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

Jun 24, 2018

$\approx 3.4022$

#### Explanation:

We have
$x \left(t\right) = \sqrt{{t}^{2} - {t}^{3}}$
then

$x ' \left(t\right) = \frac{1}{2} \cdot \frac{- 3 {t}^{2} + 2 t}{\sqrt{{t}^{2} - {t}^{3}}}$
by the chain rule

$y \left(t\right) = {t}^{3} - {t}^{2}$
then

$y ' \left(t\right) = 3 {t}^{2} - 2 t$
by the power rule

so we have to solve

${\int}_{1}^{1} \sqrt{{\left(\frac{1}{2} \cdot \frac{- 3 {t}^{2} + 2 t}{\sqrt{- {t}^{3} + {t}^{2}}}\right)}^{2} + {\left(3 {t}^{2} - 2 t\right)}^{2}} \mathrm{dt}$
we get by a numerical method

$\approx 3.4022$

Jun 25, 2018

$\frac{3 \sqrt{2}}{2} + \frac{2 \sqrt{43}}{27} + \frac{1}{4} {\sinh}^{-} 1 \left(2 \sqrt{2}\right) + \frac{1}{2} {\sinh}^{-} 1 \left(\frac{4}{\sqrt{27}}\right)$
$\approx 3.4022$

#### 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

$\mathrm{ds} = \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx} = \sqrt{1 + 4 {x}^{2}} \mathrm{dx}$

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

$x \left(t\right) = \sqrt{{t}^{2} - {t}^{3}}$

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

As can be seen clearly, the value of $x \left(t\right)$

• 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

$2 {t}_{0} - 3 {t}_{0}^{2} = 0 \implies {t}_{0} = \frac{2}{3} \implies$

${x}_{0} = \sqrt{{\left(\frac{2}{3}\right)}^{2} - {\left(\frac{2}{3}\right)}^{3}} = \sqrt{\frac{4}{27}}$

So the parabola $y = - {x}^{2}$ is traversed in three steps

• from $x = \sqrt{2}$ to $x = 0$
arc length ${L}_{1} = | {\int}_{\sqrt{2}}^{0} \sqrt{1 + 4 {x}^{2}} \mathrm{dx} |$
• from $x = 0$ to $x = {x}_{0} = \sqrt{\frac{4}{27}}$
arc length ${L}_{2} = | {\int}_{0}^{\sqrt{\frac{4}{27}}} \sqrt{1 + 4 {x}^{2}} \mathrm{dx} |$
• from $x = {x}_{0} = \sqrt{\frac{4}{27}}$ back to $x = 0$
arc length ${L}_{3} = {L}_{2}$

Since
$\int \sqrt{1 + 4 {x}^{2}} \mathrm{dx} = \frac{1}{2} x \sqrt{1 + 4 {x}^{2}} + \frac{1}{4} {\sinh}^{-} 1 \left(2 x\right)$

we have

${L}_{1} = \frac{3 \sqrt{2}}{2} + \frac{1}{4} {\sinh}^{-} 1 \left(2 \sqrt{2}\right)$
${L}_{2} = {L}_{3} = \frac{\sqrt{43}}{27} + \frac{1}{4} {\sinh}^{-} 1 \left(\frac{4}{\sqrt{27}}\right)$

Thus the total arc length is

$L = {L}_{1} + 2 {L}_{2}$
$= \frac{3 \sqrt{2}}{2} + \frac{2 \sqrt{43}}{27} + \frac{1}{4} {\sinh}^{-} 1 \left(2 \sqrt{2}\right) + \frac{1}{2} {\sinh}^{-} 1 \left(\frac{4}{\sqrt{27}}\right)$
$\approx 3.4022$