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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 k1 and k2 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 k1 = k2 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 k1 . Then the normal curvature kn(φ) in direction φ is given by

            kn(φ) = k1 cos2φ + k2 sin2φ

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 fxy 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 fx' y' = 0.

Proof. Using the chain rule of partial differentiation










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 fxy (0, 0) = 0. See Fig. 1. Now draw a line PM in the xy-plane making an angle ole9.gif 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 fx(0, 0) = fy(0, 0) = 0 (since the x and y axes lie in the tangent plane) we get

            f(x, y) = ½ (fxx x2 + fyy y2) + 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 ole11.gif , y = ξ sin ole12.gif and thus




If we set ole15.gif , in 4) we find that fxx and fyy are the curvatures k1 and k2 of the normal sections in the directions of the x and y axes. Thus 4) is Euler’s formula

            k = k1 cos2 ole16.gif + k2 sin2 ole17.gif .

The fact that k1 and k2 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 ole18.gif at which dk/d ole19.gif = 0. Differentiation shows that


and this function is equal to zero at ole21.gif = 0, π/2, π, ...

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