What is the distance between the following polar coordinates?: # (2,(9pi)/4), (14,(-3pi)/8) #

Question

What is the distance between the following polar coordinates?: #   (2,(9pi)/4), (14,(-3pi)/8)   #

Ezra Adams · Accepted Answer

The distance #~~ 14.88# When given two polar points #(r_1,theta_1)# and #(r_2,theta_2)#, the distance between the two points can be found using a variant of the the Law of Cosines: #d = sqrt(r_1^2+r_2^2-2(r_1)(r_2)cos(theta_2-theta_1)# We are given: #r_1 = 2, theta_1 = (9pi)/4, r_2 = 14, and theta_2 = (-3pi)/8# #d = sqrt(2^2+14^2-2(2)(14)cos((-3pi)/8-(9pi)/4)# #d~~14.88#

Evelyn Agard · Answer

undefined To find the distance between two polar coordinates, you can use the formula:

$$ d = \sqrt{r_1^2 + r_2^2 - 2r_1r_2\cos(\theta_2 - \theta_1)} $$

Where $ (r_1, \theta_1) $ and $ (r_2, \theta_2) $ are the polar coordinates.

For the given coordinates $ (2, \frac{9\pi}{4}) $ and $ (14, -\frac{3\pi}{8}) $:

$ r_1 = 2, \theta_1 = \frac{9\pi}{4} $

$ r_2 = 14, \theta_2 = -\frac{3\pi}{8} $

Using the formula:

$$ d = \sqrt{2^2 + 14^2 - 2(2)(14)\cos\left(-\frac{3\pi}{8} - \frac{9\pi}{4}\right)} $$

$$ d = \sqrt{4 + 196 - 56\cos\left(-\frac{15\pi}{8}\right)} $$

$$ d = \sqrt{200 - 56\cos\left(\frac{\pi}{8}\right)} $$

$$ d \approx \sqrt{200 - 56 \times 0.9239} $$

$$ d \approx \sqrt{200 - 52.0764} $$

$$ d \approx \sqrt{147.9236} $$

$$ d \approx 12.16 $$

So, the distance between the two polar coordinates is approximately $ 12.16 $.