Calculating The Directional Derivative Of F(x, Y, Z) = Xy³ + Yz³ At (2, -1, 1)

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Introduction

In multivariable calculus, the directional derivative is a fundamental concept that extends the idea of a derivative to functions of several variables. It measures the rate of change of a function along a specific direction. This article delves into the process of calculating the directional derivative of the function f(x, y, z) = xy³ + yz³ at the point (2, -1, 1) in the direction of the vector i + 2j + 2k. We will break down the steps involved, providing a comprehensive understanding of the underlying principles and calculations.

Understanding the Directional Derivative

The directional derivative is a powerful tool for analyzing how a function changes in different directions. Unlike partial derivatives, which measure the rate of change along the coordinate axes, the directional derivative allows us to explore the function's behavior along any arbitrary direction. This is particularly useful in various applications, such as physics, engineering, and computer graphics, where understanding how a quantity changes in a specific direction is crucial.

Definition and Formula

The directional derivative of a scalar function f(x, y, z) at a point P(x₀, y₀, z₀) in the direction of a unit vector u = <a, b, c> is denoted by Dᵤf(x₀, y₀, z₀) and is defined as:

Dᵤf(x₀, y₀, z₀) = ∇f(x₀, y₀, z₀) ⋅ u

where ∇f(x₀, y₀, z₀) is the gradient of f at the point P, and ⋅ denotes the dot product. The gradient, denoted by ∇f, is a vector that points in the direction of the greatest rate of increase of the function f. It is calculated as:

∇f(x, y, z) = <∂f/∂x, ∂f/∂y, ∂f/∂z>

Steps to Calculate the Directional Derivative

To calculate the directional derivative, we follow these steps:

  1. Calculate the gradient of the function: Find the partial derivatives of f with respect to each variable (x, y, and z) and form the gradient vector.
  2. Evaluate the gradient at the given point: Substitute the coordinates of the point P(x₀, y₀, z₀) into the gradient vector.
  3. Find the unit vector in the given direction: If the given direction is not a unit vector, normalize it by dividing it by its magnitude.
  4. Calculate the dot product: Take the dot product of the gradient at the point P and the unit vector. The result is the directional derivative.

Step-by-Step Calculation for f(x, y, z) = xy³ + yz³ at (2, -1, 1) in the Direction of i + 2j + 2k

Now, let's apply these steps to our specific problem: finding the directional derivative of f(x, y, z) = xy³ + yz³ at the point (2, -1, 1) in the direction of the vector i + 2j + 2k.

1. Calculate the Gradient of f(x, y, z)

The gradient of f(x, y, z) is a vector composed of its partial derivatives with respect to x, y, and z. Let's calculate these partial derivatives:

  • Partial derivative with respect to x (∂f/∂x):

    To find ∂f/∂x, we treat y and z as constants and differentiate f(x, y, z) with respect to x:

    ∂f/∂x = ∂(xy³ + yz³)/∂x = y³

  • Partial derivative with respect to y (∂f/∂y):

    To find ∂f/∂y, we treat x and z as constants and differentiate f(x, y, z) with respect to y:

    ∂f/∂y = ∂(xy³ + yz³)/∂y = 3xy² + z³

  • Partial derivative with respect to z (∂f/∂z):

    To find ∂f/∂z, we treat x and y as constants and differentiate f(x, y, z) with respect to z:

    ∂f/∂z = ∂(xy³ + yz³)/∂z = 3yz²

Now, we can construct the gradient vector:

∇f(x, y, z) = <y³, 3xy² + z³, 3yz²>

2. Evaluate the Gradient at the Point (2, -1, 1)

Next, we need to evaluate the gradient at the point (2, -1, 1). This means substituting x = 2, y = -1, and z = 1 into the gradient vector:

∇f(2, -1, 1) = <(-1)³, 3(2)(-1)² + (1)³, 3(-1)(1)²>

Simplifying this, we get:

∇f(2, -1, 1) = <-1, 3(2)(1) + 1, -3> = <-1, 7, -3>

So, the gradient of f at the point (2, -1, 1) is the vector <-1, 7, -3>. This vector points in the direction of the greatest rate of increase of the function f at this specific point.

3. Find the Unit Vector in the Direction of i + 2j + 2k

We are given the direction vector v = i + 2j + 2k, which can be written in component form as v = <1, 2, 2>. To find the directional derivative, we need a unit vector in this direction. A unit vector has a magnitude of 1. To normalize v, we divide it by its magnitude.

First, let's calculate the magnitude of v:

||v|| = √(1² + 2² + 2²) = √(1 + 4 + 4) = √9 = 3

Now, we can find the unit vector u by dividing v by its magnitude:

u = v / ||v|| = <1, 2, 2> / 3 = <1/3, 2/3, 2/3>

So, the unit vector in the direction of i + 2j + 2k is <1/3, 2/3, 2/3>. This vector has the same direction as the original vector but a magnitude of 1, making it suitable for calculating the directional derivative.

4. Calculate the Dot Product

Finally, we can calculate the directional derivative by taking the dot product of the gradient at the point (2, -1, 1) and the unit vector u:

Dᵤf(2, -1, 1) = ∇f(2, -1, 1) ⋅ u

We have already found that ∇f(2, -1, 1) = <-1, 7, -3> and u = <1/3, 2/3, 2/3>. Now, let's calculate the dot product:

Dᵤf(2, -1, 1) = <-1, 7, -3> ⋅ <1/3, 2/3, 2/3>

Dᵤf(2, -1, 1) = (-1)(1/3) + (7)(2/3) + (-3)(2/3)

Dᵤf(2, -1, 1) = -1/3 + 14/3 - 6/3

Dᵤf(2, -1, 1) = 7/3

Therefore, the directional derivative of f(x, y, z) = xy³ + yz³ at the point (2, -1, 1) in the direction of i + 2j + 2k is 7/3.

Conclusion

In this article, we have demonstrated the step-by-step process of calculating the directional derivative of a multivariable function. We first defined the directional derivative and its formula, then applied this knowledge to a specific example. By calculating the gradient, evaluating it at the given point, finding the unit vector in the specified direction, and computing the dot product, we successfully determined the directional derivative of f(x, y, z) = xy³ + yz³ at the point (2, -1, 1) in the direction of i + 2j + 2k. The result, 7/3, represents the rate of change of the function in that particular direction. Understanding the directional derivative is crucial for various applications in mathematics, physics, and engineering, providing insights into how functions change in different directions.

This comprehensive guide provides a solid foundation for understanding and calculating directional derivatives. By mastering this concept, you can gain a deeper understanding of multivariable calculus and its applications in various fields.