Question

Find the length of the curve.
x=3t² +5₁y = 2t³ +5,0 ≤t≤1

Answer 1

The length of the curve will be given by the definite integral

`\displaystyle \int_0^1 \sqrt{\left((dx)/(dt)\right)^2 + \left((dy)/(dt)\right)^2} \, dt`

From the given parametric equations, we get derivatives

`x(t) = 3t^2 + 5 \implies (dx)/(dt) = 6t`

`y(t) = 2t^3 + 5 \implies (dy)/(dt) = 6t^2`

Then the arc length integral becomes

`\displaystyle \int_0^1 √(\left(6t\right)^2 + \left(6t^2\right)^2) \, dt = \int_0^1 √(36t^2 + 36t^4) \, dt \\\\ = \int_0^1 6|t| √(1 + t^2) \, dt`

Since 0 ≤ t ≤ 1, we have |t| = t, so

`\displaystyle \int_0^1 6|t| √(1 + t^2) \, dt = 6 \int_0^1 t √(1 + t^2) \, dt`

For the remaining integral, substitute
`u = 1 + t^2` and
`du = 2t \, dt`. Then

`\displaystyle 6 \int_0^1 t √(1 + t^2) \, dt = 3 \int_1^2 √(u) \, du \\\\ = 3* \frac23 u^(3/2) \bigg|_(u=1)^2 \\\\ = 2 \left(2^(3/2) - 1^(3/2)\right) = 2^(5/2) - 2 = \boxed{4\sqrt2-2}`