Problem 19 - APEX Calculus

Question

In Exercises \(19-24,\) a function \(z=f(x, y)\) and a point \(P\) are given. (a) Find the direction of maximal increase of \(f\) at \(P\). (b) What is the maximal value of \(D_{\vec{u}} f\) at \(P ?\) (c) Find the direction of minimal increase of \(f\) at \(P\). (d) Give a direction \(\vec{u}\) such that \(D_{\vec{u}} f=0\) at \(P\). Note: these are the same functions and points as in Exercises 13 through 18 . $$ f(x, y)=-x^{2} y+x y^{2}+x y, P=(2,1) $$

Step-by-Step Solution

Answer

Direction of maximal increase: \( \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \); Maximal \( D_{\vec{u}}f: 2\sqrt{2} \); Direction of minimal increase: \( \left( \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right) \); Zero increase direction: \( \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \).

Step 1: Calculate the Gradient of the Function

The gradient of a function \( f(x, y) \) is given by \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Calculate the partial derivatives:\[ \frac{\partial f}{\partial x} = -2xy + y^2 + y \]\[ \frac{\partial f}{\partial y} = -x^2 + 2xy + x \]So, the gradient is \( abla f = ( -2xy + y^2 + y, -x^2 + 2xy + x ) \).

Step 2: Evaluate the Gradient at Point P(2,1)

Substitute \( x = 2 \) and \( y = 1 \) into the gradient to find:\[ abla f (2, 1) = (-2(2)(1) + 1^2 + 1, -(2)^2 + 2(2)(1) + 2) \]\[ = (-4 + 1 + 1, -4 + 4 + 2) \]\[ = (-2, 2) \].The gradient at \( P \) is \( abla f (2, 1) = (-2, 2) \).

Step 3: Determine the Direction of Maximal Increase

The direction of maximal increase of \( f \) at \( P \) is along the direction of the gradient vector. Hence, the direction is \( (-2, 2) \). To express as a unit vector, divide by its magnitude:Calculate magnitude:\[ \| abla f (2, 1) \| = \sqrt{(-2)^2 + (2)^2} = \sqrt{4 + 4} = \sqrt{8} = 2\sqrt{2} \]Unit vector:\[ \vec{u}_{max} = \left( -\frac{2}{2\sqrt{2}}, \frac{2}{2\sqrt{2}} \right) = \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \].This normalized vector is the direction of maximal increase.

Step 4: Compute the Maximal Value of the Directional Derivative

The maximal value of the directional derivative at \( P \) is the magnitude of the gradient at that point, which was calculated previously:\[ \text{Maximal } D_{\vec{u}} f = \| abla f (2, 1) \| = 2\sqrt{2} \].

Step 5: Determine the Direction of Minimal Increase

The direction of minimal increase (maximal decrease) of \( f \) at \( P \) is opposite to the direction of maximal increase. Thus, it is \( (2, -2) \). To express it as a unit vector, use:Unit vector:\[ \vec{u}_{min} = \left( \frac{2}{2\sqrt{2}}, -\frac{2}{2\sqrt{2}} \right) = \left( \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right) \].

Step 6: Find a Direction Where Directional Derivative is Zero

The directional derivative is zero in any direction perpendicular to the gradient. A vector perpendicular to \( (-2, 2) \) can be found by switching the components and changing one sign. Choose \( (2, 2) \) for simplicity.Make it a unit vector:\[ \vec{u}_{zero} = \left( \frac{2}{2\sqrt{2}}, \frac{2}{2\sqrt{2}} \right) = \left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \].

Key Concepts

Gradient Vector

When dealing with multivariable functions like \( f(x, y) = -x^2 y + xy^2 + xy \), the gradient vector \( abla f \) plays a vital role. It represents the direction and rate of fastest increase of the function. The gradient is a vector composed of partial derivatives, specifically \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). This requires finding the partial derivatives of the function with respect to \( x \) and \( y \).

For our function, these are calculated as follows:

• \( \frac{\partial f}{\partial x} = -2xy + y^2 + y \)
• \( \frac{\partial f}{\partial y} = -x^2 + 2xy + x \)

Evaluating at point \( P(2,1) \), we substitute to find the gradient vector: \( abla f (2, 1) = (-2, 2) \). This vector indicates both the direction and the strength of the change in the function at \( P \).

The gradient vector not only points in the direction of steepest ascent but also shows how steep that ascent is.

Maximal Increase

The notion of maximal increase refers to the highest possible increase in function value per unit distance moved from a given point, which is precisely along the gradient vector. For the function \( f(x, y) \), at point \( P(2,1) \), the direction of maximal increase is aligned with the gradient \( abla f (2, 1) = (-2, 2) \).

To express this direction as a unit vector, which is important for consistency, we normalize the gradient by dividing it by its magnitude. The unit vector \( \vec{u}_{max} \) is computed as follows:

• Magnitude of \( abla f (2, 1) = \sqrt{(-2)^2 + (2)^2} = 2\sqrt{2} \)

Giving us the normalized direction:

• \( \vec{u}_{max} = \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \)

This unit vector expresses the direction in which \( f \) increases most rapidly. The value of the directional derivative in this direction demonstrates the maximum rate of increase at point \( P \), which equals the gradient's magnitude, \( 2\sqrt{2} \).

Unit Vector

A unit vector is a vector with a magnitude (length) of one. It retains the direction of the original vector but scales it down, making it easier to handle in calculations. In the context of directional derivatives and gradients, unit vectors help in standardizing directions for comparison.

After finding the gradient vector \( abla f (2, 1) = (-2, 2) \), we need its corresponding unit vector for determining maximal and minimal changes. The normalization process involves dividing each component of the vector by its magnitude.

For instance, the unit vector for the maximal increase direction is:

• \( \vec{u}_{max} = \left( -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right) \)

Similarly, the unit vector pointing in the minimal increase (or decrease) direction is:

• \( \vec{u}_{min} = \left( \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right) \)

These unit vectors are essential for clearly conveying the direction of change without concern for magnitude.

Partial Derivatives

Partial derivatives are fundamental in understanding how functions behave with respect to their individual variables. They provide the rate of change of the function as one variable changes while others are held constant. For a function \( f(x, y) \), partial derivatives are needed to form the gradient vector:

• \( \frac{\partial f}{\partial x} \) indicates the change in \( f \) as \( x \) increases, keeping \( y \) fixed.
• \( \frac{\partial f}{\partial y} \) measures how \( f \) changes with \( y \), keeping \( x \) constant.

In this exercise, evaluating these derivatives gives us:

• \( \frac{\partial f}{\partial x} = -2xy + y^2 + y \)
• \( \frac{\partial f}{\partial y} = -x^2 + 2xy + x \)

Each derivative uncovers how the function bends or tilts along the respective axis. These derivatives not only help find the gradient but also aid in understanding the local behavior of \( f \) around a particular point.

By examining partial derivatives, we gain insight into how each variable uniquely affects the overall function, a vital step in optimization and analysis.