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

`y = - ( √(3) )/(3)x`

`{x}^(2) + {y}^(2) = 1`

Explanation:

You are correct with the graph, the angle

`(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 = (5\pi)/(6)`

`\tan( \alpha ) = \tan( (5\pi)/(6) )`

`(y)/(x) = - ( √(3) )/(3)`

`y = - ( √(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 ) = (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`