What is the arclength of #(9t^3-2t^2+15,-2t^2-12t-1)# on #t in [-2,3]#?

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Answer 1 · 2016-08-28T13:35:07

#approx 325#

Arc length is the time integral of speed, so #s = int_(-2)^3 dot s dt#

and speed #dot s = sqrt(vec v * vec v)# where #vec v# is the velocity vector

#vec v = d/dt vec r #

#= d/dt ((9t^3 - 2t^2 + 15),(-2t^2 - 12t - 1))#

#= ((27t^2 - 4t ),(-4t - 12))#

So #s = int_(-2)^3 sqrt(vec v * vec v) dt#

#= int_(-2)^3 sqrt(((27t^2 - 4t ),(-4t - 12)) * ((27t^2 - 4t ),(-4t - 12))) dt#

#= int_(-2)^3 sqrt( (27t^2 - 4t )^2 + (4t + 12)^2) dt#

#approx 325# units according to computer solution