Question

Sketch the parametric curve by plotting points.

$x=t^3-t,y=t^3+t,t\in \mathrm{ℝ}$

Short answer

The required graph is

Step-by-step solution

Given information

The parametric curve is $x=t^3-t,y=t^3+t,t\in \mathrm{ℝ}$

Calculation

Consider the parametric curves $x=t^3-t,y=t^3+t,t\in \mathrm{ℝ}$.

The objective is to sketch the parametric curve.

To draw the graph for the parametric equations assume $t=-2,-1,0,1,2$.

Substitute different t values in the parametric equations and find the values of $x,y$.

The point $(x,y)$When $t=-2$is,

$(x,y)=\left(t^3-t,t^3+t\right)$

$$\begin{gathered} \left.(x,y)=\left((-2{)}^3-(-2),(-2{)}^3+(-2)\right)\text{[since by substituting}t=-2\right] \\ (x,y)=(-8+2,-8-2) \\ (x,y)=(-6,-10)\text{simplify} \end{gathered}$$

The point $(x,y)$When $t=-1$is,

$$\begin{gathered} (x,y)=\left(t^3-t,t^3+t\right) \\ (x,y)=\left((-1{)}^3-1,(-1{)}^3+(-1)\right)[\text{since by substituting}t=-1] \\ (x,y)=(-1-1,-1-1) \\ (x,y)=(-2,-2)\text{simplify} \end{gathered}$$

The point $(x,y)$When $t=0$is,

$$\begin{gathered} (x,y)=\left(t^3-t,t^3+t\right) \\ \left.(x,y)=\left(0^3-0,0^3+0\right)\text{[since by substituting}t=0\right] \\ (x,y)=(0,0) \end{gathered}$$

Further calculation

The point $(x,y)$When $t=1$is,

$$\begin{gathered} (x,y)=\left(t^3-t,t^3+t\right) \\ (x,y)=\left(1^3-1,1^3+1\right)[\text{since by substituting}t=1] \\ (x,y)=(0,2)\text{simplify} \end{gathered}$$

The point $(x,y)$When $t=2$is,

$$\begin{gathered} (x,y)=\left(t^3-t,t^3+t\right) \\ \left.(x,y)=\left(2^3-2,2^3+2\right)\text{[since by substituting}t=2\right] \\ (x,y)=(8-2,8+2)(x,y)=(6,10)\text{simplify} \end{gathered}$$

The tabular representation of the points is as follows,

The graphical representation is shown as below,