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

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Answer 1 · 2017-05-13T04:58:36

#approx 1.64833#

The arc length for a parametric function is:
#L=int_a^b (sqrt((dx/dt)^2+(dy/dt)^2))dt#

In order to plug in values into this equation, we need to find #dx/dt# and #dy/dt# by differentiating the given function #(x(t),y(t))#:

#x(t)=(3t-2)^(1/2)#

#dx/dt=1/2(3t-2)^(-1/2)(3)=3/2(3t-2)^(-1/2)=frac{3}{2sqrt(3t-2)}#

#y(t)=(t+3)^(-1/2)#

#dy/dt=(-1/2)(t+3)^(-3/2)=frac{-1}{2(t+3)^(3/2)}#

#L=int_1^3(sqrt((frac{3}{2sqrt(3t-2)})^2+(frac{-1}{2(t+3)^(3/2)})^2))dt#

#=int_1^3(sqrt(frac{9}{4(3t-2)}+frac{1}{4(t+3)^3}))dt#

#approx 1.64833#