}-\frac{f(\theta_{n})}{f'(\theta_{n})}$$

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 $$\pi /2$$

 

[hunt_mat]

Having done the calculation myself (just out of interest), the iteration process out should use is:

$$\theta_{n+1}=\theta_{n}-\frac{\theta_{n}\cos\theta_{n} +\sin 2\theta_{n}\cos\theta_{n}+2}{\cos\theta_{n} (1+2\cos 2\theta_{n} )-\sin\theta_{n} (\theta_{n} +\sin 2\theta_{n} ))}$$

After about 3 iterations it settled down to the value of $$\theta =2.7861$$ radians.