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Computation of the principal curvatures k1 and k2

The problem of computing the values of k1 and k2 is that of finding the maxima and minima of the function


where du and dv are viewed as variables (rather than differentials). For clarity let us replace du by x and dv by y. Our problem then is to find the maxima and minima of the function


Let us also introduce the notation

            p(x, y) = Lx2 + 2Mxy + Ny2

            q(x, y) = Ex2 + 2Fxy + Gy2



The requirement for kn(x, y) to have a maxima or minima at a point (x, y) is that the following two conditions be met:


Using the formula from calculus


we compute the partial derivatives of


with respect to x and y as




which can be written more concisely as



5)        px = 2L x + 2M y

            qx = 2E x + 2F y

            py = 2M x + 2N y

            qy = 2F x + 2G y .

If we multiply equations 4) by q we obtain


Now make the substitution

            kn = p/q

and 6) becomes


7)        px - knqx = 0             py - knqy = 0

Now substituting the values of px, qx, py, qy from 5) into 7) we obtain


8)        (L - knE) x + (M - knF) y = 0

            (M - knF) x + (N - knG) y = 0

We wish to solve equations 8) for x and y. It is a homogeneous system and will have a nontrivial solution only if the determinant of the coefficients is equal to zero i.e. only if


If we expand 9) we obtain the equation


10)      (EG - F2)kn2 - (EN + GL - 2FM)kn + (LN - M2) = 0 .

Solving this equation for kn will give us two values, k1 and k2, which will represent the minimum and maximum values of the function kn(x, y) i.e. the principal curvatures.

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