Chapter 10.7, Problem 24AYU

To determine

The graph of curve of the parametric equation $x=\csc t,y=\cot t;\frac{\pi }{4}\le t\le \frac{\pi }{2}$, show its orientation. Find the rectangular equation of curve.

Answer to Problem 24AYU

Solution:

The graph of the parametric equations $x=\csc t,y=\cot t;\frac{\pi }{4}\le t\le \frac{\pi }{2}$ is,

The rectangular equation of the parametric equation $x=\csc t,y=\cot t;\frac{\pi }{4}\le t\le \frac{\pi }{2}$ is $x^2-y^2=1,1\le x\le \sqrt{2},0\le y\le 1$.

Explanation of Solution

Given information:

The parametric equation is $x=\csc t,y=\cot t;\frac{\pi }{4}\le t\le \frac{\pi }{2}$.

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 $\frac{\pi }{4}\le t\le \frac{\pi }{2}$.

$\begin{array}{cccc}t& x\left(t\right)& y\left(t\right)& \left(x,y\right)\\ \frac{\pi }{4}& \frac{2}{\sqrt{2}}& 1& \left(\frac{2}{\sqrt{2}},1\right)\\ \frac{\pi }{3}& \frac{2}{\sqrt{3}}& \frac{1}{\sqrt{3}}& \left(\frac{2}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)\\ \frac{\pi }{2}& 1& 0& \left(1,0\right)\end{array}$

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 $\frac{\pi }{4}\le t\le \frac{\pi }{2}$.

As $t$ takes values from $\frac{\pi }{4}\le t\le \frac{\pi }{2}$, the corresponding orientation is from $\left(\frac{2}{\sqrt{2}},1\right)$ to $\left(1,0\right)$

Use points in the table to sketch the curve.

The graph of the parametric equation $x=\csc t,y=\cot t;\frac{\pi }{4}\le t\le \frac{\pi }{2}$ is,

The orientation of the given curve is counter-clockwise.

Let $x=\csc t,y=\cot t$

By using Pythagorean identity $1+{\cot }^{2}t={\csc }^{2}t$,

$\Rightarrow 1+y^2=x^2$

$\Rightarrow x^2-y^2=1$

$\Rightarrow \frac{x^2}{1}-\frac{y^2}{1}=1$

$\Rightarrow x^2-y^2=1,1\le x\le \sqrt{2},0\le y\le 1$

The rectangular equation of the parametric equation $x=\csc t,y=\cot t;\frac{\pi }{4}\le t\le \frac{\pi }{2}$ is $x^2-y^2=1,1\le x\le \sqrt{2},0\le y\le 1$.