Find the length of the curve from 0 to 1

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SUMMARY

The discussion focuses on calculating the length of the curve defined by the parametric equation r(t) = <4t, t^(2) + 1/6(t)^(3)> for the interval 0 ≤ t ≤ 1. The length is derived using the formula L(t) = ∫ from 0 to 1 of √((dx/dt)² + (dy/dt)² + (dz/dt)²) dt. Participants encountered difficulties simplifying the integral ∫ from 0 to 1 of √(16 + 4t² + 1/324(t⁴)) dt, particularly in the algebraic manipulation required to extract terms from the radical.

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Homework Statement



find the length of the curve… r(t) = <4t, t^(2) + 1/6(t)^(3)> from 0≤t≤1

Homework Equations



L(t) = ∫a to b √(dx/dt)^(2) +(dy/dt)^(2) + (dz/dt)^(2))dt

The Attempt at a Solution



After taking the derivative of all components of the curve and finding the magnitude…
∫ 0 to 1 √(16+4t^(2) + 1/324(t)^(4))dt

And I can't manage to simplify any further.
I tried, doing ∫ 0 to 1 √(t)^(4) + 1296t^(2) + 5184

I don't know how to go any further in simplifying and pulling some t's out of the radical.
I know, it's some algebra, but I have no idea how to go about this from here.
 
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jorgegalvan93 said:
∫ 0 to 1 √(16+4t^(2) + 1/324(t)^(4))dt
The seem to be a few errors in the (dy/dt)2 term.
OTOH, corecting them makes the integrand look even worse.
 

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