What is the arclength of $r = - θ sin (\frac{θ}{2} - \frac{27 π}{16}) - 3 cos (\frac{θ}{2} - \frac{11 π}{16})$ $r=-thetasin(theta/2-(27pi)/16) -3cos(theta/2 -(11pi)/16)$ on $θ \in [\frac{11 π}{8}, \frac{13 π}{8}]$ $theta in [(11pi)/8,(13pi)/8]$ ?

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What is the arclength of $r = - θ sin (\frac{θ}{2} - \frac{27 π}{16}) - 3 cos (\frac{θ}{2} - \frac{11 π}{16})$ $r=-thetasin(theta/2-(27pi)/16) -3cos(theta/2 -(11pi)/16)$ on $θ \in [\frac{11 π}{8}, \frac{13 π}{8}]$ $theta in [(11pi)/8,(13pi)/8]$ ?

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Answer 1 · 2017-06-08T01:05:56

See here for a derivation of the arc length in polar coordinates.

We obtained:

$s = \int_{θ_{1}}^{θ_{2}} \sqrt{r^{2} + {(\frac{d r}{d θ})}^{2}} d θ$ $s = int_(theta_1)^(theta_2) sqrt(r^2 + ((dr)/(d theta))^2)d theta$

and used it to get about $2.734$ $2.734$ , using technology.

First, let's rewrite $r$ $r$ so that it has the same arguments... might help us later if we need to use any trig identities.

Using Wolfram Alpha, we got:

$r = 3 sin((3 π)/16 - θ/2) - θ cos((3 π)/16 - θ/2)$

using identities like $sinx = cos(x-pi/2)$ .

Then, we should first square $r$ to get:

$r^2 = 9 sin^2((3 pi)/16 - theta/2) - 6 theta sin((3 pi)/16 - theta/2)cos((3 pi)/16 - theta/2) + theta^2 cos^2((3 pi)/16 - theta/2)$

Alright, and now, the derivative with respect to $theta$ ...

$(dr)/(d theta) = 3cos((3pi)/16 - theta/2)cdot-1/2 - [theta cdot -sin((3pi)/16 - theta/2)cdot-1/2 + cos((3pi)/16 - theta/2)]$

$= -3/2cos((3pi)/16 - theta/2) - theta/2 cdot sin((3pi)/16 - theta/2) - cos((3pi)/16 - theta/2)$

$= -5/2cos((3pi)/16 - theta/2) - theta/2 sin((3pi)/16 - theta/2)$

This squared gives:

$((dr)/(d theta))^2 = [-5/2cos((3pi)/16 - theta/2) - theta/2 sin((3pi)/16 - theta/2)]^2$

$= 25/4 cos^2((3pi)/16 - theta/2) + 5/2thetasin((3pi)/16 - theta/2)cos((3pi)/16 - theta/2) + theta^2/4sin^2((3pi)/16 - theta/2)$

Well... let's see where we are now.

$s = int_(11pi//8)^(13pi//8) sqrt(9 sin^2((3 pi)/16 - theta/2) - 6 theta sin((3 pi)/16 - theta/2)cos((3 pi)/16 - theta/2) + theta^2 cos^2((3 pi)/16 - theta/2) + 25/4 cos^2((3pi)/16 - theta/2) + 5/2thetasin((3pi)/16 - theta/2)cos((3pi)/16 - theta/2) + theta^2/4sin^2((3pi)/16 - theta/2))d theta$

$= int_(11pi//8)^(13pi//8) sqrt((9 + theta^2/4)sin^2((3 pi)/16 - theta/2) - 7/2 theta sin((3 pi)/16 - theta/2)cos((3 pi)/16 - theta/2) + (theta^2 + 25/4) cos^2((3pi)/16 - theta/2))d theta$

Welp, this is as simplified as I'm willing to make it. If this was a less painful expression to evaluate, I think you could do it.

But for this one, plug it into Wolfram Alpha and save yourself some brain. I only did two steps because the first step was too long for Wolfram Alpha to accept!

$color(blue)(s ~~ 2.734)$

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What is the arclength of $r = - θ sin (\frac{θ}{2} - \frac{27 π}{16}) - 3 cos (\frac{θ}{2} - \frac{11 π}{16})$ $r=-thetasin(theta/2-(27pi)/16) -3cos(theta/2 -(11pi)/16)$ on $θ \in [\frac{11 π}{8}, \frac{13 π}{8}]$ $theta in [(11pi)/8,(13pi)/8]$ ?

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What is the arclength of r=-thetasin(theta/2-(27pi)/16) -3cos(theta/2 -(11pi)/16)r=−θsin(θ2−27π16)−3cos(θ2−11π16)r=-thetasin(theta/2-(27pi)/16) -3cos(theta/2 -(11pi)/16) on theta in [(11pi)/8,(13pi)/8]θ∈[11π8,13π8]theta in [(11pi)/8,(13pi)/8]?

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What is the arclength of $r = - θ sin (\frac{θ}{2} - \frac{27 π}{16}) - 3 cos (\frac{θ}{2} - \frac{11 π}{16})$ $r=-thetasin(theta/2-(27pi)/16) -3cos(theta/2 -(11pi)/16)$ on $θ \in [\frac{11 π}{8}, \frac{13 π}{8}]$ $theta in [(11pi)/8,(13pi)/8]$ ?