Problem 34 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Evaluate the first partial derivatives of the function at the given point. \(f(x, y, z)=x^{2} y^{2}+z^{2} ;(1,1,2)\)

Step-by-Step Solution

Answer

The first partial derivatives evaluated at point (1,1,2) are \(f_x = 2\), \(f_y = 2\), and \(f_z = 4\).

Step 1: Find the partial derivative with respect to x

Let's begin by differentiating the function with respect to x, while keeping y and z as constants. The partial derivative with respect to x is denoted as \(f_x\). \(f(x, y, z) = x^2y^2 + z^2\) Taking partial derivative, we get: \(f_x = \frac{\partial}{\partial x} (x^2y^2 + z^2)\) Here, the term \(z^2\), being independent of x, is treated as a constant and its derivative with respect to x becomes zero. So we now have: \(f_x = \frac{\partial}{\partial x} (x^2y^2) = 2xy^2\)

Step 2: Find the partial derivative with respect to y

Now let's differentiate the function with respect to y, while keeping x and z as constants. The partial derivative with respect to y is denoted as \(f_y\). Taking partial derivative, we get: \(f_y = \frac{\partial}{\partial y} (x^2y^2 + z^2)\) Here, the term \(z^2\), being independent of y, is treated as a constant and its derivative with respect to y becomes zero. So we have: \(f_y = \frac{\partial}{\partial y} (x^2y^2) = 2x^2y\)

Step 3: Find the partial derivative with respect to z

Finally, let's differentiate the function with respect to z, while keeping x and y as constants. The partial derivative with respect to z is denoted as \(f_z\). Taking partial derivative, we get: \(f_z = \frac{\partial}{\partial z} (x^2y^2 + z^2)\) Here, the term \(x^2y^2\), being independent of z, is treated as a constant and its derivative with respect to z becomes zero. So we have: \(f_z = \frac{\partial}{\partial z} (z^2) = 2z\)

Step 4: Evaluate the partial derivatives at the given point (1,1,2)

Now, we have all three first-order partial derivatives. Let's plug the given point (1,1,2) into each of them to evaluate: \(f_x(1,1,2) = 2(1)(1^2) = 2\) \(f_y(1,1,2) = 2(1^2)(1) = 2\) \(f_z(1,1,2) = 2(2) = 4\) So, the first partial derivatives evaluated at point (1,1,2) are: \(f_x = 2\) \(f_y = 2\) \(f_z = 4\)

Key Concepts

Multivariable Calculus

Multivariable calculus is an extension of calculus that deals with functions of more than one variable. Instead of examining curves, it explores surfaces and shapes in higher dimensions. This type of calculus helps us understand how changes in one variable affect another in a multi-dimensional space.

When dealing with functions of multiple variables, we often represent these functions as expressions like \( f(x, y, z) \), indicating that the function depends on several variables. This contrasts single-variable calculus, where functions depend on just one variable and describe problems generally in two-dimensions.

Some common applications of multivariable calculus include:

• Calculating the trajectory of objects in physics.
• Modeling economic systems where multiple variables interact.
• Predicting weather patterns with climate models.

Understanding multivariable calculus forms the basis of more advanced topics in mathematics and science, bridging the gap between simple functions and complex, real-world problems.

Differentiation

Differentiation in the context of multivariable functions refers to finding the derivative of a function with respect to a particular variable. This is especially crucial when a function depends on several variables, as it allows us to observe how changes in one variable impact the function's output, while keeping all other variables constant.

In multivariable calculus, we achieve this by using partial derivatives. If we consider a function \( f(x, y, z) \), where it depends on \( x, y, \) and \( z \), taking the partial derivative with respect to \( x \) would mean:

• Focusing solely on how \( x \) changes.
• Treating \( y \) and \( z \) as constants.

When differentiating multiple variables like in this exercise,

• The notation \( f_x \) or \( \frac{\partial}{\partial x}f(x, y, z) \) is commonly used to denote the partial derivative with respect to \( x \).
• Similar notation follows for partial derivatives with respect to \( y \) and \( z \).

Using differentiation, especially via partial derivatives, allows us to pinpoint the contribution of each variable to the overall function's behavior at any point.

Functions of Several Variables

Functions of several variables are important in mathematics as they allow us to represent and solve problems involving multiple inputs. Instead of having a function that depends on one variable, these functions depend on two, three, or even more variables.

The function given in the exercise, \( f(x, y, z) = x^2 y^2 + z^2 \), serves as a prime example. Here, \( x, y, \) and \( z \) are independent variables that together determine the function's output. Each variable can be thought of as a dimension in a multi-dimensional space, making these functions crucial for modeling situations where one-dimensional perspectives fall short.

What's particularly interesting about functions of several variables is how they:

• Allow complex relationships to be broken down into simpler, more understandable parts.
• Enable a clear analysis of interactions between variables by using techniques like partial differentiation.
• Provide a framework for analyzing systems in diverse fields, such as physics, engineering, and economics.

Understanding how each variable contributes to the output is central to grasping functions of several variables. This broad approach is necessary for anyone looking to solve real-world problems that cannot be reduced to single-variable scenarios.