Problem 19 - Calculus Volume 3

Question

Find the gradient vector field of each function \(f\). $$ f(x, y)=\ln \left(1+x^{2}+2 y^{2}\right) $$

Step-by-Step Solution

Answer

The gradient vector field is \( \nabla f = \left( \frac{2x}{1+x^2+2y^2}, \frac{4y}{1+x^2+2y^2} \right) \).

Step 1: Understand the Gradient Vector Field

The gradient vector field of a function \( f(x, y) \) is a vector field consisting of the partial derivatives of \( f \) with respect to each variable. The gradient is denoted as \( abla f \), and for a function \( f(x, y) \), it is given by the vector \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \).

Step 2: Compute Partial Derivative with respect to x

To find \( \frac{\partial f}{\partial x} \), treat \( y \) as a constant and differentiate \( f(x, y) = \ln(1 + x^2 + 2y^2) \) with respect to \( x \). Using the chain rule:\[\frac{\partial}{\partial x}[\ln(u)] = \frac{1}{u} \cdot \frac{\partial u}{\partial x}\]where \( u = 1 + x^2 + 2y^2 \).Thus, \( \frac{\partial u}{\partial x} = 2x \), so:\[\frac{\partial f}{\partial x} = \frac{1}{1 + x^2 + 2y^2} \cdot 2x = \frac{2x}{1 + x^2 + 2y^2}\]

Step 3: Compute Partial Derivative with respect to y

Next, compute \( \frac{\partial f}{\partial y} \) by treating \( x \) as a constant and differentiating \( f(x, y) = \ln(1 + x^2 + 2y^2) \) with respect to \( y \). Again using the chain rule:\[\frac{\partial}{\partial y}[\ln(u)] = \frac{1}{u} \cdot \frac{\partial u}{\partial y}\]where \( u = 1 + x^2 + 2y^2 \).Thus, \( \frac{\partial u}{\partial y} = 4y \), so:\[\frac{\partial f}{\partial y} = \frac{1}{1 + x^2 + 2y^2} \cdot 4y = \frac{4y}{1 + x^2 + 2y^2}\]

Step 4: Write the Gradient Vector Field

Combine the partial derivatives to write the gradient vector field:\[abla f = \left( \frac{2x}{1 + x^2 + 2y^2}, \frac{4y}{1 + x^2 + 2y^2} \right)\]This represents the gradient vector field for the function \( f(x, y) = \ln(1 + x^2 + 2y^2) \).

Key Concepts

Partial Derivatives

Partial derivatives measure how a function changes as one variable changes, while other variables are held constant. In multivariable calculus, understanding partial derivatives is key to analyzing functions with more than one variable.
Partial derivatives are denoted using the symbol \( \frac{\partial}{\partial x} \) for the derivative with respect to \( x \), and similarly for other variables like \( y \) or \( z \). When you need to find a partial derivative of a function \( f(x, y) \), you treat all other variables as constants and differentiate only with respect to the chosen variable. This isolates the effect that one variable has on the function, helping to understand a multivariable function's behavior in a specific direction.

• The partial derivative \( \frac{\partial f}{\partial x} \) shows how the function \( f \) changes as \( x \) changes, holding \( y \) constant.`
• Similarly, \( \frac{\partial f}{\partial y} \) reflects the change in \( f \) as \( y \) changes, keeping \( x \) constant.

Understanding these concepts helps identify how each variable individually influences the function’s value, allowing for easier analysis of complex surfaces or curves given by functions of two or more variables.

Chain Rule in Calculus

The chain rule is a formula for computing the derivative of a composite function. In the context of partial derivatives, it becomes extremely useful when differentiating a function like \( f(x, y) = \ln(1 + x^2 + 2y^2) \). Here, the chain rule aids in handling nested functions effectively.

When applying the chain rule to partial derivatives, you focus on differentiating the outer function first, and then multiply by the derivative of the inner function. Consider \( u = 1+x^2+2y^2 \) as the inner function, and the outer function as \( \ln(u) \).

• For \( \frac{\partial}{\partial x}[\ln(u)] \), start by differentiating \( \ln(u) \) with respect to \( u \), which is \( \frac{1}{u} \), and then multiply by \( \frac{\partial u}{\partial x} \).
• To find \( \frac{\partial}{\partial y}[\ln(u)] \), follow the same approach: differentiate \( \ln(u) \), then multiply by \( \frac{\partial u}{\partial y} \).

This method systematically tears down complex multivariable problems into simpler parts, making derivatives manageable. By understanding how to apply the chain rule, you can solve intricate function derivatives that initially seem daunting.

Multivariable Functions

Multivariable functions, as the name suggests, are functions that have more than one variable. These are incredibly common in calculus and are fundamental in fields like physics, engineering, and economics. A multivariable function takes the form \( f(x, y) \), which implies output values depending on both \( x \) and \( y \).

What's crucial with these functions is understanding how they map input pairs to an output and how changes in each variable separately affect the outcome.

• The function \( f(x, y) = \ln(1+x^2+2y^2) \) is an example where the resulting value depends on both variables \( x \) and \( y \) simultaneously.
• The value of the function is influenced by changes in either \( x \) or \( y \), or both, demonstrating different behaviors through partial derivatives.

In contexts like optimization or physics, knowing how to characterize these behaviors with multivariable functions helps to find extremes (like maxima or minima) or even solve real-world problems involving multidimensional changes.