How do I graph $r = 3 - 3\sin \theta$?

Question

How can I draw the graph of $$r = 3 - 3\sin \theta$$ in polar coordinates? I know it's a Cardioid, but I'm not sure how to get the graph step by step.

Answer 1

The cardioid cannot be represented as graph of an implicit function. However if we consider only the region in the first quadrant, we can represent the graph as the graph of an imlplicit function

we will have $sin\theta = {y\over \sqrt{x^2+y^2}}$ and so you'll get $$\sqrt{x^2+y^2} = 3 - 3{y\over \sqrt{x^2+y^2}}$$ i.e. $${x^2 + y^2\over 3} + y = \sqrt {x^2 + y^2}$$ squaring you' ll get $${{(x^2+y^2)}^2\over 9} + {2y(x^2+y^2)\over 3} = x^2$$ which you can plot using some graph plotter

Answer 2

The engineer's approach

The simplest approach (for graphing) is the following

$$x(\theta)=r(\theta)\cos(\theta)=3(1-\sin(\theta))\cos(\theta)$$ and $$y(\theta)=r(\theta)\sin(\theta)=3(1-\sin(\theta))\sin(\theta).$$ Now, list the values of $\theta$ from $0$ to $2\pi$, say, in steps of $0.05$ and put marks at the points determined by $(x(\theta),y(\theta)).$

With an Excel you can do the graph in a second:

Of course, I appreciate the other answer showing that our curve not only looks like a cardioid but it actually is a cardioid.

Answer 3

Let's look at $r=1-\sin\theta$ the curve we're looking for is obtained by a dilation of a factor $3$.

We first notice that $r(\pi-\theta)=r(\theta)$ and therefore the curve is invariant by a y-axis reflection. We can limit our study to the interval $[-\pi/2,\pi/2]$

Next we note that $r(-\pi/2)=2$, $r(0)=1$ and $r(\pi/2)=0$ and therefore the curve passes through $(-2,0)$, $(1,0)$ and $(0,0)$ and at the origin the tangent is the vertical positive semi axis.

At any point of the curve a tangent vector is $(dr/d\theta,r)$ in the local reference frame and $dr/d\theta=-\cos\theta$. This has the following consequences.

It confirms that for $\theta=\pi/2$ we have a singularity and we have a cusp point.

For $\theta=-\pi/2$ the tangent vector in the local frame is $(0,2)$ so the tangent is horizontal in the $x,y$ frame and points to the $x\geq 0$

For $\theta=0$ the tangent vector in the local frame is $(-1,1)$ and the tangent is parallel to the second diagonal $y=-x$.

Let's look for other points where the tangent is horizontal. In the local frame a tangent vector is $(-\cos\theta,1-\sin\theta)$, and it transforms into $(-\cos{2\theta}-\sin\theta,-\sin{2\theta}+\cos\theta)$. The tangent is horizontal when

$$\sin{2\theta}=\cos{\theta}=\sin{{\pi\over 2}-\theta}$$

And this is solved with $2\theta=\pi/2-\theta$ i.e. $\theta =\pi/6$ or $2\theta=-\pi/2+\theta$ i.e. $\theta=-\pi/2$ consistent with what we have already.

The tangent is vertical when

$$\cos{2\theta}=-\sin{\theta}=\cos{{\pi\over 2}+\theta}$$

And this is solved for $2\theta=\pi/2+\theta$ i.e $\theta=\pi/2$ consistent with what we had before or $2\theta=-\pi/2-\theta$ i.e $\theta=-\pi/6$.

And this is enough to draw a cardioid with the y- axis as a symmetry axis