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Polar Coordinates help

Hi guys,

Can someone explain why r=(1/1+2cosϴ) for -2/3Pi < ϴ , 2/3 Pidoes not have a pole? I was able to differentiate it so I'm quite confused. I got ϴ = 0

Thanks
Original post by ChemBoy1
Hi guys,

Can someone explain why r=(1/1+2cosϴ) for -2/3Pi < ϴ , 2/3 Pidoes not have a pole? I was able to differentiate it so I'm quite confused. I got ϴ = 0

Thanks


The pole is where r=0r=0 and clearly the function 11+2cos⁡θ\dfrac{1}{1+2\cos \theta} is never 0
Thank you very much, I seem to having difficulties with this part of the topic. It said in the book that if F(Thetha) = 0 then the theta is tangent to the graph
Original post by RDKGames
The pole is where r=0r=0 and clearly the function 11+2cos⁡θ\dfrac{1}{1+2\cos \theta} is never 0
Also if there was a situation with more than 1 trig function e..g. 2+ cos theta - cos^2 theta. How are you meant to find out the tangents to the pole?
Original post by ChemBoy1
Also if there was a situation with more than 1 trig function e..g. 2+ cos theta - cos^2 theta. How are you meant to find out the tangents to the pole?


Try to use the correct terminology. The pole is a point, so it wouldn't make sense to have something 'tangent' to it.

Here you are interested in straight lines going through the pole, which are tangent to your polar eq. r=f(θ)r=f(\theta). These lines have equation θ=α\theta = \alpha for −π<α≤π-\pi < \alpha \leq \pi (or between 0 and 2 pi, if you want to use that).
The first step is to obvious set f(θ)=0f(\theta) = 0 and solve it for the θ\theta values. But make sure that f′(θ)≠0f'(\theta) \neq 0.
For the example above, we would simply need to solve 2+cos⁡θ−cos⁡2θ=02+\cos \theta - \cos^2 \theta = 0 but note that this is just a quadratic in cos⁡θ\cos \theta which also factorises.

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