I need help figuring out the wording to this Calc 2 problem

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I need help figuring out the wording to this Calc 2 problem

Mathematics
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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 \[0 \le \theta \le \frac{ 11 }{ 8 }\pi\]
I understand what the arc length means, but the equations have two separate arcs over the interval
@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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\[\int_{a}^{b} ds\] given that ds is \[ds=\sqrt{(x')^2+(y')^2}dt\]
Where does this equation come from? Just so I have a base of understanding. I don't remember seeing anything like that in lecture
its just the result of takeing the segment of an arc ... and comparing it to a right triangle
|dw:1439771344418:dw|
as we work our way to infinitesimals we get: |dw:1439771398544:dw| and the pythag takes care of the rest
Wikipedia is your friend. https://en.wikipedia.org/wiki/Arc_length#Finding_arc_lengths_by_integrating
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
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
So that equation accounts for the two arcs?
yes, whether it makes life simpler or not ... thats to be seen :)
\[\int~ds\implies\int_{a}^{b}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}~dt\]
http://tutorial.math.lamar.edu/Classes/CalcII/ParaArcLength.aspx
\[ \begin{align*} x^\phantom{'}&=16\left(\cos(\theta)+\theta\sin(\theta)\right)\\ x'&=16\left(-\sin(\theta)+\sin(\theta)+\theta\cos(\theta)\right)\\ y^\phantom{'}&=16\left(\sin(\theta)-\theta\cos(\theta)\right)\\ y'&=16\left(\cos(\theta)-\cos(\theta)+\theta\sin(\theta)\right)\\ \end{align*} \]
The integral is much easier to evaluate once I got rid of my error lol.
heheh .. my x' got befuddled as well
I got \(x'=\sin(\theta)+\sin(\theta)+\theta\cos(\theta)\) and got a mess when evaluating the integral.
16t dt from a to b ...
maybe ... trying to do it in my head bites at my age
Just use the answer box as scrap paper lol.
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
\[ \frac{d}{dx}\cos(x)=-\sin(x) \] And product rule.
Okay, thank you so much guys, got the answer right finally!

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