What is the arclength of the polar curve $f (θ) = 2 θ sin (5 θ) - θ cot 2 θ$ $f(theta) = 2thetasin(5theta)-thetacot2theta$ over $θ \in [\frac{π}{12}, \frac{π}{3}]$ $theta in [pi/12,pi/3]$ ?

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Answer 1 · 2018-06-03T08:14:10

$\approx 2.73636$ $approx 2.73636$

From
$r (θ) = 2 θ \cdot sin (5 θ) - θ cot (2 θ)$ $r(theta)=2theta*sin(5theta)-thetacot(2theta)$

we get by differentiating with respect to $θ$ $theta$ :

$r'(theta)=10thetacos(5theta)-cot(2theta)+2thetacsc^2(2theta)+2sin(5theta)$
and our integral is given by

$int_(pi/12)^(pi/3)sqrt((2theta*sin(5theta)-thetacot(2theta))^2+(10thetacos(5theta)-cot(2theta)+2thetacsc^2(2theta)+2sin(5theta))^2)d theta$

What is the arclength of the polar curve f(theta) = 2thetasin(5theta)-thetacot2theta f(θ)=2θsin(5θ)−θcot2θf(theta) = 2thetasin(5theta)-thetacot2theta over theta in [pi/12,pi/3] θ∈[π12,π3]theta in [pi/12,pi/3] ?