Chapter 12: Q. 12 (page 975)

Show that the function $h (x, y) = x^{3} + y^{3}$ has a stationary point at the origin. Show that the discriminant det( $H_{h}$ (0, 0)) = 0. Show that there are points arbitrarily close to the origin such that h(x, y) > 0. Show that there are points arbitrarily close to the origin such that h(x, y) < 0. Explain why all this shows that h has a saddle at the origin .

Short Answer

Expert verified

$All the points near the origin of the form (- x, 0), (- x, - y) and (0, - y), with r > 0 . y > 0, h (x, y) < 0 and for the points of the form (x, 0), (x, y) and (0, y) with x > 0, y > 0, h (x, y) > 0 . Hence near the origin the function has a saddle . .$

Step by step solution

Step 1. Given

$F u n c t i o n h (x, y) = x^{3} + y^{3}$

Step 2. Calculating critical point

$Given function is h (x, y) = x^{3} + y^{3} which is polynomial and hence differentiable . The gradient of the function is \nabla h (x, y, z) = \frac{\partial h}{\partial x} i + \frac{\partial h}{\partial y} j . = 3 x^{2} i + 3 y^{2} j At the critical points the gradient of a function vanishes, that is \nabla h (x, y) = 0, T h a t i s 3 x^{2} = 0 . . . . . (1) a n d 3 y^{2} = 0 . . . . . (2) The solution of 1 and 2 is, x = 0 a n d y = 0, So the critical point is (0, 0) .$

Step 3. Discriminate of function .

$Now the second order derivative of the given function are, \frac{\partial^{2} h}{\partial x^{2}} = 6 x, \frac{\partial^{2} h}{\partial y^{2}} = 6 y, \frac{\partial^{2} h}{\partial x \partial y} = 0 Hence the discriminate of the function is, H_{h} (x, y) = \frac{\partial^{2} h}{\partial x^{2}} \frac{\partial^{2} h}{\partial y^{2}} - {(\frac{\partial^{2} h}{\partial x \partial y})}^{2} = (6 x) (6 y) - 0 = 36 x y At (0, 0) H_{h} (0, 0) = 0 . so no conclusion can be drawn using the discriminate Now all the points near the origin of the form (- x, 0), (- x, - y) and (0, - y), with r > 0 . y > 0, h (x, y) < 0 and for the points of the form (x, 0), (x, y) and (0, y) with x > 0, y > 0, h (x, y) > 0 . Hence near the origin the function has a saddle .$

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Short Answer

Step by step solution

Step 1. Given

Step 2. Calculating critical point

Step 3. Discriminate of function .

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