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

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Answer 1 · 2017-05-01T20:15:44

#"Arclength " approx 10.601267#

#f(t)=(t^2sqrt(t-1),t^2+t-1)#

Formula for parametric arclength: #color(blue)(L = int_a^b sqrt((dx/dt)^2+(dy/dt)^2)" "dt)#

First, find #f'(t)# by differentiating #f(t)#

#dx/dt=(t^2)(1/2(t-1)^(-1/2))+(2t)(t-1)^(1/2)#

#dx/dt=frac{t^2}{2sqrt(t-1)}+2tsqrt(t-1)=frac{t^2+4t(t-1)}{2sqrt(t-1)}#

#dy/dt=2t+1#

Plug in expressions for #dx/dt# and #dy/dt# into the arclength formula:
#L=int_2^3sqrt((frac{t^2+4t(t-1)}{2sqrt(t-1)})^2+(2t+1)^2)" "dt#

Use a calculator to evaluate:
#"Arclength " approx 10.601267#