Problem 29 - 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. $$ \begin{array}{l} \frac{\partial z}{\partial x}=-4, \quad \frac{\partial z}{\partial y}=9 \\ \frac{\partial x}{\partial s}=5, \quad \frac{\partial x}{\partial t}=7, \quad \frac{\partial y}{\partial s}=-2, \quad \frac{\partial y}{\partial t}=6 \end{array} $$

Step-by-Step Solution

Answer

\(\frac{\partial z}{\partial s} = -38\); \(\frac{\partial z}{\partial t} = 26\).

Step 1: Recognize the Chain Rule

To find \( \frac{\partial z}{\partial s} \) and \( \frac{\partial z}{\partial t} \), we apply the chain rule for partial derivatives. The chain rule helps us relate the derivative of \( z \) with respect to \( s \) and \( t \) through \( x \) and \( y \).

Step 2: Apply the Chain Rule for \( \frac{\partial z}{\partial s} \)

The chain rule gives us: \[\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s}\]Substituting the given values: \[\frac{\partial z}{\partial s} = (-4)(5) + (9)(-2)\]Calculating this yields:\[\frac{\partial z}{\partial s} = -20 - 18 = -38\]

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

Using the chain rule, we have: \[\frac{\partial z}{\partial t} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial t}\]Substitute the known values: \[\frac{\partial z}{\partial t} = (-4)(7) + (9)(6)\]Perform the calculations:\[\frac{\partial z}{\partial t} = -28 + 54 = 26\]

Key Concepts

Chain Rule

The chain rule is a fundamental concept in calculus, especially when dealing with functions of multiple variables. It provides a way to compute the derivative of a function based on its dependency on two or more interrelated variables. Think of it as a method of passing the derivative through a series of nested functions to determine the overall rate of change.
To understand the chain rule, consider a function \( z \) that depends on variables \( x \) and \( y \), which themselves depend on other variables like \( s \) and \( t \). The chain rule allows us to find the derivative of \( z \) with respect to \( s \) or \( t \) by taking into account how \( x \) and \( y \) change with these variables as well.

• For \( \frac{\partial z}{\partial s} \) and \( \frac{\partial z}{\partial t} \), the chain rule provides a formula that relates these derivatives to partial derivatives with respect to \( x \) and \( y \).
• This process involves multiplying the partial derivatives of \( z \) by those of \( x \) and \( y \), then summing the results.

The chain rule can be extended to situations with more variables, making it a versatile and essential tool in multivariable calculus.

Multivariable Calculus

Multivariable calculus expands upon the concepts of single-variable calculus by addressing functions that depend on more than one variable. It is a natural progression, allowing us to model and solve problems that are more complex and real-world in nature.
In this realm, functions might take the form \( f(x, y, z, ...) \), and understanding their behavior involves computing partial derivatives. Each partial derivative represents the rate of change of the function with respect to one variable while keeping the other variables constant.

• For instance, when computing \( \frac{\partial z}{\partial x} = -4 \) and \( \frac{\partial z}{\partial y} = 9 \), we observe how \( z \) changes as \( x \) and \( y \) vary individually.
• Since multiple outputs can come from changing multiple input variables, this area of calculus requires careful management of all variable interdependencies.

Applications of multivariable calculus span various fields like physics, engineering, and economics, where problems often involve multiple influencing factors. This discipline helps us optimize systems and predict changes accurately.

Derivative Computation

Computing derivatives, especially in a multivariable context, requires precision and understanding of the relationships between variables. In the given exercise, we calculated the partial derivatives \( \frac{\partial z}{\partial s} \) and \( \frac{\partial z}{\partial t} \), employing the chain rule effectively.
Let's break down the steps for computation:

• Identify the dependency of variables. Here, \( z \) depends on \( x \) and \( y \), which further depend on \( s \) and \( t \).
• Use the chain rule formula to express derivatives in terms of known partial derivatives. For instance, \( \frac{\partial z}{\partial s} = \frac{\partial z}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial s} \).
• Substitute the given values to perform the calculation: for \( \frac{\partial z}{\partial s} \), it simplified to \( (-4)(5) + (9)(-2) = -38 \).

In derivative computation, clarity about the function's variable dependencies is crucial. Correctly applying these principles allows for accurate results when assessing how changes in input variables affect an outcome. Being systematic about setting up the derivative expressions is a skill that improves with practice and foundational understanding.