Problem 44 - Calculus Early Transcendentals

Question

A quantity grows exponentially according to \(y(t)=y_{0} e^{k t} .\) What is the relationship between \(m, n,\) and \(p\) such that \(y(p)=\sqrt{y(m) y(n)} ?\)

Step-by-Step Solution

Answer

Answer: The relationship between \(m\), \(n\), and \(p\) is \(2p - m - n = 0\).

Step 1: Express y(m), y(n), and y(p) using the given equation

Use the given equation, \(y(t) = y_0e^{kt}\), to express \(y(m)\), \(y(n)\), and \(y(p)\): - \(y(m) = y_0e^{km}\) - \(y(n) = y_0e^{kn}\) - \(y(p) = y_0e^{kp}\)

Step 2: Combine the expressions of y(m), y(n), and y(p) as required

According to the requirement, we have \(y(p) = \sqrt{y(m) \cdot y(n)}\). Substitute the expressions we found above: \(y_0e^{kp} = \sqrt{y_0e^{km} \cdot y_0e^{kn}}\)

Step 3: Simplify the equation

Simplify the equation as follows: 1. Simplify the square root: \(\sqrt{y_0e^{km} \cdot y_0e^{kn}} = \sqrt{y_0^2e^{km+kn}} = y_0e^{\frac{km+kn}{2}}\) 2. Replace the square root in the equation: \(y_0e^{kp} = y_0e^{\frac{km+kn}{2}}\) 3. Divide both sides by \(y_0\): \(e^{kp} = e^{\frac{km+kn}{2}}\) 4. Since the bases are equal, we can equate the exponents: \(kp = \frac{km+kn}{2}\)

Step 4: Find the relationship between m, n, and p

From the last step, we have \(kp = \frac{km+kn}{2}\). Now, we can find the relationship between \(m\), \(n\), and \(p\): 1. Multiply both sides by 2: \(2kp = km + kn\) 2. Rearrange the terms: \(2kp - km - kn = 0\) 3. Factor out k: \(k(2p - m - n) = 0\) Since k is a constant in the exponential growth function, the relationship between \(m\), \(n\), and \(p\) is given by: \(2p - m - n = 0\)

Key Concepts

Exponential Function

An exponential function is a mathematical expression in which a constant base is raised to a variable exponent. It is represented as \(y(t) = y_0 e^{kt}\). Here, \(y(t)\) describes how a quantity changes over time, \(y_0\) is the initial quantity at time zero, and \(k\) is the growth constant that affects the rate of growth. Exponential growth occurs when the constant \(k\) is positive.

Exponential functions are crucial in modeling real-world situations where quantities grow or decay rapidly, such as population growth, radioactive decay, and interest calculations. In the context of this problem, the function models how one quantity evolves over different periods represented by \(m, n,\) and \(p\). Understanding this foundation is key to analyzing and solving more complex problems involving growth mechanisms.

Mathematical Problem Solving

Mathematical problem solving involves breaking down complex problems into manageable steps and systematically finding a solution. In this exercise, we are asked to establish a relationship between three different points in time, \(m, n,\) and \(p\), using an exponential function.

To solve such a problem, it is helpful to:

• Identify what is given and what is required.
• Express variables and equations clearly using the provided formulas.
• Substitute and simplify the equations to isolate the desired terms.

This methodical approach not only helps in solving this exercise but also serves as a blueprint for tackling other mathematical challenges. It encourages logical thinking and builds confidence in handling abstract concepts.

Algebraic Simplification

Algebraic simplification is the process of reducing complex expressions into simpler forms without changing their value. This is an essential skill in problem solving, as it allows us to make sense of equations and expressions, like those found in exponential functions.

In our step-by-step solution, we employ simplification techniques to manage the equations. We utilize exponent rules, such as:\

• The product rule: \(e^{a} \cdot e^{b} = e^{a+b}\)
• The property of equality: if \(e^{x} = e^{y}\), then \(x = y\)

By simplifying the algebraic expressions involved in the equation \(y(p) = \sqrt{y(m)\cdot y(n)}\), we identify the key relationship \(2p - m - n = 0\).

This example demonstrates the power of algebraic simplification in revealing underlying relationships and facilitating problem-solving in mathematical contexts.