When The curvature is maximum of $x^\frac{1}{2}+y^\frac{1}{2}=a^\frac{1}{2}$

QUESTION

Find Where The Curvature has an extremum ? $$x^\frac{1}{2}+y^\frac{1}{2}=a^\frac{1}{2}$$ MY APPROACH

$$x^\frac{1}{2}+y^\frac{1}{2}=a^\frac{1}{2}. . . . . (1)$$

$$\Rightarrow y^\frac{1}{2}=a^\frac{1}{2}-x^\frac{1}{2}$$ now differntiating both sides we will get: $$\frac{1}{2}{y^\frac{-1}{2}}\frac{dy}{dx}=(-1)\frac{1}{2}x^\frac{-1}{2}$$ $$\Rightarrow \frac{dy}{dx}=-(\frac{y}{x})^\frac{1}{2}. . . . . .(2)$$ Now differentiating again with respect to x again: $$\frac{d^2y}{dx^2}=-\bigg(\frac{1}{2}(\frac{y}{x})^\frac{-1}{2}.\frac{d}{dx}(\frac{y}{x})\bigg)$$ $$=-\frac{1}{2}\Bigg(\frac{x^\frac{1}{2}}{y^\frac{1}{2}}\bigg(\frac{d}{dx}(\frac{1}{x})y-\frac{dy}{dx}(\frac{1}{x})\bigg)\Bigg)$$ $$=-\frac{1}{2}\Bigg(\frac{x^\frac{1}{2}}{y^\frac{1}{2}}\bigg(\frac{-y}{x^2}-\frac{dy}{dx}(\frac{1}{x})\bigg)\Bigg)$$ now put the value of $\frac{dy}{dx}$ : $$=\frac{1}{2}\Bigg(\frac{x^\frac{1}{2}}{y^\frac{1}{2}}\bigg(\frac{y}{x^2}-(\frac{y}{x})^\frac{1}{2}\frac{1}{x}\bigg)\Bigg)$$ After simplifying i got :

$$\frac{d^2y}{dx^2}=\frac{1}{2x}\bigg(\frac{y^\frac{1}{2}}{x^2}-1\bigg). . . . .(3)$$ but from simplification of equation (2) in terms of $a$ wil result : $$\frac{dy}{dx}=1-(\frac{a}{x})^\frac{1}{2}$$ here i can easily simlify this to get $\frac{d^2y}{dx^2}$ i.e. $$\Rightarrow\frac{d^2y}{dx^2}=\frac{\sqrt a}{2x\sqrt x}. . . . .(4)$$ I dont know why i am unable to reduce (3) to (4).May be there exists some calculation error,even thats not my question. proceeding to find radius of curvature and curvature :

FROM FORMULA $$\rho=\frac{\bigg(1+(\frac{dy}{dx})^2\bigg)^\frac{3}{2}}{\frac{d^2y}{dx^2}}$$ putting the value of (2) and (4): $$\Rightarrow \rho=\frac{\bigg(1+\frac{y}{x}\bigg)^\frac{3}{2}2x\sqrt x}{\sqrt a}$$ $$\Rightarrow\rho=\frac{2(x+y)^\frac{3}{2}}{\sqrt a}$$ so curvature is $$\frac{1}{\rho}=\kappa=\frac{\sqrt a}{2(2x+a-2\sqrt a\sqrt x)^3/2}$$ [notice that i have put y in terms of a nad x]

NOW BEGINS THE PROBLEM

for being extremum $\frac{d\kappa}{dx} =0$ and i have to check the sign of $\frac{d^2\kappa}{dx^2}$ : as you can see in the image : $$\frac{d\kappa}{dx}=\frac{3\sqrt a(2-\frac{\sqrt a}{\sqrt x})}{4(2x-2\sqrt x\sqrt a+a)^\frac{3}{2}}$$ Now letting this to zero we have : $$2=\frac{\sqrt a}{\sqrt x}$$ thus i am getting $x=\frac{a}{4}$ as a critical point. BUT the answer is given as $\frac{\sqrt 2}{a}$ you can even see by inspection $x=\frac{a}{4}$ is not a critical point. please help and let me know where i have made mistake. THIS IS MY HUMBLE REQUEST.

• Why not to think about a parametrization such as $x=a \cos^4(t)$, $y=s \sin^4(t)$ ? Commented Sep 4, 2018 at 10:49

Perhaps not a complete answer, but below you will find my solution to the problem, which leads to the same answer as yours, so we are either both wrong or the answer sheet is wrong.

Parameterize your curve as $$x(t) = a\cos^4 t, \quad y(t) = a\sin^4 t.$$ Then \begin{align} \dot x(t) &= -4a\cos^3 t \sin t,\\ \ddot x(t) &= -4a(-3\cos^2t\sin^2t + \cos^4 t)\\ &= -4a\cos^2 t(\cos^2t -3\sin^2 t), \end{align} and \begin{align} \dot y(t) &= 4a\sin^3 t \cos t,\\ \ddot y(t) &= 4a(3\sin^2t\cos^2t - \sin^4 t)\\ &= 4a\sin^2 t(3\cos^2t -\sin^2 t). \end{align} Then \begin{align} \dot x \ddot y - \ddot x \dot y &= -16a^2\cos^3 t \sin^3 t (3\cos^2t -\sin^2 t) + 16a^2\sin^3 t \cos^3 t(\cos^2t -3\sin^2 t)\\ &= 16a^2\cos^3 t \sin^3 t (-4\cos^2 t - 4\sin^2t) = -64 a^2 \cos^3 t \sin^3 t, \end{align} and \begin{align} (\dot x^2 + \dot y^2 )^{3/2} &= (16a^2\cos^6t\sin^2t+16a^2\sin^6t\cos^2t)^{3/2}\\ &= 64a^3\cos^3t\sin^3t (\cos^4t+\sin^4t)^{3/2}. \end{align} This gives $$\kappa(t) = \frac{\dot x \ddot y - \ddot x \dot y}{(\dot x^2 + \dot y^2 )^{3/2}} = - \frac{1}{a(\cos^4t+\sin^4t)^{3/2}}.$$ Now \begin{align} \dot\kappa(t) &= \frac{3}{2a}\frac{-4\cos^3t\sin t + 4\sin^3 t\cos t}{(\cos^4t+\sin^4t)^{5/2}}\\ &= \frac{3}{2a}\frac{4\cos t\sin t(\sin^2t- \cos^2 t}{(\cos^4t+\sin^4t)^{5/2}}\\ &= \frac{3}{2a}\frac{-2\sin(2t)\cos(2t)}{(\cos^4t+\sin^4t)^{5/2}}\\ &= -\frac{3\sin(4t)}{2a(\cos^4t+\sin^4t)^{5/2}}. \end{align} So $\dot \kappa = 0$ whenever $\sin(4t) = 0,$ which happens for $$4t = \pi n \Leftrightarrow t = \frac{\pi}{4}n, \quad n\in \mathbb{Z}.$$ Since (I presume that) $x,y > 0$, we are considering $t \in (0, \pi/2)$, so the only valid option is $t = \pi/4$, and then $$x\left(\frac{\pi}{4}\right) = a\cos^4\left(\frac{\pi}{4}\right) = \frac{a}{\sqrt{2}^4} = \frac{a}{4},$$ which confirms your answer.

Also, the graph does seem to indicate that the curvature has an extremum at this point. Note that we are not looking for an extremum of the curve, but rather the extremum of its curvature.

• Thanks,you are right if we consider the graph st x=a/4 we see that there exists little change around its neighbourhood.So i guess it the minima. Commented Sep 4, 2018 at 11:08
• @NewBornMATH I'm glad if it helped!
– MSDG
Commented Sep 4, 2018 at 11:09
• Ya if i take x1=a/4 then at x1-delta to x1+delta for a small delta and take that small segment to join it around then i have a radius which tends to infinity thus curvature tends to zero.which proves we both are right :) Commented Sep 4, 2018 at 11:13
• @NewBornMATH It's always good to hear that!
– MSDG
Commented Sep 4, 2018 at 11:14
• According to you if x=acos^4t and y=asin^4t then x^1/2+y^1/2=a^1/2.cos^2t+a^1/2.sin^2t=2a^1/2 but it given as a^1/2. Commented Sep 4, 2018 at 11:23

the equation $$\sqrt{x}+\sqrt{y}=\sqrt{a}$$ is equivalent to $$y=x-2\sqrt{ax}+a,~~~0\leq x \leq a.$$

Hence, $$y'=1-\frac{a}{\sqrt{ax}},~~~y''=\frac{a}{2x\sqrt{ax}}.$$

Therefore, $$k=\frac{|y'|}{(1+y''^2)^{3/2}}=\frac{1}{2}\sqrt{\frac{a}{(2x-2\sqrt{ax}+a)^3}}.$$

Notice that $$2x-2\sqrt{ax}+a=2\left(\sqrt{x}-\frac{\sqrt{a}}{2}\right)^2+\frac{a}{2}\geq \frac{a}{2}$$ with the equality holding if and only if $\sqrt{x}=\dfrac{\sqrt{a}}{2}$, namely $x=\dfrac{a}{4}$. As a result, $k$ takes its maximum value $k=\dfrac{\sqrt{2}}{a}$ at $x=\dfrac{a}{4}.$

• How you can say it is maximum not minimum ? Commented Sep 4, 2018 at 11:36
• you minimize the denominator and get the maximum... Commented Sep 4, 2018 at 11:38