What is the distance between the following polar coordinates?: # (2,(13pi)/4), (1,(-pi)/8) #

1 Answer
Dec 29, 2015

#2.5555# units.

Explanation:

I will first convert these 2 polar co-ordinates into rectangular form and then use the Euclidean metric distance formula to find the distance between them.

By the conversion transformations, the polar form #z=(r,theta)# may be converted into the rectangular form #(x,y)=x+iy# as follows :
#x=rcostheta and y=rsintheta#.

Therefore in this particular case, #z_1=(2,(13pi)/4) and z_2=(1,-pi/8)# and so :
#x_1=2cos((13pi)/4)=-1.41421 and y_1=2sin((13pi)/4)=-1.41421#.

#x_2=1cos(-pi/8)=0.92388 and y_2=1sin(-pi/8)=-0.38268#.

Now assuming these 2 points are in #RR^# which is a metric space with the normal Euclidean metric, we may interpret and use this metric to find the distance between the 2 points as follows :

#d(z_1,z_2)=sqrt((x_1-x_2)^2+(y_1-y_2)^2#
#=sqrt((-1.41421-0.92388)^2+(-1.41421+0.38268)^2)#
#=2.5555#.

Note that since #RR^2# is also a complete normed space with the normal Euclidean norm, an alternate metric we could also have used is the metric induced by the norm:
#d(z_1,z_2)=||z_1-z_2||#, where #||z||=sqrt(x_1^2+x_2^2)#.
Clearly this would produce the same final result.