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tangent to a cardioid at theta=pi, r=0

Hi,

I've been asked to find the points on the cardioid:

r=a(1+cosθ)r = a(1 + cos\theta)

where the tangents are perpendicular to the initial line.

So I said:

x=rcosθx = rcos\theta

So x=a(1+cosθ)cosθx = a(1 + cos\theta ) cos\theta

⇒x=a(cosθ+cos2θ)\Rightarrow x = a(cos\theta + cos^2\theta)

⇒dxdθ=−asinθ(1+2cosθ)\Rightarrow \dfrac{dx}{d\theta} = -asin\theta(1 + 2cos\theta)

So when dxdθ=0\dfrac{dx}{d\theta} = 0

sinθ=0⇒θ=0sin\theta = 0 \Rightarrow \theta = 0 or θ=π\theta = \pi

and cosθ=−12⇒θ=±2π3cos\theta = -\frac{1}{2} \Rightarrow \theta = \pm\dfrac{2\pi}{3}

So I thought that my coordinates would be:

(2a,0)(2a, 0)

(0,π)(0, \pi)

(a2,2π3)(\dfrac{a}{2}, \dfrac{2\pi}{3})

(a2,−2π3)(\dfrac{a}{2}, -\dfrac{2\pi}{3})

However, my book doesn't agree on (0,π)(0, \pi) being a coordinate on the graph who's tangent is perpendicular to the initial line. When I look at the graph of the cardioid I agree with them, but I can't see where I've gone wrong with the maths. Is it to do with the fact that dydθ=0\dfrac{dy}{d\theta} = 0 at this point as well? Why would it have this effect?

Thank you :biggrin:
It might be because dydx\frac{dy}{dx} is undefined at that point, but if you're reading this and have a better answer, please say so because I've been wondering about this very question too!

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