Question

What is the arclength of `(3t^2,t^4-t)` on `t in [-4,1]`?

Answer 1

In parametric form, the Arc Length is given by:

`L = int_alpha^beta sqrt((dx/dt)^2+(dy/dt)^2) dt`

We have:

`{ ( x(t) = 3t^2, =>dx/dt=6t ), (y(t)=t^4-t, =>dy/dt=4t^3-1 ) :}`

So the Arc Length is given by;

`L = int_-4^1 sqrt( (6t)^2 + (4t^3-1)^2 ) dt`
`:. L = int_-4^1 sqrt( 36t^2 + 16t^6-8t^3+1 ) dt`

And this is as far as we can get analytically as there is no elementary anti-derivative and we would have to resort to a numerical solution to find the value of the definite integral