What is the arclength of #r=-cos(theta/2-(3pi)/8)/theta # on #theta in [(3pi)/4,(7pi)/4]#?

Question

1623 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-13T07:10:29

Arc length #s=0.757882# units

#r=−cos(θ/2−3π/8)/θ# on #θ∈[3π/4,7π/4]#

there is a need to determine first #(dr)/ (d theta)#

#(dr)/(d theta)=(theta*sin(theta/2-3 pi/8)(1/2)-(-cos(theta/2-3 pi/8))*1)/theta^2#

#(dr)/(d theta)=(theta/2*sin(theta/2-3 pi/8)+cos(theta/2-3 pi/8))/theta^2#

#(dr)/(d theta)=(1/(2 theta)*sin(theta/2-3 pi/8)+1/theta^2*cos(theta/2-3 pi/8))#

The formula for arc length s in Polar coordinates is

#s=int_a^b sqrt(r^2+((dr)/(d theta))^2) d theta#

#s=int_(3 pi/4)^(7 pi/4) sqrt((−cos(θ/2−3π/8)/θ)^2#+
#(1/(2 theta)*sin(theta/2-3 pi/8)+1/theta^2*cos(theta/2-3 pi/8))^2) d theta#

#s=0.757882 units