Question

Arc length of cycloidal curve?

Answer 1

Find the arclength of the cycloid curve given by (x, y) = (t−sin(t), 1−cos(t))
from the point given by t =1/2 * pi to the point given by t = 5/3 * pi
(A) 2√2 + 2√3,
(B) 2√2 + 3,
(C) 2√2 + 2,
(D) 2√3 + 2,
(E) 3√3 − 2.

Answer 2

Paul's notes are pretty good
http://tutorial.math.lamar.edu/Classes/CalcII/ParaArcLength.aspx

Answer 3

If I remember correctly the arc length is \[L = \int\limits_{a}^{b} \sqrt{\left( \frac{ dx }{ dt } \right)^2+\left( \frac{ dy }{ dt } \right)^2}dt\]

Answer 4

And you have your paramaterization already so I guess you can integrate away
(First find the derivatives)

Answer 5

i found the derivatives as dx/dt = 1-cos(t) and dy/dt = sin t
and then i applied the same formula as @iambatman found but my answer is -2, which is clearly not correct

Answer 6

Then you square the derivatives so you get, \[\left( \frac{ dx }{ dt } \right)^2 = \cos^2t-2cost+1\] and \[\left( \frac{ dy }{ dt } \right)^2 = \sin^2 t\] so your integral should look like \[\int\limits_{\pi/2}^{5 \pi/3} \sqrt{(\cos^2t-2cost+1+\sin^2t)}dt\] use the relationship \[\cos^2t+\sin^2t=1\] to simplify the integrand, and see what you get.

Answer 7

You may also need the following: \[1-cost = 2\sin^2\left( \frac{ t }{ 2 } \right)\]

Answer 8

oh i see now where i made a mistake!! i might have forgotten the (t/2) and just worked with t while plugging the numbers in!! let me re-do that