Question

How do I graph cardioid $$r=a(1-\cos \theta )$$?

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
$$r=a(1-\cos \theta )$$
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 $$0,{\displaystyle \frac{\pi }{2}},\pi ,{\displaystyle \frac{3\pi }{2}},2\pi$$.
To draw the graph let us assume value of a as $$2$$

$$\theta$$	$$r=a(1-\cos \theta )$$
$$0$$	$$0$$
$${\displaystyle \frac{\pi }{2}}$$	$$2$$
$$\pi$$	$$4$$
$${\displaystyle \frac{3\pi }{2}}$$	$$2$$
$$2\pi$$	$$0$$

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 $$r=a(1-\cos \theta )$$.

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.