Question

Find and prove a formula for the distance between the

points $(r_1,{\theta }_1)$ and $(r_2,{\theta }_1)$ when they are plotted in a polar

coordinate system.

Short answer

The required answer is

$d=\sqrt{r^2{}_1+r_2^2-2r_1{r}_2\cos \left({\theta }_2-{\theta }_1\right)}$

Step-by-step solution

The objective is to find the distance between the points.

Assume that $\left(r_1,{\theta }_1\right)\left(r_2,{\theta }_2\right)$are the polar coordinates of the points $P,Q$

The points are $r_1=r_1\cos {\theta }_1,{\theta }_1=r_1\sin {\theta }_1$and $r_2=r_2\cos {\theta }_2,{\theta }_2=r_2\sin {\theta }_2$

The distance formula is as follows:

$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

The distance is calculated by substituting the coordinates.

$$\begin{gathered} d=\sqrt{{\left(r_2\cos {\theta }_2-r_1\cos {\theta }_1\right)}^2+{\left(r_2\sin {\theta }_2,-r_1\sin {\theta }_1\right)}^2} \\ \Rightarrow d=\sqrt{{r_2}^2{\cos }^2{\theta }_2-2r_1{r}_2\cos {\theta }_2+{r_1}^2{\cos }^2{\theta }_1+({r_2}^2{\sin }^2{\theta }_2-2r_1{r}_2\sin {\theta }_1\sin {\theta }_2+{r_1}^2{\sin }^2\theta } \\ \Rightarrow d=\sqrt{{r_1}^2+{r_2}^2-2r_1{r}_2(\sin {\theta }_1\sin {\theta }_2+\cos {\theta }_1{\cos }_2)} \\ \Rightarrow d=\sqrt{{r_1}^2+{r_2}^2-2r_1{r}_2(\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2} \end{gathered}$$

Find the distance between the points in the polar form 

On simplification,

$d=\sqrt{r_1^2+r_2^2-2r_1{r}_2\cos \left({\theta }_2-{\theta }_1\right)}\left[\sin ce\left.\cos {\theta }_1\cos {\theta }_2+\sin {\theta }_1\sin {\theta }_2=\cos \left({\theta }_2-{\theta }_1\right)\right]\right.$

The distance between the points in the polar form is:

$d=\sqrt{r_1^2+r_2^2-2r_1{r}_2\cos \left({\theta }_2-{\theta }_1\right)}$

Hence the required answer is

$d=\sqrt{r^2{}_1+r_2^2-2r_1{r}_2\cos \left({\theta }_2-{\theta }_1\right)}$