Find the exact length of the curve.

Question

x = 5cost - cos5t, y = 5sint - sin5t,
0 < t < pi

Answer 1

Good old Pythagoras's theorem says that:

ds^2 = dx^2 + dy^2

We can write:

dx = [-5 sin(t) + 5 sin(5t)] dt

dy = [5 cos(t) - 5 cos(t)] dt

ds^2 = dx^2 + dy^2 =

(using sin^2(x) + cos^2(x) = 1 twice for x = t and x = 5t)

25 [2 - 2 sin(t) sin(5t) - 2 cos(t) cos(5t)] dt^2=

50 [1 - cos(5t-t)] dt^2=

50 [1 - cos(4t] dt^2

Then using cos(2x) = 1-2 sin^2(x), this becomes:

ds^2 = 100 sin^2(2t) dt^2 ----->

ds = 10 |sin(2t)| dt


The integral from 0 to pi is twice the integral from 0 to pi/2, which is an interval where sin(2t) is positive:

s = 20 Integral from 0 to pi/2 of sin(2t) dt =

-10 [cos(pi) - cos(0)] = 20

Answer 2

ds = sqrt (dx^2 + dy^2)

dx = -5sin t +5 sin5t

dx^2= 25sin^2 t -50sint sin5t+25sin^2 5t

dy = 5 cos t -5 cos 5t

dy^2= 25cos^2 t-50costcos5t +25 cos^2 5t

luckily sin^2 + cos^2 = 1
dx^2+dy^2=50-50(sin t sin 5t+costcos 5t)

but sin a sin b + cos a cos b = cos(a-b)
= 50 - 50 cos 4 t = 50 (1 - cos 4t)

so
50 integral 0 to pi of (1-cos 4t)^.5 dt

yuuk ! better check my arithmetic.