91Ó°ÊÓ

In multivariable calculus, critical points are locations on a graph where the function appears to have a completely flat tangent plane, meaning there's neither a slope nor a peak. They can be thought of as the equivalent of turning points in single-variable calculus. To find them, you must set the first partial derivatives of the function to zero.

For the function given, the partial derivatives are:

• \(\frac{\partial f}{\partial x} = 2x(y+1)^3\)
• \(\frac{\partial f}{\partial y} = 3x^2(y+1)^2 + 2y\)

Setting these to zero helps identify potential critical points, which are coordinates where there is no change in the direction of the tangent plane. In this exercise,

• For \(x\), either \(x=0\) or \((y+1)^3=0\) holds;
• For \(y\), solving the equation given by \(x=0\) results in \(y=0\).

This process identified (0,0) as the critical point.

Partial derivatives are a key tool in multivariable calculus. They measure how a function changes as each variable changes separately, while keeping the others constant. This idea is similar to taking normal derivatives in single-variable calculus, but now each variable can change individually.

For example, for the function \(f(x, y)=x^{2}(y+1)^{3}+y^{2}\), its rate of change with respect to \(x\) is expressed by \(\frac{\partial f}{\partial x} = 2x(y+1)^3\), showing how \(f\) changes as \(x\) shifts.

Similarly, \(\frac{\partial f}{\partial y} = 3x^2(y+1)^2 + 2y\) tells us about changes in \(f\) when \(y\) changes. Both derivatives need to be computed and evaluated to find critical points, as we do at points where these derivatives equal zero.

The second derivative test helps classify the nature of critical points in a multivariable function. It uses second-order partial derivatives to tell us whether a critical point is a local minimum, maximum, or saddle point.

For the function at hand, we calculate the relevant second derivatives:

• \(f_{xx} = 2(y+1)^3\)
• \(f_{yy} = 6x^2(y+1) + 2\)
• \(f_{xy} = 6x(y+1)^2\)

To apply the second derivative test, we use the Hessian determinant \(D\), given by:\[D = f_{xx}f_{yy} - (f_{xy})^2\]At the critical point (0,0), this simplifies to \(D = 4 > 0\), combined with \(f_{xx} > 0\), confirming a local minimum.

The Hessian determinant is a critical component for analyzing the curvature at critical points of a function involving multiple variables. This determinant helps to determine if a critical point is a local minimum, a local maximum, or a saddle point.

It is computed from the matrix of second partial derivatives of the function. For our example, the Hessian matrix is:\[\begin{bmatrix} f_{xx} & f_{xy} \ f_{xy} & f_{yy} \end{bmatrix}\]The Hessian determinant \(D\) is calculated as \(f_{xx}f_{yy} - (f_{xy})^2\). At the critical point (0,0) in this exercise,

• The determinant was found to be 4, a positive value which is key to confirming the presence of a local minimum when combined with \(f_{xx} > 0\).

This positive determinant implies that the function curve opens upwards, making our critical point a local minimum, although not necessarily a global one.