What is the arclength of $r = cos θ - 3 sin 2 θ$ $r=costheta-3sin2theta$ on $θ \in [- \frac{π}{4}, π]$ $theta in [-pi/4,pi]$ ?

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Answer 1 · 2018-06-03T07:50:12

$\approx 16.7916$ $approx 16.7916$

From
$r (θ) = cos (θ) - 3 sin (3 θ)$ $r(theta)=cos(theta)-3sin(3theta)$
we get by differentiating

$r'(theta)=-sin(theta)-6cos(2theta)$
so we get

$int_(-pi/4)^pisqrt((cos(theta)-3*sin(2theta))^2+(-sin(theta)-6cos(2theta))^2)d theta$

By a numerical method we obtain

$16.7916$

What is the arclength of r=costheta-3sin2theta r=cosθ−3sin2θr=costheta-3sin2theta on theta in [-pi/4,pi]θ∈[−π4,π]theta in [-pi/4,pi]?