Chapter 9, Problem 38CLT

a.

To determine

Draw the graph of the parametric equation.

Explanation of Solution

Given:

It is given in the question that the equations are $x=4\ln t,y=\frac{1}{2}{t}^2$ .

Graph:

  

Interpretation: Here, the two parametric equations are $x=4\ln t,y=\frac{1}{2}{t}^2$ , put that equation in the parametric graphing calculator, it gives the required graph in the proper way.

b.

To determine

Draw the graph by taking a points of $(x,y)$ .

Explanation of Solution

Given:

It is given in the question that the equations are $x=4\ln t,y=\frac{1}{2}{t}^2$ .

Graph:

  

Interpretation:

Since, find the points which satisfies the given equation $x=4\ln t,y=\frac{1}{2}{t}^2$ ,

Now,

Put $t=1$ ,

  $x=4\ln 1=0.000;y=\frac{1}{2}{(1)}^2=0.500$

Put $t=2$ ,

  $x=4\ln 2=2.773;y=\frac{1}{2}{(2)}^2=2$

Put $t=3$ ,

  $x=\cos 3\pi /4=-0.707;y=2{(\sin 3\pi /4)}^2=1$

Put $t=4$ ,

  $x=4\ln 4=5.545;y=\frac{1}{2}{(4)}^2=8$

Put $t=5$ ,

  $x=4\ln 5=6.438;y=\frac{1}{2}{(5)}^2=12.5$

So, the points that obtained are $(0,0.5),(2.7,2),(4.39,4.50),(5.45,8),(6.43,12.5)$ .Put that point in the graphing calculator, and verify it is the same as it comes in (a).

c.

To determine

Find the rectangular equation by eliminating the parameters.

Answer to Problem 38CLT

The rectangular equation is $y=\frac{1}{2}\sqrt{e^x}$ .

Explanation of Solution

Given: It is given in the question that the equations are $x=4\ln t,y=\frac{1}{2}{t}^2$ .

Concept Used:

In this, use the concept of eliminating the parameters and put that in the another equation and get the rectangular equations.

Calculation:

Here, the equations are $x=4\ln t,y=\frac{1}{2}{t}^2$ .

  $\begin{array}{l}x=4\ln t\\ x=\ln t^4(Powerproperty)\\ Exponenting,\\ t^4=e^x\\ t^2=e^{x/2}\end{array}$

Now, substitute the term in other equation:

  $\begin{array}{l}y=\frac{1}{2}{t}^2\\ y=\frac{1}{2}\times e^{x/2}\\ y=\frac{1}{2}\sqrt{e^x}\end{array}$

Conclusion:

The equation is $y=\frac{1}{2}\sqrt{e^x}$ .