Problem 23 - The Calculus with Analytic Geometry

Question

If \(f\) is a differentiable function of the variable \(u\), let \(u=b x-a y\) and prove that \(z=f(b x-a y)\) satisfies the equation \(a(\partial z / \partial x)+b(\partial z / \partial y)=0\), where \(a\) and \(b\) are constants.

Step-by-Step Solution

Answer

By applying the chain rule and substituting the partial derivatives into the equation, it can be shown that the equation holds true.

Step 1 - Expressing the function in terms of u

Let us start by defining the function as dependent on a new variable. Given that $$u = bx - ay$$ and that the function can be written as $$z = f(u) = f(bx - ay)$$.

Step 2 - Finding the partial derivative of z with respect to x

To find $$\frac{\text{d} z}{\text{d} x}$$ we use the chain rule. Since $$z = f(u)$$, we have $$\frac{\text{d} z}{\text{d} x} = \frac{\text{d} f(u)}{\text{d} u} \frac{\text{d} u}{\text{d} x}$$. Now calculate $$\frac{\text{d} u}{\text{d} x}$$ if $$u = bx - ay$$ thus $$\frac{\text{d} u}{\text{d} x} = b$$. Therefore, $$\frac{\text{d} z}{\text{d} x} = f'(u) \times b$$ or more generally: $$\frac{\text{d} z}{\text{d} x} = b f'(u)$$.

Step 3 - Finding the partial derivative of z with respect to y

Similarly, we apply the chain rule to find $$\frac{\text{d} z}{\text{d} y}$$. Using $$z = f(u)$$ again, we have $$\frac{\text{d} z}{\text{d} y} = \frac{\text{d} f(u)}{\text{d} u} \frac{\text{d} u}{\text{d} y}$$. Now calculate $$\frac{\text{d} u}{\text{d} y}$$ if $$u = bx - ay$$ thus $$\frac{\text{d} u}{\text{d} y} = -a$$. Therefore, $$\frac{\text{d} z}{\text{d} y} = f'(u) \times (-a)$$ or $$\frac{\text{d} z}{\text{d} y} = -a f'(u)$$.

Step 4 - Substituting the derivatives into the given equation

Next, substitute $$\frac{\text{d} z}{\text{d} x}$$ and $$\frac{\text{d} z}{\text{d} y}$$ into the equation $$a \frac{\text{d} z}{\text{d} x} + b \frac{\text{d} z}{\text{d} y} = 0$$. Thus, we have $$a (b f'(u)) + b (-a f'(u)) = 0$$.

Step 5 - Simplifying the expression

Simplify the equation to verify that it holds: $$ab f'(u) - ab f'(u) = 0$$. Clearly, this simplifies to $$0 = 0$$ which confirms the equation is satisfied.

Step 6: Conclusion

Thus, we have proven that $$z = f(b x - a y)$$ satisfies $$a \frac{\text{d} z}{\text{d} x} + b \frac{\text{d} z}{\text{d} y} = 0$$.

Key Concepts

chain rule

The chain rule is a fundamental concept in calculus for finding the derivative of composite functions. When you have a function, say \(z = f(u)\), and \(u\) itself is a function of \(x\) and \(y\) (e.g., \(u = bx - ay\)), the chain rule allows us to differentiate \(z\) with respect to \(x\) or \(y\) by breaking it down into simpler parts. This is how it works:
To find \(\frac{\text{d} z}{\text{d} x}\), we compute:
\[ \frac{\text{d} z}{\text{d} x} = \frac{\text{d} f(u)}{\text{d} u} \frac{\text{d} u}{\text{d} x} \]
Similarly, to find \( \frac{\text{d} z}{\text{d} y} \), we use:
\[ \frac{\text{d} z}{\text{d} y} = \frac{\text{d} f(u)}{\text{d} u} \frac{\text{d} u}{\text{d} y} \]
In both cases, the goal is to first differentiate \(f(u)\) with respect to \(u\), and then \(u\) with respect to either \(x\) or \(y\). This method simplifies our computations and keeps things organized.

partial derivatives

Partial derivatives are used when dealing with functions of multiple variables. They measure how the function changes as each individual variable changes, holding the other variables constant.
In our exercise, we have \( z = f(bx - ay) \). To find the partial derivative of \(z\) with respect to \(x\) (notated as \( \frac{\text{d} z}{\text{d} x} \)):
1. Use the chain rule to differentiate \(z\) as \( z = f(u)\), where \( u = bx - ay\).
2. This gives us \( \frac{\text{d} z}{\text{d} x} = f'(u) \times b \).
Similarly, to find the partial derivative of \(z\) with respect to \(y\) ( \( \frac{\text{d} z}{\text{d} y} \) ):
1. Again use the chain rule, resulting in \( \frac{\text{d} z}{\text{d} y} = f'(u) \times (-a) \).
Partial derivatives give us critical information about how the function \(z\) behaves by isolating the effects of each independent variable.

differentiable functions

A differentiable function is a function that has a derivative at each point in its domain. This means the function's graph has a tangible slope and is smooth without breaks or corners.
In our example, we start with \( f(u) \), which is a differentiable function of \(u\). This is crucial because it allows us to apply the chain rule and find the partial derivatives correctly. These are key to solving our main equation.
Being differentiable assures that \( f(u) \) behaves nicely under differentiation, meaning it has continuous partial derivatives, which is a critical assumption when dealing with complex functions in higher dimensions.

proofs in calculus

Proofs in calculus provide a rigorous method to verify that mathematical statements are true. In the given exercise, we proved that if \( z = f(bx - ay) \), then it satisfies \( a \frac{\text{d} z}{\text{d} x} + b \frac{\text{d} z}{\text{d} y} = 0 \).
Here’s a breakdown of the proof:
- Reframe the function \( z \) in terms of a new variable \( u = bx - ay \).
- Compute the partial derivatives \( \frac{\text{d} z}{\text{d} x} \) and \( \frac{\text{d} z}{\text{d} y} \) using the chain rule.
- Substitute these derivatives into the given equation.
- Simplify the equation to show it holds true.
This step-by-step verification confirms the mathematical relationship and strengthens our understanding of the function’s behavior.