Converting Cartesian $(4,-3)$ to Polar Coordinates:

Question

Given $0 \le \phi < 2\pi$, and $r \le 0$, convert the cartesian coordinate $(4, -3)$ to polar coordinates

I'll use a more standard conversion between these 2 coordinate systems to show how I attempt these questions. For example, Converting $(3, \sqrt{3})$ to polar coordinate

I found the value $r = -6 $ such that it satisfies $r \le 0$, and the point is in the 1st quadrant which meant that its angle when $r \ge 0$ is between $0$ and $\frac{\pi}{2}$. Thus, to find the equivalent point with $r \le 0$, I use the ray that points in the opposite direction which means adding $\pi$ to the 1st quadrant angle.

From here, I find the angle $\phi$ between $\pi$ and $\frac{3\pi}{2}$ such that $\tan \phi = \frac{3\sqrt{3}}{3}$

Now, coming to the question I posted,

I found that $r=-5$ to satisfy $r \le 0$.

Now, to find the $\phi$ portion from polar coordinate, I understand that $(4,-3)$ is in the 4th quadrant.

Now, my teacher told us that because $\frac{-3}{4} $ is not a known unit circle value, the angle will need to be written in terms of the inverse tangent function, $\tan^ {-1} \left(\frac{-3}{4}\right)$.

Why is solving this so different? What does it mean by "not a known unit circle value"?

So, I just carried on doing with what he told us. Knowing that the range of inverse tangent is between $\pi/2$ and $\frac{\pi}{2}$. And since the point lies in the 4th quadrant, $\tan^ {-1} \left(\frac{-3}{4}\right)$ is an angle between $\frac{\pi}{2}$ and $0$ and I will then need to add $\pi$ to the quadrant $4$ angle to obtain the $\phi$ portion of polar coordinate.

However, I did all these with zero understanding. How do I understand or visualise this?

Answer 1

Well it's known, namely its $\tan^{-1}(-3/4)$. But what your teacher means is that the angle not a multiple of $\pi/6$.

You said you want to understand polar coordinates. The key here is Pythagorean theorem (so "polar coordinates" serves as an answer to "when do we ever use this"). In polar coordinates, you are allowed to move in two direction, namely a radial direction $r$ and an angular direction $\theta$. How does movement in these two directions compare to the standard $x,y$ ones? If we're at a point $p=(x,y)$ notice the distance from the radius $r$ is the length of line segment from the origin, and furthermore it's the hypothenuse of the right triangle, where the other two sides are given by projecting $p$ onto the x and y axes. Therefore, we have the formula $$r=\sqrt{x^2+y^2}.$$ What about the angular coordinate? Again, consider this right triangle, and let's now call $\theta$ the angle measured from the positive $x$-axis in the counterclockwise direction (this is just convention). In this case, notice that by definition, $\tan\theta=y/x$. Taking inverses solves for $\theta$, namely $$\theta=\tan^{-1}(y/x).$$

This solves $r,\theta$ in terms of $x,y$. In calculus, it turns out that it is important to solve for $x,y$ in terms of $r,\theta$. In that case, the formulas are given by $$x=r\cos\theta,$$ $$y=r\sin\theta.$$ Can you see why that's the case? Hint: consider, once again, right triangles.

Answer 2

You are asking for a visualization. Here is one :

where we will concentrate on $A'$ the symmetric point of $A$ with respect to $O$.

• First of all, the module is length($OA$)=length($OA'$)=$r=\sqrt{3^2+4^2}=5$.

• Then, the polar angle $\varphi$. As you have to turn according to the positive (= anticlockwise) way starting from unit vector $\vec{OI}$ to final vector $\vec{OA'}$ (green angle) (in compliance with constraint $\varphi \in [0, 2 \pi)$). This green angle can be decomposed into a first $\frac{\pi}{2}$ turn to which one adds the angle $\tan^{-1}\frac{BA'}{BO}=\tan^{-1}\frac43$.

So the final answer is

$$[r,\varphi]=[-5,\frac{\pi}{2}+\tan^{-1}\frac43]$$

Remark: Don't pay a special attention to the angle values indicated in degrees : they are just rounded values to the nearest integer degree...