What is the arc length of the curve given by # x = 1 + 3t^2, y = 4 + 2t^3# on # t in [0, 1]#?

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Answer 1 · 2018-06-14T04:11:05

#L=4sqrt2-2# units.

#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#