# 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#