What is the arc length of the curve given by x = 1 + 3t^2, y = 4 + 2t^3 on t in [0, 1]?
1 Answer
Jun 14, 2018
Explanation:
x=1+3t^2
x'=6t
y=4+2t^3
y'=6t^2
The arc length is given by:
L=int_0^1sqrt(36t^2+36t^4)dt
Simplify:
L=6int_0^1tsqrt(1+t^2)dt
Integrate directly:
L=2[(1+t^2)^(3/2)]_0^1
Insert the limits:
L=4sqrt2-2