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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?

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    $\begingroup$ If it helps, Wolfram Alpha agrees with you upon some simple experimentation (I integrated from $0\to5$ and compared your formula's answer to Wolfram's answer and they agreed). However, I do not own Wolfram Pro and a general form integral took the poor computer too long to calculate, so my one experiment may not be proof that your answer is correct. That being said, mathematically, it does look right to me! $\endgroup$
    – FShrike
    Oct 4, 2021 at 9:18
  • $\begingroup$ @FShrike That was a great idea. I checked my solution against Wolfram and it seems good. But my concern is regarding the legitimacy of the math I've done to arrive at that solution. $\endgroup$ Oct 4, 2021 at 12:40

1 Answer 1

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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 $a=2$):

enter image description here

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.

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  • $\begingroup$ I did a check using Wolfram Alpha and my answer matches with what I've derived. But my concern is whether the math that I've done is legit. wolframalpha.com/input/… $\endgroup$ Oct 4, 2021 at 12:32
  • $\begingroup$ I have corrected an error of mine giving a formula closer to yours, but definitely not the same. $\endgroup$
    – Jean Marie
    Oct 4, 2021 at 12:52
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    $\begingroup$ That's a good catch. I should have not said '1st Quadrant'. I meant the 'one-half'. Let me correct that. $\endgroup$ Oct 4, 2021 at 13:00
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    $\begingroup$ Your answer is correct. It works for all values of $a$. Mine doesn't and also your observation is on point that $L\rightarrow 0$ as $a\rightarrow\infty$ $\endgroup$ Oct 4, 2021 at 13:19
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    $\begingroup$ My genius brain made an integration mistake. I've fixed it now and it is same as yours. Thank you so much for all the help. $\endgroup$ Oct 4, 2021 at 13:35

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