Problem 14 - Advanced Engineering Mathematics

Question

Find the directional derivative of the given function at the given point in the indicated direction. $$ f(x, y)=\frac{x y}{x+y} ;(2,-1), 6 \mathbf{i}+8 \mathbf{j} $$

Step-by-Step Solution

Answer

The directional derivative is 3.8.

Step 1: Understand the Directional Derivative

The directional derivative of a function gives the rate at which the function changes in a specific direction. For a function \( f(x,y) \), the directional derivative at a point \((x_0, y_0)\) in the direction of a vector \( \mathbf{v} \) is given by \( D_\mathbf{u}f(x_0, y_0) = abla f(x_0, y_0) \cdot \mathbf{u} \), where \( \mathbf{u} \) is a unit vector in the direction of \( \mathbf{v} \).

Step 2: Compute the Gradient

First, we need to find the gradient of the function \( f(x, y) \). The gradient \( abla f(x, y) \) is computed as follows:\[ abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \]Calculate the partial derivatives, where:- \( \frac{\partial f}{\partial x} = \frac{y(x+y) - xy}{(x+y)^2} = \frac{y^2}{(x+y)^2} \)- \( \frac{\partial f}{\partial y} = \frac{x(x+y) - xy}{(x+y)^2} = \frac{x^2}{(x+y)^2} \)Thus:\[ abla f(x, y) = \left( \frac{y^2}{(x+y)^2}, \frac{x^2}{(x+y)^2} \right) \].

Step 3: Evaluate the Gradient at the Given Point

Substitute the point \((2, -1)\) into the gradient \( abla f(x, y) \):\[ abla f(2, -1) = \left( \frac{(-1)^2}{(2-1)^2}, \frac{2^2}{(2-1)^2} \right) = (1, 4) \].

Step 4: Normalize the Direction Vector

The direction vector given is \( 6 \mathbf{i} + 8 \mathbf{j} \). First, find its magnitude:\[ \| \mathbf{v} \| = \sqrt{6^2 + 8^2} = \sqrt{100} = 10 \].Then, convert \( \mathbf{v} \) into a unit vector \( \mathbf{u} \):\[ \mathbf{u} = \frac{1}{10} (6, 8) = (0.6, 0.8) \].

Step 5: Calculate the Directional Derivative

Now, compute the directional derivative using the gradient and the unit vector:\[ D_\mathbf{u}f(2, -1) = abla f(2, -1) \cdot \mathbf{u} = (1, 4) \cdot (0.6, 0.8) = 1 \times 0.6 + 4 \times 0.8 = 3.8 \].

Key Concepts

Understanding Gradient Vector

The gradient vector of a function plays a crucial role in determining how the function behaves at any point. For a two-variable function like \( f(x, y) \), the gradient is a vector representing the function's steepest ascent \[ abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \] This means the direction of this vector shows where the function increases most rapidly. The gradient not only indicates the path of fastest increase but its magnitude shows the rate at which the function increases. When calculating the gradient at any given point, such as \((2, -1)\), you substitute these coordinates into the equations for the partial derivatives. This gives us an insight into both the direction and speed of change at that specific point.

Partial Derivatives Simplified

Partial derivatives represent the fundamental building blocks of the gradient. They measure how the function changes as you move along one axis while keeping the other constant. For instance, in the function \( f(x, y) = \frac{xy}{x+y} \), the partial derivative with respect to \( x \) is calculated by assessing the function's change when \( y \) is constant. To find these derivatives for our function, we used the quotient rule. The result:

• Partial derivative with respect to \( x \), \( \frac{\partial f}{\partial x} = \frac{y^2}{(x+y)^2} \)
• Partial derivative with respect to \( y \), \( \frac{x^2}{(x+y)^2} \)

These derivatives are essential for forming the gradient vector, which we then use to explore directions of change, establishing the basis for directional derivatives.

The Role of a Unit Vector

Unit vectors are vital for accurately determining the direction vector in which you want to measure changes in a function. A unit vector is a vector with a magnitude of one, ensuring the direction is maintained without exaggerating the scale of its impact. In our exercise, we started with the vector \( 6\mathbf{i} + 8\mathbf{j} \). The first step was calculating its magnitude, which was 10, allowing us to transform it into a unit vector: \[ \mathbf{u} = \frac{1}{10} (6, 8) = (0.6, 0.8) \] By converting to a unit vector, we could accurately measure how the function's value changes in the desired direction without introducing distortions due to varying vector lengths.

Exploring Rate of Change

The rate of change in a specific direction of a function, known as the directional derivative, tells us how quickly the function is increasing or decreasing. To determine this for a point like \((2, -1)\) in our function, we calculate its directional derivative using the formula: \[ D_\mathbf{u}f(x_0, y_0) = abla f(x_0, y_0) \cdot \mathbf{u} \] This means multiplying the components of the gradient vector by the unit vector, effectively focusing on the function's rate of change in the specified direction. For our exercise, the directional derivative calculation gave us a rate of 3.8 in the direction of \( \mathbf{u} = (0.6, 0.8) \). This reveals how quickly the function's value increases in the desired direction from the given point.