What is the arc length of $f(t)=(3te^t,t-e^t)$ over $t in [2,4]$ ?

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Answer 1 · 2017-05-15T12:41:30

$612.530$ (3dp)

We have:

$f(t) = (3te^t, t-e^t )$ where $t in [2,4]$

The parametric arc-length is given by:

$L = int_(alpha)^(beta) \ sqrt((dx/dt)^2 + (dy/dt)^2 ) \ dt$

We can differentiate the parameters:

$x(t) = 3te^t => dx/dt = 3te^t + 3e^t$

$y(t) = t-e^t => dy/dt = 1-e^t$

Then the arc-length is given by:

$L = int_2^4 \ sqrt( (3te^t + 3e^t)^2 + (1-e^t)^2 ) \ dt$

This integral dos not have a trivial anti-derivative, and so is evacuated using numerical methods to give:

$L = 612.530$ (3dp)

What is the arc length of f(t)=(3te^t,t-e^t) f(t)=(3te^t,t-e^t) over t in [2,4]t in [2,4]?