Q.7 - 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. 

$x=e^t,y=\ln t,t>0$

Step-by-Step Solution

Answer

The curve rises and goes to the right.

Step: 1 Given information

Consider the parametric equations $x=e^t,y=\ln t,t>0$

Step 2: Calculation

Consider the parametric equations $x=e^{\text{'}},y=\ln t,t>0$

The goal is to examine the parametric equations' direction of motion.

Take the function $x=e^{\text{'}}.$

Differentiate with respect to $\mathit{t}.$

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

$x^{\text{'}}\left(t\right)=e^{\text{'}}>0$

Step 3 Further calculation

Take the function now.$y=\ln t$.

different in terms of$t$.

$$\begin{gathered} y^3\left(t\right)=\frac{d}{dt}\left(\ln t\right) \\ {y}^3\left(t\right)=\frac{1}{t}>0 \end{gathered}$$

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

The value of $t$, the derivatives of the functions $x\left(t\right)$, $y\left(t\right)$are also growing. As a response, the curve goes up and to the right throughout its motion.