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Chapter 13: Problem 30

Find the critical points and test for relative extrema. List the critical points for which the Second-Partials Test fails. $$ f(x, y)=\left(x^{2}+y^{2}\right)^{2 / 3} $$

Short Answer

Expert verified
The only critical point of the function is at (0,0), but it is not defined by the Second-Partials Test due to the undefined nature of the second derivatives at this point.

Step by step solution

01

Find the Partial Derivatives

The first partial derivatives of the function with respect to x and y are calculated as follows: \[f_x=\frac{4}{3}x\left(x^{2}+y^{2}\right)^{-1 / 3}\] and \[f_y=\frac{4}{3}y\left(x^{2}+y^{2}\right)^{-1 / 3}\].
02

Find the Critical Points

The critical points occur where both partial derivatives are equal to 0 or are undefined. Setting both \(f_x\) and \(f_y\) to 0: \[\frac{4}{3}x\left(x^{2}+y^{2}\right)^{-1 / 3}=0\] and \[\frac{4}{3}y\left(x^{2}+y^{2}\right)^{-1/3}=0\]. The only solution is at (x, y) = (0, 0), which is a critical point.
03

Apply the Second-Partials Test

The second partials test requires the computation of the second derivatives of the function, including \(f_{xx}, f_{yy},\) and \(f_{xy}\). These are: \[f_{xx}=\frac{4}{3}(1-\frac{x^{2}}{x^{2}+y^{2}})\left(x^{2}+y^{2}\right)^{-1 / 3}, f_{yy}=\frac{4}{3}(1-\frac{y^{2}}{x^{2}+y^{2}})\left(x^{2}+y^{2}\right)^{-1 / 3},\] and \(f_{xy}=\frac{4}{3}(-\frac{xy}{x^{2}+y^{2}})\left(x^{2}+y^{2}\right)^{-1 / 3}\). The discriminant D at a point (x, y) is defined as \(D = f_{xx}f_{yy} - (f_{xy})^{2}\). At the critical point (0,0), these are undefined, and hence the Second-Partials Test fails. Therefore, the critical point (0,0) is undefined for the Second-Partials Test.

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