Why does $r = \cos \theta$ produce a circle?


Solution 1

$$r = \cos \theta$$

$$\sqrt{x^2 + y^2} = \frac{x}{\sqrt{x^2 + y^2}}$$

$$x^2 + y^2 = x$$

$$x^2 - x + \frac{1}{4} + y^2 = \frac{1}{4}$$

$$\left(x - \frac{1}{2}\right)^2 + (y - 0)^2 = \left(\frac{1}{2}\right)^2$$

This is the equation for a circle in Cartesian coordinates $(x, y)$ with center $({1 \over 2}, 0)$ and radius $1 \over 2$.

Solution 2

A bit different then the other answers: $$\begin{align} r &= \cos(\theta) \quad &\Rightarrow \\ r^2 &= r\cos(\theta) \quad&\Rightarrow\\ x^2 + y^2 &= x. \end{align} $$ (And then finish as in the other answers.)

Solution 3

No, there aren't several values of $r$ associated with each value of $\theta$. Remember that the rays are being issued from the origin, not from the center of the circle.

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Solution 4

It is not clear to see actually, because $r=\cos \theta$ is in polar coordinates. If you make a change of coordinates, to rectangular for example this is most clear. Take $x=r\cos \theta$ and $y=r\sin \theta$ then you have $r^2=r\cos\theta$ or $$x^2+y^2=x$$ or $$(x-\frac{1}{2})^2+y^2=\frac{1}{4}$$ which describes a circle.

Solution 5

To see why the graph displays a function, ignoring the origin for the moment, note that the "vertical line test" you may be thinking of is inappropriate for polar functions. Each value of $\theta$ specifies a line through the origin, and it is this line which should only pass through one point on the graph. However this test is still inappropriate, because there are multiple values of $\theta$ which specify the same line through the origin (specifically, $\theta+\pi n$ for every integer $n$), so it is possible for a graph of a polar function to have many values passing through the same line through the origin.

In this particular case, every nonvertical line through the origin passes through two points of the graph: one on the origin, and the other elsewhere. It is the other point that the equation $r = \cos \theta$ specifies. On the vertical line, $\theta = \pi/2+\pi n$, we have $\cos\theta=0$, so this gives the origin.

As for why this gives a circle, the other answers have given algebraic proofs for this. I'm partial to the one Thomas and Luis gave, which multiplies by $r$ first, to give $r^2 = r\cos \theta$, myself. ;)


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Updated on November 06, 2020


  • user1251385
    user1251385 about 3 years

    I am trying to do a double integral over the following region in polar coordinates:

    enter image description here

    I know that the limits of integration are:

    $$\theta = -\frac{\pi}{2} \quad \to \quad \theta = \frac{\pi}{2} \\ r = 0 \quad \to \quad r = \cos \theta$$

    However, I don't understand how $r = 0 \quad \to \quad r = \cos \theta$ works. Cosine is a function (not just a relation) meaning that it has only one value of $r$ for every value of $\theta$. However, it seems like the graph $r = \cos \theta$ has two values of $r$ for every value of $\theta$. Why does $r = \cos \theta$ produce a circle?

    • Ben Grossmann
      Ben Grossmann almost 10 years
      Note that as you go from $0$ to $2\pi$, the bit from $\pi$ to $2\pi$ retraces what you have between $0$ and $\pi$.
    • user1251385
      user1251385 almost 10 years
      @Omnomnomnom If I plot the bit just from $\pi$ to $2\pi$, I get the entire graph: wolframalpha.com/input/…
    • Valent
      Valent almost 10 years
      Apparently all had the same idea to answer.
  • abiessu
    abiessu almost 10 years
    (Where $r=\sqrt{x^2+y^2}$ and $\cos\theta=\frac x{\sqrt{x^2+y^2}}$, using the translation from polar to rectangular coordinates)