What is the distance between #(2 , (5 pi)/8 )# and #(3 , (1 pi )/3 )#?

Question

What is the distance between #(2   , (5   pi)/8    )# and #(3  , (1   pi )/3   )#?

Quinn Anderson · Accepted Answer

The distance between those two coordinates is #sqrt(13 - 12 cos((7pi)/24)) ~~2.39#.

You can use the law of cosines to do that.

Let me illustrate why:

Polar coordinates #(r, theta)# are defined by the radius #r# and the angle #theta#.

Imagine lines leading from the pole to your respective polar coordinates. Those lines represent two sides of a triangle with lengths #A = 3# and #B = 2#. The distance between those two coordinates being the third side, #C#.

Furthermore, the angle between #A# and #B# can be computed as the difference between the two angles of your polar coordinates:

#gamma = (5pi)/8 - pi/3 = (7pi)/24#
Thus, the length of the side #C# can be found with the help of law of cosines on that triangle:

#C^2 = A^2 + B^2 - 2AB cos(gamma)#

#= 3^2 + 2^2 - 2 * 3 * 2 * cos((7pi)/24)#
#= 13 - 12 cos((7pi)/24)#

#=> C = sqrt(13 - 12 cos((7pi)/24)) ~~2.39#

Brooklyn Anderson · Answer

undefined To find the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ in a two-dimensional coordinate system, we use the distance formula:

$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$

Substituting the given coordinates:

$$ d = \sqrt{(3 - 2)^2 + \left(\frac{\pi}{3} - \frac{5\pi}{8}\right)^2} $$

$$ d = \sqrt{1^2 + \left(\frac{8\pi - 15\pi}{24}\right)^2} $$

$$ d = \sqrt{1 + \left(\frac{-7\pi}{24}\right)^2} $$

$$ d = \sqrt{1 + \frac{49\pi^2}{576}} $$

$$ d = \sqrt{\frac{576 + 49\pi^2}{576}} $$

$$ d = \frac{\sqrt{576 + 49\pi^2}}{24} $$

Therefore, the distance between the points $(2, \frac{5\pi}{8})$ and $(3, \frac{\pi}{3})$ is $ \frac{\sqrt{576 + 49\pi^2}}{24} $ units.