Problem 42 - Advanced Engineering Mathematics

Question

Use the Chain Rule to find the indicated partial derivatives. $$ z=\frac{x-y}{x+y} ; x=\frac{u}{v}, y=\frac{v^{2}}{u} ; \frac{\partial z}{\partial u}, \frac{\partial z}{\partial v} $$

Step-by-Step Solution

Answer

\( \frac{\partial z}{\partial u} = \frac{2y}{v(x+y)^2} + \frac{2xv^2}{u^2(x+y)^2} \); \( \frac{\partial z}{\partial v} = -\frac{2yu}{v^2(x+y)^2} - \frac{4xv}{u(x+y)^2} \).

Step 1: Identify the Functions

We have the function \( z = \frac{x-y}{x+y} \) which depends on variables \( x \) and \( y \). Furthermore, \( x \) and \( y \) are functions of \( u \) and \( v \) given by \( x = \frac{u}{v} \) and \( y = \frac{v^2}{u} \). We need to find \( \frac{\partial z}{\partial u} \) and \( \frac{\partial z}{\partial v} \). This implies using the Chain Rule since \( z \) is not directly a function of \( u \) and \( v \).

Step 2: Express z in terms of u and v

Substitute the expressions for \( x \) and \( y \) into \( z \):\[z = \frac{\frac{u}{v} - \frac{v^2}{u}}{\frac{u}{v} + \frac{v^2}{u}}\]Simplify the expression by finding a common denominator for the terms in the numerator and the denominator:

Step 3: Simplify the Expression

Simplify the expression for \( z \):\[z = \frac{\frac{u^2 - v^3}{uv}}{\frac{u^2 + v^3}{uv}}\]This simplifies to:\[z = \frac{u^2 - v^3}{u^2 + v^3}\]

Step 4: Apply the Chain Rule for Partial Derivatives

The chain rule in this multivariable context is \[\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial u},\] \[\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial v}.\] We first need the partial derivatives \( \frac{\partial z}{\partial x} \) and \( \frac{\partial z}{\partial y} \).

Step 5: Compute Partial Derivatives with respect to x and y

\( z \) is a function of \( x \) and \( y \) by:\[z = \frac{x-y}{x+y}\]For \( \frac{\partial z}{\partial x} \):\[\frac{\partial z}{\partial x} = \frac{(1)(x+y) - (x-y)(1)}{(x+y)^2} = \frac{2y}{(x+y)^2}\]For \( \frac{\partial z}{\partial y} \):\[\frac{\partial z}{\partial y} = \frac{(x-y)(-1) - (-1)(x+y)}{(x+y)^2} = \frac{-2x}{(x+y)^2}\]

Step 6: Compute Partial Derivatives of x and y with respect to u and v

Differentiate \( x = \frac{u}{v} \) and \( y = \frac{v^2}{u} \):For \( \frac{\partial x}{\partial u} \):\[\frac{\partial x}{\partial u} = \frac{1}{v} \quad \text{(holding v constant)}\]For \( \frac{\partial x}{\partial v} \):\[\frac{\partial x}{\partial v} = -\frac{u}{v^2} \quad \text{(holding u constant)}\]For \( \frac{\partial y}{\partial u} \):\[\frac{\partial y}{\partial u} = -\frac{v^2}{u^2}\]For \( \frac{\partial y}{\partial v} \):\[\frac{\partial y}{\partial v} = \frac{2v}{u}\]

Step 7: Compute \( \frac{\partial z}{\partial u} \) and \( \frac{\partial z}{\partial v} \)

Substitute the previously found derivatives into the chain rule expressions:\(\frac{\partial z}{\partial u} = \frac{2y}{(x+y)^2} \cdot \frac{1}{v} + \frac{-2x}{(x+y)^2} \cdot \left(-\frac{v^2}{u^2}\right)\)evalutes to:\(\frac{\partial z}{\partial u} = \frac{2y}{v(x+y)^2} + \frac{2xv^2}{u^2(x+y)^2}\)\(\frac{\partial z}{\partial v} = \frac{2y}{(x+y)^2} \cdot \left(-\frac{u}{v^2}\right) + \frac{-2x}{(x+y)^2} \cdot \frac{2v}{u}\)evaluetes to:\(\frac{\partial z}{\partial v} = -\frac{2yu}{v^2(x+y)^2} - \frac{4xv}{u(x+y)^2}\)

Key Concepts

Partial Derivatives

Understanding partial derivatives is crucial in analyzing functions that depend on more than one variable. When a function has several variables, its rate of change can vary depending on which variable changes while keeping others constant. These specific rates of change are called partial derivatives. In simpler terms:

• The partial derivative with respect to a variable measures how the function changes as that variable changes.
• Other variables are held constant during this change.

In the provided exercise, we are determining how the function \( z \) changes with respect to variables \( u \) and \( v \). We find \( \frac{\partial z}{\partial u} \) and \( \frac{\partial z}{\partial v} \) using the chain rule. This process involves computing how \( z \) reacts when only \( u \) or \( v \) changes. These derivatives help us to understand the sensitivity of the function \( z \) to changes in these underlying parameters.

Multivariable Functions

A multivariable function is one where the output depends on more than one input variable. This is the case with the function \( z = \frac{x-y}{x+y} \), where \( z \) depends on two intermediates, \( x \) and \( y \), which are themselves functions of \( u \) and \( v \).Key aspects of multivariable functions include:

• The ability to describe complex real-world phenomena where multiple factors affect outcomes.
• They may be expressed in terms of other functions, as seen with the substitutions \( x = \frac{u}{v} \) and \( y = \frac{v^2}{u} \).

These types of functions are widely used in engineering, physics, and other sciences to model situations where outcomes are affected by several factors simultaneously. Understanding how to manipulate and derive such functions is essential for solving complex problems in engineering and the sciences.

Mathematical Simplification

Mathematical simplification is an important technique in calculus and algebra that involves rewriting expressions in a simpler, often more manageable form. This process can make it easier to apply mathematical theorems or to integrate and differentiate complicated expressions.Here's how simplification plays a role:

• The original expression for \( z \) was \( \frac{\frac{u}{v} - \frac{v^2}{u}}{\frac{u}{v} + \frac{v^2}{u}} \). By finding a common denominator, this was simplified to \( z = \frac{u^2 - v^3}{u^2 + v^3} \).
• Such simplification can reduce computation errors and make it easier to visualize or further analyze the mathematical model.

In the context of our exercise, simplifying \( z \) allowed for more straightforward application of the chain rule and easier computation of the desired partial derivatives. This step is often crucial in making complex calculus problems more manageable.

Derivatives in Engineering Mathematics

Derivatives are an essential component of engineering mathematics, providing information about the rate and direction of changes. In many engineering problems, we deal with systems that are affected by multiple variables – a classic case in point for using multivariable calculus. Here’s why derivatives are especially useful:

• They help predict how small changes in one aspect of the system affect the overall system behavior.
• In engineering, this can mean the difference between optimal performance and suboptimal or even unsafe conditions.

The application of partial derivatives enables engineers to fine-tune systems for performance by understanding sensitivities and response characteristics. By using tools like the chain rule to find these derivatives, engineers can solve complex problems involving multivariable interactions efficiently and effectively. This skill is essential for innovating and improving systems and processes across various fields of engineering.