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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

Thus

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

where

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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