Chapter 8, Problem 7T

To determine

To sketch: The graph of the parametric curve $x=3\sin t+3y=2\cos t\left(0\le t\le \pi \right)$

Explanation of Solution

Given: The parametric curve $x=3\sin t+3y=2\cos t\left(0\le t\le \pi \right)$

First make table t , x and y and plot the points on graph.

Taking some random value of t between $\left(0\le t\le \pi \right)$

  $\begin{array}{ccc}t& x& y\\ 0& 3& 2\\ \frac{\pi }{4}& 5.12& 1.41\\ \frac{\pi }{2}& 6& 0\\ \frac{3\pi }{4}& 5.12& -1.41\\ \pi & 3& -2\end{array}$

Plot the points.

  

To determine

To obtain: The parametric curve into rectangular coordinate.

Answer to Problem 7T

Rectangular form: ${\left(\frac{x-3}{3}\right)}^2+{\left(\frac{y}{2}\right)}^2=1$

Explanation of Solution

Given: The parametric curve $x=3\sin t+3y=2\cos t\left(0\le t\le \pi \right)$

The parametric curve, $x=3\sin t+3y=2\cos t\left(0\le t\le \pi \right)$

First isolate the variable t

  $\begin{array}{l}x=3\sin t+3\Rightarrow \sin t=\frac{x-3}{3}\mathrm{...}\left(i\right)\\ y=2\cos t\Rightarrow \cos t=\frac{y}{2}\mathrm{...}\left(ii\right)\end{array}$

Using ( i ) and (ii)

  $\begin{array}{c}{\sin }^{2}t+{\cos }^{2}t=1\\ {\left(\frac{x-3}{3}\right)}^2+{\left(\frac{y}{2}\right)}^2=1\\ \frac{{\left(x-3\right)}^2}{9}+\frac{y^2}{4}=1\end{array}$