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

Question

Evelyn Agard · Accepted Answer

The formula for the arclength #L# is

#L=int_a^b sqrt ((dx/dt)^2+(dy/dt)^2) dt#

Your parametric equations are

#x=2t^2-t and y= t^4-t#, so

#dx/dt= 4t-1 and dy/dt= 4t^3-1#.

With an interval of #[a,b] = [-4,1]#, this makes

#L=int_-4^1sqrt((4t-1)^2+(4t^3-1)^2) dt#

The inside, #(4 t - 1)^2 + (4 t^3 - 1)^2#, simplifies to #16 t^6-8 t^3+16 t^2-8 t+2#, but this doesn't make the indefinite integral any easier.

And your numerical integral is approximately 266.536.

Charles Arocha · Answer

undefined To find the arc length of the curve $ (2t^2 - t, t^4 - t) $ on the interval $ [-4, 1] $, we'll use the formula for arc length:

$$ L = \int_{a}^{b} \sqrt{{(dx/dt)^2 + (dy/dt)^2}} dt $$

Where:
$ (dx/dt) $ is the derivative of the x-component of the curve with respect to $ t $
$ (dy/dt) $ is the derivative of the y-component of the curve with respect to $ t $
$ a $ and $ b $ are the endpoints of the interval

First, we find the derivatives:

$ dx/dt = 4t - 1 $

$ dy/dt = 4t^3 - 1 $

Now, we substitute these derivatives into the formula and integrate over the interval $[-4, 1]$:

$$ L = \int_{-4}^{1} \sqrt{{(4t - 1)^2 + (4t^3 - 1)^2}} dt $$

This integral can be solved numerically.