Problem 282 - Calculus 3

Question

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

Step-by-Step Solution

Answer

The gradient of \(f\) at \(P\) in the direction \(\mathbf{u}\) is \(-\frac{158}{351}\).

Step 1: Find the Gradient of f

First, let's determine the gradient \( abla f \) of the function \( f(x, y, z) = \ln(x^2 + 2y^2 + 3z^2) \). The gradient is the vector of partial derivatives:\[ abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \]- **Partial derivative with respect to \( x \):** \[ \frac{\partial f}{\partial x} = \frac{1}{x^2 + 2y^2 + 3z^2} \cdot (2x) \]- **Partial derivative with respect to \( y \):** \[ \frac{\partial f}{\partial y} = \frac{1}{x^2 + 2y^2 + 3z^2} \cdot (4y) \]- **Partial derivative with respect to \( z \):** \[ \frac{\partial f}{\partial z} = \frac{1}{x^2 + 2y^2 + 3z^2} \cdot (6z) \]Thus, the gradient is:\[ abla f = \left( \frac{2x}{x^2 + 2y^2 + 3z^2}, \frac{4y}{x^2 + 2y^2 + 3z^2}, \frac{6z}{x^2 + 2y^2 + 3z^2} \right) \]

Step 2: Evaluate the Gradient at Point P

Substitute the coordinates of point \( P(2, 1, 4) \) into the gradient \( abla f \):- Substitute \( x = 2 \), \( y = 1 \), and \( z = 4 \): - Calculate \( x^2 + 2y^2 + 3z^2 = 2^2 + 2(1^2) + 3(4^2) = 4 + 2 + 48 = 54 \). - Then, \( \frac{2x}{54} = \frac{2(2)}{54} = \frac{4}{54} = \frac{2}{27} \). - \( \frac{4y}{54} = \frac{4(1)}{54} = \frac{4}{54} = \frac{2}{27} \). - \( \frac{6z}{54} = \frac{6(4)}{54} = \frac{24}{54} = \frac{4}{9} \).So, \( abla f(2, 1, 4) = \left( \frac{2}{27}, \frac{2}{27}, \frac{4}{9} \right) \).

Step 3: Find the Directional Derivative

Now, find the directional derivative of \( f \) at \( P \) in the direction of \( \mathbf{u} \) by using the dot product of \( abla f(2, 1, 4) \) and \( \mathbf{u} \):\( \mathbf{u} = \frac{-3}{13} \mathbf{i} - \frac{4}{13} \mathbf{j} - \frac{12}{13} \mathbf{k} \).The directional derivative \( D_{\mathbf{u}}f \) is given by:\[ D_{\mathbf{u}}f = abla f(2, 1, 4) \cdot \mathbf{u} = \left( \frac{2}{27}, \frac{2}{27}, \frac{4}{9} \right) \cdot \left( \frac{-3}{13}, \frac{-4}{13}, \frac{-12}{13} \right) \]Calculate the dot product:- \( \frac{2}{27} \times \frac{-3}{13} = -\frac{6}{351} \).- \( \frac{2}{27} \times \frac{-4}{13} = -\frac{8}{351} \).- \( \frac{4}{9} \times \frac{-12}{13} = -\frac{48}{117} \).Sum these:\( D_{\mathbf{u}}f = -\frac{6}{351} - \frac{8}{351} - \frac{48}{117} = -\frac{14}{351} - \frac{144}{351} = -\frac{158}{351} \).

Step 4: Simplify the Directional Derivative Result

Simplify the directional derivative result \(-\frac{158}{351}\). Since \(158\) and \(351\) do not have common factors other than \(1\), \(-\frac{158}{351}\) is already in its simplest form.

Key Concepts

Partial Derivatives

Partial derivatives are fundamental in understanding how a multivariable function changes. These derivatives break down how the function behaves as each independent variable shifts, one at a time. For a function like our given \( f(x, y, z) = \ln(x^2 + 2y^2 + 3z^2) \), we examine how changes in \( x \), \( y \), and \( z \) individually affect the entire function.

To find these, you temporarily fix all other variables and differentiate with respect to the one you’re focusing on.

• For \( \frac{\partial f}{\partial x} \), consider \( y \) and \( z \) constant, which results in \( \frac{2x}{x^2 + 2y^2 + 3z^2} \).
• When finding \( \frac{\partial f}{\partial y} \), \( x \) and \( z \) are fixed, leading to \( \frac{4y}{x^2 + 2y^2 + 3z^2} \).
• Lastly, for \( \frac{\partial f}{\partial z} \), \( x \) and \( y \) remain unchanged, yielding \( \frac{6z}{x^2 + 2y^2 + 3z^2} \).

Collectively, these results form a vector known as the gradient, which represents the direction of greatest increase of the function. This vector has important implications in vector calculus.

Directional Derivative

The directional derivative extends the concept of partial derivatives by considering the rate of change of the function not just along the axes, but in any specified direction. This measure is quite critical because it tells how much the function increases or decreases as we move in a particular direction.

Here's how it's tackled: Given a unit vector \( \mathbf{u} \), we calculate the dot product of the gradient vector and \( \mathbf{u} \). In the problem:

• The gradient at point \( P \) was found to be \( abla f = \left( \frac{2}{27}, \frac{2}{27}, \frac{4}{9} \right) \).
• The vector \( \mathbf{u} = \frac{-3}{13} \mathbf{i} - \frac{4}{13} \mathbf{j} - \frac{12}{13} \mathbf{k} \).

To find the directional derivative, compute:

• The dot product results in \(-\frac{6}{351} - \frac{8}{351} - \frac{48}{117} = -\frac{158}{351}\).

This value indicates how quickly \( f \) changes as we move from point \( P \) in the direction of \( \mathbf{u} \). It combines both the gradient's magnitude and the tilt introduced by \( \mathbf{u} \). This concept is key when interfering with fields like physics and engineering, where path direction matters.

Vector Calculus

Vector calculus is a branch of mathematics focused on vector fields and their derivatives and integrals. It's crucial for understanding how physical forces and quantities behave in space.

The gradient and directional derivative, as seen in this exercise, showcase just two of vector calculus's tools. The gradient:

• Is a vector representation of the partial derivatives and heads toward the greatest increase of a function.
• Enables us to evaluate the change of the multivariable function as we explored with \( abla f = \left( \frac{2x}{x^2 + 2y^2 + 3z^2}, \frac{4y}{x^2 + 2y^2 + 3z^2}, \frac{6z}{x^2 + 2y^2 + 3z^2} \right) \).

The directional derivative:

• Provides insight into how the function evolves as we move along any arbitrary path determined by a vector \( \mathbf{u} \).
• Is calculated by dotting the gradient with \( \mathbf{u} \), leading to our specific example result of \(-\frac{158}{351}\).

Vector calculus lays the groundwork for advanced studies in electromagnetism, gravitation, and fluid dynamics—fields where vector quantities are at play. Its principles are not just abstract, but applicable in engineering and the physical sciences, aiding in the design of everything from bridges to circuit boards.