mathproject >> 7.10. Geometric Properties of Differentiable Functions

Now we are going to study two geometric behaviours in more detail, monotony and curvature. It will turn out that monotony is related to the first derivative and curvature to the second.

Definition: A function $f : A \to ℝ$ is called

increasing (or monotonic increasing) on $B \subset A$ , if the implication

x < y \Rightarrow f (x) \leq f (y)

[7.10.1]

decreasing (or monotonic decreasing) on $B \subset A$ , if the implication

x < y \Rightarrow f (x) \geq f (y)

[7.10.2]

holds for all $x, y \in B$ .

f is called strictly increasing on B, if the conclusion in [7.10.1] could be tightened to $f (x) < f (y)$ . The property strictly decreasing is introduced appropriately.

Usually the term "on B" is only used if $A \neq B$ .

[1] and [2] however allow to restate the monotony condition in terms of difference quotient functions:

Proposition: A function $f : A \to ℝ$ is

monotonic increasing on $B \subset A$ , if and only if any two different points $x, y \in B$ satisfy

\frac{f (y) - f (x)}{y - x} \geq 0

[7.10.3]

monotonic deacreasing on $B \subset A$ , if and only if any two different points $x, y \in B$ satisfy

\frac{f (y) - f (x)}{y - x} \leq 0

[7.10.4]

Proof: As both assertions are similar to prove, we only show the first one.

" $\Rightarrow$ ": Let f be increasing. From [1] we see that for $x \neq y$ both differences, $y - x$ and $f (y) - f (x)$ , have the same sign, so that the quotient $\frac{f (y) - f (x)}{y - x}$ is always positive.

" $\Leftarrow$ ": If now $y - x > 0$ , then $f (y) - f (x) \geq 0$ because otherwise we would have $\frac{f (y) - f (x)}{y - x} < 0$ in contrast to the premise.

For differentiable functions on intervals the above result will lead to the monotony test, a method to characterise the monotony of f by its derivative behaviour. The monotony test puts the first derivative into a geometric perspective.

Proposition (monotony test): Let f be differentiable on an interval I, i.e. $f \in D^{1} (I)$ . Then we have

f is increasing on I $\Leftrightarrow f^{'} (x) \geq 0 for all x \in I$

[7.10.5]

f is decreasing on I $\Leftrightarrow f^{'} (x) \leq 0 for all x \in I$

[7.10.6]

Proof: Again we only deal with 1.

" $\Rightarrow$ ": If f is increasing we know from [7.10.3] that

m_{x} (y) = \frac{f (y) - f (x)}{y - x} \geq 0

for all

y \in I \ {x}

According to [6.9.4] we thus get $f^{'} (x) = \lim_{y \to x} m_{x} (y) \geq 0$ .

" $\Leftarrow$ ": Now take $x, y \in I$ such that $x < y$ . Due to the mean value theorem [7.9.5] there is an $\tilde{x} \in] x, y [$ satisfying

f (y) = f (x) + \underset{> 0}{\underset{︸}{(y - x)}} \cdot \underset{\geq 0}{\underset{︸}{f^{'} (\tilde{x})}} \geq f (x)

Note that the direction "

\Leftarrow

" provided the typical context for an application of the mean value theorem. Further, the details in the proof show that

The monotony now widens the options to confirm a local extreme point and this will result in a further sufficient criterion (cf. the necessary criterion [7.9.2] and the sufficient criterion [7.9.17] for

C^{n + 1}

-functions). As a start we observe that a transition point between two different monotony domains has to be an extreme point.

Proposition: $f : A \to ℝ$ has a local extremum at $a \in A$ if there is an $ε > 0$ such that f has a different monotony behaviour on the relative semi-neighbourhoods

A \cap] a - ε, a]

and

A \cap [a, a + ε [

[7.10.7]

The reverse does not hold.

Proof: We assume f is increasing on $A \cap] a - ε, a]$ and decreasing on $A \cap [a, a + ε [$ . That means for all $x \in$ $A_{a, ε}$
i

$A_{a, ε} = A \cap] a - ε, a + ε [$

:

\begin{array}{l} f (x) \leq f (a),  if x \leq a \\ f (a) \geq f (x),  if x \geq a \end{array}

which actually says that f has a local maximum at a.

A counter example will prove that the reverse is false. The indicator function $χ_{ℚ}$
i

$χ_{ℚ} (x) = {\begin{array}{l} 1, if x \in ℚ \\ 0, if x \notin ℚ \end{array}$
has a global minimum at 0, but fails to be monotone on any interval.

A new sufficient criterion ("

f^{'}

changes the sign at its zero a") for differentiable functions on intervals is now available.

Proposition: A function $f \in D^{1} (I)$ has a local extremum at $a \in I$ if there is an $ε > 0$ such that $f^{'}$ has a different sign on the semi-neighbourhoods

I \cap] a - ε, a]

and

I \cap [a, a + ε [

[7.10.8]

The reverse is not true.

Proof: If the sign of $f^{'}$ differs on the intervals $I \cap] a - ε, a]$ and $I \cap [a, a + ε [$ , let's say $f^{'} (x) \geq 0$ for all $x \in I \cap] a - ε, a]$ and $f^{'} (x) \leq 0$ for all $x \in I \cap] a, a + ε]$ , f has a different monotony behaviour on these intervals according to [7.10.5./6.]. Due to [7.10.7] f thus has a local extremum at a.

The function $f : ℝ \to ℝ$ defined by

f (x) ≔ {\begin{array}{l} 0,  if x = 0 \\ (x \cdot \sin x^{- 1})^{2},  if x \neq 0 \end{array}

shows that the reverse of [7.10.8] does not hold: f is differentiable with

f^{'} (x) = {\begin{array}{l} \lim_{y \to 0} \frac{(y \cdot \sin y^{- 1})^{2}}{y} = \lim_{y \to 0} y \cdot \sin^{2} y^{- 1} = 0,  if x = 0 (note: \sin^{2} is bounded!) \\ 2 (x \cdot \sin x^{- 1}) (\sin x^{- 1} - x^{- 1} \cdot \cos x^{- 1}) = 2 x \cdot \sin^{2} x^{- 1} - 2 \cdot \sin x^{- 1} \cdot \cos x^{- 1},  if x \neq 0 \end{array}

and has a global minimum at 0. But $f^{'}$ changes the sign arbitrary often in each semi-neighbourhood of 0. We show this exemplarily for a semi-neighbourhood to the right. To that end we calculate for an arbitrary $n \in ℕ^{*}$ the values $f^{'} ((n π + \frac{π}{4})^{- 1})$ and $f^{'} ((n π + 3 \frac{π}{4})^{- 1})$ using the addition theorems for sin and cos [4.3.*]. With

\begin{array}{l} \sin (n π + \frac{π}{4}) = \underset{= 0}{\underset{︸}{\sin (n π)}} \cdot \cos \frac{π}{4} + \cos (n π) \cdot \underset{= \frac{\sqrt{2}}{2}}{\underset{︸}{\sin \frac{π}{4}}} = \cos (n π) \cdot \frac{1}{2} \sqrt{2} \\ \cos (n π + \frac{π}{4}) = \cos (n π) \cdot \underset{= \frac{\sqrt{2}}{2}}{\underset{︸}{\cos \frac{π}{4}}} - \underset{= 0}{\underset{︸}{\sin (n π)}} \cdot \sin \frac{π}{4} = \cos (n π) \cdot \frac{1}{2} \sqrt{2} \\ \sin (n π + 3 \frac{π}{4}) = \underset{= 0}{\underset{︸}{\sin (n π)}} \cdot \cos 3 \frac{π}{4} + \cos (n π) \cdot \underset{= \frac{\sqrt{2}}{2}}{\underset{︸}{\sin 3 \frac{π}{4}}} = \cos (n π) \cdot \frac{1}{2} \sqrt{2} \\ \cos (n π + 3 \frac{π}{4}) = \cos (n π) \cdot \underset{= - \frac{\sqrt{2}}{2}}{\underset{︸}{\cos 3 \frac{π}{4}}} - \underset{= 0}{\underset{︸}{\sin (n π)}} \cdot \sin 3 \frac{π}{4} = - \cos (n π) \cdot \frac{1}{2} \sqrt{2} \end{array}

and $(\cos (n π) \cdot \frac{1}{2} \sqrt{2})^{2} = \frac{1}{2}$ we now calculate

\begin{array}{l} f^{'} ((n π + \frac{π}{4})^{- 1}) = 2 (n π + \frac{π}{4})^{- 1} \cdot \frac{1}{2} - 2 \cdot \frac{1}{2} = (n π + \frac{π}{4})^{- 1} - 1 < 0 \\ f^{'} ((n π + 3 \frac{π}{4})^{- 1}) = 2 (n π + 3 \frac{π}{4})^{- 1} \cdot \frac{1}{2} + 2 \cdot \frac{1}{2} = (n π + \frac{π}{4})^{- 1} + 1 > 0 \end{array}

We turn to a second geometric aspect which is illustrated by the two functions below.

Definition: A function $f : A \to ℝ$ is called

convex (or curved to the left or said to have a positive curvature) on $B \subset A$ if

f (x) \leq f (a) + \frac{f (b) - f (a)}{b - a} (x - a) for all x \in] a, b [\cap B

[7.10.9]

concave (or curved to the right or said to have a negative curvature) on $B \subset A$ if

f (x) \geq f (a) + \frac{f (b) - f (a)}{b - a} (x - a) for all x \in] a, b [\cap B

[7.10.10]

holds for all $a, b \in B$ such that $a < b$ .

If [7.10.9] even allows $<$ instead of $\leq$ , f is called strictly convex on B. strictly concave is defined appropriately.

We omit the the phrase "on B" if $A = B$ .

We now concentrate on convex functions. All the results are transferable to concave functions as well, and, with only rare exceptions, even to the strict case.

as an equivalent condition for the convexity of f. Considering further that the lines

are the same, i.e. that [7.10.9] is replaceable by

f (x) \leq f (b) + \frac{f (b) - f (a)}{b - a} (x - b)

, we get another convexity test (note:

x - b < 0

!):

Finally, combining [4] and [5] results in a third, more versatile convexity condition.

Proposition: $f : A \to ℝ$ is convex on $B \subset A$ if and only if for any $a, b \in B$ such that $a < b$ the following condition holds:

\frac{f (x) - f (a)}{x - a} \leq \frac{f (b) - f (x)}{b - x} for all x \in] a, b [\cap B

[7.10.11]

Proof: For all $x \in] a, b [\cap B$ the following inequalities are equivalent:

\begin{matrix} \frac{f (x) - f (a)}{x - a} & \leq & \frac{f (b) - f (x)}{b - x} \\ \Leftrightarrow & f (x) (\frac{1}{x - a} + \frac{1}{b - x}) & \leq & \frac{f (a)}{x - a} + \frac{f (b)}{b - x} \\ \Leftrightarrow & f (x) (\underset{= b - a}{\underset{︸}{b - x + x - a}}) & \leq & f (a) (\underset{= b - a + a - x}{\underset{︸}{b - x}}) + f (b) (x - a) \\ = & f (a) (b - a) + (f (b) - f (a)) (x - a) \\ \Leftrightarrow & f (x) & \leq & f (a) + \frac{f (b) - f (a)}{b - a} (x - a) \end{matrix}

Now we are prepared, at least for differentiable functions on intervals, to prove some more comfortable convexity criteria.

Proposition:

If $f \in D^{1} (I)$ , then:

f is convex

\Leftrightarrow f^{'}

is monotonic increasing

[7.10.12]

If $f \in D^{2} (I)$ , then:

f is convex

\Leftrightarrow {f^{'}}^{'} (x) \geq 0 for all x \in I

[7.10.13]

Proof:

1. ► For the direction " $\Rightarrow$ " we take two points a and b of I such that $a < b$ . Due to [4] and [5] we have for all $x \in] a, b [$ :

m_{a} (x) = \frac{f (x) - f (a)}{x - a} \leq \frac{f (b) - f (a)}{b - a} \leq \frac{f (x) - f (b)}{x - b} = m_{b} (x)

As $f^{'} (a) = \lim_{x \to a} m_{a} (x) = \lim_{x \to a} m_{a} |] a, b [(x)$ and $f^{'} (b) = \lim_{x \to b} m_{b} (x) = \lim_{x \to b} m_{b} |] a, b [(x)$ (see [6.9.1] for details) we thus know:

f^{'} (a) \leq \frac{f (b) - f (a)}{b - a} \leq f^{'} (b)

which in fact proves that $f^{'}$ is increasing.

" $\Leftarrow$ ": Now we take a fixed point $x \in] a, b [$ , $a, b \in I$ . According to the mean value theorem [7.9.4] we find two points ${\tilde{x}}_{1}, {\tilde{x}}_{2}$ nested $a < {\tilde{x}}_{1} < x < {\tilde{x}}_{2} < b$ , such that, due to the monotony of $f^{'}$ , the following estimate holds:

\frac{f (x) - f (a)}{x - a} = f^{'} ({\tilde{x}}_{1}) \leq f^{'} ({\tilde{x}}_{2}) = \frac{f (b) - f (x)}{b - x}

[7.10.11] now guarantees the convexity of f.

2. ► follows immediately from 1. due to [7.10.5].

[7.10.7] illustrated the role of transition points between two different monotony domains. We should pay attention to transition points between two different curvature domains as well.

[6] and [7] make it easy to find necessary existence criteria for inflection points.

Proposition: If $f : I \to ℝ$ has an inflection point at $a \in I$ we have for any

$D^{1}$ -function:

f^{'}

has a local extremum at a.

[7.10.15]

$D^{2}$ -function:

{f^{'}}^{'} (a) = 0

[7.10.16]

In both cases the reverse does not hold.

Proof:

1. ► According to [6] the monotony behaviour of $f^{'}$ changes at a so that $f^{'}$ has a local extremum at a due to [7.10.7].

We need a counter example to prove that the reverse is not valid. We turn back to the differentiable, thus continuous function $f : ℝ \to ℝ$ introduced in [7.10.8] by

f (x) ≔ {\begin{array}{l} 0,  if x = 0 \\ (x \cdot \sin x^{- 1})^{2},  if x \neq 0 \end{array}

As f is continuous [8.1.5] guarantees that there will be a differentiable function $g : ℝ \to ℝ$ such that $g^{'} = f$ . We read the properties of f proved in [7.10.8] now like this:

$g^{'}$ has a global minimum at 0.
${g^{'}}^{'}$ does'nt change its sign at 0 so that g has no inflection point at 0 due to [7].

2. ► is an immediate consequence of [7.9.2] due to 1.

The non-constant power function $X^{4}$ is convex on the whole of $ℝ$ according to [7.10.13] ( $(X^{4} {)^{'}}^{'} (x) = 12 x^{2} \geq 0$ for all x! ) thus has no inflection point at all. Nevertheless we have $(X^{4} {)^{'}}^{'} (0) = 0$ .

We now get a sufficient criterion for

C^{n + 2}

-functions with the help of Taylor's formula [7.9.16]. The approach will be parallel to [7.9.17].

Proposition (sufficient criterion for $\underline{C^{n + 2}}$ -functions): Take $f \in C^{n + 2} (I)$ and an interior point a of I such that

{f^{'}}^{'} (a) = \dots = f^{(n + 1)} (a) = 0 \land f^{(n + 2)} (a) \neq 0

[7.10.17]

Then we have:

If n + 2 is odd f has an inflection point at a.
If n + 2 is even f has no inflection point at a.

Proof: We apply Taylor's formula to the $C^{n}$ -function ${f^{'}}^{'}$ . Assuming ${f^{'}}^{'} (a) > 0$ will provide, due to continuity, an $ε > 0$ such that ${f^{'}}^{'} (x) > 0$ for all $x \in I_{a, ε}$ . For each of these x there is now an $\tilde{x}$ in between x and a such that

{f^{'}}^{'} (x) = \sum_{i = 0}^{n - 1} \frac{f^{(i + 2)} (a)}{i!} {(x - a)}^{i} + \frac{f^{(n + 2)} (\tilde{x})}{n!} {(x - a)}^{n} = \underset{> 0}{\underset{︸}{\frac{f^{(n + 2)} (\tilde{x})}{n!}}} {(x - a)}^{n}

${f^{'}}^{'}$ thus changes its sign at a iff n is odd. Due to [7] this is the assertion.

We use the terms "convex" and "concave" to describe the quality of a curvature. Deciding if a function is curved to the left or curved to the right however does not provide any information of the curvature's quantity, i.e. we are unable to measure how small or how big the curvature at a certain point is.

It is only with circles that we have an idea how to measure curvature, namely by considering the radius: As the curvature looks the smaller the bigger the radius is, we will introduce the reciprocal of the radius as a psychological sound measure for the curvature of a circle.

If we succeed now to find an appropriate circle, the so called circle of curvature, that nestles perfectly to a given function f at a suitable point a we could take its radius as the radius of curvature of f at a.

The sketch to the left shows the circle of curvature with respect to the reciprocal function

\frac{1}{X}

at 1. The distance between its centre (2,2) and its boundary point (1,1), i.e. the radius of curvature calculates to

\sqrt{2}

in this case.

But how to construct such a ideal circle, how to find its centre and its radius? The first idea is to look for it among those circles that are attached vertically to f. Thus we will concentrate on circles through (a, f(a)) that have its centre on the normal

n_{a}

, i.e. on the perpendicular to the tangent

t_{a}

To pick the "right" one out of those we need to know the local geometry of the graph, so that the values of f in the vicinity of a are involved. When we constructed the tangent we used secants, auxiliary lines that resulted from considering additional points (x, f(x)). Analog to that we will introduce "secant circles" that have (x, f(x)) as an additional point. Subsequently we will calculate their centres as the insection points of the normal and the perpendicular bisector of the secant determined by a and x. The following interactive applet illustrates the concept quite convincing.

We now consider

C^{2}

-functions at points a with a definite curvature. We will thus subsequently exclude inflection points and will assume that

{f^{'}}^{'} (a) \neq 0

. Further we will markedly benefit from the vector notation in the calculations ahead. Tangent, normal and perpendicular bisector are thus presented like this:

Proposition: If $f \in C^{2} (I)$ and a is an interior point of I such that ${f^{'}}^{'} (a) \neq 0$ there are, without restriction, two points $\tilde{x}, \tilde{\tilde{x}}$ in between x and a for every $x \in I \ {a}$ such that the lines $n_{a}$ and $s_{x}$ intersect solely at

M_{a} (x) ≔ \frac{1}{2} (\begin{matrix} x + a \\ f (x) + f (a) \end{matrix}) - \frac{1 + f^{'} (a) f^{'} (\tilde{x})}{{f^{'}}^{'} (\tilde{\tilde{x}})} (\begin{matrix} \frac{f (x) - f (a)}{x - a} \\ - 1 \end{matrix})

[7.10.18]

Proof: As ${f^{'}}^{'}$ is continuous we know that ${f^{'}}^{'} (x) \neq 0$ for all x in an appropriate neighbourhood $I_{a, ε}$ of a. Without restriction we assume $I_{a, ε} = I$ .

To calculate the intersection we now use a method outlined in 9.9. For an arbitrary vector

\vec{z} = \frac{1}{2} (\begin{matrix} x + a \\ f (x) + f (a) \end{matrix}) + α (\begin{matrix} f (x) - f (a) \\ a - x \end{matrix}) \in s_{x}

the following equivalence accordingly holds:

\begin{array}{l} \vec{z} \in n_{a} & \Leftrightarrow (\begin{matrix} f^{'} (a) \\ - 1 \end{matrix}) \vec{y} = (\begin{matrix} \frac{x + a}{2} - a + α (f (x) - f (a)) \\ \frac{f (x) + f (a)}{2} - f (a) + α (a - x) \end{matrix}) is solvable \\ \underset{I + f^{'} (a) II}{\Leftrightarrow} (\begin{matrix} 0 \\ - 1 \end{matrix}) \vec{y} = (\begin{matrix} \frac{x - a}{2} + α (f (x) - f (a)) + f^{'} (a) (\frac{f (x) - f (a)}{2} + α (a - x)) \\ \frac{f (x) - f (a)}{2} + α (a - x) \end{matrix}) is solvable \\ \Leftrightarrow α (f (x) - f (a) + f^{'} (a) (a - x)) + \frac{x - a}{2} + f^{'} (a) \frac{f (x) - f (a)}{2} = 0 [ 8 ] \end{array}

Using Taylor's theorem [7.9.16] we find two points $\tilde{x}, \tilde{\tilde{x}}$ in between x and a such that

$f (x) - f (a) = f^{'} (\tilde{x}) (x - a)$
$f (x) - f (a) - f^{'} (a) (x - a) = \frac{1}{2} {f^{'}}^{'} (\tilde{\tilde{x}}) {(x - a)}^{2}$

which allows to extend the equivalence [8] by

\begin{array}{l} α \frac{1}{2} {f^{'}}^{'} (\tilde{\tilde{x}}) {(x - a)}^{2} + \frac{x - a}{2} + f^{'} (a) \frac{f^{'} (\tilde{x}) (x - a)}{2} = 0 \\ \Leftrightarrow & α {f^{'}}^{'} (\tilde{\tilde{x}}) (x - a) + 1 + f^{'} (a) f^{'} (\tilde{x}) = 0 \\ \Leftrightarrow & α = - \frac{1 + f^{'} (a) f^{'} (\tilde{x})}{{f^{'}}^{'} (\tilde{\tilde{x}}) (x - a)}, \end{array}

so that finally the unique solution is calculated to

\begin{array}{l} \vec{z} & = \frac{1}{2} (\begin{matrix} x + a \\ f (x) + f (a) \end{matrix}) - \frac{1 + f^{'} (a) f^{'} (\tilde{x})}{{f^{'}}^{'} (\tilde{\tilde{x}}) (x - a)} (\begin{matrix} f (x) - f (a) \\ a - x \end{matrix}) \\ = \frac{1}{2} (\begin{matrix} x + a \\ f (x) + f (a) \end{matrix}) - \frac{1 + f^{'} (a) f^{'} (\tilde{x})}{{f^{'}}^{'} (\tilde{\tilde{x}})} (\begin{matrix} \frac{f (x) - f (a)}{x - a} \\ - 1 \end{matrix}) \end{array}

Now we are able to analyse the limit behaviour of the intersection points $M_{a} (x)$ which in fact are the centres of the secant circles. If x converges to a the intermediate points $\tilde{x}$ and $\tilde{\tilde{x}}$ are forced to do the same. Now that f is differentiable at a and $f, f^{'}, {f^{'}}^{'}$ are continuous at that point, the function $M_{a}$ has a limit at a:

\lim_{x \to a} M_{a} (x) = (\begin{matrix} a \\ f (a) \end{matrix}) - \frac{1 + {(f^{'} (a))}^{2}}{{f^{'}}^{'} (a)} (\begin{matrix} f^{'} (a) \\ - 1 \end{matrix})

Obviously this limit is a point on the normal $n_{a}$ and its distance to the boundary point (a, f(a)) calculates to

| \lim_{x \to a} M_{a} (x) - (\begin{matrix} a \\ f (a) \end{matrix}) | = | \frac{1 + {(f^{'} (a))}^{2}}{{f^{'}}^{'} (a)} | \cdot | (\begin{matrix} f^{'} (a) \\ - 1 \end{matrix}) | = | \frac{1 + {(f^{'} (a))}^{2}}{{f^{'}}^{'} (a)} | \sqrt{1 + {(f^{'} (a))}^{2}}

With these results we have achieved our aim: To describe curvature in quantitative terms.

Definition: For any $f \in C^{2} (I)$ and an interior point a of I such that ${f^{'}}^{'} (a) \neq 0$ the circle of curvature of f with respect to a is defined by its

centre of curvature

M (a) ≔ (\begin{matrix} a \\ f (a) \end{matrix}) - \frac{1 + {(f^{'} (a))}^{2}}{{f^{'}}^{'} (a)} (\begin{matrix} f^{'} (a) \\ - 1 \end{matrix})

[7.10.19]

and its

radius of curvature

r (a) ≔ \frac{{\sqrt{1 + {(f^{'} (a))}^{2}}}^{3}}{| {f^{'}}^{'} (a) |}

[7.10.20]

The number $k (a) = \frac{1}{r (a)} = \frac{| {f^{'}}^{'} (a) |}{{\sqrt{1 + {(f^{'} (a))}^{2}}}^{3}}$ is called the curvature of f at a.

In case ${f^{'}}^{'} (a) = 0$ we additionally set $r (a) ≔ \infty$ and $k (a) ≔ 0$ .

As an example we calculate the curvature data of the standard parabola and of a semi-circle with radius r centered at the origin. With the latter we expect all radii of curvature to coincide with r and all centres of curvature to be (0,0).

Example:

For $f = X^{2}, f^{'} = 2 X, {f^{'}}^{'} = 2 and a \in ℝ$ we get:

$M (a) = (\begin{matrix} a \\ a^{2} \end{matrix}) - \frac{1 + 4 a^{2}}{2} (\begin{matrix} 2 a \\ - 1 \end{matrix}) = (\begin{matrix} - 4 a^{3} \\ \frac{1}{2} + 3 a^{2} \end{matrix})$

$r (a) = \frac{{\sqrt{1 + 4 a^{2}}}^{3}}{2}$

$k (a) = \frac{2}{{\sqrt{1 + 4 a^{2}}}^{3}}$
For $f = \sqrt{r^{2} - X^{2}}, f^{'} = - \frac{X}{\sqrt{r^{2} - X^{2}}}, {f^{'}}^{'} = - \frac{r^{2}}{{\sqrt{r^{2} - X^{2}}}^{3}} and a \in] - r, r [$ we initially have

$\frac{1 + \frac{a^{2}}{r^{2} - a^{2}}}{- \frac{r^{2}}{{\sqrt{r^{2} - a^{2}}}^{3}}} = - \frac{{\sqrt{r^{2} - a^{2}}}^{3} + a^{2} \sqrt{r^{2} - a^{2}}}{r^{2}} = - \sqrt{r^{2} - a^{2}} \frac{{\sqrt{r^{2} - a^{2}}}^{2} + a^{2}}{r^{2}} = - \sqrt{r^{2} - a^{2}}$ .

and from that we calculate

$M (a) = (\begin{matrix} a \\ \sqrt{r^{2} - a^{2}} \end{matrix}) + \sqrt{r^{2} - a^{2}} (\begin{matrix} - \frac{a}{\sqrt{r^{2} - a^{2}}} \\ - 1 \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$

$r (a) = \frac{{\sqrt{1 + \frac{a^{2}}{r^{2} - a^{2}}}}^{3}}{\frac{r^{2}}{{\sqrt{r^{2} - a^{2}}}^{3}}} = \frac{{\sqrt{r^{2} - a^{2} + a^{2}}}^{3}}{r^{2}} = r$

$k (a) = \frac{1}{r}$

7.9. 7.11.