Problem 32 - Advanced Engineering Mathematics

Question

If \(f(x, y)=x^{2}+x y+y^{2}-x\), find all points where \(D_{\mathrm{u}} f(x, y)\) in the direction of \(\mathbf{u}=(1 / \sqrt{2})(\mathbf{i}+\mathbf{j})\) is zero.

Step-by-Step Solution

Answer

All points \((x, y)\) on the line \(y = \frac{1}{3} - x\) where the directional derivative of \(f(x, y)\) is zero in direction \(\mathbf{u}\).

Step 1: Compute the Gradient

The given function is \( f(x, y) = x^2 + xy + y^2 - x \). To find the gradient \( abla f(x, y) \), we calculate the partial derivatives: \(abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).\ Compute \( \frac{\partial f}{\partial x} = 2x + y - 1 \) and \( \frac{\partial f}{\partial y} = x + 2y \). Thus, the gradient vector is \( abla f(x, y) = (2x + y - 1, x + 2y) \).

Step 2: Direction of the Derivative

We need to find where the directional derivative \( D_{\mathbf{u}} f(x, y) \) is zero in the direction of \( \mathbf{u} = \frac{1}{\sqrt{2}}(\mathbf{i} + \mathbf{j}) \). The formula for the directional derivative is \( D_{\mathbf{u}} f(x, y) = abla f(x, y) \cdot \mathbf{u} \).

Step 3: Compute the Dot Product

Substitute \( abla f(x, y) \) and \( \mathbf{u} \) into the dot product:\[D_{\mathbf{u}} f(x, y) = (2x + y - 1, x + 2y) \cdot \frac{1}{\sqrt{2}}(1, 1)\]This results in:\[D_{\mathbf{u}} f(x, y) = \frac{1}{\sqrt{2}} \left( (2x + y - 1) + (x + 2y) \right)\]

Step 4: Simplify the Expression

Simplify the expression obtained from the dot product:\[D_{\mathbf{u}} f(x, y) = \frac{1}{\sqrt{2}} (3x + 3y - 1)\]

Step 5: Set the Expression to Zero

For the directional derivative to be zero, we set the expression equal to zero:\[\frac{1}{\sqrt{2}} (3x + 3y - 1) = 0\]Simplify this:\[3x + 3y - 1 = 0\]

Step 6: Solve for Points

Re-arrange the simplified equation to solve for \(x\) and \(y\):\[3x + 3y = 1\]Divide both sides by 3 to solve for \(y\):\[y = \frac{1}{3} - x\]The solutions are all points \((x, y)\) on the line given by \(y = \frac{1}{3} - x\).

Key Concepts

Gradient Vector

The gradient vector is a central concept in understanding how functions change. It represents the rate and direction of fastest increase of a function. For a function of two variables, like our example function \(f(x, y) = x^2 + xy + y^2 - x\), the gradient is a vector field formed by the partial derivatives with respect to each variable.
This vector is denoted by \(abla f(x, y)\), and is calculated as:

• The partial derivative with respect to \(x\), noted as \(\frac{\partial f}{\partial x}\), yielding \(2x + y - 1\).
• The partial derivative with respect to \(y\), noted as \(\frac{\partial f}{\partial y}\), giving \(x + 2y\).

Thus, the gradient vector of the function is \(abla f(x, y) = (2x + y - 1, x + 2y)\).
This vector points in the direction of the greatest rate of increase of the function at any given point \(x, y\).

Partial Derivatives

Partial derivatives are foundational for functions of multiple variables. They measure how the function changes as one of the variables is varied, while keeping others constant.
In our exercise, we have the function \(f(x, y) = x^2 + xy + y^2 - x\), and we find its partial derivatives as follows:

• \(\frac{\partial f}{\partial x}\): This measures the change of \(f\) along \(x\), resulting in \(2x + y - 1\).
• \(\frac{\partial f}{\partial y}\): This depicts the change of \(f\) in the direction of \(y\), resulting in \(x + 2y\).

The resulting partial derivatives are used to form the gradient vector, which helps in understanding how the function behaves in its two-variable domain.

Dot Product

The dot product plays a crucial role in calculating the directional derivative, which describes the rate of change of the function in a specified direction.
For our function \(f(x, y)\), the directional derivative in the direction of vector \(\mathbf{u}\) is given by \(D_{\mathbf{u}} f(x, y) = abla f(x, y) \cdot \mathbf{u}\).
With \(abla f(x, y) = (2x + y - 1, x + 2y)\) and \(\mathbf{u} = \frac{1}{\sqrt{2}}(1, 1)\), the dot product is computed as:

• \(D_{\mathbf{u}} f(x, y) = \frac{1}{\sqrt{2}} ((2x + y - 1) + (x + 2y))\).
• This simplifies to \(\frac{1}{\sqrt{2}} (3x + 3y - 1)\).

The presence of the dot product allows us to find when the directional derivative equals zero, revealing critical points.

Functions of Two Variables

A function of two variables, such as \(f(x, y) = x^2 + xy + y^2 - x\), maps ordered pairs \((x, y)\) to a real number. These functions are often visualized as surfaces in a three-dimensional space.
Understanding such functions involves examining their changes along the \(x\) and \(y\) axes separately, using tools like partial derivatives and gradients. In our solution, we investigate where the directional derivative in a given direction is zero.
This tells us where the function remains constant along that direction. By solving \(3x + 3y - 1 = 0\), we find that all points \((x, y)\) on the line \(y = \frac{1}{3} - x\) have this property. Thus, the function does not change in the direction of \(\mathbf{u}\) along these points.