What is the arclength of the polar curve $f(theta) = -3sin(theta)-2sec^2theta$ over $theta in [0,pi/8]$ ?

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Answer 1 · 2018-07-15T20:16:32

$approx 1.827918426331$

We are using the formula

$s=int_alpha^betasqrt(r^2+((d r)/(d theta))^2)d theta$
$r(theta)=-3sin(theta)-2sec^2(theta)$

$r'(theta)=-3cos(theta)-4sec^2(theta)tan(theta)$

so we have to integrate

$int_0^(pi/8)sqrt((-3sin(theta)-2sec^2(theta))^2+(-3cos(theta)-4sec^2(theta)tan(theta))^2)d theta$

by a numerical method we get $approx 1.827918426331$

What is the arclength of the polar curve f(theta) = -3sin(theta)-2sec^2theta f(theta) = -3sin(theta)-2sec^2theta over theta in [0,pi/8] theta in [0,pi/8] ?