Chapter 10, Problem 12CT

To determine

The graph of the curve of a parametric equation $x=3t-2$, $y=1-\sqrt{t};0\le t\le 9$, show its orientation. Find the rectangular equation of curve of the parametric equations.

Answer to Problem 12CT

Solution:

The graph of the parametric equations $x=3t-2$, $y=1-\sqrt{t};0\le t\le 9$ is

  

The rectangular equation of the parametric equation $x=3t-2,y=1-\sqrt{t};0\le t\le 9$ is $y=1-\sqrt{\frac{x+2}{3}},-2\le x\le 25$ .

Explanation of Solution

Given information:

The parametric equation is $x=3t-2,y=1-\sqrt{t};0\le t\le 9$

To draw the given parametric equation, plug some values of $t$ in the given equation and find few points on the curve.

The interval for $t$ is $0\le t\le 9$

  $\begin{array}{cccc}t& x\left(t\right)& y\left(t\right)& \left(x,y\right)\\ 0& -2& 1& \left(-2,1\right)\\ 1& 1& 0& \left(1,0\right)\\ 4& 10& -1& \left(10,-1\right)\\ 9& 25& -2& \left(25,-2\right)\end{array}$

The orientation is curve traced out in a certain direction by the corresponding succession of points $\left(x,y\right)$ .

The arrow shows the orientation along the curve as $t$ varies from $0\le t\le 9$ .

As $t$ takes values from $0\le t\le 9$, the corresponding orientation is from $\left(-2,1\right)$ to $\left(25,-2\right)$ .

Use points in the table to sketch the curve.

The graph of the parametric equation $x=3t-2,y=1-\sqrt{t};0\le t\le 9$ is

  

Let $x=3t-2$ --------- (1)

  $y=1-\sqrt{t}$ ----------- (2)

Solve equation (1) for variable $t$,

  $\Rightarrow 3t=x+2$

  $\Rightarrow t=\frac{x+2}{3}$

Substitute this result for $t$ into equation (2),

  $\Rightarrow y=1-\sqrt{\frac{x+2}{3}},-2\le x\le 25$

The rectangular equation of the parametric equation $x=3t-2,y=1-\sqrt{t};0\le t\le 9$ is $y=1-\sqrt{\frac{x+2}{3}},-2\le x\le 25$ .