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Tangents with polar equations

Hi,

I've got the polar equation:

r=a(1+cosθ)r = a(1+cos\theta)

I've been asked to find the coordinates of the points on this equation where the tangents are parallel to the initial line.

So I did this:

y=rsinθ=a(sinθ+sinθcosθ)y = rsin\theta = a(sin\theta + sin\theta cos\theta)

⇒dydθ=a(cosθ+cos2θ−sin2θ)\Rightarrow \dfrac{dy}{d\theta} = a(cos\theta + cos^2\theta - sin^2\theta)

So if I let dydθ=0\dfrac{dy}{d\theta} = 0 to find the tangent to the initial line, I get:

a(cosθ+cos2θ−sin2θ)=0a(cos\theta + cos^2\theta - sin^2\theta) = 0

⇒2cos2θ+cosθ−1=0\Rightarrow 2cos^2\theta + cos\theta - 1 = 0

⇒(2cosθ−1)(cosθ+1)=0\Rightarrow (2cos\theta - 1)(cos\theta + 1) = 0

So:

cosθ=12⇒θ=±π3cos\theta = \frac{1}{2} \Rightarrow \theta = \pm \dfrac{\pi}{3} and so r=3a2 r = \dfrac{3a}{2}

or:

cosθ=−1⇒θ=±πcos\theta = -1 \Rightarrow \theta = \pm \pi and so r=0 r = 0

And so the tangents to the initial line are at:

(3a2,±π3)(\dfrac{3a}{2}, \pm\dfrac{\pi}{3}) and (0,±π)(0, \pm \pi)

However my book says the tangents are at:

(3a2,±π3)(\dfrac{3a}{2}, \pm\dfrac{\pi}{3}) and (0,π)(0, \pi)

How come I can't have θ=−π\theta = -\pi? Is this true for all polar equations?

Thank you :biggrin:
Original post by so it goes


How come I can't have θ=−π\theta = -\pi? Is this true for all polar equations?



Hint: What's the domain of theta?

If in doubt see here
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so it goes
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Original post by ghostwalker
Hint: What's the domain of theta?

If in doubt see here


Right, thank you, I think I get it. So because my r is 0, I just express it as (0,π)(0, \pi) because that's basically the same as (0,−π)(0, -\pi)?

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