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

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Answer 1 · 2015-11-28T06:28:25

The formula for the arclength #L# is

#L=int_a^b sqrt ((dx/dt)^2+(dy/dt)^2) dt#

Your parametric equations are

#x=2t^2-t and y= t^4-t#, so

#dx/dt= 4t-1 and dy/dt= 4t^3-1#.

With an interval of #[a,b] = [-4,1]#, this makes

#L=int_-4^1sqrt((4t-1)^2+(4t^3-1)^2) dt#

The inside, #(4 t - 1)^2 + (4 t^3 - 1)^2#, simplifies to #16 t^6-8 t^3+16 t^2-8 t+2#, but this doesn't make the indefinite integral any easier.

And your numerical integral is approximately 266.536.