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.
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 \(\dfrac{x^{2}}{4}+y^{2}=1\).
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 \[ x( t) =2\sin ( \pi t) \qquad y( t) =\cos ( \pi t)\;0\leq t\leq 2 \]
The period of a sinusoidal graph is discussed in Section P. 6, p. 50.