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

Apr 14, 2017

$11.927$

#### Explanation:

The arc length formula for a parametric equation is:

A=int_a^bsqrt((dx/dt)^2+(dy/dt)^2

We need to take the derivative of $\left({e}^{2 t} - t , t - \frac{t}{e} ^ \left(t - 1\right)\right)$ which is in the form $\left(x \left(t\right) , y \left(t\right)\right)$ in order to get $\frac{\mathrm{dx}}{\mathrm{dt}}$ and $\frac{\mathrm{dy}}{\mathrm{dt}}$.

For $\frac{\mathrm{dx}}{\mathrm{dt}}$, use chain rule:

$\frac{\mathrm{dx}}{\mathrm{dt}} = 2 {e}^{2 t} - 1$

For $\frac{\mathrm{dy}}{\mathrm{dt}}$, use quotient rule. It may be easier to write $\frac{t}{e} ^ \left(t - 1\right)$ as $e \frac{t}{e} ^ t$ and then ignore the $e$ since it's a constant:

$\frac{\mathrm{dy}}{\mathrm{dt}} = 1 - e \frac{1 \left({e}^{t}\right) - \left({e}^{t}\right) t}{e} ^ \left(2 t\right) = 1 - \left(1 - t\right) {e}^{1 - t}$

Now plug into the formula and solve:

$A = {\int}_{-} {1}^{1} \sqrt{{\left(2 {e}^{2 t} - 1\right)}^{2} + {\left(1 - \left(t - 1\right) {e}^{t + 1}\right)}^{2}} = 11.927$