# What is the arc length of the curve given by r(t)=(4t,3t-6) in the interval t in [0,7]?

Nov 25, 2016

35

#### Explanation:

The Arc Length of $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$ is given by:

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

Let $x \left(t\right) = 4 t \implies \frac{\mathrm{dx}}{\mathrm{dt}} = 4$
And, $y \left(t\right) = 3 t - 6 \implies \frac{\mathrm{dy}}{\mathrm{dt}} = 3$

So,

$L = {\int}_{0}^{7} \sqrt{{\left(4\right)}^{2} + {\left(3\right)}^{2}} \mathrm{dt}$
$\therefore L = {\int}_{0}^{7} \sqrt{16 + 9} \mathrm{dt}$
$\therefore L = {\int}_{0}^{7} \sqrt{25} \mathrm{dt}$
$\therefore L = 5 {\int}_{0}^{7} \mathrm{dt}$
$\therefore L = 5 {\left[t\right]}_{0}^{7}$
$\therefore L = 5 \left(7 - 0\right)$
$\therefore L = 35$