Find the arc length on the interval for t between 0 and 1 inclusive for the curve described with the parametric equations: x = 1 + 3t^2, y = 2t^3 + 4. (20 points)

Question

A) 2 times the square root of 2 minus 1
B) 4 times the square root of 2 minus 2
C) 2 times the square root of 2 minus 2
D) the square root of 2 minus 2

Answer 1

s = ∫[0,1] √(36t^2 + 36t^4) dt
= ∫[0,1] 6t√(1 + t^2) dt
Now let u = 1+t^2 and du = 2t dt, giving
s = ∫[1,2] 3√u du
now finish it off

Answer 2

Is the integral from 1 to 2 or 2 to 1

Answer 3

come on, think a bit. You had
∫ from 0 to 1 dt
u = 1+t^2, so now you have
∫ from 1 to 2 du