What is the arclength of the polar curve $f (θ) = 2 cos (9 θ) - 3 sin θ$ $f(theta) = 2cos(9theta)-3sintheta$ over $θ \in [0, \frac{π}{3}]$ $theta in [0,pi/3]$ ?

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Answer 1 · 2018-07-15T16:27:54

$\approx 13.20973$ $approx 13.20973$

We use the formula

$s = \int_{α}^{β} \sqrt{r^{2} + {(\frac{d r}{d θ})}^{2}} d θ$ $s=int_alpha^beta\sqrt(r^2+((dr)/(d theta))^2)d theta$
where

$r (θ) = 2 cos (9 θ) - 3 sin (θ)$ $r(theta)=2cos(9theta)-3sin(theta)$
$r'(theta)=-18sin(9theta)-3cos(theta)$

so we have to integrate

$int_0^(pi/3)sqrt(173-160cos(18 theta)+60sin(8theta)+48sin(10theta))d theta$
by a numerical method we get

$approx 13.20973$

What is the arclength of the polar curve f(theta) = 2cos(9theta)-3sintheta f(θ)=2cos(9θ)−3sinθf(theta) = 2cos(9theta)-3sintheta over theta in [0,pi/3] θ∈[0,π3]theta in [0,pi/3] ?