Question

help please with 7b

Answer 1

Your polar Coordinates are in a form \(\color{black}{\displaystyle \left(r,\,\theta \right) }\).
In this case, your angle is \(\color{black}{\displaystyle 2\pi/3 }\), and the length is \(\color{black}{\displaystyle 4 }\).

Answer 2

oh yeah i got that one already

Answer 3

hmmm is there a formula for that ?

Answer 4

Suppose \(\color{black}{\displaystyle (x,y)=(\alpha,\beta) }\).
We know that \(\color{black}{\displaystyle r^2=x^2+y^2 }\), from the Pyth. Trm.
Therefore, \(\color{black}{ r=\sqrt{\alpha^2+\beta^2} }\).

You know that \(\color{black}{\displaystyle \tan \theta =x/y }\).
So, in case your points are of a form \(\color{black}{\displaystyle (x,y)=(-\alpha,\alpha)}\),
you have \(\color{black}{\displaystyle \tan \theta =-1 }\), and knowing that the point is in
the 2nd quadrant you add \(\color{black}{\displaystyle \pi }\).

So, you get \(\color{black}{\displaystyle \theta=(-\pi/4)\color{blue}{+\pi} }\),
\(\color{black}{\displaystyle \theta=3\pi/4 }\).

Answer 5

So basically, any points in a form \(\color{black}{\displaystyle (x,y)=(-\alpha,\alpha)}\),
Will be converted as
*) \(\color{black}{ r=\sqrt{(-\alpha)^2+\alpha^2}=\sqrt{2}\alpha \ }\).
*) \(\color{black}{ \theta=3\pi/4 \ }\).

So, it would be \(\color{black}{ (\sqrt{{\tiny~}2}\alpha,\,\,3\pi/4)}\).

Answer 6

When you say any points does that apply numbers? Or radians

Answer 7

Numbers. I am converting cartesian to polar.

Answer 8

Oh okay