What is the arclength of $r = 2 sin (θ - \frac{3 π}{16}) + 9 θ$ $r=2sin(theta-(3pi)/16) + 9theta$ on $θ \in [\frac{- 5 π}{16}, \frac{9 π}{16}]$ $theta in [(-5pi)/16,(9pi)/16]$ ?

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Answer 1 · 2017-03-24T11:09:08

Store $r = 2 sin (θ - \frac{3 π}{16}) + 9 θ$ $r=2sin(theta-(3pi)/16)+9theta$ in your calculator.

Find the derivative $\frac{d r}{d θ} = 2 cos (θ - \frac{3 π}{16}) + 9$ $(dr)/(d theta)=2cos(theta-(3pi)/16)+9$ and store this separately.

Plug these into the integral that gives the arc length for polar equations:

$s=int_((-5pi)/16)^((9pi)/16)sqrt(r^2+((dr)/(d theta))^2)d theta=37.1478...$

What is the arclength of r=2sin(theta-(3pi)/16) + 9thetar=2sin(θ−3π16)+9θr=2sin(theta-(3pi)/16) + 9theta on theta in [(-5pi)/16,(9pi)/16]θ∈[−5π16,9π16]theta in [(-5pi)/16,(9pi)/16]?