Question

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

Answer 1

`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 `(e^(2t)-t,t-t/e^(t-1))` which is in the form `(x(t),y(t))` in order to get `dx/dt` and `dy/dt`.

For `dx/dt`, use chain rule:

`dx/dt=2e^(2t)-1`

For `dy/dt`, use quotient rule. It may be easier to write `t/e^(t-1)` as `et/e^t` and then ignore the `e` since it's a constant:

`dy/dt=1-e(1(e^t)-(e^t)t)/e^(2t)=1-(1-t)e^(1-t)`

Now plug into the formula and solve:

`A=int_-1^1 sqrt((2e^(2t)-1)^2 +(1-(t-1)e^(t+1))^2)=11.927`