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Computation of the principal curvatures k_{1} and k_{2}

The problem of computing the values of k_{1} and k_{2} 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) = Lx^{2} + 2Mxy + Ny^{2}

q(x, y) = Ex^{2} + 2Fxy + Gy^{2}

Thus

The requirement for k_{n}(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) p_{x} = 2L x + 2M y

q_{x} = 2E x + 2F y

p_{y} = 2M x + 2N y

q_{y} = 2F x + 2G y .

If we multiply equations 4) by q we obtain

Now make the substitution

k_{n} = p/q

and 6) becomes

7) p_{x }- k_{n}q_{x} = 0 p_{y }- k_{n}q_{y} = 0

Now substituting the values of p_{x}, q_{x}, p_{y}, q_{y} from 5) into 7) we obtain

8) (L - k_{n}E) x + (M - k_{n}F) y = 0

(M - k_{n}F) x + (N - k_{n}G) 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 - F^{2})k_{n}^{2} - (EN + GL - 2FM)k_{n} + (LN - M^{2}) = 0 .

Solving this equation for k_{n} will give us two values, k_{1} and k_{2}, which will represent the minimum
and maximum values of the function k_{n}(x, y) i.e. the principal curvatures.

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