Problem 64 - Calculus Early Transcendentals

Question

Power functions and percent change Suppose that \(z=f(x, y)=x^{a} y^{b},\) where \(a\) and \(b\) are real numbers. Let \(d x / x, d y / y,\) and \(d z / z\) be the approximate relative (percent) changes in \(x, y,\) and \(z,\) respectively. Show that \(d z / z=a(d x) / x+b(d y) / y ;\) that is, the relative changes are additive when weighted by the exponents \(a\) and \(b\)

Step-by-Step Solution

Answer

Question: Show that the approximate relative (percent) changes in \(x\), \(y\), and \(z\) are additive when weighted by the exponents \(a\) and \(b\) for the power function \(z = f(x, y) = x^a y^b\), where \(a\) and \(b\) are real numbers. Answer: We have shown that the approximate relative changes in \(x\), \(y\), and \(z\) follow the equation \(\frac{dz}{z} = a\left(\frac{dx}{x}\right) + b\left(\frac{dy}{y}\right)\) which demonstrates that the changes are additive when weighted by the exponents \(a\) and \(b\).

Step 1: Differentiating the function

First, we need to differentiate the function \(z=f(x,y)=x^a y^b\) with respect to x and y variables using partial derivatives: $$\frac{\partial z}{\partial x}= a x^{a-1}y^b$$ $$\frac{\partial z}{\partial y}= b x^ay^{b-1}$$

Step 2: Finding the approximate relative changes

Now, we need to find the approximate relative changes: $$\frac{dz}{z} = \frac{\frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y}dy}{f(x,y)} = \frac{a x^{a-1}y^bdx + b x^ay^{b-1}dy}{x^ay^b}$$

Step 3: Simplifying the expression

We can simplify the expression \(\frac{dz}{z}\) further by dividing by \(x^ay^b\): $$\frac{dz}{z} = \frac{a dx}{x} + \frac{b dy}{y}$$

Step 4: Concluding the solution

Therefore, we've shown that the approximate relative (percent) changes in \(x\), \(y\), and \(z\) are additive when weighted by the exponents \(a\) and \(b\): $$\frac{dz}{z}= a\left(\frac{dx}{x}\right) + b\left(\frac{dy}{y}\right)$$

Key Concepts

Partial Derivatives

Partial derivatives are a crucial concept in multivariable calculus. They allow us to understand the rate of change of a multivariable function concerning each variable individually, keeping other variables constant. When dealing with functions like \(z = f(x, y) = x^a y^b\), it is important to calculate how changes in \(x\) and \(y\) individually affect \(z\).

To find the partial derivatives of this function:

• The partial derivative with respect to \(x\), denoted \(\frac{\partial z}{\partial x}\), measures how much \(z\) changes as \(x\) changes, while keeping \(y\) constant. Here, it is found by differentiating \(f\) as if \(y\) were a constant: \(\frac{\partial z}{\partial x} = a x^{a-1} y^b\).
• Similarly, the partial derivative with respect to \(y\), \(\frac{\partial z}{\partial y}\), shows how \(z\) changes as \(y\) changes, with \(x\) constant: \(\frac{\partial z}{\partial y} = b x^a y^{b-1}\).

Partial derivatives provide essential insight into the behavior and interactions of variables within a multivariable function, forming the foundation for concepts like gradients and differentials.

Power Functions

Power functions such as \(z = x^a y^b\) involve variables raised to constant exponents. They are frequently encountered in mathematical modeling of real-world phenomena, ranging from physics to economics. Understanding these functions involves examining how changing one variable impacts the entire expression.

In a power function, each variable \(x\) and \(y\) is raised to an exponent \(a\) and \(b\), respectively. These exponents dictate the function's growth rate relative to each variable. For instance:

• If \(a > 1\), the function grows faster with changes in \(x\) compared to linear growth.
• If \(a < 1\), the growth rate is slower compared to linear.

Power functions demonstrate the concept of scaling, where certain changes in variables lead to proportional changes in the function's output, making them essential in understanding complex systems.

Relative Change

Relative change quantifies how much a variable changes in comparison to its original value. It is often expressed as a percentage, making it a useful measure in many scientific and mathematical contexts.

In the context of the problem, relative change is applied to the function \(z = x^a y^b\) to determine how much \(z\) changes in response to changes in \(x\) and \(y\). This is expressed as:

• The relative change in \(z\) is \( \frac{dz}{z} \), describing the change in \(z\) per original value of \(z\).
• Similarly, \( \frac{dx}{x} \) and \( \frac{dy}{y} \) represent the relative changes in \(x\) and \(y\).

The equation \( \frac{dz}{z} = a\left(\frac{dx}{x}\right) + b\left(\frac{dy}{y}\right) \) illustrates how the relative changes in \(x\) and \(y\) contribute to the relative change in \(z\), weighted by their respective exponents \(a\) and \(b\). This relationship showcases the additive nature of relative changes in power functions.

Differential Calculus

Differential Calculus helps us understand how functions change. It deals with the concept of derivatives, which represent an instantaneous rate of change. For functions of several variables, such as \(z = x^a y^b\), differential calculus provides the tools to analyze how small changes impact the function's output.

The differentiation process can break down complex multivariable functions into more manageable parts through:

• Partial derivatives, determining how each variable individually affects the function's rate of change.
• Differentials, which use derivatives to estimate changes in the function value for small changes in inputs.

In the exercise, differential calculus is used to express the change in \(z\) through \( \frac{dz}{z} = a\left(\frac{dx}{x}\right) + b\left(\frac{dy}{y}\right) \), highlighting how weighted changes in inputs contribute to overall change. This mathematical tool can model and solve problems involving dynamic systems in science, engineering, and finance.