Problem 3 - Thomas Calculus

Question

Find the gradient fields of the functions $$ g(x, y, z)=e^{z}-\ln \left(x^{2}+y^{2}\right) $$

Step-by-Step Solution

Answer

The gradient field is \( \nabla g = \left( -\frac{2x}{x^2+y^2}, -\frac{2y}{x^2+y^2}, e^z \right) \).

Step 1: Understanding the Gradient

The gradient of a function is a vector field that consists of all its partial derivatives with respect to its variables. For a function \( g(x, y, z) \), the gradient is denoted as \( abla g \) and is calculated as \( abla g = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y}, \frac{\partial g}{\partial z} \right) \).

Step 2: Calculate Partial Derivative with Respect to x

To find \( \frac{\partial g}{\partial x} \), treat \( y \) and \( z \) as constants. The function involves the term \(-\ln(x^2 + y^2)\) whose derivative is obtained using the chain rule: \( \frac{d}{dx}[ -\ln(x^2 + y^2)] = -\frac{1}{x^2 + y^2} \cdot 2x = -\frac{2x}{x^2 + y^2} \). Hence, \( \frac{\partial g}{\partial x} = -\frac{2x}{x^2 + y^2} \).

Step 3: Calculate Partial Derivative with Respect to y

For \( \frac{\partial g}{\partial y} \), treat \( x \) and \( z \) as constants. Similarly to \( x \), use the chain rule for the term \(-\ln(x^2 + y^2)\): \( \frac{d}{dy}[ -\ln(x^2 + y^2)] = -\frac{1}{x^2 + y^2} \cdot 2y = -\frac{2y}{x^2 + y^2} \). Thus, \( \frac{\partial g}{\partial y} = -\frac{2y}{x^2 + y^2} \).

Step 4: Calculate Partial Derivative with Respect to z

For \( \frac{\partial g}{\partial z} \), treat \( x \) and \( y \) as constants. The term \( e^z \) differentiates to \( e^z \). Therefore, \( \frac{\partial g}{\partial z} = e^z \).

Step 5: Forming the Gradient Field

Combine the partial derivatives obtained to express the gradient field: \( abla g = \left( -\frac{2x}{x^2+y^2}, -\frac{2y}{x^2+y^2}, e^z \right) \). This vector represents the gradient of the function \( g(x, y, z) \).

Key Concepts

Partial Derivatives

Partial derivatives are crucial when dealing with functions of multiple variables. Imagine a surface defined by a function in three dimensions. Each point on this surface changes based on its position. The partial derivative answers how the function changes as we move along one direction, while keeping the other variables constant.

For example, suppose we have a function, like the one in the exercise:

• g(x, y, z) = e^{z} - \ln(x^{2} + y^{2})

To find how this function changes with respect to just one variable, say x, we need \( \frac{\partial g}{\partial x} \). It essentially tells us the slope of the function in the x direction, considering y and z are kept constant.

This process is repeated for all variables involved, providing insights into how each variable specifically impacts the entire function. This insight is foundational when compiling the entire gradient of a multi-variable function.

Vector Field

A vector field is an assignment of a vector to every point in a subset of space. Think of it like a map, where each point shows the direction and magnitude of some value. In the context of gradients, each point in the space gets a vector that tells us how steeply the function increases in each direction.

The step-by-step solution forms the gradient field by combining the partial derivatives. For the function g(x, y, z), the gradient field is:

• \( abla g = \left( -\frac{2x}{x^2+y^2}, -\frac{2y}{x^2+y^2}, e^z \right) \)

This collection of vectors at each point helps understand the overall change of the function in space.

When visualizing a vector field, arrows can represent each vector, showing both the direction and the speed of change the function is undergoing at each point in the field. This way, you get a tangible sense of the function's behavior over its entire domain.

Chain Rule

The Chain Rule is an essential tool for finding derivatives of composite functions. It acts like a connection, allowing you to compute complex derivatives by breaking them down into simpler parts.

In the provided exercise, the Chain Rule is used to compute the partial derivatives of terms like \(-\ln(x^2 + y^2)\). Here's how it works:

• First, identify the inner function, like u = x^2 + y^2.
• Differentiate u with respect to x or y, yielding simple derivatives like 2x for x.
• Then, differentiate the outer function, -\ln(u), with respect to u.

Multiply these results, and you get the derivative of the original function with respect to each variable.

The Chain Rule allows mathematicians to handle these derivatives systematically, making complex derivative computations manageable. Without this rule, dealing with derivative problems involving nested functions could become cumbersome.