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

Question

Charles Anctil · Accepted Answer

#approx 3.4022#

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#

William Abdalla · Answer

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

Isaiah Allen · Answer

undefined The arc length of $f(t) = (\sqrt{t^2-t^3}, t^3-t^2)$ on $t$ in $[-1,1]$ can be found using the formula for arc length:

$$ L = \int_{-1}^{1} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt $$

Where $ \frac{dx}{dt} $ and $ \frac{dy}{dt} $ are the derivatives of $ x $ and $ y $ with respect to $ t $, respectively. In this case,

$$ \frac{dx}{dt} = \frac{t - \frac{3}{2}t^2}{\sqrt{t^2 - t^3}} $$
$$ \frac{dy}{dt} = 3t^2 - 2t $$

Substituting these derivatives into the arc length formula and evaluating the integral over the interval $[-1,1]$ will give the arc length of the curve.