Vtyjkyuk

QUESTION

Find Where The Curvature has an extremum ?
$$x^frac12+y^frac12=a^frac12$$
MY APPROACH

$$x^frac12+y^frac12=a^frac12. . . . . (1)$$

$$Rightarrow y^frac12=a^frac12-x^frac12$$
now differntiating both sides we will get:
$$frac12y^frac-12fracdydx=(-1)frac12x^frac-12$$
$$Rightarrow fracdydx=-(fracyx)^frac12. . . . . .(2)$$
Now differentiating again with respect to x again:
$$fracd^2ydx^2=-bigg(frac12(fracyx)^frac-12.fracddx(fracyx)bigg)$$
$$=-frac12Bigg(fracx^frac12y^frac12bigg(fracddx(frac1x)y-fracdydx(frac1x)bigg)Bigg)$$
$$=-frac12Bigg(fracx^frac12y^frac12bigg(frac-yx^2-fracdydx(frac1x)bigg)Bigg)$$
now put the value of $fracdydx$ :
$$=frac12Bigg(fracx^frac12y^frac12bigg(fracyx^2-(fracyx)^frac12frac1xbigg)Bigg)$$
After simplifying i got :

$$fracd^2ydx^2=frac12xbigg(fracy^frac12x^2-1bigg). . . . .(3)$$
but from simplification of equation (2) in terms of $a$ wil result :
$$fracdydx=1-(fracax)^frac12$$
here i can easily simlify this to get $fracd^2ydx^2$ i.e.
$$Rightarrowfracd^2ydx^2=fracsqrt a2xsqrt x. . . . .(4)$$
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
$$rho=fracbigg(1+(fracdydx)^2bigg)^frac32fracd^2ydx^2$$
putting the value of (2) and (4):
$$Rightarrow rho=fracbigg(1+fracyxbigg)^frac322xsqrt xsqrt a$$
$$Rightarrowrho=frac2(x+y)^frac32sqrt a$$
so curvature is
$$frac1rho=kappa=fracsqrt a2(2x+a-2sqrt asqrt x)^3/2$$
[notice that i have put y in terms of a nad x]

NOW BEGINS THE PROBLEM

for being extremum
$fracdkappadx =0 $ and i have to check the sign of $fracd^2kappadx^2$ :

as you can see in the image :
$$fracdkappadx=frac3sqrt a(2-fracsqrt asqrt x)4(2x-2sqrt xsqrt a+a)^frac32$$
Now letting this to zero we have :
$$2=fracsqrt asqrt x$$
thus i am getting $x=fraca4$ as a critical point.
BUT the answer is given as $fracsqrt 2a$
you can even see by inspection $x=fraca4$ is not a critical point.

please help and let me know where i have made mistake.
THIS IS MY HUMBLE REQUEST.

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
$$ x(t) = acos^4 t, quad y(t) = asin^4 t. $$
Then
beginalign
dot x(t) &= -4acos^3 t sin t,\
ddot x(t) &= -4a(-3cos^2tsin^2t + cos^4 t)\
&= -4acos^2 t(cos^2t -3sin^2 t),
endalign
and
beginalign
dot y(t) &= 4asin^3 t cos t,\
ddot y(t) &= 4a(3sin^2tcos^2t - sin^4 t)\
&= 4asin^2 t(3cos^2t -sin^2 t).
endalign
Then
beginalign
dot x ddot y - ddot x dot y
&= -16a^2cos^3 t sin^3 t (3cos^2t -sin^2 t) + 16a^2sin^3 t cos^3 t(cos^2t -3sin^2 t)\
&= 16a^2cos^3 t sin^3 t (-4cos^2 t - 4sin^2t) = -64 a^2 cos^3 t sin^3 t,
endalign
and
beginalign
(dot x^2 + dot y^2 )^3/2
&= (16a^2cos^6tsin^2t+16a^2sin^6tcos^2t)^3/2\
&= 64a^3cos^3tsin^3t (cos^4t+sin^4t)^3/2.
endalign
This gives
$$ kappa(t) = fracdot x ddot y - ddot x dot y(dot x^2 + dot y^2 )^3/2 = - frac1a(cos^4t+sin^4t)^3/2.$$
Now
beginalign
dotkappa(t) &= frac32afrac-4cos^3tsin t + 4sin^3 tcos t(cos^4t+sin^4t)^5/2\
&= frac32afrac4cos tsin t(sin^2t- cos^2 t(cos^4t+sin^4t)^5/2\
&= frac32afrac-2sin(2t)cos(2t)(cos^4t+sin^4t)^5/2\
&= -frac3sin(4t)2a(cos^4t+sin^4t)^5/2.
endalign
So $dot kappa = 0$ whenever $sin(4t) = 0,$ which happens for
$$ 4t = pi n Leftrightarrow t = fracpi4n, quad nin mathbbZ. $$
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
$$ xleft(fracpi4right) = acos^4left(fracpi4right) = fracasqrt2^4 = fraca4, $$
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.

the equation $$sqrtx+sqrty=sqrta$$ is equivalent to $$y=x-2sqrtax+a,~~~0leq x leq a.$$

Hence, $$y'=1-fracasqrtax,~~~y''=fraca2xsqrtax.$$

Therefore, $$k=fracy'(1+y''^2)^3/2=frac12sqrtfraca(2x-2sqrtax+a)^3.$$

Notice that $$2x-2sqrtax+a=2left(sqrtx-fracsqrta2right)^2+fraca2geq fraca2$$ with the equality holding if and only if $sqrtx=dfracsqrta2$, namely $x=dfraca4$. As a result, $k$ takes its maximum value $k=dfracsqrt2a$ at $x=dfraca4.$