# What is the arclength of f(t) = (sqrt(t-2),t^2) on t in [2,3]?

Feb 19, 2017

$L = {\int}_{2}^{3} \sqrt{\frac{1}{4 \left(t - 2\right)} + 4 {t}^{2}} \mathrm{dt} \approx 5.1927$

#### Explanation:

The arc length of the parametric function $f \left(t\right) = \left(x \left(t\right) , y \left(t\right)\right)$ on $t \in \left[a , b\right]$ is given through:

$L = {\int}_{a}^{b} \sqrt{{\left(x ' \left(t\right)\right)}^{2} + {\left(y ' \left(t\right)\right)}^{2}} \mathrm{dt}$

We see that:

$x \left(t\right) = \sqrt{t - 2} = {\left(t - 2\right)}^{\frac{1}{2}}$

Then:

$x ' \left(t\right) = \frac{1}{2} {\left(t - 2\right)}^{- \frac{1}{2}} \frac{d}{\mathrm{dt}} \left(t - 2\right) = \frac{1}{2 \sqrt{t - 2}}$

And:

$y \left(t\right) = {t}^{2} \text{ "=>" } y ' \left(t\right) = 2 t$

Combining these, we see that:

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

$L = {\int}_{2}^{3} \sqrt{\frac{1}{4 \left(t - 2\right)} + 4 {t}^{2}} \mathrm{dt} \approx 5.1927$

This can't be integrated by hand.