Question

Use the results of Exercise 3 to analyze the direction of motion for the parametric curves given by the equations in Exercises 5–8

$x=t^3-t,y=t^3+t^{3}t\in \mathrm{ℝ}$

Short answer

The curve moves up and to the right.

Step-by-step solution

Step: 1 Given information

The equations of parameters$x=t^3-t,y=t^3+t^{3}t\in \mathrm{ℝ}$

Calculation

Consider the parametric equations $x=t^3-t,y=t^3+t,t>0$.

The objective is to analyze the direction of motion for the parametric equations.

Take the function $x=t^3-t$.

Differentiate with respect to t.

Then, $x^1\left(t\right)=\frac{d}{dt}\left(t^3-t\right)$

$$\begin{gathered} x^1\left(t\right)=\frac{d}{dt}{t}^3-\frac{d}{dt}t \\ {x}^1\left(t\right)=3t^2-1 \end{gathered}$$

Now take the function $y=t^3+t$.

Differentiate with respect to t.

$y^{\text{'}}\left(t\right)=\frac{d}{dt}\left(t^3+t\right)$

$y^{\text{'}}\left(t\right)=3t^2+1$

Thus, $x^1\left(t\right)>0$and $y^1\left(t\right)>0$.

Further the derivatives of the functions $x\left(t\right),y\left(t\right)$are increasing for the value of t.

As a result, the curve moves up and to the right in its motion.