What is the arclength of #f(t) = (t-2,1/(t-2))# on #t in [0,1]#?

Question

1008 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-16T21:42:22

Use the equation:

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

#L ≈ 1.13209#

The arc length, L, for parametric equations is:

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

#dx/dt = 1#
#dy/dt = -(t - 2)^-2#
#a = 0#
#b = 1#

#L = int_0^1sqrt((1)^2 + (-(t - 2)^-2)^2)dt#

#L = int_0^1sqrt(1 + (t - 2)^-4)dt#

#L ≈ 1.13209#