Question

What is the formula for the distance between two polar coordinates?

Answer 1

Hint: We know that distance between two points in Cartesian coordinates is given by $D=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}$ . We can form this formula into polar form by replacing $x=r\cos \theta $ and $y=r\sin \theta $ . We can further simplify the equation by using $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$ and hence, we will get the required formula.

Complete step-by-step solution:
We know that the distance between any two points $\left( {{x}_{1}},{{y}_{1}} \right)$ and $\left( {{x}_{2}},{{y}_{2}} \right)$ in the Cartesian plane is given by
$D=\sqrt{{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}+{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}}...\left( i \right)$
This is commonly known as the Distance formula.
We know that we can express any point in Cartesian system into polar coordinates by substituting
$x=r\cos \theta $ and $y=r\sin \theta $ .
So, let us assume that ${{x}_{1}}={{r}_{1}}\cos {{\theta }_{1}}$ and ${{y}_{1}}={{r}_{1}}\sin {{\theta }_{1}}$.
And similarly let us assume that ${{x}_{2}}={{r}_{2}}\cos {{\theta }_{2}}$ and ${{y}_{2}}={{r}_{2}}\sin {{\theta }_{2}}$.
So, now we can write
${{x}_{2}}-{{x}_{1}}={{r}_{2}}\cos {{\theta }_{2}}-{{r}_{1}}\cos {{\theta }_{1}}$
We can take squares on both sides of the equation, and hence we will get
${{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}={{\left( {{r}_{2}}\cos {{\theta }_{2}}-{{r}_{1}}\cos {{\theta }_{1}} \right)}^{2}}$
We know that the identity for the expansion of square of difference is given as
\[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]
By using this identity, we can expand the right hand side (RHS) of the equation as,
\[{{\left( {{x}_{2}}-{{x}_{1}} \right)}^{2}}={{r}_{2}}^{2}{{\cos }^{2}}{{\theta }_{2}}+{{r}_{1}}^{2}{{\cos }^{2}}{{\theta }_{1}}-2{{r}_{1}}{{r}_{2}}\cos {{\theta }_{1}}\cos {{\theta }_{2}}...\left( ii \right)\]
Similarly, we can write
${{y}_{2}}-{{y}_{1}}={{r}_{2}}\sin {{\theta }_{2}}-{{r}_{1}}\sin {{\theta }_{1}}$
We can take squares on both sides of the equation, and hence we get
${{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}={{\left( {{r}_{2}}\sin {{\theta }_{2}}-{{r}_{1}}\sin {{\theta }_{1}} \right)}^{2}}$
We know that the identity for the expansion of the square of difference is given as
\[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\]
By using this identity, we can expand the right hand side (RHS) of our equation as,
\[{{\left( {{y}_{2}}-{{y}_{1}} \right)}^{2}}={{r}_{2}}^{2}{{\sin }^{2}}{{\theta }_{2}}+{{r}_{1}}^{2}{{\sin }^{2}}{{\theta }_{1}}-2{{r}_{1}}{{r}_{2}}\sin {{\theta }_{1}}\sin {{\theta }_{2}}...\left( iii \right)\]
Using the values from equation (ii) and equation (iii) and putting them in equation (i), we get
$D=\sqrt{\left( {{r}_{2}}^{2}{{\cos }^{2}}{{\theta }_{2}}+{{r}_{1}}^{2}{{\cos }^{2}}{{\theta }_{1}}-2{{r}_{1}}{{r}_{2}}\cos {{\theta }_{1}}\cos {{\theta }_{2}} \right)+\left( {{r}_{2}}^{2}{{\sin }^{2}}{{\theta }_{2}}+{{r}_{1}}^{2}{{\sin }^{2}}{{\theta }_{1}}-2{{r}_{1}}{{r}_{2}}\sin {{\theta }_{1}}\sin {{\theta }_{2}} \right)}$
We can regroup the terms present here as,
$D=\sqrt{{{r}_{1}}^{2}\left( {{\sin }^{2}}{{\theta }_{2}}+{{\cos }^{2}}{{\theta }_{2}} \right)+{{r}_{2}}^{2}\left( {{\sin }^{2}}{{\theta }_{1}}+{{\cos }^{2}}{{\theta }_{1}} \right)-2{{r}_{1}}{{r}_{2}}\left( \cos {{\theta }_{1}}\cos {{\theta }_{2}}+\sin {{\theta }_{1}}\sin {{\theta }_{2}} \right)}$
We all know very well that ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ . Using this identity, we get,
$D=\sqrt{{{r}_{1}}^{2}+{{r}_{2}}^{2}-2{{r}_{1}}{{r}_{2}}\left( \cos {{\theta }_{1}}\cos {{\theta }_{2}}+\sin {{\theta }_{1}}\sin {{\theta }_{2}} \right)}$
We are aware of the trigonometric identity $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$.
Using this identity, we can write
$D=\sqrt{{{r}_{1}}^{2}+{{r}_{2}}^{2}-2{{r}_{1}}{{r}_{2}}\cos \left( {{\theta }_{1}}-{{\theta }_{2}} \right)}$
This is the required formula for the calculation of distance between any two points in polar coordinates.

Note: We must remember that in polar coordinates, the angle $\theta $ is always expressed in radians. So, in this question the angles ${{\theta }_{1}}\text{ and }{{\theta }_{2}}$ are in radians and not degrees. We must remember these formulae in polar form as well, to solve the questions easily.