# What is the arc length of r(t)=(t,t,t) on tin [1,2]?

Apr 12, 2018

$\sqrt{3}$

#### Explanation:

We seek the arc length of the vector function:

$\boldsymbol{\underline{r} \left(t\right)} = \left\langlet , t , t\right\rangle$ for $t \in \left[1 , 2\right]$

Which we can readily evaluated using:

$L = {\int}_{\alpha}^{\beta} \setminus | | \boldsymbol{\underline{r '} \left(t\right)} | | \setminus \mathrm{dt}$

So we calculate the derivative, $\boldsymbol{\underline{r '} \left(t\right)}$:

$\boldsymbol{\underline{r} ' \left(t\right)} = \left\langle1 , 1 , 1\right\rangle$

Thus we gain the arc length:

$L = {\int}_{1}^{2} \setminus | | \left\langle1 , 1 , 1\right\rangle | | \setminus \mathrm{dt}$

$\setminus \setminus = {\int}_{1}^{2} \setminus \sqrt{{1}^{1} + {1}^{2} + {1}^{2}} \setminus \mathrm{dt}$

$\setminus \setminus = {\int}_{1}^{2} \setminus \sqrt{3} \setminus \mathrm{dt}$

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

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

$\setminus \setminus = \sqrt{3}$

This trivial result should come as no surprise as the given original equation is that of a straight line.