Question

In problem graph the curve whose parametric equations are given and show its orientation. Find the rectangular equation of each curve. Verify your graph using a graphing utility.

$x=t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+1,y=\sqrt{t};t\ge 0.$

Short answer

The answer is$x=y^3+1.$

Step-by-step solution

Step 1. Given information.

$x=t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+1,y=\sqrt{t}$is given.

Step 2. Calculation.

Graph the curve defined by the parametric equations -

$t$	$x$	$y$	$(x,y)$
$0$	$0$	$0$	$(0,0)$
$4$	$9$	$2$	$(9,2)$

We find the rectangular equation by eliminating the parameter t from the parametric equations.

$$\begin{gathered} y=\sqrt{t} \\ t=y^2 \end{gathered}$$

Put the value of t in equation-

role="math" localid="1648404939468" $$\begin{gathered} x=t^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+1 \\ x={\left(y^2\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}+1 \\ x=y^3+1 \end{gathered}$$

Thus, the equation isrole="math" localid="1648404969030" $x=y^3+1.$