# What is the distance between the following polar coordinates?:  (1,(3pi)/4), (5,(5pi)/8)

Mar 21, 2018

Distance$\approx 4.094$

#### Explanation:

Call the two points A and B:

$A = \left({r}_{a} , {\theta}_{a}\right) = \left(1 , \frac{3 \pi}{4}\right)$

$B = \left({r}_{b} , {\theta}_{b}\right) = \left(5 , \frac{5 \pi}{8}\right)$

Convert them to rectangular form:

$A = \left({x}_{a} , {y}_{a}\right)$

${x}_{a} = {r}_{a} \cos \left({\theta}_{a}\right) = 1 \cdot \cos \left(\frac{3 \pi}{4}\right) = - \frac{\sqrt{2}}{2}$

${y}_{a} = {r}_{a} \sin \left({\theta}_{a}\right) = 1 \cdot \sin \left(\frac{3 \pi}{4}\right) = \frac{\sqrt{2}}{2}$

$A = \left(- \frac{\sqrt{2}}{2} , \frac{\sqrt{2}}{2}\right) \approx \left(- 0.707 , 0.707\right)$

$B = \left({x}_{b} , {y}_{b}\right)$

${x}_{b} = {r}_{b} \cos \left({\theta}_{b}\right) = 5 \cdot \cos \left(\frac{5 \pi}{8}\right) = - 5 \frac{\sqrt{2 - \sqrt{2}}}{2}$

${y}_{b} = {r}_{b} \sin \left({\theta}_{b}\right) = 5 \cdot \sin \left(\frac{5 \pi}{8}\right) = 5 \frac{\sqrt{2 + \sqrt{2}}}{2}$

$B = \left(- 5 \frac{\sqrt{2 - \sqrt{2}}}{2} , 5 \frac{\sqrt{2 + \sqrt{2}}}{2}\right) \approx \left(- 1.913 , 4.619\right)$

Now apply the Pythagorean Theorem to find the length of the line segment $\overline{A B}$.

$| | \overline{A B} | | = \sqrt{{\left({x}_{b} - {x}_{a}\right)}^{2} + {\left({y}_{b} - {y}_{a}\right)}^{2}}$

$| | \overline{A B} | | \approx \sqrt{{\left(\left(- 1.913\right) - \left(- 0.707\right)\right)}^{2} + {\left(4.619 - 0.707\right)}^{2}}$

$| | \overline{A B} | | \approx \sqrt{{\left(- 1.206\right)}^{2} + {\left(3.912\right)}^{2}} = \sqrt{1.455 + 15.306}$

$| | \overline{A B} | | \approx \sqrt{16.761} \approx 4.094$