Points of inflection in a implicite curve

**scifo** · July 21st 2010, 09:27 PM

Hi!

You can say like finding the eventual points of inflection of a curve given in implicit form f (x, y)=0 and that not it can or is difficult to explicit in form y=y (x) or x=x (y),
how for example
x^5 + y^5 + x^2*y - x*y^2 = 0 ?

P.S. I hope it is the correct subforum ...

**chisigma** · July 21st 2010, 09:59 PM

Given a function $y(x)$ expressed in the form $f(x,y)=0$ , if certain conditions are satisfied, the derivative exists and is...

$\displaystyle y^{'} (x) = - \frac{f_{x}^{'} (x,y)}{f_{y}^{'} (x,y)}$ (1)

Now if further conditions are satisfied it exists also the second derivative and is...

$\displaystyle y^{''} (x) = - \frac{f_{xx}^{''}\ f_{y}^{' 2} - 2 f_{xy}^{''}\ f_{x}^{'}\ f_{y}^{'} + f_{yy}^{''}\ f_{x}^{'2}}{f_{y}^{'3}}$ (2)

At this point the inflection points of $y(x)$ are found in standard way imposing that $y^{''} (x)=0$ ...

Kind regards

$\chi$ $\sigma$

**HallsofIvy** · July 22nd 2010, 12:08 AM

chisigma, I am confused as to why you have both the $f'$ that is normally used with ordinary derivatives and $f_x$ and $f_y$ for partial derivatives.

I would say that if f(x,y)= 0, then $f_x+ f_y\frac{dy}{dx}= 0$ so that $\frac{dy}{dx}= -\frac{f_x}{f_y}$ .

Then, using the quotient rule, $\frac{d^2y}{dx^2}=-\frac{f_{xx}f_y- f_xf_{yy}}{(f_y)^2}$ . That will be 0 when the numerator is 0: when $f_{xx}f_y= f_xf_{yy}$ .

**scifo** · July 22nd 2010, 12:50 AM

Thanks for your explanation.
Therefore x,y is a point of inflection when
simply fxx*fy = fx*fyy ?

**chisigma** · July 22nd 2010, 12:19 PM

Originally Posted by HallsofIvy

... I would say that if f(x,y)= 0, then $f_x+ f_y\frac{dy}{dx}= 0$ so that $\frac{dy}{dx}= -\frac{f_x}{f_y}$ . Then, using the quotient rule, $\frac{d^2y}{dx^2}=-\frac{f_{xx}f_y- f_xf_{yy}}{(f_y)^2}$ . That will be 0 when the numerator is 0: when $f_{xx}f_y= f_xf_{yy}$ .

The matter is a little different: starting from...

$f(x,y(x))=0$ (1)

... deriving both terms respect to x You obtain...

$\displaystyle f_{x} + f_{y} \ y^{'} = 0 \rightarrow y^{'} = - \frac{f_{x}}{f_{y}}$ (2)

Now if You derive both terms of (2) respect to x You obtain...

$\displaystyle f_{xx} + 2\ f_{xy}\ y^{'} + f_{yy}\ y^{'2} + f_{y}\ y^{''} = 0$ (3)

... that setting $y^{'} = - \frac{f_{x}}{f_{y}}$ becomes...

$\displaystyle y^{''} = - \frac{f_{xx}\ f_{y}^{2} - 2\ f_{xy}\ f_{x}\ f_{y} + f_{yy}\ f_{x}^{2}}{f_{y}^{3}}$ (4)

Now the point of inflexion of y(x), if they exist, are the points in the x y plane for which the relations...

$f = 0$

$f_{xx}\ f_{y}^{2} - 2\ f_{xy}\ f_{x}\ f_{y} + f_{yy}\ f_{x}^{2} =0$ (5)

... are both satisfied with the condition that $f_{y} \ne 0$ ...

Kind regards

$\chi$ $\sigma$

**HallsofIvy** · July 23rd 2010, 02:29 AM

Thanks, that was a lot clearer.

**Ackbeet** · July 23rd 2010, 02:47 AM

You've stated necessary but not sufficient conditions for points to be points of inflection. Technically, the second derivative being zero does not imply that you have a point of inflection. You also need the lowest-order non-zero derivative to be of odd order. So, check out the first derivative at any candidate points. If it's nonzero at a candidate point, then it's a point of inflection. If the first derivative is zero at a candidate point, then check the third derivative (oh, joy!). Alternatively, a point of inflection is a point where the second derivative changes sign. So, if you find all the zeros of the second derivative, you should be able, by plugging in appropriate x-values, to see whether the second derivative changes sign at its zeros or not.

**scifo** · July 25th 2010, 07:55 PM

Thank you very much for your explanations about point of inflection.

But I have a doubt...
For the quintic
F = y(20^4 - 60y^2 + 39) + 20x^2y(5x^2 + 6y^2 - 7) + x = 0
I have 13 points of inflection or 14?
The 14° point is infinite orizzontal point ?

　
I have the graphic of this curve
F = 0
and its intersections with the curve
FxxFy^2-2FxyFxFy+FyyFx^2 = 0 ,
but I don't know how to send you (please, without the http:\!)　
Help me, please!

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