What is the distance between #(4 ,( 9 pi)/8 )# and #(-1 ,( 3 pi )/2 )#?

Question

What is the distance between #(4  ,(  9  pi)/8   )# and #(-1  ,( 3 pi )/2       )#?

Josiah Anderson · Accepted Answer

#d~~4.48#

Each point on a polar plane is represented by the ordered pair #(r,theta)#.

So lets call the coordinates of #P_1# as #(r_1,theta_1)# and coordinates of #P_2# as #(r_2,theta_2)# . To find the distance between two points on a polar plane use the formula #d=sqrt((r_1) ^2+(r_2)^2-2r_1r_2cos(theta_2-theta_1))#

Thererfore using the points #(4,(9pi)/8)# and #(-1,(3pi)/2)#, and the formula

#d=sqrt((r_1) ^2+(r_2)^2-2r_1r_2cos(theta_2-theta_1))#

we have

#d=sqrt((4)^2+(-1)^2-2*4*-1cos((3pi)/2-(9pi)/8))#

#:. d~~4.48#

Brooklyn Anderson · Answer

undefined To find the distance between two points in polar coordinates, we can use the formula:

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

Given the points $(4, \frac{9\pi}{8})$ and $(-1, \frac{3\pi}{2})$, we have:

$$ r_1 = 4, \quad r_2 = -1, \quad \theta_1 = \frac{9\pi}{8}, \quad \theta_2 = \frac{3\pi}{2} $$

$$ \text{Distance} = \sqrt{(-1)^2 + (4)^2 - 2(-1)(4) \cos(\frac{3\pi}{2} - \frac{9\pi}{8})} $$

$$ = \sqrt{1 + 16 + 8\cos(\frac{3\pi}{2} - \frac{9\pi}{8})} $$

$$ = \sqrt{17 + 8\cos(\frac{12\pi - 9\pi}{8})} $$

$$ = \sqrt{17 + 8\cos(\frac{3\pi}{8})} $$

$$ = \sqrt{17 + 8 \left(\frac{\sqrt{2 + \sqrt{2}}}{2}\right)} $$

$$ = \sqrt{17 + 4\sqrt{2 + \sqrt{2}}} $$

$$ \approx \sqrt{22.657} $$

$$ \approx 4.757 $$