Question

What is the arclength of `f(t) = (t-sqrt(t^2+2),t+te^(t-2))` on `t in [-1,1]`?

Answer 1

Arc length `s=3.2352212144206" "`units

Explanation:

The formula to obtain the arc length `s`

`s=int_(t_1)^(t_2) sqrt((dx/dt)^2+(dy/dt)^2)*dt`

From the given `f(t)=(t-sqrt(t^2+2), t+t*e^(t-2))`

We have `x=t-sqrt(t^2+2)" "`and `y= t+t*e^(t-2)`

We need to obtain the derivatives `(dx)/dt` and `(dy)/dt`

`(dx)/dt=1-t/sqrt(t^2+2)" "` and `(dy)/dt=1+t*e^(t-2)+e^(t-2)`

We solve now the arclength `s`

`s=int_(t_1)^(t_2) sqrt((dx/dt)^2+(dy/dt)^2)*dt`

`s=int_(-1)^(1) sqrt((1-t/sqrt(t^2+2))^2+(1+t*e^(t-2)+e^(t-2))^2)*dt`

The integral is complicated and requires the use of calculator or Simpson's Rule formula

`s=int_(-1)^(1) sqrt((1-t/sqrt(t^2+2))^2+(1+t*e^(t-2)+e^(t-2))^2)*dt`

`s=3.2352212144206" "`units

God bless....I hope the explanation is useful.