What is the distance between the following polar coordinates?: # (2,(3pi)/4), (1,(15pi)/8) #

Question

What is the distance between the following polar coordinates?: #   (2,(3pi)/4), (1,(15pi)/8)   #

Emilia Aguilar · Accepted Answer

#D=sqrt(5+4cos(pi/8))~~2.9488#

We know that ,

#"Distance between Polar Co-ordinates:"A(r_1,theta_1)and B(r_2,theta_2) # is

#color(red)(D=sqrt(r_1^2+r_2^2-2r_1r_2cos(theta_1-theta_2))...to(I)#

We have , #P_1(2,(3pi)/4) and P_2(1,(15pi)/8)#.

So , #r_1=2 , r_2=1 , theta_1=(3pi)/4 and theta_2=(15pi)/8#

#=>theta_1-theta_2=(3pi)/4-(15pi)/8=(6pi-15pi)/8=(-9pi)/8#

#=>cos(theta_1-theta_2)=cos((-9pi)/8)# #=>cos(theta_1-theta_2)=cos((9pi)/8)to[becausecos(-theta)=costheta]# #=>cos(theta_1-theta_2)=cos(pi+pi/8)=-cos(pi/8)#

#"Using : " color(red)((I)# we get

#D=sqrt(2^2+1^2-2(2)(1)(-cos(pi/8))#

#=>D=sqrt(4+1+4*cos(pi/8))#

#=>D=sqrt(5+4cos(pi/8))#

#=>D~~2.9488#

Isabella Ackerman · Answer

#sqrt(2+sqrt(2-sqrt(2)))#

It is nontrivial to determine polar distances (since drawing a line in polar coordinates isn't very easy), so we should convert to Cartesian coordinates.

The first point is moderately easy: #x = rcostheta = 2 * -sqrt(2)/2 = -sqrt(2)# #y = rsintheta = 2 * sqrt(2)/2 = sqrt(2)#

The second point is a little harder, but we only need to use half angle identities. Let #theta = (15pi)/4 equiv -(pi)/4#. Therefore, #cos(theta) = sqrt(2)/2# and

#cos(theta/2) = pm sqrt((1+costheta)/2) = pm sqrt(2+sqrt2)/2# #sin(theta/2) = pm sqrt(1-costheta)/2 = pm sqrt(2-sqrt2)/2#

Since #(15pi)/8# is in the 4th quadrant, its cosine is positive and its sine is negative. Therefore,

#x = r cos((15pi)/8) = sqrt(2+sqrt2)/2# #y = r sin((15pi)/8) = sqrt(2-sqrt2)/2#

This allows us to use Pythagorean theorem in order to find the distance between the two points:

#Delta x = sqrt(2+sqrt2)/2 + sqrt2# #Delta y = sqrt(2-sqrt2)/2 - sqrt2#

Pulling out a factor of #sqrt(2)# from each, then squaring and adding them, #ds^2 = 2[1 + sqrt2/2 + 1 + 1 - sqrt2/2 + 1 + 2sqrt(1+sqrt2/2) - 2 sqrt(1 - sqrt2/2)] # #= 2[4 + 2(sqrt(1+sqrt2/2) - sqrt(1-sqrt2/2))] # Using the cosine sum formula with #pi/8# and #pi/4#, we get #= 8 (1 + cos ((3pi)/8) ) = 4 cos^2((3pi)/16)#

Therefore the total distance is #D = 2 cos((3pi)/16) = sqrt(2 + sqrt(2 - sqrt(2))) approx 1.66#

Ezra Adams · Answer

#2.949\ ext{unit}# The distance between the points #(r_1, heta_1)\equiv(2, {3\pi}/4)# & #(r_2, heta_2)\equiv(1, {15\pi}/8)# is given by the formula as follows #\sqrt{r_1^2+r_2^2-2r_1r_2\cos( heta_1- heta_2)}# #=\sqrt{2^2+1^2-2(2)(1)\cos({3\pi}/4-{15\pi}/8)}# #=\sqrt{5-4\cos({9\pi}/8)}# #=\sqrt{5+4\cos({\pi}/8)}# #=\sqrt{\frac{7\sqrt2+2}{\sqrt2}}# #=2.949\ ext{unit}#

Emilia Aguilar · Answer

undefined To find the distance between the polar coordinates $ (2, \frac{3\pi}{4}) $ and $ (1, \frac{15\pi}{8}) $:

1. Convert each polar coordinate to Cartesian coordinates using the formulas:
   $$ x = r \cos(\theta) $$
   $$ y = r \sin(\theta) $$

2. For $ (2, \frac{3\pi}{4}) $:
   $$ x_1 = 2 \cos\left(\frac{3\pi}{4}\right) $$
   $$ y_1 = 2 \sin\left(\frac{3\pi}{4}\right) $$

3. For $ (1, \frac{15\pi}{8}) $:
   $$ x_2 = 1 \cos\left(\frac{15\pi}{8}\right) $$
   $$ y_2 = 1 \sin\left(\frac{15\pi}{8}\right) $$

4. Calculate the distance between the two Cartesian points using the distance formula:
   $$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$