# What is the distance between the following polar coordinates?:  (21,(4pi)/3), (11,(11pi)/8)

Jan 5, 2017

$10.20 \setminus \text{units}$ (2dp)

#### Explanation:

There are two basic methods, which invoke either working in polar coordinates or converting to Cartesian coordinates. The polar coordinate equation is a little harder to remember but easier to use, whereas the rectangular formula is easy to remember (it just relies upon Pythagoras) but invokes more work due to the polar to Cartesian conversion

Polar Coordinates
The distance between two polar coordinates $\left({r}_{1} , {\theta}_{1}\right)$ and $\left({r}_{2} , {\theta}_{2}\right)$ is given by;

$d = \sqrt{\left({r}_{1}^{2} + {r}_{2}^{2} - 2 {r}_{1} r\right) 2 \cos \left({\theta}_{2} - {\theta}_{1}\right)}$

So for polar coordinates $\left(21 , \frac{4 \pi}{3}\right)$ and $\left(11 , \frac{11 \pi}{8}\right)$ we have:

d=sqrt(21^2+11^2-2*21*11cos((11pi)/8-(4pi)/3)
$\setminus \setminus = \sqrt{441 + 121 - 462 \cos \left(\frac{\pi}{24}\right)}$
$\setminus \setminus = \sqrt{103.9524 . .}$
$\setminus \setminus = 10.1957 \ldots$

Cartesian (Rectangular) Coordinates
The distance between two Cartesian coordinates $\left({x}_{1} , {y}_{1}\right)$ and $\left({x}_{2} , {y}_{2}\right)$ is given by:

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

The Cartesian equivalent of the polar coordinate $\left(r , \theta\right)$ is $\left(r \cos \theta , r \sin \theta\right)$, So:

$\left(21 , \frac{4 \pi}{3}\right) \rightarrow \left(21 \cos \left(\frac{4 \pi}{3}\right) , 21 \sin \left(\frac{4 \pi}{3}\right)\right)$
$\text{ } = \left(- 10.5 , - 18.1865 \ldots\right)$

$\left(11 , \frac{11 \pi}{8}\right) \rightarrow \left(11 \cos \left(\frac{11 \pi}{8}\right) , 11 \sin \left(\frac{11 \pi}{8}\right)\right)$
$\text{ } = \left(- 4.2095 \ldots , - 10.1626 \ldots\right)$

So s the distance between these coordinates is:

$d = \sqrt{{\left(- 10.5 - \left(- 4.2095 \ldots\right)\right)}^{2} + {\left(- 18.1865 \ldots - \left(- 10.1626 \ldots\right)\right)}^{2}}$
$\setminus \setminus = \sqrt{{\left(- 6.2904 \ldots\right)}^{2} + {\left(- 8.0238\right)}^{2}}$
$\setminus \setminus = \sqrt{39.5701 \ldots + 64.3823 \ldots}$
$\setminus \setminus = \sqrt{103.9524 \ldots}$
$\setminus \setminus = 10.1957 \ldots$, as above