Problem 30 - Calculus Early Transcendentals: Pearson New International Edition

Question

Show that if \(w=f(r-s, s-t, t-r)\) then $$ \frac{\partial w}{\partial r}+\frac{\partial w}{\partial s}+\frac{\partial w}{\partial t}=0 $$

Step-by-Step Solution

Answer

The sum of the partial derivatives is zero: \(\frac{\partial w}{\partial r} + \frac{\partial w}{\partial s} + \frac{\partial w}{\partial t} = 0\).

Step 1: Identify the Given Function

We have the function defined as \(w = f(r-s, s-t, t-r)\). This indicates that \(w\) depends on three new variables, which are \(u = r-s\), \(v = s-t\), and \(w = t-r\). These are transformations of the original variables \(r\), \(s\), and \(t\).

Step 2: Use Chain Rule for Partial Derivatives

To find the partial derivatives of \(w\) with respect to \(r\), \(s\), and \(t\), we use the chain rule for partial derivatives. Specifically, \[ \frac{\partial w}{\partial r} = \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial r} + \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial r} + \frac{\partial f}{\partial w} \cdot \frac{\partial w}{\partial r} \].

Step 3: Find the Partial Derivative \(\frac{\partial w}{\partial r}\)

Using the chain rule, find the derivative: - \(\frac{\partial u}{\partial r} = 1\) (since \(u = r-s\))- \(\frac{\partial v}{\partial r} = 0\) (since \(v = s-t\) is independent of \(r\))- \(\frac{\partial w}{\partial r} = -1\) (since \(w = t-r\)) This gives:\[ \frac{\partial w}{\partial r} = \frac{\partial f}{\partial u} - \frac{\partial f}{\partial w} \].

Step 4: Find the Partial Derivative \(\frac{\partial w}{\partial s}\)

Using the chain rule, the partial derivatives with respect to \(s\) are:- \(\frac{\partial u}{\partial s} = -1\) (since \(u = r-s\))- \(\frac{\partial v}{\partial s} = 1\) (since \(v = s-t\))- \(\frac{\partial w}{\partial s} = 0\) (since \(w = t-r\))This gives:\[ \frac{\partial w}{\partial s} = \frac{\partial f}{\partial v} - \frac{\partial f}{\partial u} \].

Step 5: Find the Partial Derivative \(\frac{\partial w}{\partial t}\)

Using the chain rule, calculate:- \(\frac{\partial u}{\partial t} = 0\) (since \(u = r-s\))- \(\frac{\partial v}{\partial t} = -1\) (since \(v = s-t\))- \(\frac{\partial w}{\partial t} = 1\) (since \(w = t-r\))Therefore,\[ \frac{\partial w}{\partial t} = \frac{\partial f}{\partial w} - \frac{\partial f}{\partial v} \].

Step 6: Sum the Partial Derivatives

Sum the expressions obtained for the partial derivatives: \[ \frac{\partial w}{\partial r} + \frac{\partial w}{\partial s} + \frac{\partial w}{\partial t} = (\frac{\partial f}{\partial u} - \frac{\partial f}{\partial w}) + (\frac{\partial f}{\partial v} - \frac{\partial f}{\partial u}) + (\frac{\partial f}{\partial w} - \frac{\partial f}{\partial v}) \].

Step 7: Simplify the Expression

Upon simplification, notice that all terms cancel each other:- \(\frac{\partial f}{\partial u}\) cancels out- \(\frac{\partial f}{\partial v}\) cancels out- \(\frac{\partial f}{\partial w}\) cancels outThe sum simplifies to zero:\[ \frac{\partial w}{\partial r} + \frac{\partial w}{\partial s} + \frac{\partial w}{\partial t} = 0 \].

Key Concepts

Chain Rule

The chain rule is a fundamental tool in calculus, especially when dealing with composite functions. When we have a function composed of other functions, like in the problem where \( w = f(u, v, w) \), it simplifies the process of differentiation. Here, the chain rule tells us how to differentiate \( w \) with respect to the original variables \( r, s, \) and \( t \). Instead of differentiating directly, we break it down into parts by calculating derivatives of intermediate variables (\( u, v, \) and \( w \) ) first.

Using the chain rule, we take the derivative of \( w \) with respect to each new variable it depends on, and then multiply by the derivative of that intermediate variable with respect to the original variable. This approach helps in systematically solving complex differentiation problems.

In this exercise, for the partial derivative \( \frac{\partial w}{\partial r} \), we applied the chain rule as follows:

• Find \( \frac{\partial f}{\partial u} \cdot \frac{\partial u}{\partial r}, \) \( \frac{\partial f}{\partial v} \cdot \frac{\partial v}{\partial r}, \) and \( \frac{\partial f}{\partial w} \cdot \frac{\partial w}{\partial r}. \)

This rule allows us to differentiate within a structured framework, leading to the correct calculation of derivatives.

Function Transformation

Transformations are crucial when dealing with functions as they allow us to simplify a problem by changing the perspective or the parameters involved. In the original exercise, function transformation occurs when we define new variables: \( u = r-s, v = s-t, \text{ and } w = t-r. \)

This transformation helps in revealing inherent symmetries and simplifies computation. By expressing the function \( \text{f} \) in terms of \( u, v, \text{ and } w, \) our problem becomes easier to manage.

Each of the new variables is a linear combination of the original ones, making it straightforward to calculate their partial derivatives. This approach often uncovers hidden relationships that are not immediately apparent, showing how different functions relate to each other after a transformation.

Variable Substitution

Variable substitution is a key strategy in calculus to simplify problems by switching from one set of variables to another. This technique is especially useful when the direct approach seems complex or impractical.

In this exercise, we substitute \( r, s, \text{ and } t \) with \( u, v, \text{ and } w, \) using:

• \( u = r - s \),
• \( v = s - t \),
• \( w = t - r \).

This substitution changes the form of the function \( w \), making it a function of \( u, v, \text{ and } w \) instead. By doing so, we can exploit new variable dependencies and relationships. It leads us through the partial derivatives more smoothly and often simplifies the final calculations significantly. Variable substitution helps to convert a challenging problem into a more tractable one.

Calculus Problem Solving

Solving calculus problems often involves a suite of techniques and strategies to arrive at a solution methodically. In this exercise, we explored how different calculus methods, like the chain rule and variable substitution, come together.

First, we identified how the function \( w = f(u, v, w) \) depends on the variables \( r, s, \text{ and } t. \) Then, by formulating their relationships and applying transformation, we redefined the problem. From there, the strategy included:

• Applying the chain rule to find each partial derivative accurately.
• Summing the calculated derivatives which revealed the relation \( \frac{\partial w}{\partial r} + \frac{\partial w}{\partial s} + \frac{\partial w}{\partial t} = 0 \).

This systematic approach is how calculus helps solve complex problems. By breaking problems into smaller, manageable parts, calculus offers not just a solution, but insight into the nature of mathematical relationships.