The conditions for parameterisation

Question

I have proved that $\gamma(t) = (1-cost, tant-sint)$ satisfies the equation for the conchoid $(x-1)^2(x^2+y^2)=x^2$. But is there any reason why this is not a parameterisation? How do I have to restrict the parameter $t$ to get a parameterisation for each branch of the curve?

The graph of the conchoids is produced here, which is the union of two disjoint connected curves.

What I can think about is regarding the domain of $t$. In this problem, considering $sint$, $cost$ and $tant$, $t$ shouldn't be equal to $\frac{\pi}{2}+k\pi$, where $k$ is an integer. Hence, neither $sint$ nor $cost$ can reach their maxima and minima. So $1-cost≠1$, while $tant-sint$ tends to be infinity and negative infinity from different directions, which seems satisfying the graph of the given conchoid.

I am a bit confused here. Thanks in advance. :)

Answer 1

Usually, by a plane curve we mean a continuous function $\gamma \colon I \to \mathbb R^2$ where $I \subseteq \mathbb R$ is an interval.

If $\gamma$ is defined by $\gamma(t) = (1 - \cos t, \tan t - \sin t)$, there are two natural choices for the interval:

• if we choose $I_1 = \left ( - \frac \pi 2, \frac \pi 2 \right )$ we obtain the branch of the conchoid containing the singular point $(0, 0)$, because as $t$ goes from $- \frac \pi 2$ to $0$ the abscissa $(1 - \cos t)$ goes from $1$ to $0$, and then as $t$ goes from $0$ to $\frac \pi 2$ the abscissa goes from $0$ to $1$;
• if we choose $I_2 = \left ( \frac \pi 2, \frac 3 2 \pi \right )$ by a similar reasoning we obtain the other branch.

Any other interval having length $\pi$ will give you the same points given by $I_1$ or $I_2$ because $\gamma$ has period $2 \pi$ and as you already said $\frac \pi 2 + k \pi$ must be excluded from the domain for any $k \in \mathbb Z$.