Problem 1 - Thomas Calculus

Question

Find the gradient fields of the functions in Exercises \(1-4\) $$f(x, y, z)=\left(x^{2}+y^{2}+z^{2}\right)^{-1 / 2}$$

Step-by-Step Solution

Answer

The gradient field is \( \nabla f = -x (x^2 + y^2 + z^2)^{-3/2}, -y (x^2 + y^2 + z^2)^{-3/2}, -z (x^2 + y^2 + z^2)^{-3/2} \).

Step 1: Identify the Function

The function provided is \( f(x, y, z) = (x^2 + y^2 + z^2)^{-1/2} \). We need to find its gradient field. This function represents the reciprocal of the square root of the sum of squares of \( x \), \( y \), and \( z \).

Step 2: Understand the Gradient

The gradient of a function \( f(x, y, z) \) is a vector composed of partial derivatives: \( abla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) \). It points in the direction of greatest increase of the function.

Step 3: Calculate Partial Derivative with Respect to x

Calculate \( \frac{\partial f}{\partial x} \):Using the chain rule, \( \frac{d}{dx} (x^2 + y^2 + z^2)^{-1/2} = \frac{d}{du} u^{-1/2} \cdot \frac{d}{dx} (x^2 + y^2 + z^2) \), where \( u = x^2 + y^2 + z^2 \).The chain rule gives: \( -\frac{1}{2} (x^2 + y^2 + z^2)^{-3/2} \times 2x = -x (x^2 + y^2 + z^2)^{-3/2} \).

Step 4: Calculate Partial Derivative with Respect to y

Calculate \( \frac{\partial f}{\partial y} \) similarly as in Step 3:\( \frac{d}{dy} (x^2 + y^2 + z^2)^{-1/2} = -y (x^2 + y^2 + z^2)^{-3/2} \).

Step 5: Calculate Partial Derivative with Respect to z

Calculate \( \frac{\partial f}{\partial z} \) similarly:\( \frac{d}{dz} (x^2 + y^2 + z^2)^{-1/2} = -z (x^2 + y^2 + z^2)^{-3/2} \).

Step 6: Combine the Partial Derivatives into the Gradient

The gradient field \( abla f \) is:\[ abla f = \left( -x (x^2 + y^2 + z^2)^{-3/2}, -y (x^2 + y^2 + z^2)^{-3/2}, -z (x^2 + y^2 + z^2)^{-3/2} \right) \].

Key Concepts

Partial Derivatives

In the fascinating realm of calculus, partial derivatives allow us to understand how a function changes with respect to each of its variables independently. When a function, such as \( f(x, y, z) \), contains multiple variables, partial derivatives measure the rate of change of the function when only one variable changes at a time, while others are held constant. This is crucial in multivariable calculus.
To compute a partial derivative, like \( \frac{\partial f}{\partial x} \), we treat all variables except \( x \) as constants and use the same differentiation techniques as in single-variable calculus. For instance, by applying the chain rule on our function \( f(x, y, z) = (x^2 + y^2 + z^2)^{-1/2} \), we find that \( \frac{\partial f}{\partial x} = -x (x^2 + y^2 + z^2)^{-3/2} \). Similarly, we can calculate \( \frac{\partial f}{\partial y} \) and \( \frac{\partial f}{\partial z} \).
Knowing how to calculate partial derivatives is a powerful skill as it lets us analyze how each variable influences a function, a critical process in modeling real-world phenomena.

Multivariable Calculus

Multivariable calculus extends the concepts of single-variable calculus to functions with two or more variables. This branch of mathematics is invaluable in fields like physics, engineering, and economics, where functions of multiple inputs are common.
In multivariable calculus, we explore functions like \( f(x, y, z) \) that depend on several variables. This makes it possible to understand complex systems with multiple interacting components by looking at how each variable affects the output.

• We use concepts such as partial derivatives to examine how a function changes with specific variables.
• The gradient, divergence, and curl are additional tools from multivariable calculus that help describe changing fields and flows in space.

True to the foundational principles of calculus, multivariable calculus continues to explore rates of change and the concept of limits, but in a multidimensional space. The techniques developed in multivariable calculus are foundational to mastering further mathematical and scientific studies.

Gradient Vector

The gradient vector is an essential concept in multivariable calculus which combines all partial derivatives of a function into a single vector. For a function \( f(x, y, z) \), the gradient \( abla f \) points in the direction of steepest ascent. It’s like a compass guiding you to the highest elevation point on a hill.
In our exercise, the gradient of \( f(x, y, z) = (x^2 + y^2 + z^2)^{-1/2} \) is given by \( abla f = \left( -x (x^2 + y^2 + z^2)^{-3/2}, -y (x^2 + y^2 + z^2)^{-3/2}, -z (x^2 + y^2 + z^2)^{-3/2} \right) \). Each component of this vector is a partial derivative, representing the rate of change of the function along the \( x \), \( y \), and \( z \) axes, respectively.
The gradient not only indicates the direction of the greatest increase but also its magnitude, showing how steep the slope is. This directionality and magnitude make the gradient a vital tool in optimization problems, as it informs us of how to adjust variables to increase or decrease function values effectively.