Question

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

Answer 1

`= 2 sqrt(3)`

Explanation:

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)`