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

L=4sqrt2-2 units.

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