# What is the distance between the following polar coordinates?:  (6,pi/3), (0,pi/2)

Jan 18, 2018

6.

#### Explanation:

In cartesian coordinates $\left(x , y\right)$, the distance $\left(d\right)$ between two points $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by:

${d}^{2} = {\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}$.

Now let us transform that to polar coordinates. In this new system, we know that $x = \rho \cos \theta$ and $y = \rho \sin \theta$. Then:

${d}^{2} = {\left({\rho}_{2} \cos {\theta}_{2} - {\rho}_{1} \cos {\theta}_{1}\right)}^{2} + {\left({\rho}_{2} \sin {\theta}_{2} - {\rho}_{1} \sin {\theta}_{1}\right)}^{2}$.

With a few algebric steps, we can rewrite the above expression as:

${d}^{2} = {\rho}_{2}^{2} \left({\cos}^{2} {\theta}_{2} + {\sin}^{2} {\theta}_{2}\right) + {\rho}_{1}^{2} \left({\cos}^{2} {\theta}_{1} + {\sin}^{2} {\theta}_{1}\right) - 2 {\rho}_{1} {\rho}_{2} \left(\cos {\theta}_{1} \cos {\theta}_{2} + \sin {\theta}_{1} \sin {\theta}_{2}\right)$ .

Now, by using the trigonometric relations, we obtain the following expression for $d$:

d = sqrt(rho_1^2 + rho_2^2 - 2rho_1rho_2cos(theta_2 - theta_1).

Then, applying our points $\left({\rho}_{1} , {\theta}_{1}\right) = \left(6 , \frac{\pi}{3}\right)$ and $\left({\rho}_{2} , {\theta}_{2}\right) = \left(0 , \frac{\pi}{2}\right)$:

$d = \sqrt{36 + 0 - 0} = 6$.