What is the distance between the following polar coordinates?: # (3,(15pi)/8), (9,(-2pi)/8) #

Question

What is the distance between the following polar coordinates?: #   (3,(15pi)/8), (9,(-2pi)/8)   #

Christopher Agresta · Accepted Answer

Distance is #6.33 #unit

Polar coordinates of point A is #r_1=3.0 , theta_1=(15pi)/8=337.5^0# Polar coordinates of point B is #r_2=9, theta_2=-(2pi)/8 , theta_2=-45^0=315^0# Cartesian coordinates of point A is #x_1=r_1 cos theta_1 # or # x_1= 3.0 cos 337.5 ~~ 2.77, y_1=r_1 sin theta # or #y_1=3.0 sin 337.5 ~~-1.15 :. # Cartesian coordinates of point A is #(x_1,y_1) or (2.77, -1.15)# Cartesian coordinates of point B is #x_2=r_2 cos theta_2 # or # x_2= 9.0 cos 315 = 6.36, y_2=r_2 sin theta_2 # or #y_2=9.0 sin 315=-6.36 #. Cartesian coordinates of point B is #(x_2,y_2) or (6.36,-6.36)# Distance between them #D= sqrt ((x_1-x_2)^2+(y_1-y_2)^2# #D= sqrt ((2.77-6.36)^2+(-1.15+6.36)^2) ~~ 6.33#unit [Ans]

Aurora Alvarado · Answer

undefined The distance between the polar coordinates $ (3, \frac{15\pi}{8}) $ and $ (9, \frac{-2\pi}{8}) $ can be found using the formula:

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

Substituting the given values:

$$ r_1 = 3, \ r_2 = 9, \ \theta_1 = \frac{15\pi}{8}, \ \theta_2 = \frac{-2\pi}{8} $$

$$ \text{Distance} = \sqrt{3^2 + 9^2 - 2(3)(9)\cos\left(\frac{-2\pi}{8} - \frac{15\pi}{8}\right)} $$

$$ \text{Distance} = \sqrt{9 + 81 - 54\cos\left(\frac{-2\pi - 15\pi}{8}\right)} $$

$$ \text{Distance} = \sqrt{90 - 54\cos\left(\frac{-17\pi}{8}\right)} $$

Since $ \cos(\frac{-17\pi}{8}) = \cos(\frac{\pi}{8}) $, we can rewrite:

$$ \text{Distance} = \sqrt{90 - 54\cos\left(\frac{\pi}{8}\right)} $$

Now, calculate the value of $ \cos\left(\frac{\pi}{8}\right) $ and then substitute it back into the formula to find the distance.