# What is the arclength of (t,t-1) on t in [-7,1]?

Jan 5, 2017

$8 \sqrt{2}$

#### Explanation:

The Arc Length for a Parametric Curve is given by

$L = {\int}_{\alpha}^{\beta} \sqrt{{\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)}^{2} + {\left(\frac{\mathrm{dy}}{\mathrm{dt}}\right)}^{2}} \setminus \mathrm{dt}$

So in this problem we have

$x = t \implies \frac{\mathrm{dx}}{\mathrm{dt}} = 1$
$y = t - 1 \implies \frac{\mathrm{dy}}{\mathrm{dt}} = 1$

So the Arc Length is;

$L = {\int}_{-} {7}^{1} \sqrt{{\left(1\right)}^{2} + {\left(1\right)}^{2}} \setminus \mathrm{dt}$
$\setminus \setminus \setminus = \sqrt{2} \setminus {\int}_{-} {7}^{1} \setminus \mathrm{dt}$
$\setminus \setminus \setminus = \sqrt{2} \setminus {\left[t \textcolor{w h i t e}{\int}\right]}_{-} {7}^{1} \setminus \mathrm{dt}$
$\setminus \setminus \setminus = \sqrt{2} \setminus \left\{\left(1\right) - \left(- 7\right)\right\}$
$\setminus \setminus \setminus = 8 \sqrt{2}$

Additionally, If we look at the actual graph of the parametric curve:

we can see that the equations represent a straight line, so in fact we can easily calculate the are length (coloured blue) from a triangle using Pythagoras:

$L = \sqrt{{\left(1 - \left(- 7\right)\right)}^{2} + {\left(0 - 8\right)}^{2}}$
$\setminus \setminus \setminus = \sqrt{{8}^{2} + {8}^{2}}$
$\setminus \setminus \setminus = \sqrt{64 + 64}$
$\setminus \setminus \setminus = \sqrt{128}$
$\setminus \setminus \setminus = 8 \sqrt{2}$, as above

So, Maths Works!