Length of the curve $\displaystyle 3ay^2=x(x-a)^2$

Question

I'm working on finding out the length of the curve $$ 3ay^2=x(x-a)^2 \tag{1} $$

I ran into a small problem, but was able to end up with an answer that looks right but I'm not entirely sure about it. Here's my approach:

The length of a curve $f(x)$ between $x=a$ and $x=b$ where $b>a$, is given by $$ L=\int_a^b\sqrt{1+\big[f'(x)\big]^2}dx $$ where $f'(x)$ is continuous in $[a,b]$. In our case

$$ L=\int_0^x\sqrt{1+\big[y'\big]^2}dt \tag{2} $$

Curve $(1)$ is symmetric about the $x$-axis. Hence I'll just work on the length of the part that is above the $x$-axis and then double that to obtain the entire length. $$ y=(x-a)\sqrt{\frac{x}{3a}} \text{ (One half of the curve, the other half being the negative multiple)} $$ Note that $(1)$ is defined for $x\geq0$ and $a>0$. That is why I've taken the limits of integration as $0$ to $x$.

Therefore, $$ y'=\frac{3x-a}{\sqrt{12ax}} $$

But $y'$ is not continuous at $x=0$ and hence I can't plug this into $(2)$. So, let me calculate the length from $h>0$ to any $x$. Therefore, we have $$ \begin{aligned} L &= \int_h^x\sqrt{1+\big[y'\big]^2}dt \\ &= \int_h^x\sqrt{1+\left(\frac{3t-a}{\sqrt{12at}}\right)^2}dt \\ &= \int_h^x \frac{3t+a}{\sqrt{12at}} dt \\ &= \frac{x^{\frac{3}{2}}-h^{\frac{3}{2}}+ax^{\frac{1}{2}}-ah^{\frac{1}{2}}}{\sqrt{3a}} \end{aligned} $$

Now, since I want the length in $[0,x]$, I'll just take the limit $h\rightarrow0$, which yields $$ \boxed{L=(x+a)\sqrt{\frac{x}{3a}}} $$ The length of the curve in question would be double the above.

I don't see any problem with what I've done. Is this good?

Answer 1

Cubic curve with implicit equation:

$$3ay^2=x(x-a)^2\tag{1}$$

has an "alpha" shape with a double point $D(a,0)$ as can be seen on this Desmos figure (in the case $a=2$):

This curve can be described in an alternative way using the following parametric representation :

$$\begin{cases}x&=&3am^2\\y&=&am(3m^2-1)\end{cases}\tag{2}$$

Explanation: parameter $m$ has the following geometrical interpretation : it is the slope of a variable line (represented in blue on the figure) passing through double point $D$ with equation:

$$y=m(x-a)\tag{3}$$

Plugging (3) into (1) gives (2), after simplification (end of explanation of (2)).

The length of the curve between parameters values $m_1$ and $m_2$ is (using a classical formula):

$$L=\int_{m_1}^{m_2}\sqrt{x'(m)^2+y'(m)^2}dm$$

$$L=\int_{m_1}^{m_2}\sqrt{(6am)^2+|9am^2-a|^2}dm$$

$$L=a\int_{m_1}^{m_2}\sqrt{(6m)^2+(9m^2-1)^2}dm$$

$$L=a\int_{m_1}^{m_2}\sqrt{81m^4+18m^2+1}dm$$

$$L=a\int_{m_1}^{m_2}(9m^2+1)dm$$

$$L=a[m(3m^2+1)]_{m_1}^{m_2},\tag{4}$$

plainly. It remains to convert (4) in terms of variable $x$ where

$$x=3am^2 \iff m=\sqrt{\frac{x}{3a}}$$

(if we consider only positive slopes).

(4) gives

$$L=(x+a)\sqrt{\frac{x}{3a}}$$

a result which is now the same as yours.

Remarks:

1. we are here in an exceptional case where we have an analytic formula for the arc length... These exceptional cases for cubic curves have been studied here using cubic Bezier curves techniques.

2. The parameterization using the slope of a line passing through the double point is classical.