Cartesian to polar conversions

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Homework Statement



find polar coordinates of the points whose cartesian coordinates are given.

Homework Equations



heres the point: (3sqrt(3), 3)

The Attempt at a Solution



well i know that r^2 = (sqrt(a^2 + b^2))
so the answer here is : 6

and if we use tan(theta) = o/a = 3/(3sqrt(3)
so the answer here is: 1/sqrt(3)

so theta is pi/6

so how do i know its pi/6?
how do i convert to get this answer with pi? radians?

help please.
 
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rcmango said:

Homework Statement



find polar coordinates of the points whose cartesian coordinates are given.

Homework Equations



heres the point: (3sqrt(3), 3)

The Attempt at a Solution



well i know that r^2 = (sqrt(a^2 + b^2))
so the answer here is : 6

and if we use tan(theta) = o/a = 3/(3sqrt(3)
so the answer here is: 1/sqrt(3)
The right hand side of that equation, not the "answer" (to what question?!), is [itex]\frac{1}{\sqrt{3}}[/itex]. In general, if you know what [itex]tan(\theta)[/itex] is, you can find [itex]\theta[/itex] by using the arctan function- perhaps on a calculator. Here, you are probably expected to know that [itex]sin(\pi/6)= \frac{1}{2}[/itex] and that [itex]cos(\pi/6)= \frac{\sqrt{3}}{2}[/itex] so that [itex]tan(\pi/6)= \frac{1}{\sqrt{3}}[/itex].

so theta is pi/6

so how do i know its pi/6?
how do i convert to get this answer with pi? radians?

help please.

As benorin said- in polar coordinates the angle is always in radians. As a rule, the only time you use degrees is when the problem specifically involves angle that are given in degrees.