# What is the arclength of (t^2-t-1,9t^2-2/t) on t in [1,4]?

Jul 17, 2018

$\approx 137.033$

#### Explanation:

Let $x \left(t\right) = {t}^{2} - t - 1$ then

$x ' \left(t\right) = 2 t - 1$

$y \left(t\right) = 9 {t}^{2} - \frac{2}{t}$

$y ' \left(t\right) = 18 t + \frac{2}{t} ^ 2$
then we have to calculate

${\int}_{1}^{4} \sqrt{{\left(2 t - 1\right)}^{2} + {\left(18 t + \frac{2}{t} ^ 2\right)}^{2}} \mathrm{dt}$

by a numerical Methode we get $\approx 137.033$
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