arclength of <1,t^2,t^3> from 0≤t≤1?
idk what im doing wrong.. i know the answer but cant seem to get it...
i know length is integral of the magnitude of the derivative
lr'(t)l =sqrt((4t^2)+(9t^4))
but i cant seemto be able to work it from there...
thanks for the help
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You want the integral of sqrt(4t^2 + 9t^4)dt = t*sqrt(4 + 9t^2)dt from t = 0 to t = 1.
Let u = 4 + 9t^2, so so du/18 = tdt.
I know you can finish it.