anonymous
  • anonymous
I need help figuring out the wording to this Calc 2 problem
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Consider the parametric equation \[x = 16(\cos \theta + \theta \sin \theta)\] \[y = 16( \sin \theta - \theta \cos \theta)\] What is the length of the curve for \[0 \le \theta \le \frac{ 11 }{ 8 }\pi\]
anonymous
  • anonymous
I understand what the arc length means, but the equations have two separate arcs over the interval
anonymous
  • anonymous
@mathstudent55 you seem like the most qualified person answering right now, I just need to understand what it's asking for, I can do the math myself

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amistre64
  • amistre64
\[\int_{a}^{b} ds\] given that ds is \[ds=\sqrt{(x')^2+(y')^2}dt\]
anonymous
  • anonymous
Where does this equation come from? Just so I have a base of understanding. I don't remember seeing anything like that in lecture
amistre64
  • amistre64
its just the result of takeing the segment of an arc ... and comparing it to a right triangle
amistre64
  • amistre64
|dw:1439771344418:dw|
amistre64
  • amistre64
as we work our way to infinitesimals we get: |dw:1439771398544:dw| and the pythag takes care of the rest
thomas5267
  • thomas5267
Wikipedia is your friend. https://en.wikipedia.org/wiki/Arc_length#Finding_arc_lengths_by_integrating
amistre64
  • amistre64
x=16(cos(t) +t sin(t)) x' = 16(cos(t)+sin(t) +t cos(t)) dt y=16(sin(t) -t cos(t)) y'=16(cos(t) -cos(t) +t sin(t)) dt
amistre64
  • amistre64
if we can eliminate the parameter, and make y a function of x, then x'=1 ... but thats not always a simple case is it
anonymous
  • anonymous
So that equation accounts for the two arcs?
amistre64
  • amistre64
yes, whether it makes life simpler or not ... thats to be seen :)
amistre64
  • amistre64
\[\int~ds\implies\int_{a}^{b}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}~dt\]
amistre64
  • amistre64
http://tutorial.math.lamar.edu/Classes/CalcII/ParaArcLength.aspx
thomas5267
  • thomas5267
\[ \begin{align*} x^\phantom{'}&=16\left(\cos(\theta)+\theta\sin(\theta)\right)\\ x'&=16\left(-\sin(\theta)+\sin(\theta)+\theta\cos(\theta)\right)\\ y^\phantom{'}&=16\left(\sin(\theta)-\theta\cos(\theta)\right)\\ y'&=16\left(\cos(\theta)-\cos(\theta)+\theta\sin(\theta)\right)\\ \end{align*} \]
thomas5267
  • thomas5267
The integral is much easier to evaluate once I got rid of my error lol.
amistre64
  • amistre64
heheh .. my x' got befuddled as well
thomas5267
  • thomas5267
I got \(x'=\sin(\theta)+\sin(\theta)+\theta\cos(\theta)\) and got a mess when evaluating the integral.
amistre64
  • amistre64
16t dt from a to b ...
amistre64
  • amistre64
maybe ... trying to do it in my head bites at my age
thomas5267
  • thomas5267
Just use the answer box as scrap paper lol.
anonymous
  • anonymous
Thomas, dumb question but could you walk me through the derivation of x real quick? I know the 16 can just get pulled out first
thomas5267
  • thomas5267
\[ \frac{d}{dx}\cos(x)=-\sin(x) \] And product rule.
anonymous
  • anonymous
Okay, thank you so much guys, got the answer right finally!

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