Chapter 10.7, Problem 24AYU

In Problems 7-26, graph the curve whose parametric equations are given and show its orientation. Find the rectangular equation of each curve. Verify your graph using a graphing utility.

$x=\csc t$, $y=\cot t$; $\frac{\pi }{4}\le t\le \frac{\pi }{2}$

To determine

a.

To find: The rectangular equation of the curve.

Answer to Problem 24AYU

$x^2-y^2=1;\text{ 1 }\le x<\sqrt{2},0\le y\le 1$

Explanation of Solution

Given:

$x=\csc t,y=\cot t;\frac{\pi }{4}\le t\le \frac{\pi }{2}$

Calculation:

$x=\csc t,y=\cot t;\frac{\pi }{4}\le t\le \frac{\pi }{2}$

$x^2={\csc }^{2}t$

$x^2=1+{\cot }^{2}t$

$x^2=1+y^2$

$x^2-y^2=1;\text{ 1 }\le x<\sqrt{2},0\le y\le 1$

To determine

b.

To graph: $x=\csc t,y=\cot t;\frac{\pi }{4}\le t\le \frac{\pi }{2}$ verify with the graph of rectangular equation of the curve.

Answer to Problem 24AYU

Explanation of Solution

Given:

$x=\csc t,y=\cot t;\frac{\pi }{4}\le t\le \frac{\pi }{2}$

Graph:

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

The graph of $x^2-y^2=1;\text{ 1 }\le x<\sqrt{2},0\le y\le 1$ is plotted.

Graph 1 represents the graph of $x=\csc t,y=\cot t;\frac{\pi }{4}\le t\le \frac{\pi }{2}$ plotted in parametric mode and the orientation is shown.

The rectangular equation of the curve is calculated. Graph 2 represents the curve of the rectangular equation $x^2-y^2=1,\text{ 1 }\le x<\sqrt{2};0\le y\le 1$. Both the equation represents the same graph and orientation. Hence it is verified.