Problem 30 - APEX Calculus

Question

In Exercises \(27-30\), find \(\frac{d z}{d t},\) or \(\frac{\partial z}{\partial s}\) and \(\frac{\partial z}{\partial t},\) using the supplied information. $$ \begin{array}{l} \frac{\partial z}{\partial x}=2, \quad \frac{\partial z}{\partial y}=1 \\ \frac{\partial x}{\partial s}=-2, \quad \frac{\partial x}{\partial t}=3, \quad \frac{\partial y}{\partial s}=2, \quad \frac{\partial y}{\partial t}=-1 \end{array} $$

Step-by-Step Solution

Answer

\( \frac{\partial z}{\partial s} = -2 \); \( \frac{\partial z}{\partial t} = 5 \).

Step 1: Using the Chain Rule Formula

To find \( \frac{\partial z}{\partial s} \), we employ the chain rule for partial derivatives: \( \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s} \). Insert the provided values: \( \frac{\partial z}{\partial x} = 2, \frac{\partial x}{\partial s} = -2, \frac{\partial z}{\partial y} = 1, \frac{\partial y}{\partial s} = 2 \).

Step 2: Calculate \( \frac{\partial z}{\partial s} \)

Substitute the values from Step 1 into the formula: \[ \frac{\partial z}{\partial s} = (2)(-2) + (1)(2) = -4 + 2 = -2 \].

Step 3: Using the Chain Rule Formula for \( \frac{\partial z}{\partial t} \)

To find \( \frac{\partial z}{\partial t} \), apply the chain rule: \( \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t} \). Use the given values: \( \frac{\partial z}{\partial x} = 2, \frac{\partial x}{\partial t} = 3, \frac{\partial z}{\partial y} = 1, \frac{\partial y}{\partial t} = -1 \).

Step 4: Calculate \( \frac{\partial z}{\partial t} \)

Insert the values into the formula: \[ \frac{\partial z}{\partial t} = (2)(3) + (1)(-1) = 6 - 1 = 5 \].

Key Concepts

Partial Derivatives

In multivariable calculus, partial derivatives represent the rate of change of a function with respect to one of its variables, while keeping other variables constant. This is crucial when dealing with functions that depend on multiple variables. For example, in the function \( f(x, y) \), \( \frac{\partial f}{\partial x} \) indicates how \( f \) changes as \( x \) changes, with \( y \) remaining fixed. This concept is similar to regular derivatives, but it applies specifically to functions with more than one variable.

Key notations to be familiar with include:

• \( \frac{\partial z}{\partial x} \): the partial derivative of \( z \) with respect to \( x \)
• \( \frac{\partial z}{\partial y} \): the partial derivative of \( z \) with respect to \( y \)

Partial derivatives are not only a method for finding rates of change but also a foundational component in optimization problems and gradient calculations in fields like physics and engineering.

In our original exercise, understanding the values \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) helps us use the chain rule to find further derivatives like \( \frac{\partial z}{\partial s} \) and \( \frac{\partial z}{\partial t} \). This process articulates how changes in variables \( s \) and \( t \) indirectly influence \( z \).

Multivariable Calculus

Understanding multivariable calculus is essential for dealing with systems or phenomena where numerous factors interact simultaneously. It extends single-variable calculus concepts to functions of multiple variables. For example, a function \( z = f(x, y) \) describes a surface in three-dimensional space, where \( x \) and \( y \) serve as independent variables.

In multivariable calculus, you often encounter tasks such as finding maximum or minimum values of functions, integrating over regions, and understanding vector fields. A frequent necessity is determining how a function changes when multiple variables are involved. This is where partial derivatives come into play, offering insights into these multidimensional changes.

Concepts associated with multivariable calculus not only aid in solving academic exercises but also provide tools for practical applications in physics, economics, and engineering among many other fields. In the original exercise, the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \) are used to analyze the behavior of the function as different circumstances in \( x \) and \( y \) affect \( z \). This reflects the utility of multivariable thinking in assessing complex systems.

Differentiation Techniques

Differentiation is a fundamental concept in calculus, and it becomes more complex yet more powerful when extended to functions with multiple variables. Techniques like the chain rule are crucial differentiation tools, especially when working across different variables.

The chain rule for partial derivatives allows us to compute derivatives of composite functions efficiently. Specifically, it helps in finding how a dependent variable \( z \) changes with respect to independent variables \( s \) and \( t \) through intermediary variables \( x \) and \( y \).

• To compute \( \frac{\partial z}{\partial s} \), the formula is: \( \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial s} \).
• For \( \frac{\partial z}{\partial t} \), use: \( \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial t} \).

These formulas show how components of change interact when variables are interrelated. Practicing these methods is key in mastering multivariable calculus and applying it to various scientific and engineering problems. Understanding these differentiation techniques ensures that students can handle complex functions and systems effectively.