Problem 14 - Advanced Engineering Mathematics

Question

In Problems, 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 \( \frac{19}{5} \).

Step 1: Understand the Problem

We need to find the directional derivative of the function \( f(x, y) = \frac{xy}{x+y} \) at the point \((2, -1)\) in the direction of the vector \( \mathbf{v} = 6\mathbf{i} + 8\mathbf{j} \). The directional derivative tells us how the function changes as we move in a specific direction.

Step 2: Compute the Gradient

First, we find the gradient of \( f(x, y) \), which is the vector of partial derivatives: \( abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \). Calculate \( \frac{\partial f}{\partial x} = \frac{y(x+y) - xy}{(x+y)^2} = \frac{y^2}{(x+y)^2} \) and \( \frac{\partial f}{\partial y} = \frac{x(x+y) - xy}{(x+y)^2} = \frac{x^2}{(x+y)^2} \). 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 Given Point

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

Step 4: Normalize the Direction Vector

The direction vector is \( \mathbf{v} = 6\mathbf{i} + 8\mathbf{j} \). First, find its magnitude: \( \| \mathbf{v} \| = \sqrt{6^2 + 8^2} = 10 \). Then, normalize \( \mathbf{v} \) by dividing it by its magnitude, resulting in the unit vector \( \mathbf{u} = \left( \frac{6}{10}, \frac{8}{10} \right) = \left( \frac{3}{5}, \frac{4}{5} \right) \).

Step 5: Calculate the Directional Derivative

The directional derivative in the direction of \( \mathbf{u} \) is given by the dot product \( abla f(2, -1) \cdot \mathbf{u} = (1, 4) \cdot \left( \frac{3}{5}, \frac{4}{5} \right) \). This equals \( 1 \times \frac{3}{5} + 4 \times \frac{4}{5} = \frac{3}{5} + \frac{16}{5} = \frac{19}{5} \).

Step 6: Conclusion

Thus, the directional derivative of \( f \) at \( (2, -1) \) in the direction of the vector \( 6\mathbf{i} + 8\mathbf{j} \) is \( \frac{19}{5} \).

Key Concepts

Gradient

The gradient, denoted by \( abla f \), is a vector that consists of the partial derivatives of a function. It provides a direction of the steepest ascent of a function. The gradient is extremely useful in vector calculus for analyzing how functions change. In context of the given problem, the function is \( f(x, y) = \frac{xy}{x+y} \), and its gradient at any point \((x, y)\) can be expressed as:

• \(\frac{\partial f}{\partial x}\): The derivative of \( f \) with respect to \( x \).
• \(\frac{\partial f}{\partial y}\): The derivative of \( f \) with respect to \( y \).

By computing these partial derivatives, we form the gradient vector: \( abla f(x,y) = \left( \frac{y^2}{(x+y)^2}, \frac{x^2}{(x+y)^2} \right) \). The gradient tells us how small changes in \( x \) and \( y \) affect the function \( f \). Evaluating this gradient at a specific point allows us to determine the rate and direction of change of the function at that point. This is crucial in finding the directional derivative.

Partial Derivatives

Partial derivatives represent how a function changes as one specific variable is changed, while all other variables are held constant. Essentially, they measure the sensitivity of the function to change in one of its input variables.
In the exercise, the function \( f(x, y) = \frac{xy}{x+y} \) involves two variables, \( x \) and \( y \). Therefore, we compute two partial derivatives:

• \(\frac{\partial f}{\partial x} = \frac{y^2}{(x+y)^2}\): This tells us how \( f \) changes as \( x \) changes.
• \(\frac{\partial f}{\partial y} = \frac{x^2}{(x+y)^2}\): This tells us how \( f \) changes as \( y \) changes.

These derivatives are assembled into the gradient, which gives us a complete picture of how the function \( f \) changes in response to small variations in its input variables, enabling the calculation of the directional derivative.

Vector Normalization

Vector normalization is the process of converting a vector into a unit vector. A unit vector has a magnitude of one and keeps the same direction. This is important when finding directional derivatives because a direction should not depend on scale.
For the direction vector \( \mathbf{v} = 6\mathbf{i} + 8\mathbf{j} \) in the exercise, we start with calculating its magnitude as:\[\| \mathbf{v} \| = \sqrt{6^2 + 8^2} = 10\]To normalize \( \mathbf{v} \), we divide each component by this magnitude:

• Unit vector: \( \mathbf{u} = \left( \frac{6}{10}, \frac{8}{10} \right) = \left( \frac{3}{5}, \frac{4}{5} \right) \)

This normalized vector \( \mathbf{u} \) is then used to calculate the directional derivative, ensuring that we’re considering just the direction and not the scale of movement.

Dot Product

The dot product is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. In the context of directional derivatives, it combines the gradient vector and a direction vector.
To get the directional derivative, we calculate the dot product of the gradient at a certain point and the normalized direction vector. This is done by multiplying corresponding components and summing them:

• If \( abla f(2,-1) = (1, 4) \) and \( \mathbf{u} = \left( \frac{3}{5}, \frac{4}{5} \right) \)
• Directional derivative: \(1 \times \frac{3}{5} + 4 \times \frac{4}{5} = \frac{3}{5} + \frac{16}{5} = \frac{19}{5}\)

This calculated value, \( \frac{19}{5} \), indicates how fast and in what direction the function \( f \) increases when moving in the specified direction.