What is the arclength of $(2t^3-t^2+1,2t^2-t+4)$ on $t in [-2,1]$ ?

Question

1515 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-06T00:02:52

$L=int_(-2)^1sqrt((6t^2-2t)^2+(4t-1)^2)dtapprox24.465$

The arc length of the parametric curve $(x(t),y(t))$ is found for $tin[a,b]$ through:

$L=int_a^bsqrt((dx/dt)^2+(dy/dx)^2)dt$

So we first need to find the derivatives:

$x=2t^3-t^2+1$

$dx/dt=6t^2-2t$

$y=2t^2-t+4$

$dy/dt=4t-1$

Then:

$L=int_(-2)^1sqrt((6t^2-2t)^2+(4t-1)^2)dt$

This cannot be integrated by hand, but in a calculator we see that:

$L=int_(-2)^1sqrt((6t^2-2t)^2+(4t-1)^2)dtapprox24.465$

What is the arclength of (2t^3-t^2+1,2t^2-t+4)(2t^3-t^2+1,2t^2-t+4) on t in [-2,1]t in [-2,1]?