What is the arclength of #f(t) = (t,t,t)# on #t in [1,3]#?

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Answer 1 · 2016-06-26T10:08:02

#= 2 sqrt(3)#

that's a straight line so the Pythagorean answer is simply

#sqrt{(3-1)^2+(3-1)^2+(3-1)^2} = 2 sqrt(3)#

if you want to do it using calculus then

#ds = sqrt{dot x ^2 + dot y ^2 + dot z ^2} \ dt= sqrt(3) \ dt#

[where the dot, #dot#, denotes #d/(dt}#]

so you have

#S = int \ ds = int_{t=1}^3 \ sqrt(3) \ dt#

# = sqrt(3) \ [t]_{1}^3 = 2 sqrt(3)#