Problem 2 - Calculus Early Transcendentals

Question

Suppose \((0,2)\) is a critical point of a function \(g\) with continuous second derivatives. In each case, what can you say about \(g\) ? (a) $$g_{x x}(0,2)=-1, \quad g_{x},(0,2)=6, \quad g_{y}(0,2)=1$$ (b) $$g_{x x}(0,2)=-1, \quad g_{x},(0,2)=2, \quad g_{y}(0,2)=-8$$ (c) $$g_{x x}(0,2)=4, \quad g_{x y}(0,2)=6, \quad g_{y y}(0,2)=9$$

Step-by-Step Solution

Answer

For all parts, there is an inconsistency as stated, but in (c) the result is inconclusive from the second derivative test.

Step 1: Analyze given information

We know \((0,2)\) is a critical point of the function \(g\). This means that both the first partial derivatives \(g_x(0,2)\) and \(g_y(0,2)\) should be zero at the critical point for typical definitions unless explicitly given as non-zero.

Step 2: Confirm conditions for part (a)

Part (a) provides \(g_{xx}(0,2)=-1\), \(g_{x}(0,2)=6\), and \(g_{y}(0,2)=1\). However, for a critical point, \(g_{x}(0,2)\) and \(g_{y}(0,2)\) should be zero. Therefore, the given information seems inconsistent with \((0,2)\) as a critical point.

Step 3: Confirm conditions for part (b)

Part (b) provides \(g_{xx}(0,2)=-1\), \(g_{x}(0,2)=2\), and \(g_{y}(0,2)=-8\). Again, \(g_{x}(0,2)\) and \(g_{y}(0,2)\) should be zero at a critical point. The inconsistency here also suggests an issue with the statement of it being a critical point.

Step 4: Confirm conditions for part (c)

Part (c) provides \(g_{xx}(0,2)=4\), \(g_{xy}(0,2)=6\), and \(g_{yy}(0,2)=9\). We can use the second derivative test because we have enough information on second partial derivatives.

Step 5: Apply the Second Derivative Test for part (c)

Calculate the discriminant \(D = g_{xx}(0,2)g_{yy}(0,2) - (g_{xy}(0,2))^2\) which equals \(4 \cdot 9 - 6^2 = 36 - 36 = 0\). The discriminant \(D = 0\) indicates the second derivative test is inconclusive. The critical point could be a saddle point or a point of inflection.

Key Concepts

Second Derivative Test

The second derivative test is a tool used to classify critical points of multivariable functions. When you have a critical point, this test helps determine whether it is a local maximum, local minimum, or saddle point.

For a function with continuous second derivatives:

• The critical point \( (x_0, y_0) \) is first confirmed by ensuring \( g_x(x_0, y_0) = 0 \) and \( g_y(x_0, y_0) = 0 \).
• The Hessian matrix is formed using the second partial derivatives: \([g_{xx}(x_0,y_0), g_{xy}(x_0,y_0), g_{xy}(x_0,y_0), g_{yy}(x_0,y_0)]\).
• The discriminant, \( D = g_{xx}(x_0, y_0)g_{yy}(x_0, y_0) - (g_{xy}(x_0, y_0))^2 \), is then calculated to determine the nature of the critical point.

If \( D > 0 \) and \( g_{xx}(x_0, y_0) > 0 \), the point is a local minimum. If \( D > 0 \) and \( g_{xx}(x_0, y_0) < 0 \), it's a local maximum. \( D < 0 \) indicates a saddle point. However, if \( D = 0 \), as in part (c) of our exercise, the test is inconclusive.

Partial Derivatives

Partial derivatives involve taking the derivative of a multivariable function with respect to one variable at a time, while treating all other variables as constants. This concept is crucial in determining critical points of such functions.

In our exercise, partial derivatives are represented by \( g_x(x, y) \) and \( g_y(x, y) \), which denote the rate of change of \( g \) with respect to \( x \) and \( y \) respectively. For a point \( (x_0, y_0) \) to be critical, these must equal zero:

• \( g_x(x_0, y_0) = 0 \)
• \( g_y(x_0, y_0) = 0 \)

This means at the critical point, the function does not increase or decrease along the directions of either of the axes. The second partial derivatives, \( g_{xx}, g_{yy}, \) and \( g_{xy} \), are then used in further analysis, such as applying the second derivative test.

Saddle Points

A saddle point in the context of multivariable calculus is a type of critical point. Unlike maxima or minima, which are extremum points, saddle points have distinctive characteristics.

Saddle points are characterized by their nature of being neither a local minimum nor maximum. This means:

• The surface slopes away in different directions from the saddle point.
• Saddle points are usually identified when \( D < 0 \) during a second derivative test. However, if \( D = 0 \), as given in one part of our exercise, the second derivative test cannot confirm a saddle point conclusively – it could also be an inflection point, requiring further analysis.

Saddle points are important in understanding the topography of a function's graph, as they represent points of critical but changing behavior. Recognizing and calculating saddle points can provide insight into the overall structure and behavior of the function's surface.