Q.5 - Calculus

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. 

$x=t^2,y=t^3,t\in \mathrm{ℝ}$

Step-by-Step Solution

Answer

As a consequence, the curve's motion is up and to the right when $t>0$, and up and to the left when $t<0$

For this reason, the correct answer is up and to the left.$t<0$,up and to the right when $t>0$.

Step 1 : Given information

Consider the equations that are parametric. $x=t^2,y=t^3,t\in \mathrm{ℝ}$

step 2: calculation

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

Take the function$x=t^2$

Differentiate with respect to$t$.

Then width="93" style="max-width: none; vertical-align: -47px;" $$\begin{gathered} x^1\left(t\right)=\frac{d}{dt}\left(t^2\right) \\ =2t \\ {x}^{\text{'}}\left(t\right)>0 \end{gathered}$$ 

step 3 : further calculation

Now take the function $y=t^3$

Differentiate with respect to t

$$\begin{gathered} y^{\text{'}}\left(t\right)=\frac{d}{dx}\left(3t^2\right) \\ =6t \\ {y}^1\left(t\right)=6t>0 \end{gathered}$$

Thus, $x^{\text{'}}\left(t\right)>0$and $y^{\text{'}}\left(t\right)>0.$