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## Summary:

- critical points of a function on "critical lines"

## Main Question or Discussion Point

Hi all,

I have recently faced some problem about distances between two curves, and (re?)"discovered" an interesting point that I would like to share with you.

In the following, we consider a function of two variables ##f(x,y)##, but it should be clear that the definitions and the result is valid for functions of more variables.

First, let me set few definitions:

Let ##f(x,y)## be a differentiable function and ##(x_0, y_0)## be a point in its domain. A

Similarly,

If furthermore the point ##(x_0, y_0)## is such that

$$ {df (x, \varphi(x))\over dx}\bigg|_{x=x_0} = 0$$

(resp. ## {df(\psi(y), y)\over dy}\bigg|_{y = y_0} = 0##), then we say that ##(x_0, y_0)## is a critical point of ##f## on the critical partial line ##\varphi## (resp ##\psi##).

With these definitions, we have the following theorem:

Before I give the (very simple proof), let me give an application:

Let ##y = f(x)## and ## y = g(x)## be two curves defined in open intervals. Assume that ##P_1 = (a_1, f(a_1)) ## is a point of curve ##f## that minimize (resp. maximize) the distance ##d((x_1, f(x_1), g)##, and ##P_2= (a_2, g(a_2))## is a point of ##g## that minimize (resp. maximize) the distance ##d((x_2, g(x_2)), f)##.

Then ##(a_1, a_2)## is a critical point of the distance function

$$h(x_1, x_2) = (x_1 - x_2)^2 + (f(x_1) - g(x_2))^2. $$

Furthermore, the straight line between ##P_1## and ##P_2## is orthogonal to both curves (this is the not so easy point).

$$\min_{x_2} h(a_1, x_2).$$ A point of minimum ##a_2## exists and must fulfill

$${\partial h (a_1, a_2)\over \partial x_2} = 0.$$

Moreover, the straight line from ##(a_1, f(a_1))## to ##(a_2, g(a_2))##

is orthogonal to curve ##g##, as is well known.

By the implicit function theorem, there is a function ##\varphi(x_1)## defined in the neighborhood of ##a_1## such that

$${\partial h (x_1, \varphi(x_1))\over \partial x_2} = 0.$$

So ##\varphi## is a critical partial line of ##f## with respect to ##x_2##.

Thus, the problem reduces to find a point ##a_1## that is a critical point of ##h## on ##\varphi##.

By the theorem, ##(a_1, a_2= \varphi(a_1))## must be a critical point of ##f##.

Furthermore, we have seen that the straight line ##(a_1, f(a_1))-(a_2, g(a_2))## is orthogonal to ##g##. By symmetry (which is not evident a priori), taking a critical partial line of ##f## with respect to ##x_1##, it is also orthogonal to ##f## (q.e.d).

Assume for example that ##\varphi## is a critical partial line of ##f## with respect to ##y## at ##(x_0, y_0)##.

We have

$${df (x, \varphi(x))\over dx}\bigg|_{x=x_0} = {\partial f(x_0, \varphi(x_0))\over \partial x} + \varphi'(x_0) {\partial f(x_0, \varphi(x_0))\over \partial y}. $$

Hence the condition

$${\partial f(x_0, y_0)\over \partial x} = 0 \ {\rm and} \ {\partial f(x_0, y_0)\over \partial y} = 0$$

is equivalent to the condition

$${df (x, \varphi(x))\over dx}\bigg|_{x=x_0} = 0 \ {\rm and} \ {\partial f(x_0, \varphi(x_0))\over \partial y} = 0 ,$$

but from the hypothese,

$$ {\partial f(x_0, \varphi(x_0))\over \partial y} = 0 $$

since ##\varphi## is a critical line of ##f## at ##(x_0, y_0)##, with respect to ##y## (q.e.d.)

I have recently faced some problem about distances between two curves, and (re?)"discovered" an interesting point that I would like to share with you.

In the following, we consider a function of two variables ##f(x,y)##, but it should be clear that the definitions and the result is valid for functions of more variables.

First, let me set few definitions:

Let ##f(x,y)## be a differentiable function and ##(x_0, y_0)## be a point in its domain. A

*critical partial line of f*in the neighborhood of ##(x_0, y_0)##, with respect to the variable ##y##, is a function ##\varphi(x)## defined in the neighborhood of ##x_0## such that ##y_0 = \varphi(x_0)## and $$ {\partial f \over \partial y} (x, \varphi(x)) = 0.$$Similarly,

*a critical partial line of f*in the neighborhood of ##(x_0, y_0)## with respect to the variable ##x##, is a function ##\psi(y)## defined in the neighborhood of ##y_0## such that ##x_0 = \psi(y_0)## and $$ {\partial f \over \partial x} (\psi(y), y) = 0.$$If furthermore the point ##(x_0, y_0)## is such that

$$ {df (x, \varphi(x))\over dx}\bigg|_{x=x_0} = 0$$

(resp. ## {df(\psi(y), y)\over dy}\bigg|_{y = y_0} = 0##), then we say that ##(x_0, y_0)## is a critical point of ##f## on the critical partial line ##\varphi## (resp ##\psi##).

With these definitions, we have the following theorem:

*The critical points of ##f## on its critical partial lines are the critical points of ##f##. In particular, they do not depend on the variable with respect to which the critical partial line refers to.*Before I give the (very simple proof), let me give an application:

Let ##y = f(x)## and ## y = g(x)## be two curves defined in open intervals. Assume that ##P_1 = (a_1, f(a_1)) ## is a point of curve ##f## that minimize (resp. maximize) the distance ##d((x_1, f(x_1), g)##, and ##P_2= (a_2, g(a_2))## is a point of ##g## that minimize (resp. maximize) the distance ##d((x_2, g(x_2)), f)##.

Then ##(a_1, a_2)## is a critical point of the distance function

$$h(x_1, x_2) = (x_1 - x_2)^2 + (f(x_1) - g(x_2))^2. $$

Furthermore, the straight line between ##P_1## and ##P_2## is orthogonal to both curves (this is the not so easy point).

**Proof**: The distance of a point ##(a_1, f(a_1))## to curve ##g## is given by$$\min_{x_2} h(a_1, x_2).$$ A point of minimum ##a_2## exists and must fulfill

$${\partial h (a_1, a_2)\over \partial x_2} = 0.$$

Moreover, the straight line from ##(a_1, f(a_1))## to ##(a_2, g(a_2))##

is orthogonal to curve ##g##, as is well known.

By the implicit function theorem, there is a function ##\varphi(x_1)## defined in the neighborhood of ##a_1## such that

$${\partial h (x_1, \varphi(x_1))\over \partial x_2} = 0.$$

So ##\varphi## is a critical partial line of ##f## with respect to ##x_2##.

Thus, the problem reduces to find a point ##a_1## that is a critical point of ##h## on ##\varphi##.

By the theorem, ##(a_1, a_2= \varphi(a_1))## must be a critical point of ##f##.

Furthermore, we have seen that the straight line ##(a_1, f(a_1))-(a_2, g(a_2))## is orthogonal to ##g##. By symmetry (which is not evident a priori), taking a critical partial line of ##f## with respect to ##x_1##, it is also orthogonal to ##f## (q.e.d).

**Proof of the theorem:**Assume for example that ##\varphi## is a critical partial line of ##f## with respect to ##y## at ##(x_0, y_0)##.

We have

$${df (x, \varphi(x))\over dx}\bigg|_{x=x_0} = {\partial f(x_0, \varphi(x_0))\over \partial x} + \varphi'(x_0) {\partial f(x_0, \varphi(x_0))\over \partial y}. $$

Hence the condition

$${\partial f(x_0, y_0)\over \partial x} = 0 \ {\rm and} \ {\partial f(x_0, y_0)\over \partial y} = 0$$

is equivalent to the condition

$${df (x, \varphi(x))\over dx}\bigg|_{x=x_0} = 0 \ {\rm and} \ {\partial f(x_0, \varphi(x_0))\over \partial y} = 0 ,$$

but from the hypothese,

$$ {\partial f(x_0, \varphi(x_0))\over \partial y} = 0 $$

since ##\varphi## is a critical line of ##f## at ##(x_0, y_0)##, with respect to ##y## (q.e.d.)

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