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

##### 1 Answer

Nov 12, 2016

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