## anonymous one year ago I need help figuring out the wording to this Calc 2 problem

1. 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 $0 \le \theta \le \frac{ 11 }{ 8 }\pi$

2. anonymous

I understand what the arc length means, but the equations have two separate arcs over the interval

3. 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

4. amistre64

$\int_{a}^{b} ds$ given that ds is $ds=\sqrt{(x')^2+(y')^2}dt$

5. 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

6. amistre64

its just the result of takeing the segment of an arc ... and comparing it to a right triangle

7. amistre64

|dw:1439771344418:dw|

8. amistre64

as we work our way to infinitesimals we get: |dw:1439771398544:dw| and the pythag takes care of the rest

9. thomas5267

10. 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

11. 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

12. anonymous

So that equation accounts for the two arcs?

13. amistre64

yes, whether it makes life simpler or not ... thats to be seen :)

14. amistre64

$\int~ds\implies\int_{a}^{b}\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}~dt$

15. amistre64
16. 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*}

17. thomas5267

The integral is much easier to evaluate once I got rid of my error lol.

18. amistre64

heheh .. my x' got befuddled as well

19. thomas5267

I got $$x'=\sin(\theta)+\sin(\theta)+\theta\cos(\theta)$$ and got a mess when evaluating the integral.

20. amistre64

16t dt from a to b ...

21. amistre64

maybe ... trying to do it in my head bites at my age

22. thomas5267

Just use the answer box as scrap paper lol.

23. 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

24. thomas5267

$\frac{d}{dx}\cos(x)=-\sin(x)$ And product rule.

25. anonymous

Okay, thank you so much guys, got the answer right finally!