Problem 8 - Calculus with Concepts in Calculus

Question

Find the directional derivative of \(f\) at the point \(P\) in the direction of a. $$ f(x, y, z)=3 x-2 y+4 z ; P=(1,-1,2) ; \mathbf{a}=\mathbf{i}+\mathbf{j}+\mathbf{k} $$

Step-by-Step Solution

Answer

The directional derivative is \( \frac{5\sqrt{3}}{3} \).

Step 1: Find the Gradient of f

The gradient of the function \( f(x, y, z) = 3x - 2y + 4z \) is \( abla f(x, y, z) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \). Calculate each partial derivative: \( \frac{\partial f}{\partial x} = 3 \), \( \frac{\partial f}{\partial y} = -2 \), \( \frac{\partial f}{\partial z} = 4 \). Hence, \( abla f = (3, -2, 4) \).

Step 2: Evaluate the Gradient at the Point P

Substitute the coordinates of \( P = (1, -1, 2) \) into the gradient calculated in Step 1. Since the gradient itself does not change with these variables, \( abla f(P) = (3, -2, 4) \).

Step 3: Normalize the Direction Vector a

The direction vector \( \mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k} \) has components \( (1, 1, 1) \). Normalize it by dividing by its magnitude. Calculate the magnitude: \( \| \mathbf{a} \| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \). The normalized vector is \( \mathbf{u} = \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \).

Step 4: Calculate the Directional Derivative

The directional derivative of \( f \) at point \( P \) in the direction of \( \mathbf{a} \) is given by the dot product \( abla f(P) \cdot \mathbf{u} \). Compute this dot product: \( (3, -2, 4) \cdot \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) = 3 \times \frac{1}{\sqrt{3}} - 2 \times \frac{1}{\sqrt{3}} + 4 \times \frac{1}{\sqrt{3}} \). This results in \( \frac{5}{\sqrt{3}} \).

Step 5: Simplify the Directional Derivative

Simplify the expression \( \frac{5}{\sqrt{3}} \) by rationalizing the denominator: \( \frac{5}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{3} \). This is the final value of the directional derivative.

Key Concepts

Gradient

The gradient is a vector that points in the direction of the greatest rate of increase of a function. It helps us understand how a function changes at any given point. For a function of several variables, such as \( f(x, y, z) \), the gradient is denoted as \( abla f \) and is composed of partial derivatives. When dealing with a function like \( f(x, y, z) = 3x - 2y + 4z \), the gradient \( abla f \) is calculated by finding the partial derivatives with respect to each variable:- Partial with respect to \( x \): \( \frac{\partial f}{\partial x} = 3 \)- Partial with respect to \( y \): \( \frac{\partial f}{\partial y} = -2 \)- Partial with respect to \( z \): \( \frac{\partial f}{\partial z} = 4 \)These partial derivatives are combined to form the gradient vector: \( abla f = (3, -2, 4) \).The gradient is useful because it tells us both the direction and rate of the fastest increase of the function. It is a fundamental concept used throughout multivariable calculus and physics.

Partial Derivative

Partial derivatives are the building blocks of the gradient. They measure how a function changes as each individual variable is varied, holding the others constant. This is useful in multivariable calculus, where functions depend on more than one variable.For a function \( f(x, y, z) \), we compute its partial derivatives to see how \( f \) changes as \( x \,\), \( y \), and \( z \) change independently:

• \( \frac{\partial f}{\partial x} \): Change in \( f \) when only \( x \) changes
• \( \frac{\partial f}{\partial y} \): Change in \( f \) when only \( y \) changes
• \( \frac{\partial f}{\partial z} \): Change in \( f \) when only \( z \) changes

In our example, the partial derivatives were straightforward to compute: \( \frac{\partial f}{\partial x} = 3 \), \( \frac{\partial f}{\partial y} = -2 \), \( \frac{\partial f}{\partial z} = 4 \).Understanding these helps decode complex relationships within functions and provides the information needed to find the gradient.

Vector Normalization

Vector normalization is the process of scaling a vector so that it has a length, or magnitude, of 1. This transformed vector is called a unit vector. Unit vectors are crucial because they provide direction but not magnitude, making them ideal for situations where the direction is needed without the influence of magnitude.Given the vector \( \mathbf{a} = \mathbf{i} + \mathbf{j} + \mathbf{k} \), the process of normalization involves dividing each component of the vector by its magnitude. First, calculate the magnitude of \( \mathbf{a} \):\[ \| \mathbf{a} \| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3} \]Next, divide each component by this magnitude to get the normalized vector \( \mathbf{u} \):\[ \mathbf{u} = \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \]Using normalized vectors allows us to focus purely on the direction they provide, which is useful in calculating directional derivatives.

Dot Product

The dot product is a mathematical operation that takes two equal-length sequences of numbers (usually vectors) and returns a single number. This operation is significant because it combines vectors and can determine angles or project one vector onto another. For two vectors \( \mathbf{a} = (a_1, a_2, a_3) \) and \( \mathbf{b} = (b_1, b_2, b_3) \), the dot product is:\[ \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \]In the context of directional derivatives, the dot product helps combine the gradient and the direction to find the rate of change of a function in a specific direction. Here, we took the dot product of the gradient \( abla f(P) = (3, -2, 4) \) and the unit vector \( \mathbf{u} = \left( \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right) \):\[ 3 \times \frac{1}{\sqrt{3}} - 2 \times \frac{1}{\sqrt{3}} + 4 \times \frac{1}{\sqrt{3}} = \frac{5}{\sqrt{3}} \]This result gives the directional derivative, representing how quickly the function changes at point \( P \) in the specified direction.