# What is the distance between the following polar coordinates?:  (2,(13pi)/4), (1,(-pi)/8)

Dec 29, 2015

$2.5555$ units.

#### Explanation:

I will first convert these 2 polar co-ordinates into rectangular form and then use the Euclidean metric distance formula to find the distance between them.

By the conversion transformations, the polar form $z = \left(r , \theta\right)$ may be converted into the rectangular form $\left(x , y\right) = x + i y$ as follows :
$x = r \cos \theta \mathmr{and} y = r \sin \theta$.

Therefore in this particular case, ${z}_{1} = \left(2 , \frac{13 \pi}{4}\right) \mathmr{and} {z}_{2} = \left(1 , - \frac{\pi}{8}\right)$ and so :
${x}_{1} = 2 \cos \left(\frac{13 \pi}{4}\right) = - 1.41421 \mathmr{and} {y}_{1} = 2 \sin \left(\frac{13 \pi}{4}\right) = - 1.41421$.

${x}_{2} = 1 \cos \left(- \frac{\pi}{8}\right) = 0.92388 \mathmr{and} {y}_{2} = 1 \sin \left(- \frac{\pi}{8}\right) = - 0.38268$.

Now assuming these 2 points are in RR^ which is a metric space with the normal Euclidean metric, we may interpret and use this metric to find the distance between the 2 points as follows :

d(z_1,z_2)=sqrt((x_1-x_2)^2+(y_1-y_2)^2
$= \sqrt{{\left(- 1.41421 - 0.92388\right)}^{2} + {\left(- 1.41421 + 0.38268\right)}^{2}}$
$= 2.5555$.

Note that since ${\mathbb{R}}^{2}$ is also a complete normed space with the normal Euclidean norm, an alternate metric we could also have used is the metric induced by the norm:
$d \left({z}_{1} , {z}_{2}\right) = | | {z}_{1} - {z}_{2} | |$, where $| | z | | = \sqrt{{x}_{1}^{2} + {x}_{2}^{2}}$.
Clearly this would produce the same final result.