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

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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

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