Find polar coordinates $(r, \theta)$ of the point, where $r > 0$ and $0 \leq \theta < 2\pi$

Question

Given these Cartesian coordinates: $(2,-3)$

This is my fourth problem of this type, I solved the other 3, but this one has weird numbers and I don't know what to do.

$$\tan\theta = -\frac{3}{2}$$

what would $\theta$ be? The number isn't convenient, if that makes sense. I don't know how to calculate it. I tried entering that into google and it just said it was $.98$ radians. That didn't work as the answer.

As for finding $r$, since I need $(r,\theta)$

I did $$x^2 + y^2 = r^2\implies 4 + 9 = r^2\Longrightarrow r^2=13\Longrightarrow r=\sqrt{13}$$

I'm pretty sure that's right but I can't check it because I can't find the accompanying theta value to submit the answer.

Answer 1

When working with polar coordinates you must use two relations:
$$r = \sqrt{x^2 + y^2}$$ $$\theta = \arctan(y/x)$$ In the case on hand, these relations give us the point: $$(r,\theta)=(\sqrt{13}, -56.3º)$$ You can check its correctness by the relation: $$(x,y)=(r*\cos(\theta), r*\sin(\theta))$$ You can always calculate the $arctan(y/x)$ without the sign and add the necessary quantity to displace the angle to the quadrant it's placed on; in our case, you just need to change the sign of the angle in question.

Hope this helps.

Answer 2

You correctly found the value of $r$. It is also true that $\tan\theta = -3/2$. One possible value of $\theta$ is $$\theta = \arctan\left(-\frac{3}{2}\right)$$ However, $$\arctan x: (-\infty, \infty) \to \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$ and $$-\frac{3}{2} < 0 \implies \arctan\left(-\frac{3}{2}\right) \in \left(-\frac{\pi}{2}, 0\right)$$ so simply taking the arctangent of $-3/2$ does not yield a value for $\theta$ in the desired interval. To find $\theta \in [0, 2\pi)$ such that $\tan\theta = -3/2$, we can use the periodicity of the tangent function. Since $f(x) = \tan x$ has period $\pi$, the general solution to the equation $\tan\theta = -3/2$ is $$\theta = \arctan\left(-\frac{3}{2}\right) + n\pi, n \in \mathbb{Z}$$ What you need to do is determine the values of $n \in \mathbb{Z}$ such that $$\theta = \arctan\left(-\frac{3}{2}\right) + n\pi \in [0, 2\pi)$$

Answer 3

The two angles are $\tan^{-1} 1.5 = -56.3, 180 -56.3 $ plus $2 k \pi$