Question

Sketch a graph of the polar equation. how do I do the second one?
Express the equation in rectangular coordinates. (Use the variables x and y.) And the other one. on the next picture.

Answer 1

Answer:

$y = - \frac{ \sqrt{3} }{3}x$

${x}^{2} + {y}^{2} = 1$

Step-by-step explanation:

You are correct with the graph, the angle

$\frac{5\pi}{6}$

should lies in the second quadrant so the polar graph will lie in the second quadrant then go to the fourth quadrant. It would be a line not a circle.

For the equation, use the tangent equation because it has y and x in it

$\alpha = \frac{5\pi}{6}$

$\tan( \alpha ) = \tan( \frac{5\pi}{6} )$

$\frac{y}{x} = - \frac{ \sqrt{3} }{3}$

$y = - \frac{ \sqrt{3} }{3} x$

For the next picture, the graph is correct,

We know that

$\cos {}^{2} (3t) + \sin {}^{2} ( {}^{}3t ) = 1$

${x}^{2} + {y}^{2} = 1$

Alternate Way:

$x = \cos(3t)$

$\frac{ \cos {}^{ - 1} (x) }{3} = t$

$y = \sin(3( \frac{ \cos {}^{ - 1} (x) }{3} )$

$y = \sin( \cos {}^{ - 1} (x) )$

Here we must imagine a triangle such that an angle has a hypotenuse 1, and a leg x, by definition the missing side is

$\sqrt{1 - {x}^{2} }$

by using the Pythagorean theorem.

$\cos( \alpha ) = \frac{x}{1}$

so

$\sin( \alpha ) = \frac{1 - \sqrt{ {x}^{2} } }{1}$

$y = \sqrt{1 - {x}^{2} }$

${y}^{2} = 1 - {x}^{2}$

${x}^{2} + {y }^{2} = 1$