Question

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

Answer 1

`approx 3.4022`

Explanation:

We have
`x(t)=sqrt(t^2-t^3)`
then

`x'(t)=1/2*(-3t^2+2t)/sqrt(t^2-t^3)`
by the chain rule

`y(t)=t^3-t^2`
then

`y'(t)=3t^2-2t`
by the power rule

so we have to solve

`int_1^1sqrt((1/2*(-3t^2+2t)/sqrt(-t^3+t^2))^2+(3t^2-2t)^2)dt`
we get by a numerical method

`approx 3.4022`

Answer 2

`(3sqrt2)/2+(2sqrt{43})/27 + 1/4sinh^-1(2sqrt2)+1/2sinh^-1(4/sqrt27)`
`~~ 3.4022`

Explanation:

It is easy to see that the curve traced out in this case is part of a parabola

`y=-x^2`

Thus, the infinitesimal arc-length between two neighboring points on this curve is given simply by

`ds = sqrt{1+(dy/dx)^2}dx = sqrt{1+4x^2}dx`

The only real problem in calculating the total arc length is that as `t` ranges from -1 to +1 different parts of this curve is traversed multiple times. To see this, take a look at the graph of

`x(t) = sqrt{t^2-t^3}`

graph{sqrt(x^2-x^3) [-1.1, 1.1, -0.5, 1.5]}

As can be seen clearly, the value of `x(t)`

• changes monotonously from `sqrt 2` to 0 as `t` goes from -1 to 0.
• After this, it increases from 0 to some `0 < x _0 < 1` as `t` increases from 0 to some `0< t_0< 1`
• and then returns from `x_0` back to 0 as `t` goes from `t_0` to 1.

It is easy to see that `t_0` satisfies

`2t_0-3t_0^2 = 0 implies t_0 = 2/3 implies`

`x_0 = sqrt{(2/3)^2-(2/3)^3} = sqrt{4/27}`

So the parabola `y=-x^2` is traversed in three steps

• from `x=sqrt 2` to `x=0`
  arc length `L_1=| int_sqrt2^0 sqrt{1+4x^2}dx|`
• from `x=0` to `x=x_0=sqrt{4/27}`
  arc length `L_2=| int_0^{sqrt{4/27}} sqrt{1+4x^2}dx|`
• from `x=x_0=sqrt{4/27}` back to `x=0`
  arc length `L_3=L_2`

Since
`int sqrt{1+4x^2} dx = 1/2xsqrt{1+4x^2}+1/4sinh^-1(2x)`

we have

`L_1 = (3sqrt2)/2+1/4sinh^-1(2sqrt2)`
`L_2 = L_3 = sqrt43/27+ 1/4sinh^-1(4/sqrt27)`

Thus the total arc length is

`L = L_1+2L_2`
`= (3sqrt2)/2+(2sqrt{43})/27 + 1/4sinh^-1(2sqrt2)+1/2sinh^-1(4/sqrt27)`
`~~ 3.4022`