Q. 70.

Question

Find and prove a formula for the distance between the

points $(r_{1}, θ_{1})$ and $(r_{2}, θ_{1})$ when they are plotted in a polar

coordinate system.

Step-by-Step Solution

Verified

Answer

The required answer is

d = \sqrt{r^{2} {}_{1}+ r_{2}^{2} - 2 r_{1} r_{2} \cos (θ_{2} - θ_{1})}

1Step 1: The objective is to find the distance between the points.

Assume that $(r_{1}, θ_{1}) (r_{2}, θ_{2})$ are the polar coordinates of the points $P, Q$

The points are $r_{1} = r_{1} \cos θ_{1}, θ_{1} = r_{1} \sin θ_{1}$ and $r_{2} = r_{2} \cos θ_{2}, θ_{2} = r_{2} \sin θ_{2}$

The distance formula is as follows:

$d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

The distance is calculated by substituting the coordinates.

$d = \sqrt{{(r_{2} \cos θ_{2} - r_{1} \cos θ_{1})}^{2} + {(r_{2} \sin θ_{2}, - r_{1} \sin θ_{1})}^{2}} \Rightarrow d = \sqrt{{r_{2}}^{2} \cos^{2} θ_{2} - 2 r_{1} r_{2} \cos θ_{2} + {r_{1}}^{2} \cos^{2} θ_{1} + ({r_{2}}^{2} \sin^{2} θ_{2} - 2 r_{1} r_{2} \sin θ_{1} \sin θ_{2} + {r_{1}}^{2} \sin^{2} θ} \Rightarrow d = \sqrt{{r_{1}}^{2} + {r_{2}}^{2} - 2 r_{1} r_{2} (\sin θ_{1} \sin θ_{2} + \cos θ_{1} \cos_{2})} \Rightarrow d = \sqrt{{r_{1}}^{2} + {r_{2}}^{2} - 2 r_{1} r_{2} (\cos θ_{1} \cos θ_{2} + \sin θ_{1} \sin θ_{2}}$

2Step 2: Find the distance between the points in the polar form

On simplification,

$d = \sqrt{r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} \cos (θ_{2} - θ_{1})} [\sin c e \cos θ_{1} \cos θ_{2} + \sin θ_{1} \sin θ_{2} = \cos (θ_{2} - θ_{1})]$