Problem 21 - Calculus with Concepts in Calculus

Question

Find \(\partial w / \partial u\) and \(\partial w / \partial v\). $$ w=\frac{y z}{x^{2}+x y} ; x=u^{2}, y=v^{2}, z=u^{2}-v^{2} $$

Step-by-Step Solution

Answer

Use the chain rule, given substitutions, to find \( \partial w/\partial u \) and \( \partial w/\partial v \).

Step 1: Understand the given function and variables

We are given that \( w = \frac{yz}{x^2 + xy} \), with \( x = u^2 \), \( y = v^2 \), and \( z = u^2 - v^2 \). Our task is to find the partial derivatives \( \partial w / \partial u \) and \( \partial w / \partial v \).

Step 2: Calculate \( \partial w / \partial u \) using chain rule

We use the chain rule: \[ \frac{\partial w}{\partial u} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial u} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial u}. \]Calculate the partial derivatives of \( w \) with respect to \( x, y, \) and \( z \), and plug in the expressions for \( x, y, \) and \( z \). Then, find \( \frac{\partial x}{\partial u} = 2u \), \( \frac{\partial y}{\partial u} = 0 \), and \( \frac{\partial z}{\partial u} = 2u \).

Step 3: Express \( w \) in terms of \( u \) and \( v \)

Replace \( x, y, \) and \( z \) in the expression for \( w \) by \( u^2, v^2, \) and \( u^2 - v^2 \), respectively. Then simplify:\[ w = \frac{(v^2)(u^2 - v^2)}{(u^2)^2 + (u^2)(v^2)} = \frac{v^2(u^2 - v^2)}{u^4 + u^2v^2}. \]

Step 4: Calculate \( \partial w / \partial v \) using chain rule

Similarly, use the chain rule for \( \partial v \):\[ \frac{\partial w}{\partial v} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial v} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial v}. \]Calculate \( \frac{\partial x}{\partial v} = 0 \), \( \frac{\partial y}{\partial v} = 2v \), and \( \frac{\partial z}{\partial v} = -2v \) and substitute these values into the expression.

Key Concepts

chain rule

The chain rule is a fundamental concept in calculus, essential for finding the derivative of a composite function. When dealing with functions of multiple variables, we need this tool to differentiate terms with interconnected variables. In our problem, we're given a function \(w = \frac{yz}{x^2 + xy}\). Here, \(w\) depends on \(x, y\), and \(z\), which in turn are functions of \(u\) and \(v\). The chain rule helps us find how \(w\) changes with respect to \(u\) and \(v\) by considering how \(w\) changes due to changes in \(x, y\), and \(z\), which are influenced by changes in \(u\) and \(v\). This interaction is shown in the formula for the partial derivatives:

• \(\frac{\partial w}{\partial u} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial u} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial u} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial u}\)
• \(\frac{\partial w}{\partial v} = \frac{\partial w}{\partial x}\frac{\partial x}{\partial v} + \frac{\partial w}{\partial y}\frac{\partial y}{\partial v} + \frac{\partial w}{\partial z}\frac{\partial z}{\partial v}\)

As we derive each part, the rule allows us to keep track of how each variable directly and indirectly affects \(w\). This becomes particularly critical in multivariable calculus scenarios like ours.

multivariable calculus

Multivariable calculus extends the principles of calculus to functions of more than one variable. Unlike single-variable calculus, where we deal with functions depending on just one variable, multivariable calculus explores how functions behave when influenced by several inputs.
In our given problem, \(w\) is expressed in terms of three variables \(x, y,\) and \(z\), which are themselves functions of \(u\) and \(v\). This kind of problem involves understanding not just how the final output responds to changes in input values, but also how intermediate variables interconnect.
Applications of multivariable calculus appear in fields like physics for describing systems with multiple factors, or in engineering to analyze problems where different parameters influence outcomes concurrently. By understanding partial derivatives, we delve into how one specific input affects a function while others are held constant, a cornerstone concept in multivariable analysis.

calculus problems

Calculus problems involving partial derivatives require a methodical approach to unravel how changes in inputs affect a given function. When calculating partial derivatives, we focus on one variable at a time while treating other variables as constants.
For our problem, computing \( \frac{\partial w}{\partial u} \) and \( \frac{\partial w}{\partial v} \) illustrates this. Using the expressions for \(x = u^2, y = v^2,\) and \(z = u^2 - v^2\), we transform the complex problem into manageable parts. This step-by-step breakdown makes it more feasible to see how changes in \(u\) and \(v\) specifically alter \(w\).

• Identify and isolate the variable of interest.
• Treat other variables as constants temporarily.
• Use the chain rule to account for all influences.

This structured approach is beneficial for tackling calculus problems with multiple interconnected variables.

expressions simplification

Expressions simplification is a pivotal technique in calculus that involves rewriting mathematical expressions in a form that's easier to manage and understand. It involves applying algebraic rules to combine terms, factor out common elements, or reduce fractions.
In our example, simplifying \(w = \frac{(v^2)(u^2 - v^2)}{(u^2)^2 + (u^2)(v^2)}\) to a more straightforward form is necessary for ease of calculation and interpretation. Simplification can reveal underlying patterns or elements of a problem and is often a prerequisite before applying further calculus techniques, like differentiation or integration.
The major steps in simplifying an expression include:

• Combine like terms to reduce complexity.
• Factor expressions to uncover common patterns.
• Reduce fractions to their simplest form.

By ensuring that functions and expressions are at their simplest, we prepare the terrain for accurate and efficient problem-solving.