Chapter 10.7, Problem 14AYU

To determine

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

Answer to Problem 14AYU

Solution:

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

The rectangular equation of the parametric equation $x=2t-4,y=4t^2;-\infty <t<\infty$ is $y={\left(x+4\right)}^2$.

Explanation of Solution

Given information:

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

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 $-\infty <t<\infty$

$\begin{array}{cccc}t& x\left(t\right)=2t-4& y\left(t\right)=4t^2& \left(x,y\right)\\ -3& -10& 36& \left(-10,36\right)\\ -2& -8& 16& \left(-8,16\right)\\ -1& -6& 4& \left(-6,4\right)\\ 0& -4& 0& \left(-4,0\right)\\ 1& -2& 4& \left(-2,4\right)\\ 2& 0& 16& \left(0,16\right)\\ 3& 2& 36& \left(2,36\right)\end{array}$

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 $-\infty <t<\infty$, the corresponding orientation is from $\left(-10,36\right)$ to $\left(2,36\right)$

Use points in the table to sketch the curve.

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

Now, let $x=2t-4$ ——– (1)

$y=4t^2$ ——– (2)

Solve equation (1) for $t$

$\Rightarrow x+4=2t$

$\Rightarrow t=\frac{x+4}{2}$

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

$\Rightarrow y=4{\left(\frac{x+4}{2}\right)}^2$

$\Rightarrow y={\left(x+4\right)}^2$

The rectangular equation of the parametric equations $x=4t-2,y=1-t;-\infty <t<\infty$ is $y={\left(x+4\right)}^2$.