how the hell do you solve something like this? x = 2cos(t), y= 2t + 2sin(t), 0<t<pi/2?

Question

will the 2cos and the 2sin will be 1 after they get plugged and simplified in the parametric curve equation ?? squared root of dx squared plus dy squared???

Answer 1

what do you want to do? once that is clear, then it is easier to explain how to solve it.
If you want to find the LENGTH of the curve
x = 2cos(t), y= 2t + 2sin(t), for 0 one needs:
dx = -2sint dt
dy = (2 +2cost )dt
length = integral from 0 to pi/2 sqrt( (-2sint)^2 + (2+2cost)^2 ) dt
= integral from 0 to pi/2 sqrt( 4sin^2 t + 4 +8 cost + 4cos^2 t ) dt
= integral from 0 to pi/2 sqrt( 4 + 4 +8 cost ) dt
= integral from 0 to pi/2 sqrt( 8+8 cost ) dt .
now remember that cos (2x) + 1 = 2 cos^2 x,
so in this case 8+8 cost =8( 1 + cost) = 8 (2 cos^2 (t/2) ) = 16 cos^2 (t/2)
so the integral becomes:
integral from 0 to pi/2 sqrt( 8+8 cost ) dt
= integral from 0 to pi/2 sqrt( 16 cos^2 (t/2) ) dt
= integral from 0 to pi/2 4 cos (t/2) dt
= 8 sin(t/2) from 0 to pi/2
= 8( sin (pi/4) - sin (0/2) ) = 8 sqrt(2) /2 = 4 sqrt(2).

Answer 2

Well, there are 3 equations and three unknowns. Looks like x and y will be ranges, just as t is. But why go into derivatives (dx and dy)? Unless I am missing something, x varies from 2cos(0) through 2cos(pi/2), and y varies from 2sin(0) through pi+2sin(pi/2).