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

2 Answers
Jun 24, 2018

#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#

Jun 25, 2018

# (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#