Question

How do I sketch the cardioid whose equation is `r=2-3sintheta`?

Answer 1

Please see below.

Explanation:

Start by plugging in values from `0` to `2pi` for `theta` and compute `r` and plot the value on the coordinate system. You will soon see your graph taking shape.

For example:

if `theta=0` then `r=2-3sin(0)=2-3*(0)=2-0=2`

If `theta=(pi/6)` then `r=2-3sin(pi/6)=2-3*(1/2)=2-3/2=1/2`

If `theta=(pi/4)` then

`r=2-3sin(pi/4)=2-3(sqrt2/2)=(4-3sqrt2)/2=-.12`

If `theta=pi/2` then

`r=2-3sin(pi/2)=2-3*(1)=2-3=-1`

And so on and so forth.

You can see the points on the graph for the above values of `r`

The graph is: