I have this question and it seems like I am stuck somewhere. Could someone help me with this? It involves parametric curves with the parameter of the curve in terms of $\pi$.
Find the cartesian equation of the curve, sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases. [Verify using Mathematica]
i). $\;\;\;x = \sin \frac{1}{2}\theta, \;\; y = \cos \frac{1}{2}\theta, \;\; -\pi \leq \theta \leq \pi.$
ii). $\;\;x = \frac{1}{2}\cos \theta, \;\; y = 2 \sin \theta, \;\; 0 \leq \theta \leq \pi.$
Here is one approach for part (i):
$x^{2}=\sin^{2}{\frac{\theta}{2}}$, $y^{2}=\cos^{2}{\frac{\theta}{2}}$
$x^{2}+y^{2}=\sin^{2}{\frac{\theta}{2}}+\cos^{2}{\frac{\theta}{2}}=1$
When $\theta=-\pi$, $x=\sin{\frac{-\pi}{2}}=-1$ and when $\theta=\pi$, $x=\sin{\frac{\pi}{2}}=1$ so the domain is $[-1,1]$
For part (ii), we need $\sin{x}$ and $\cos{x}$ to both have the same coefficient, so start by writing
$2x=\cos{\theta}$ and $\frac{y}{2}=\sin{\theta}$
Then $(2x)^{2}+(\frac{y}{2})^{2}=\sin^{2}{\theta}+\cos^{2}{\theta}=1$
So $4x^{2}+\frac{y^{2}}{4}=1$
The domain can be determined from the equation for $x$ in a similar way to part (i).
This is only one possible approach. There are others.
I will leave the direction of the curve and the verification using Mathematica up to you, since this looks a lot like homework...
(Hint though for Mathematica, the Manipulate command can be very useful here...)