What is the arclength of f(t) = (t-7,t+7) on t in [1,3]?

Jan 30, 2017

$2 \sqrt{2}$

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 (using the product rule):

$x = t - 7 \implies \frac{\mathrm{dx}}{\mathrm{dt}} = 1$
$y = t + 7 \implies \frac{\mathrm{dy}}{\mathrm{dt}} = 1$

So the Arc Length is;

$L = {\int}_{1}^{3} \sqrt{{\left(1\right)}^{2} + {\left(1\right)}^{2}} \setminus \mathrm{dt}$
$\setminus \setminus = {\int}_{1}^{3} \sqrt{2} \setminus \mathrm{dt}$
$\setminus \setminus = \sqrt{2} {\int}_{1}^{3} \setminus \mathrm{dt}$
$\setminus \setminus = \sqrt{2} {\left[t\right]}_{1}^{3}$
$\setminus \setminus = \sqrt{2} \left(3 - 1\right)$
$\setminus \setminus = 2 \sqrt{2}$

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:

$L = \sqrt{{\left(- 6 - \left(- 4\right)\right)}^{2} + {\left(10 - 8\right)}^{2}}$
$\setminus \setminus \setminus = \sqrt{4 + 4}$
$\setminus \setminus \setminus = \sqrt{8}$
$\setminus \setminus \setminus = 2 \sqrt{2}$, as above

So, Maths Works!