What is the arclength of $r = \frac{5}{2} sin (\frac{θ}{8} + \frac{3 π}{16}) - \frac{θ}{4}$ $r=5/2sin(theta/8+(3pi)/16) -theta/4$ on $θ \in [\frac{- 3 π}{16}, \frac{7 π}{16}]$ $theta in [(-3pi)/16,(7pi)/16]$ ?

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What is the arclength of $r = \frac{5}{2} sin (\frac{θ}{8} + \frac{3 π}{16}) - \frac{θ}{4}$ $r=5/2sin(theta/8+(3pi)/16) -theta/4$ on $θ \in [\frac{- 3 π}{16}, \frac{7 π}{16}]$ $theta in [(-3pi)/16,(7pi)/16]$ ?

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Answer 1 · 2016-05-13T16:59:37

$s = 2.72416$ $s = 2.72416$

Explanation:

In polar coordinates $d s^{2} = r {(θ)}^{2} + {(\frac{d r (θ)}{d θ})}^{2}$ $ds^2 = r(theta)^2+((d r(theta))/(d theta))^2$
here $r (θ) = \frac{5}{5} sin (\frac{θ}{8} + \frac{3 π}{16}) - \frac{θ}{4}$ $r(theta)=5/5sin(theta/8+(3pi)/16)-theta/4$
$\frac{d r (θ)}{d θ} = - \frac{1}{4} + \frac{5}{16} cos (\frac{3 π}{16} + \frac{θ}{8})$ $(dr(theta))/(d theta) =-1/4+5/16cos((3pi)/16+theta/8)$ doing
$\int_{\frac{- 3 π}{16}}^{\frac{7 π}{16}} \sqrt{r {(θ)}^{2} + {(\frac{d r (θ)}{d θ})}^{2}} d θ = 2.72416$ $int_((-3pi)/16)^((7pi)/16)sqrt(r(theta)^2+((d r(theta))/(d theta))^2)d theta = 2.72416$

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What is the arclength of $r = \frac{5}{2} sin (\frac{θ}{8} + \frac{3 π}{16}) - \frac{θ}{4}$ $r=5/2sin(theta/8+(3pi)/16) -theta/4$ on $θ \in [\frac{- 3 π}{16}, \frac{7 π}{16}]$ $theta in [(-3pi)/16,(7pi)/16]$ ?

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What is the arclength of $r = \frac{5}{2} sin (\frac{θ}{8} + \frac{3 π}{16}) - \frac{θ}{4}$ $r=5/2sin(theta/8+(3pi)/16) -theta/4$ on $θ \in [\frac{- 3 π}{16}, \frac{7 π}{16}]$ $theta in [(-3pi)/16,(7pi)/16]$ ?