Question

What is the polar form of `(16,-4`?

Answer 1

`( 4sqrt(17) , -0,245 )`

Explanation:

The quickest method is to use the formula:

`(x , y ) = ( sqrt(x^2+y^2), arctan(y/x))`

Below is an alternative method.

`x=rcostheta`

`y=rsintheta`

`:.`

`16=rcostheta` [ 1 ]

`-4=rsintheta` [ 2 ]

Squaring both equations:

`256=r^2cos^2(theta)`

`16=r^2sin^2(theta)`

Adding both equations:

`272=r^2cos^2(theta)+r^2sin^2(theta)`

`272=r^2(cos^2(theta)+sin^2(theta))`

`272=r^2(1)=>r=+-sqrt(272)=+-4sqrt(17)`

We only need the positive root, negative radii can exist, and a point can be represented by many different polar coordinates, unlike Cartesian coordinates which are unique.

`theta=arctan(y/x)=arctan(-4/16)=arctan(-1/4)=-0.24498`

Polar coordinate:

`( 4sqrt(17) , -0,245 )`