Question

What is the polar form of `( -4,32 )`?

Answer 1

`(r,theta)=(4sqrt65,1.69515132)`

Explanation:

The rectangular point `(-4,32)` is in the form `(x,y)`.

Polar points are in the form `(r,theta)`. See the attached image for what this means:

To find `r`, we effectively need to find the hypotenuse of the right triangle with legs of `x` and `y`.

Thus, `r=sqrt(x^2+y^2)`. Here, this becomes

`r=sqrt((-4)^2+(32)^2)=sqrt(4^2+32^2)=sqrt(4^2+4^2(8^2))=sqrt(4^2(1+8^2))=4sqrt65`

Even though the point `(-4,32)` is in Quadrant `"II"`, the value of `r` is a magnitude and is still positive.

To find `theta`, we first need to write some statement involving `theta` given the information that we know.

Looking at the image, we have `theta`, the side opposite `theta`, and the side adjacent to `theta` in a right triangle. Thus, we can say:

`tantheta="opposite"/"adjacent"=y/x`

Solving for `theta`:

`theta=tan^-1(y/x)`

Using our known values:

`theta=tan^-1(32/(-4))=tan^-1(-8)=-1.44644133`

Note, however, that this is a negative value and that `-1.44644133gt-pi/2`, so this is really an angle in Quadrant `"IV"`.

To find the value of this angle in Quadrant `"II"`, we know that it will be `pi` minus the magnitude of the angle we determined.

That is,

`theta=pi-1.44644133=1.69515132`

So, our point is:

`(r,theta)=(4sqrt65,1.69515132)`