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

Jun 26, 2016

$= 2 \sqrt{3}$

#### Explanation:

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

$\sqrt{{\left(3 - 1\right)}^{2} + {\left(3 - 1\right)}^{2} + {\left(3 - 1\right)}^{2}} = 2 \sqrt{3}$

if you want to do it using calculus then

$\mathrm{ds} = \sqrt{{\dot{x}}^{2} + {\dot{y}}^{2} + {\dot{z}}^{2}} \setminus \mathrm{dt} = \sqrt{3} \setminus \mathrm{dt}$

[where the dot, dot, denotes $\frac{d}{\mathrm{dt}}$]

so you have

$S = \int \setminus \mathrm{ds} = {\int}_{t = 1}^{3} \setminus \sqrt{3} \setminus \mathrm{dt}$

$= \sqrt{3} \setminus {\left[t\right]}_{1}^{3} = 2 \sqrt{3}$