Chapter 10.7, Problem 7AYU

In Problems $7-26,$ graph the plane curve whose parametric equations are given, and show its orientation. Find the rectangular equation of each curve.

$x(t)=3t+2,y(t)=t+1;0\le t\le 4$

To determine

The graph of the curve of a parametric equation $x=3t+2,y=t+1;0\le t\le 4$, show its orientation. Find the rectangular equation of curve of the parametric equations.

Answer to Problem 7AYU

Solution:

The graph of the parametric equations $x=3t+2,y=t+1;0\le t\le 4$ is

The rectangular equation of the parametric equation $x=3t+2,y=t+1;0\le t\le 4$ is $x-3y+1=0$.

Explanation of Solution

Given information:

The parametric equation $x=3t+2,y=t+1;0\le t\le 4$

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 $0\le t\le 4$

$\begin{array}{cccc}t& x=3t+2& y=t+1& \left(x,y\right)\\ 0& 2& 1& \left(2,1\right)\\ 1& 5& 2& \left(5,2\right)\\ 2& 8& 3& \left(8,3\right)\\ 3& 11& 4& \left(11,4\right)\\ 4& 14& 5& \left(14,5\right)\end{array}$

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 $0\le t\le 4$.

As $t$ takes values from $0\le t\le 4$, the corresponding orientation is from $\left(2,1\right)$ to $\left(14,5\right)$.

Use points in the table to sketch the curve.

The graph of the parametric equation $x=3t+2,y=t+1;0\le t\le 4$ is,

Let $x=3t+2$ and $y=t+1$

$\Rightarrow x-2=3t$ and $t=y-1$

By substituting the value $t=y-1$,

$\Rightarrow x-2=3\left(y-1\right)$

$\Rightarrow x-2=3y-3$

$\Rightarrow x-3y=2-3$

$\Rightarrow x-3y=-1$

$\Rightarrow x-3y+1=0$

The rectangular equation of the parametric equation $x=3t+2,y=t+1;0\le t\le 4$ is $x-3y+1=0$.