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

Nov 28, 2015

The formula for the arclength $L$ is

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

$x = 2 {t}^{2} - t \mathmr{and} y = {t}^{4} - t$, so

$\frac{\mathrm{dx}}{\mathrm{dt}} = 4 t - 1 \mathmr{and} \frac{\mathrm{dy}}{\mathrm{dt}} = 4 {t}^{3} - 1$.

With an interval of $\left[a , b\right] = \left[- 4 , 1\right]$, this makes

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

The inside, ${\left(4 t - 1\right)}^{2} + {\left(4 {t}^{3} - 1\right)}^{2}$, simplifies to $16 {t}^{6} - 8 {t}^{3} + 16 {t}^{2} - 8 t + 2$, but this doesn't make the indefinite integral any easier.

And your numerical integral is approximately 266.536.