Chapter 10.7, Problem 8AYU

To determine

The graph of the curve of a parametric equation $x=t-3,y=2t+4;0\le t\le 2$, show its orientation. Find the rectangular equation of curve of the parametric equations.

Answer to Problem 8AYU

Solution:

The graph of the parametric equations $x=t-3,y=2t+4;0\le t\le 2$ is

The rectangular equation of the parametric equation $x=t-3,y=2t+4;0\le t\le 2$ is $2x-y+10=0$.

Explanation of Solution

Given information:

The parametric equation $x=t-3,y=2t+4;0\le t\le 2$

Explanation:

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 $0\le t\le 2$

$\begin{array}{cccc}t& x=t-3& y=2t+4& \left(x,y\right)\\ 0& -3& 4& \left(-3,4\right)\\ 1& -2& 6& \left(-2,6\right)\\ 2& -1& 8& \left(-1,8\right)\end{array}$

The orientation is the 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 $0\le t\le 2$.

As $t$ takes values from $0\le t\le 2$, the corresponding orientation is from $\left(-3,4\right)$ to $\left(-1,8\right)$.

Use points in the table to sketch the curve.

The graph of the parametric equation $x=t-3,y=2t+4;0\le t\le 2$ is

Let $x=t-3$ and $y=2t+4$

$\Rightarrow y-4=2t$ and $t=x+3$

By substituting the value of $t=x+3$,

$\Rightarrow y-4=2\left(x+3\right)$

$\Rightarrow y-4=2x+6$

$\Rightarrow 2x-y=-10$

$\Rightarrow 2x-y+10=0$

The rectangular equation of the parametric equation $x=t-3,y=2t+4;0\le t\le 2$ is $y-2x-10=0$.