Find parametric equations that trace out the ellipse $$\begin{equation*} \frac{x^{2}}{4}+y^{2}=1 \end{equation*}$$

where the parameter $t$ is time (in seconds) and:

(a) The motion around the ellipse is counterclockwise, begins at the point $(2,0)$, and requires $1$ second for a complete revolution.

(b) The motion around the ellipse is clockwise, begins at the point $(0,1)$ , and requires $2$ seconds for a complete revolution.

Solution (a) Figure 8 shows the graph of the ellipse $\dfrac{ x^{2}}{4}+y^{2}=1$.

$\dfrac{x^{2}}{4}+y^{2}=1$

Since the motion begins at the point $( 2,0)$, we want $x=2$ and $y=0$ when $t=0$. We let $$\begin{equation*} x( t) =2\cos ( \omega t) \qquad\hbox{and}\qquad y( t) =\sin ( \omega t) \end{equation*}$$

for some constant $\omega$. This choice for $x=x( t)$ and $y=y( t)$ satisfies the equation $\dfrac{x^{2}}{4}+y^{2}=1$ and also satisfies the requirement that when $t=0$, then $x=2$ and $y=0$.

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For the motion to be counterclockwise, as $t$ increases from $0$, the value of $x$ must decrease and the value of $y$ must increase. This requires $\omega > 0$. [Do you see why? If $\omega > 0$, then $x=2\cos ( \omega t)$ is decreasing when $t > 0$ is near zero, and $y=\sin ( \omega t)$ is increasing when $t > 0$ is near zero.]

Finally, since $1$ revolution takes $1$ second, the period is $\dfrac{2\pi }{ \omega }=1$, so $\omega =2\pi$. Parametric equations that satisfy the conditions given in (a) are $$\begin{equation*} x( t) =2\cos ( 2\pi t) \qquad y( t) =\sin ( 2\pi t)\;0\leq t\leq 1 \end{equation*}$$

(b) Figure 9 shows the graph of the ellipse $\dfrac{x^{2}}{4}+y^{2}=1$.

$\dfrac{x^{2}}{4}+y^{2}=1$

Since the motion begins at the point $( 0,1)$, we want $x=0$ and $y=1$ when $t=0$. We let $$\begin{equation*} x( t) =2\sin ( \omega t) \qquad \hbox{and}\qquad y( t) =\cos ( \omega t) \end{equation*}$$

for some constant $\omega$. This choice for $x=x( t)$ and $y=y( t)$ satisfies the equation $\dfrac{x^{2}}{4}+y^{2}=1$ and also satisfies the requirement that when $t=0$, then $x=2\sin 0=0$ and $y=\cos 0=1$.

For the motion to be clockwise, as $t$ increases from $0$, the value of $x$ must increase and the value of $y$ must decrease. This requires that $\omega > 0$.

Finally, since $1$ revolution takes $2$ seconds, the period is $\dfrac{2\pi }{ \omega }=2$, or $\omega =\pi$. Parametric equations that satisfy the conditions given in (b) are $$x( t) =2\sin ( \pi t) \qquad y( t) =\cos ( \pi t)\;0\leq t\leq 2$$