Chapter 10.7, Problem 16AYU

To determine

The graph of the curve of the parametric equation $x\left(t\right)=e^t,y\left(t\right)=e^{-t};t\ge 0$, show its orientation. Find the rectangular equation of curve of the parametric equation.

Answer to Problem 16AYU

Solution:

The graph of the parametric equations $x\left(t\right)=e^t,y\left(t\right)=e^{-t};t\ge 0$ is

The rectangular equation of the parametric equation $x\left(t\right)=e^t,y\left(t\right)=e^{-t};t\ge 0$ is $xy=1$.

Explanation of Solution

Given information:

The parametric equation, $x\left(t\right)=e^t,y\left(t\right)=e^{-t};t\ge 0$

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 $t\ge 0$

$\begin{array}{cccc}t& x\left(t\right)=e^t& y\left(t\right)=e^{-t}& \left(x,y\right)\\ 0& 1& 1& \left(1,1\right)\\ 1& \text{2}\text{.718}& \text{0}\text{.368}& \left(\text{2}\text{.718},\text{0}\text{.368}\right)\\ 2& \text{7}\text{.389}& \text{0}\text{.135}& \left(\text{7}\text{.389},\text{0}\text{.135}\right)\\ 3& \text{20}\text{.086}& \text{0}\text{.05}& \left(\text{20}\text{.086},\text{0}\text{.05}\right)\\ 4& \text{54}\text{.598}& \text{0}\text{.018}& \left(\text{54}\text{.598},\text{0}\text{.018}\right)\end{array}$

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 as $t$ varies from $t\ge 0$.

As $t$ takes values from $t\ge 0$, the corresponding orientation is from $\left(1,1\right)$ to $\left(\text{54}\text{.59815},\text{0}\text{.01831}\right)$.

Use points in the table to sketch the curve.

The graph of the parametric equation $x\left(t\right)=e^t,y\left(t\right)=e^{-t};t\ge 0$ is

Let $x\left(t\right)=e^t$ and $y\left(t\right)=e^{-t}$

By simplifying,

$\ln \left(x\left(t\right)\right)=t\left(\ln e\right)$

$\Rightarrow \ln \left(x\left(t\right)\right)=t$

$\Rightarrow \ln \left(x\left(t\right)\right)=t$

$\Rightarrow t=\ln \left(x\left(t\right)\right)$

By substituting the value of $t=\ln \left(x\left(t\right)\right)$ in the equation $y\left(t\right)=e^{-t}$

$\Rightarrow y\left(t\right)=e^{-\left(\ln (x(t))\right)}$

$\Rightarrow y=e^{\ln {(x)}^{-1}}$

$\Rightarrow y=x^{-1}$

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

$\Rightarrow xy=1$

The rectangular equation of the parametric equation $x\left(t\right)=e^t,y\left(t\right)=e^{-t};t\ge 0$ is $xy=1$.