What is the distance between the following polar coordinates?: # (2,(10pi)/3), (14,(-31pi)/8) #

Question

What is the distance between the following polar coordinates?: #   (2,(10pi)/3), (14,(-31pi)/8)   #

Ezekiel Alexander · Accepted Answer

#14.39color(white)(88)#units

First we need to convert the polar coordinates to Cartesian coordinates. We can do this by using the following: #x =rcos(theta)# #y=rsin(theta)# #:.# #x=2cos((10pi)/3)=-1# #y=2sin((10pi)/3)=-sqrt(3)# Cartesian coordinate: #( -1 , -sqrt(3) )# #x=14cos((-31pi)/8)=12.93# #y=14sin((-31pi)/8)=-5.36# Cartesian coordinate: #( 12.93 , -5.36 )# Next we use the distance formula: #d = sqrt((x_2-x_1)^2+(y_2-y_1)^2)# #d=sqrt((-1-12.93)^2+(-sqrt(3)+5.36)^2)=14.39color(white)(88)#units All results to 2 .d.p.

Charles Arocha · 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 $r_1$ and $r_2$ are the radii, and $\theta_1$ and $\theta_2$ are the angles in radians.

For the given coordinates:
$r_1 = 2$, $\theta_1 = \frac{10\pi}{3}$, $r_2 = 14$, $\theta_2 = \frac{-31\pi}{8}$

$$d = \sqrt{2^2 + 14^2 - 2(2)(14)\cos\left(\frac{-31\pi}{8} - \frac{10\pi}{3}\right)}$$

$$d = \sqrt{4 + 196 - 56\cos\left(\frac{-31\pi}{8} - \frac{10\pi}{3}\right)}$$

$$d = \sqrt{200 - 56\cos\left(\frac{-31\pi}{8} - \frac{10\pi}{3}\right)}$$

$$d \approx \sqrt{200 - 56\cos\left(-\frac{17\pi}{24}\right)}$$

$$d \approx \sqrt{200 - 56\cos\left(-\frac{17}{24}\cdot\pi\right)}$$

$$d \approx \sqrt{200 - 56\cos\left(-\frac{17}{24}\cdot\frac{180}{\pi}\cdot\pi\right)}$$

$$d \approx \sqrt{200 - 56\cos(-127.5^\circ)}$$

$$d \approx \sqrt{200 - 56\cos(232.5^\circ)}$$

$$d \approx \sqrt{200 - 56(-0.7071)}$$

$$d \approx \sqrt{200 + 39.5716}$$

$$d \approx \sqrt{239.5716}$$

$$d \approx 15.48$$

Therefore, the distance between the given polar coordinates is approximately $15.48$.