# What is the arclength of (2t^3-t^2+1,2t^2-t+4) on t in [-2,1]?

##### 1 Answer
Feb 6, 2017

$L = {\int}_{- 2}^{1} \sqrt{{\left(6 {t}^{2} - 2 t\right)}^{2} + {\left(4 t - 1\right)}^{2}} \mathrm{dt} \approx 24.465$

#### Explanation:

The arc length of the parametric curve $\left(x \left(t\right) , y \left(t\right)\right)$ is found for $t \in \left[a , b\right]$ through:

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

So we first need to find the derivatives:

$x = 2 {t}^{3} - {t}^{2} + 1$

$\frac{\mathrm{dx}}{\mathrm{dt}} = 6 {t}^{2} - 2 t$

$y = 2 {t}^{2} - t + 4$

$\frac{\mathrm{dy}}{\mathrm{dt}} = 4 t - 1$

Then:

$L = {\int}_{- 2}^{1} \sqrt{{\left(6 {t}^{2} - 2 t\right)}^{2} + {\left(4 t - 1\right)}^{2}} \mathrm{dt}$

This cannot be integrated by hand, but in a calculator we see that:

$L = {\int}_{- 2}^{1} \sqrt{{\left(6 {t}^{2} - 2 t\right)}^{2} + {\left(4 t - 1\right)}^{2}} \mathrm{dt} \approx 24.465$