Question

How do you graph $r=2\sin \theta$ ?

Answer 1

Hint: The given equation is in the polar form of a conic section. To graph this equation on a cartesian plane we have to convert this equation into standard form or rectangular form. To convert this equation we can use the following conversions,
 $$\begin{gathered} x=r\cos \theta \\ y=r\sin \theta \end{gathered}$$
where $\theta$ is the angle that the line joining origin and the general point makes with the x-axis and $r$ is the magnitude or the distance of the point from origin given as $r^2=x^2+y^2$ .

Complete step by step solution:
We have been given to graph the equation $r=2\sin \theta$ .
Since this equation includes the angle $\theta$ and the magnitude $r$ , this is in the polar form. To graph this equation we will first convert it into standard form or rectangular form such that we get an equation in terms of $x$ and $y$ . We can use $x=r\cos \theta ,y=r\sin \theta and{r}^2=x^2+y^2$ .
From $y=r\sin \theta$ , we have $\sin \theta ={\displaystyle \frac{y}{r}}$ .
From $r^2=x^2+y^2$ , we have $r=\sqrt{x^2+y^2}$
Thus the given equation becomes,
 $$\begin{gathered} r=2\sin \theta \\ \Rightarrow \sqrt{x^2+y^2}=2{\displaystyle \frac{y}{r}}={\displaystyle \frac{2y}{\sqrt{x^2+y^2}}} \\ \Rightarrow x^2+y^2=2y \end{gathered}$$
We have the coefficient of $$x^2$$ is equal to the coefficient of $$y^2$$. So this equation will represent a circle in the cartesian plane.
We can simplify the equation as,
 $$\begin{gathered} x^2+y^2=2y \\ \Rightarrow x^2+y^2-2y=0 \\ \Rightarrow x^2+y^2-2y+1=1 \\ \Rightarrow x^2+{\left(y-1\right)}^2=1^2 \end{gathered}$$
Thus the circle is centered at the point $\left(0,1\right)$ and the radius is $1$ .
This can be drawn on the graph as follows,

Hence, this is the graph of the given equation. Here $r$ is the distance of the general point on the curve from origin and $\theta$ is the angle that the line joining the origin and the general point will make with the x-axis.

Note: We converted the given polar form of the equation into the rectangular form to graph the equation in a cartesian plane. The rectangular form is given in terms of $x$ and $y$ . After the conversion we have to determine what type of curve this equation represents or else we have to find the critical points using the derivatives.