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:

$L=4sqrt2-2$

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