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Monday, January 11, 2016

Differential Geometry and Its Application - Exercise 3.1.7

Problem:

Show that the principal curvatures are given in terms of $ K $ and $ H $ by

$ k_1 = H + \sqrt{H^2 - K} $ and $ k_2 = H - \sqrt{H^2 - K} $.

Solution:

We know $ H = \frac{k_1 + k_2}{2} $ and $ K = k_1 k_2 $. Therefore we can form the quadratic equation $ x^2 - 2Hx + K $ so that the roots are $ k_1 $ and $ k_2 $

Now using the quadratic formula, we get

$ k = \frac{2H \pm \sqrt{4H^2 - 4K}}{2} = H \pm \sqrt{H^2 - K} $, this is exactly what we needed.

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