Problem 282 - Calculus Volume 3

Question

For the following exercises, find the gradient. Find the gradient of \(f(x, y, z)\) at \(P\) and in the direction \(\quad\) of \(f(x, y, z)=\ln \left(x^{2}+2 y^{2}+3 z^{2}\right), P(2,1,4), \quad \mathbf{u}=\frac{-3}{13} \mathbf{i}-\frac{4}{13} \mathbf{j}-\frac{12}{13} \mathbf{k}\).

Step-by-Step Solution

Answer

The directional derivative of the function is \( \frac{-158}{351} \).

Step 1: Find the partial derivatives

To find the gradient of the function \( f(x, y, z) = \ln(x^2 + 2y^2 + 3z^2) \), we need to compute the partial derivatives with respect to each variable: \( x, y,\) and \( z \).

Step 2: Partial derivative with respect to x

\[ \frac{\partial f}{\partial x} = \frac{2x}{x^2 + 2y^2 + 3z^2} \]

Step 3: Partial derivative with respect to y

\[ \frac{\partial f}{\partial y} = \frac{4y}{x^2 + 2y^2 + 3z^2} \]

Step 4: Partial derivative with respect to z

\[ \frac{\partial f}{\partial z} = \frac{6z}{x^2 + 2y^2 + 3z^2} \]

Step 2: Evaluate the gradients at the point P

Substitute the values for \( x = 2, y = 1, z = 4 \) into the partial derivatives to find the gradient at \( P(2, 1, 4) \).

Step 6: Gradient at x

Substituting into \( \frac{\partial f}{\partial x} \):\[ \frac{2(2)}{2^2 + 2(1)^2 + 3(4)^2} = \frac{4}{4 + 2 + 48} = \frac{4}{54} = \frac{2}{27} \]

Step 7: Gradient at y

Substituting into \( \frac{\partial f}{\partial y} \):\[ \frac{4(1)}{2^2 + 2(1)^2 + 3(4)^2} = \frac{4}{54} = \frac{2}{27} \]

Step 8: Gradient at z

Substituting into \( \frac{\partial f}{\partial z} \):\[ \frac{6(4)}{2^2 + 2(1)^2 + 3(4)^2} = \frac{24}{54} = \frac{4}{9} \]

Step 3: Form the gradient vector

Combine the components to form the gradient vector at \( P(2, 1, 4) \):\[ abla f(2, 1, 4) = \langle \frac{2}{27}, \frac{2}{27}, \frac{4}{9} \rangle \]

Step 4: Find the directional derivative

The directional derivative in the direction of \( \mathbf{u} = \frac{-3}{13} \mathbf{i} - \frac{4}{13} \mathbf{j} - \frac{12}{13} \mathbf{k} \) is calculated by taking the dot product of the gradient vector with \( \mathbf{u} \).

Step 11: Compute the directional derivative

\[ abla f(2, 1, 4) \cdot \mathbf{u} = \langle \frac{2}{27}, \frac{2}{27}, \frac{4}{9} \rangle \cdot \left(\frac{-3}{13}, \frac{-4}{13}, \frac{-12}{13}\right) \]\[ = \frac{2}{27} \cdot \left(\frac{-3}{13}\right) + \frac{2}{27} \cdot \left(\frac{-4}{13}\right) + \frac{4}{9} \cdot \left(\frac{-12}{13}\right) \]

Step 12: Simplify the directional derivative

Calculate each term:\[ \frac{-6}{351} + \frac{-8}{351} + \frac{-48}{117} = \frac{-14}{351} + \frac{-144}{351} \]Combine terms:\[ \frac{-158}{351} \]

Key Concepts

Partial Derivatives

Partial derivatives are a cornerstone of calculus, especially when dealing with functions of several variables. In simple terms, they represent the rate of change of a function with respect to one of the variables, while all other variables are held constant. This allows us to dissect complex multivariable functions by looking at how they change piece by piece.

For a function of the form \( f(x, y, z) = \ln(x^2 + 2y^2 + 3z^2) \), we find the partial derivatives with respect to \( x \), \( y \), and \( z \). The process is similar to finding ordinary derivatives for single-variable functions.

• The derivative with respect to \( x \) is \( \frac{2x}{x^2 + 2y^2 + 3z^2} \).
• The derivative with respect to \( y \) is \( \frac{4y}{x^2 + 2y^2 + 3z^2} \).
• The derivative with respect to \( z \) is \( \frac{6z}{x^2 + 2y^2 + 3z^2} \).

At point \( P(2, 1, 4) \), we calculate each of these derivatives by substituting the values of \( x \), \( y \), and \( z \) to evaluate the specific rates of changes at that point.

Gradient Vector

Once we have the partial derivatives, we can compile them into a gradient vector. The gradient vector provides a direction in which the function increases most steeply at a given point. This vector consists of all the partial derivatives of the function, giving us a comprehensive view of how the function changes in any direction.

For the function \( f(x, y, z) \) at point \( P(2, 1, 4) \), the gradient vector is formed by combining the evaluated partial derivatives:
\[ abla f(2, 1, 4) = \left\langle \frac{2}{27}, \frac{2}{27}, \frac{4}{9} \right\rangle \]

This vector indicates that at point \( P \), the function increases most quickly in the direction of the vector \( \left\langle \frac{2}{27}, \frac{2}{27}, \frac{4}{9} \right\rangle \). The length, or magnitude, of this vector shows how steeply the function's value changes.

Directional Derivative

The directional derivative extends the concept of the gradient vector to a specific direction. It calculates the rate of change of the function in the direction of a given vector, \( \mathbf{u} \). This is useful when you are interested in how a function behaves not just generally, but specifically in a chosen path.

To find the directional derivative of \( f \) at point \( P(2, 1, 4) \) in the direction of \( \mathbf{u} = \frac{-3}{13} \mathbf{i} - \frac{4}{13} \mathbf{j} - \frac{12}{13} \mathbf{k} \), we perform a dot product between the gradient vector and \( \mathbf{u} \):
\[ abla f(2, 1, 4) \cdot \mathbf{u} = \left\langle \frac{2}{27}, \frac{2}{27}, \frac{4}{9} \right\rangle \cdot \left(\frac{-3}{13}, \frac{-4}{13}, \frac{-12}{13}\right) \]

This calculation breaks into simpler components, leading to the result \( \frac{-158}{351} \). This number tells us that the function decreases in the direction of \( \mathbf{u} \) at point \( P \).

Multivariable Functions

Multivariable functions are functions that rely on more than one variable. Instead of just \( f(x) \), we say \( f(x, y, z) \) to reflect the many inputs that affect the function's output. These functions are extremely useful for modeling real-world situations where multiple factors play a role.

They allow for more complex and realistic representations compared to single-variable functions, making them essential in fields like physics, engineering, and economics. For instance, the function \( f(x, y, z) = \ln(x^2 + 2y^2 + 3z^2) \) represents a relationship influenced by three different variables \( x \), \( y \), and \( z \).

When working with multivariable functions, calculating quantities such as partial derivatives and directional derivatives helps us understand how changes in the variables impact the function overall. This understanding is crucial when optimizing functions or predicting behavior based on variable shifts.