Question

How do I graph cardioid \[r=a\left( 1-\cos \theta \right)\]?

Answer 1

Hint: We will use polar coordinates concept to draw the graph. Here we have to generate the table for different known values of \[\theta \] to find r. As we already know that the polar coordinate graph will contain points as \[(r,\theta )\] we will plot the graph. The value of \[a\] can be any random variable.

Complete step by step answer:
Before going to draw the graph lets know about cardioids.
A cardioid is a plane curve traced by a point on the perimeter of a circle that is rolling around a fixed circle of the same radius. It can also be defined as an epicycloid having a single cusp. It is also a type of sinusoidal spiral, and an inverse curve of the parabola with the focus as the center of inversion
For horizontal cardioid the equation will look like
\[r=a\pm a\cos \theta \]
For vertical cardioid the equation will be like
\[r=a\pm a\sin \theta \]
Given equation is
\[r=a\left( 1-\cos \theta \right)\]
We can see it is in the form of a horizontal equation. So we will get a horizontal cardioid from our equation.
So to plot a graph first we have to create a table for the values \[(r,\theta )\] .
For this we will take the \[\theta \] values in the range between \[0\] and \[2\pi \].
\[0\le \theta <2\pi \]
So now will find \[r\] values for these \[\theta \] values \[0,\dfrac{\pi }{2},\pi ,\dfrac{3\pi }{2},2\pi \].
To draw the graph let us assume value of a as \[2\]

\[\theta \]	\[r=a\left( 1-\cos \theta \right)\]
\[0\]	\[0\]
\[\dfrac{\pi }{2}\]	\[2\]
\[\pi \]	\[4\]
\[\dfrac{3\pi }{2}\]	\[2\]
\[2\pi \]	\[0\]

So using these values we can draw the graph of the given equation.
The graph that obtained is

This is the cardioid graph of the equation \[r=a\left( 1-\cos \theta \right)\].

Note: We can see the change in the area and perimeter of the graph by changing the values of \[a\]. We can find area, perimeter, arc length and tangential angle. But if the value of \[a\] we have chosen is not a circle radius then the envelope formed is a limacon.