Chapter 9.4, Problem 10E

a.

To determine

To create a table of $x-$ and $y-$ values using the $t=-\frac{\pi }{2},-\frac{\pi }{4},0,\frac{\pi }{2},\frac{\pi }{4}$ for the given parametric equations.

Answer to Problem 10E

  $\begin{array}{cccccc}t& -\frac{\pi }{2}& -\frac{\pi }{4}& 0& \frac{\pi }{4}& \frac{\pi }{2}\\ x& 0& 2& 4& 2& 0\\ y& -4& -2\sqrt{2}& 0& 2\sqrt{2}& 4\end{array}$

Explanation of Solution

Given: The parametric equations $x=4{\cos }^{2}t$ and $y=4\sin t$ .

Calculation:

When $t=-\frac{\pi }{2}$ , $x=4{\cos }^{2}t=4{\cos }^2\left(-\frac{\pi }{2}\right)=4{\left(0\right)}^2=0,y=4\sin t=4\sin \left(-\frac{\pi }{2}\right)=4\left(-1\right)=-4$ .

When $t=-\frac{\pi }{4}$ , $x=4{\cos }^{2}t=4{\cos }^2\left(-\frac{\pi }{4}\right)=4{\left(\frac{\sqrt{2}}{2}\right)}^2=2,y=4\sin t=4\sin \left(-\frac{\pi }{4}\right)=4\left(-\frac{\sqrt{2}}{2}\right)=-2\sqrt{2}$ .

When $t=0$ , $x=4{\cos }^{2}t=4{\cos }^2\left(0\right)=4{\left(1\right)}^2=4,y=4\sin t=4\sin \left(0\right)=4\left(0\right)=0$ .

When $t=\frac{\pi }{2}$ , $x=4{\cos }^{2}t=4{\cos }^2\left(\frac{\pi }{2}\right)=4{\left(0\right)}^2=0,y=4\sin t=4\sin \left(\frac{\pi }{2}\right)=4\left(1\right)=4$ .

When $t=\frac{\pi }{4}$ , $x=4{\cos }^{2}t=4{\cos }^2\left(\frac{\pi }{4}\right)=4{\left(\frac{\sqrt{2}}{2}\right)}^2=2,y=4\sin t=4\sin \left(\frac{\pi }{4}\right)=4\left(\frac{\sqrt{2}}{2}\right)=2\sqrt{2}$ .

So, the table becomes as shown below.

  $\begin{array}{cccccc}t& -\frac{\pi }{2}& -\frac{\pi }{4}& 0& \frac{\pi }{4}& \frac{\pi }{2}\\ x& 0& 2& 4& 2& 0\\ y& -4& -2\sqrt{2}& 0& 2\sqrt{2}& 4\end{array}$

b.

To determine

To plot the points $\left(x,y\right)$ generated in part (a) and graph the parametric equation. Also, describe the orientation of the curve.

Explanation of Solution

Given: The parametric equations $x=4{\cos }^{2}t$ and $y=4\sin t$ .

Plot:

From part (a), the table is given below.

  $\begin{array}{cccccc}t& -\frac{\pi }{2}& -\frac{\pi }{4}& 0& \frac{\pi }{4}& \frac{\pi }{2}\\ x& 0& 2& 4& 2& 0\\ y& -4& -2\sqrt{2}& 0& 2\sqrt{2}& 4\end{array}$

The points become $\left(0,-4\right),\left(2,-2\sqrt{2}\right),\left(4,0\right),\left(2,2\sqrt{2}\right),\left(0,4\right)$ .

Now, plot these points on the coordinate axes as shown below.

  

Now, join the points to graph the parametric equations.

  

The curve is moving anticlockwise.

c.

To determine

To graph the parametric equations by using graphing utility.

Explanation of Solution

Given: The parametric equations $x=4{\cos }^{2}t$ and $y=4\sin t$ .

Graph:

The graph of the parametric equations using graphing utility can be given as shown below.

  

d.

To determine

To find the rectangular equation by eliminating the parameter $t$ . Then, sketch its graph and describe how the graph differs from the graphs in part (b) and (c).

Answer to Problem 10E

  $y^2=16-4x$

Explanation of Solution

Given: The parametric equations $x=\sqrt{t}$ and $y=2-t$ .

Calculation:

For the given parametric equations, $x=4{\cos }^{2}t\Rightarrow {\cos }^{2}t=\frac{x}{4}$ and $y=4\sin t\Rightarrow \sin t=\frac{y}{4}$ .

Now,

  $\begin{array}{c}{\cos }^{2}t+{\sin }^{2}t=1\\ \frac{x}{4}+{\left(\frac{y}{4}\right)}^2=1\\ \frac{x}{4}+\frac{y^2}{16}=1\\ 4x+y^2=16\\ y^2=16-4x\end{array}$

So, the rectangular equation of the given parametric equations is $y^2=16-4x$ .

The graph of this equation can be given as shown below.

  

The graph is same as obtained in part (b) and (c).