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

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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#