# What is the arc length of f(t)=(3t-4,t^3-2t)  over t in [-1,2]?

Oct 23, 2016

Integration by WolframAlpha
$L = 28.7405$

#### Explanation:

$\frac{\mathrm{dx}}{\mathrm{dt}} = 3$

$\frac{\mathrm{dy}}{\mathrm{dt}} = 3 {t}^{2} - 2$

Let L = the arclength

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

Substituting in our values:

$L = {\int}_{-} {1}^{2} \sqrt{{\left(9\right)}^{2} + {\left(3 {t}^{2} - 2\right)}^{2}} \mathrm{dt}$

$L = 28.7405$