How to show that the limaçon has only two vertices.

Question

Question: Show that the limaçon has only two vertices.

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I researched what is limaçon. And I reached the following result;

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Note that I only know that The limaçon is the parametrized curve $$\gamma(t) = ((1 + 2\cos t)\cos t, (1 + 2\cos t)\sin t)$$ for $t \in \Bbb R$.

After this, I calculated $$\dot{\gamma(t)}= (-\sin t -4\cos t \sin t, \cos t +2\cos(2t)).$$

Now I need to find the points at which $\dot{\gamma (t)}=0$ .

How can I find these points? And after I find, what do I need to do in order to conclude that it has only two vertices.

I will be happy if someone show this.

Answer 1

Given the definition Ted mentioned in the comments, the vertices of a parameterised curve are the points of a curve at which the derivative of the curvature is zero (ie a critical point of curvature), and so your first action should be to calculate the curvature of $\gamma(t)$. You can do this by one of several ways, either from the definition of curvature, by reparametising the curve in terms of arc length, or by using any of the various other formulae for curvature in terms of a parametisation. I would personally use $$\kappa(t)=\frac{\|\gamma'(t)\times\gamma''(t)\|}{\|\gamma'(t)\|^3}.$$

Now you need to show that $\kappa(t)$ has two distinct critical points $t_1$ and $t_2$, at which $\kappa'(t_i)=0$.