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Prove: Euler’s theorem. At each point P on a surface there exist two particular directions such that

1. They are mutually perpendicular.

2. The curvatures k_{1} and k_{2} of the normal sections in these directions are the smallest and the
largest values of the curvatures of all normal sections at point P (in the particular case where k_{1}
= k_{2} the curvature of all sections is the same, as for example, on a sphere).

3. Let φ be the angle, in the tangent plane, measured counterclockwise from the direction of
minimum curvature k_{1} . Then the normal curvature k_{n}(φ) in direction φ is given by

k_{n}(φ) = k_{1} cos^{2}φ + k_{2} sin^{2}φ

Proof. For the proof of Euler’s theorem we need the following lemma.

Lemma. If the function f(x, y) has continuous second derivatives at a given point P, then there exists a rotation of the axes

x = x' cos α - y' sin α

y = x' sin α + y' cos α

that will reduce the mixed derivative f_{xy} at point P to zero.

Stated a little differently the lemma says that if the function z = f(x, y) has continuous second
derivatives at a given point P then there exists some angle α by which the coordinate system can
be rotated about the z axis that will make f_{x' y'} = 0.

Proof. Using the chain rule of partial differentiation

Now

so

or

From 3) it readily follows that for a value of α satisfying

End of proof.

Proof of Euler’s theorem. Consider the surface S given by the equation z = f(x, y) in which the
coordinate system has its origin at the point of interest P with its x and y axes lying in tangent
plane T and so chosen as to make f_{xy} (0, 0) = 0. See Fig. 1. Now draw a line PM in the xy-plane
making an angle
with the x axis and consider the normal section L in the direction of this
straight line. The curvature of L at point P, taking sign into account, is given by

where f(x, y) is the distance of a point on L to line PM.. Expanding f(x, y) by Taylor’s formula
and noting that f_{x}(0, 0) = f_{y}(0, 0) = 0 (since the x and y axes lie in the tangent plane) we get

f(x, y) = ½ (f_{xx} x^{2} + f_{yy }y^{2}) + ε

where ε is a higher order infinitesimal that approaches zero as x and y approach zero. For a particular point on L, x = ξ cos , y = ξ sin and thus

or

If we set
, in 4) we find that f_{xx} and f_{yy} are the curvatures k_{1} and k_{2} of the normal
sections in the directions of the x and y axes. Thus 4) is Euler’s formula

k = k_{1 }cos^{2}
+ k_{2 }sin^{2}
.

The fact that k_{1} and k_{2} are the minimal and maximal curvatures also follows from the formula.
The maxima and minima of the function are found by finding the values of
at which dk/d
=
0. Differentiation shows that

and this function is equal to zero at = 0, π/2, π, ...

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