What is the arclength of $r = sin (θ^{2} - \frac{7 π}{6}) + 2 θ cos (θ - \frac{3 π}{8})$ $r=sin(theta^2-(7pi)/6) +2thetacos(theta -(3pi)/8)$ on $θ \in [\frac{7 π}{8}, π]$ $theta in [(7pi)/8,pi]$ ?

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Answer 1 · 2016-01-13T07:51:23

The arc length $s = 2.092001$ $s=2.092001$ units

The given polar equation:

$r=sin(θ^2−(7π)/6)+2θ*cos(θ−(3π)/8)$

The first derivative $(dr)/ (d theta)$ is needed in the formula for arc length $s$ :

$(dr)/(d theta)=cos(theta^2-(7 pi)/6)(2 theta)+2(1*cos(theta-(3 pi)/8)+ theta(-sin(theta-(3 pi)/8))*1)$

$(dr)/(d theta)=2 thetacos(theta^2-(7 pi)/6)+2 cos(theta-(3 pi)/8)-2 theta sin(theta-(3 pi)/8)$

The formula for arc length $s$ :

$s=int_a^b sqrt((sin(θ^2−(7π)/6)+2θ*cos(θ−(3π)/8))^2+ (2 thetacos(theta^2-(7 pi)/6)+2 cos(theta-(3 pi)/8)-2 theta sin(theta-(3 pi)/8))^2 d theta$

$s=2.09201$ units

What is the arclength of r=sin(theta^2-(7pi)/6) +2thetacos(theta -(3pi)/8)r=sin(θ2−7π6)+2θcos(θ−3π8)r=sin(theta^2-(7pi)/6) +2thetacos(theta -(3pi)/8) on theta in [(7pi)/8,pi]θ∈[7π8,π]theta in [(7pi)/8,pi]?