What is the distance between the following polar coordinates?: # (3,(-7pi)/12), (2,(7pi)/8) #

Question

What is the distance between the following polar coordinates?: #   (3,(-7pi)/12), (2,(7pi)/8)   #

Evelyn Agard · Accepted Answer

#sqrt(13-12cos((35pi)/24)) approx 3.81658411497#

Say you have two polar coordinates #(r_1, theta_1)# and #(r_2, theta_2)#. In this case make a triangle with vertices at the origin and each point. Around the vertex of the origin we have two sides of lengths #r_1# and #r_2#, and we have angle #|theta_1 - theta_2|#. By the law of cosines, the third side is #sqrt(r_1^2+r_2^2-2r_1r_2cos(theta_1-theta_2))#

To find our distance we just then have:

#sqrt(13-12cos((35pi)/24)) approx 3.81658411497#

Evelyn Agard · Answer

undefined To find the distance between two polar coordinates $ (r_1, \theta_1) $ and $ (r_2, \theta_2) $, you can use the formula:

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

Given the polar coordinates $ (3, -\frac{7\pi}{12}) $ and $ (2, \frac{7\pi}{8}) $, plug the values into the formula:

$$ r_1 = 3, \, \theta_1 = -\frac{7\pi}{12}, \, r_2 = 2, \, \theta_2 = \frac{7\pi}{8} $$

$$ \text{Distance} = \sqrt{3^2 + 2^2 - 2(3)(2)\cos\left(\frac{7\pi}{8} - \left(-\frac{7\pi}{12}\right)\right)} $$

$$ \text{Distance} = \sqrt{9 + 4 - 12\cos\left(\frac{7\pi}{8} + \frac{7\pi}{12}\right)} $$

$$ \text{Distance} = \sqrt{13 - 12\cos\left(\frac{21\pi}{24} + \frac{14\pi}{24}\right)} $$

$$ \text{Distance} = \sqrt{13 - 12\cos\left(\frac{35\pi}{24}\right)} $$

$$ \text{Distance} = \sqrt{13 - 12\cos\left(\frac{11\pi}{24}\right)} $$

$$ \text{Distance} \approx \sqrt{13 - 12\cos\left(\frac{11\pi}{24}\right)} $$

$$ \text{Distance} \approx \sqrt{13 - 12\left(\frac{\sqrt{3}}{2}\right)} $$

$$ \text{Distance} \approx \sqrt{13 - 6\sqrt{3}} $$

So, the distance between the given polar coordinates is approximately $ \sqrt{13 - 6\sqrt{3}} $.