Chapter 10.7, Problem 26AYU

To determine

The curve of the parametric equation $x=t^2,y=\ln t;t>0$, show its orientation. Find the rectangular equation of curve.

Answer to Problem 26AYU

Solution:

The graph of the parametric equations $x=t^2,y=\ln t;t>0$ is,

The rectangular equation of the parametric equation $x=t^2,y=\ln t;t>0$ is $y=\frac{1}{2}\ln x$.

Explanation of Solution

Given information:

The parametric equation is $x=t^2,y=\ln (t)$ ;$t>0$

Explanation:

To draw the given parametric equation, plug some values of $t$ in the given equation and find few points on the curve.

The interval for $t$ is $t>0$

$\begin{array}{cccc}t& x=t^2& y=\ln (t)& \left(x,y\right)\\ 1& 1& 0& \left(1,0\right)\\ 2& 4& \ln (2)& \left(4,\ln (2)\right)\\ 3& 9& \ln (3)& \left(9,\ln (3)\right)\end{array}$

The orientation is the curve traced out in a certain direction by the corresponding succession of points $\left(x,y\right)$.

The arrow shows the orientation along the curve for $t>0$

As $t$ takes values from $1$, the corresponding orientation started is from $\left(1,0\right)$ and continued till $\infty$

Use points in the table to sketch the curve.

The graph of the parametric equation $x=t^2,y=\ln (t)$ ;$t>0$ is

The orientation is from $\left(1,0\right)$ and continued till $\infty$

Let $x=t^2$ and $y=\ln (t)$

$y=\ln (t)\Rightarrow e^y=t$

By substituting the value of $t=e^y$ in $x=t^2$

$\Rightarrow x={\left(e^y\right)}^2=e^{2y}$

$\Rightarrow x=e^{2y}$

$\Rightarrow \ln x=\ln e^{2y}$

$\Rightarrow \ln x=2y$

$\Rightarrow y=\frac{1}{2}\ln x$

The rectangular equation of the parametric equation $x=t^2,y=\ln (t)$ ;$t>0$ is $y=\frac{1}{2}\ln x$