Chapter 10.7, Problem 11AYU

To determine

The graph of the curve of a parametric equation $x=t^2+4,y=t^2-4;-\infty <t<\infty$, show its orientation. Find the rectangular equation of curve.

Answer to Problem 11AYU

Solution:

The graph of the parametric equations $x=t^2+4,y=t^2-4;-\infty <t<\infty$ is

The rectangular equation of the parametric equation $x=t^2+4,y=t^2-4;-\infty <t<\infty$ is $y=x-8$.

Explanation of Solution

Given information:

The parametric equation, $x=t^2+4,y=t^2-4;-\infty <t<\infty$

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 $-\infty <t<\infty$.

$\begin{array}{cccc}t& x\left(t\right)=t^2+4& y\left(t\right)=t^2-4& \left(x,y\right)\\ -2& 8& 0& \left(8,0\right)\\ -1& 5& -3& \left(5,-3\right)\\ 0& 4& -4& \left(4,-4\right)\\ 1& 5& -3& \left(5,-3\right)\\ 3& 13& 5& \left(13,5\right)\end{array}$

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 $-\infty <t<\infty$.

As $t$ takes values from $-\infty <t<0$, the corresponding orientation is from $\left(8,0\right)$ to $\left(4,-4\right)$ and for $0<t<\infty$, the corresponding orientation is from $\left(5,-3\right)$ to $\left(13,5\right)$.

Use points in the table to sketch the curve.

The graph of the parametric equation $x=t^2+4,y=t^2-4;-\infty <t<\infty$ is

Now, let $x=t^2+4$ and $y=t^2-4$

By solving both the equations,

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

$\Rightarrow x-4=y+4$

$\Rightarrow y=x-8$

The rectangular equation of the parametric equation $x=t^2+4,y=t^2-4;-\infty <t<\infty$ is $y=x-8$.