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Let f be a multivariable function defined by f(x, y) = x^3 y – x^2 y^2, where x and y are real numbers. Choose a specific point to analyze and complete the following parts:

• Part A: Explain how to find the direction of maximum increase for f at your chosen point, showing all required work.
• Part B: Explain how to find the direction of maximum decrease for f at your chosen point, showing all required work.
• Part C: Explain how to find the equation of the tangent plane to f at your chosen point, showing all required work.
• Part D: Explain how to find the equation of the normal line to f at your chosen point, showing all required work.
• Part E: Demonstrate that the second derivative test for local extreme values of f is inconclusive for all points on the y-axis.

Paper For Above instruction

The multivariable function under consideration is f(x, y) = x^3 y – x^2 y^2, which entails understanding its gradient, tangent plane, normal line, and second derivative behavior at specified points. To analyze these aspects, we first select a point in the domain where the function's properties are to be examined. For simplicity and clarity, we choose the point (1, 1), which is a common choice for such analysis, but the process applies to other points as well.

Part A: Direction of Maximum Increase

The direction of maximum increase of a multivariable function at a specific point is given by the gradient vector of the function, ∇f(x, y). The gradient vector points in the direction where the function increases most rapidly, and its magnitude corresponds to the rate of this increase. To find this direction at the point (1, 1), we first compute the partial derivatives of f with respect to x and y.

The partial derivatives are:

• f_x(x, y) = ∂f/∂x = 3x^2 y – 2x y^2
• f_y(x, y) = ∂f/∂y = x^3 – 2x^2 y

Evaluating these at (1, 1):

• f_x(1, 1) = 3(1)^2(1) – 2(1)(1)^2 = 3 – 2 = 1
• f_y(1, 1) = (1)^3 – 2(1)^2(1) = 1 – 2 = -1

The gradient vector at (1, 1) is thus:

∇f(1, 1) = 1, -1.

The direction of maximum increase is along this gradient vector. Typically, we represent this as a unit vector in that direction:

u = (1/√2, -1/√2).

Therefore, the direction of maximum increase of f at (1, 1) is along the vector (1, -1), normalized to a unit vector.

Part B: Direction of Maximum Decrease

The direction of maximum decrease is directly opposite to the gradient vector. Since the gradient at (1, 1) is (1, –1), the direction of maximum decrease is along the vector:

−∇f(1, 1) = (−1, 1).

As with the previous part, we can consider the unit vector in this direction:

u = (−1/√2, 1/√2).

In conclusion, the direction of maximum decrease at (1, 1) is given by the vector (−1, 1), pointing in the direction of the steepest descent of the function.

Part C: Equation of the Tangent Plane

The tangent plane to the surface z = f(x, y) at the point (x₀, y₀, z₀) is given by:

z = z₀ + f_x(x₀, y₀)(x − x₀) + f_y(x₀, y₀)(y − y₀).

First, evaluate f at (1, 1):

f(1, 1) = (1)^3(1) – (1)^2(1)^2 = 1 – 1 = 0.

Using the previously computed partial derivatives:

• f_x(1, 1) = 1
• f_y(1, 1) = -1

Thus, the equation of the tangent plane at (1, 1, 0) is:

z = 0 + 1(x − 1) – 1(y − 1)

which simplifies to:

z = (x − 1) - (y − 1) = x - y.

Part D: Equation of the Normal Line

The normal line to the surface at a point is determined by the gradient vector at that point. The parametric equations of the normal line passing through (x₀, y₀, z₀) are:

• x = x₀ + t·f_x(x₀, y₀)
• y = y₀ + t·f_y(x₀, y₀)
• z = z₀ + t·|∇f(x₀, y₀)|

Using the values at (1, 1, 0):

• x = 1 + t·1 = 1 + t
• y = 1 + t·(–1) = 1 - t
• z = 0 + t·√(1^2 + (–1)^2) = 0 + t·√2

Therefore, the parametric equations of the normal line are:

x = 1 + t, y = 1 - t, z = t·√2.

Part E: Inconclusiveness of the Second Derivative Test on the y-axis

The second derivative test in multiple variables involves examining the Hessian matrix of second derivatives to determine the nature of critical points. For the function f(x, y), the Hessian matrix is:

H = |f_xx f_xy|

|f_yx f_yy|.

Compute the second derivatives:

• f_xx = ∂^2f/∂x^2 = 6x y – 2 y^2
• f_yy = ∂^2f/∂y^2 = –2 x^2
• f_xy = f_yx = ∂^2f/∂x∂y = 3x^2 – 2x y

On the y-axis, where x = 0, the second derivatives simplify to:

• f_xx(0, y) = 0 – 2 y^2 = –2 y^2
• f_yy(0, y) = –2 * 0 = 0
• f_xy(0, y) = 3 0 – 2 0 * y = 0

The Hessian determinant D is:

D = f_xx * f_yy – (f_xy)^2 = (–2 y^2)(0) – (0)^2 = 0 – 0 = 0.

The second derivative test states that if D > 0 and f_xx > 0, then there is a local minimum; if D > 0 and f_xx

This demonstrates that the second derivative test cannot classify the nature of critical points along the entire y-axis, as the Hessian determinant is zero everywhere on that line.

Conclusion

In summary, the analysis at the point (1, 1) reveals the directions of steepest ascent and descent, the tangent plane, and the normal line to the surface defined by f(x, y). The second derivative test fails to provide conclusive information along the y-axis, emphasizing the need for alternative methods to analyze critical points in such regions. These techniques are fundamental tools in multivariable calculus for understanding the behavior of complex functions in multiple dimensions.

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