Problem 5 - APEX Calculus

Question

In Exercises \(5-8,\) find the total differential \(d z\). $$ z=x \sin y+x^{2} $$

Step-by-Step Solution

Answer

\[ dz = (\sin y + 2x) \, dx + (x \cos y) \, dy \]

Step 1: Identify Variables and Function

The given function is \( z = x \sin y + x^2 \). Here, \( z \) is a function of two variables, \( x \) and \( y \).

Step 2: Compute Partial Derivatives

To find the total differential \( dz \), we need the partial derivatives of \( z \) with respect to \( x \) and \( y \).- The partial derivative of \( z \) with respect to \( x \) is \( \frac{\partial z}{\partial x} = \sin y + 2x \).- The partial derivative of \( z \) with respect to \( y \) is \( \frac{\partial z}{\partial y} = x \cos y \).

Step 3: Formulate the Total Differential

The total differential \( dz \) is given by the formula:\[ dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy \]Substitute the values from Step 2:\[ dz = (\sin y + 2x) \, dx + (x \cos y) \, dy \]

Key Concepts

Partial Derivatives

When dealing with functions of multiple variables, understanding partial derivatives becomes essential. A partial derivative measures how a function changes as one variable changes while keeping other variables constant. For the function \( z = x \sin y + x^2 \), we can find the partial derivative of \( z \) with respect to \( x \), treating \( y \) as a constant. This gives us \( \frac{\partial z}{\partial x} = \sin y + 2x \). Similarly, the partial derivative with respect to \( y \) is found by treating \( x \) as a constant, resulting in \( \frac{\partial z}{\partial y} = x \cos y \).

• Partial derivatives are denoted using the symbol \( \partial \).
• They provide a way to explore the behavior of a multivariable function along individual dimensions.

This ability to isolate variables and study their effects individually is a powerful tool in calculus, especially when working with complex systems.

Multivariable Calculus

Multivariable calculus extends the concepts of calculus to functions of more than one variable. In the given problem, we are dealing with a function \( z = x \sin y + x^2 \), which illustrates a typical scenario in multivariable calculus where a single output depends on multiple inputs.

• Multivariable functions are often written in the form \( f(x, y, z, \ldots) \), depending on how many variables are involved.
• This area of calculus studies limits, continuity, and differentiability in higher dimensions.

For instance, to understand how changes in \( x \) and \( y \) affect \( z \), we use tools like partial derivatives and total differentials. These allow us to predict the behavior of \( z \) based on small changes in its variables. As you become more comfortable with these concepts, you'll be able to handle more complex functions and systems common in fields such as physics and engineering.

Differential Calculus

Differential calculus focuses on understanding how functions change, primarily through derivatives. In its multivariable form, such as with the total differential, it provides an insight into the overall change of a function based on its input changes. The total differential \( dz \) of a function \( z \) with variables \( x \) and \( y \) is calculated as:\[ dz = \frac{\partial z}{\partial x} \, dx + \frac{\partial z}{\partial y} \, dy\]This formula uses the partial derivatives of \( z \) with respect to each variable to predict the change in \( z \) for small changes in \( x \) and \( y \). In practice:

• The term \( \frac{\partial z}{\partial x} \, dx \) describes how \( z \) changes as \( x \) changes while \( y \) is constant.
• The term \( \frac{\partial z}{\partial y} \, dy \) describes how \( z \) changes as \( y \) changes while \( x \) is constant.

Using these concepts, you can find how functions behave dynamically, which is crucial for modeling and solving real-world problems that involve change.