Problem 63 - Calculus and its applications

Question

Quantity \(Q_{1}\) grows exponentially with a doubling time of 1 yr. Quantity \(Q_{2}\) grows exponentially with a doubling time of 2 yr. If the initial amounts of \(Q_{1}\) and \(Q_{2}\) are the same, how long will it take for \(Q_{1}\) to be twice the size of \(Q_{2} ?\)

Step-by-Step Solution

Answer

It will take 2 years for \(Q_1\) to be twice the size of \(Q_2\).

Step 1: Identify Initial Conditions

Both quantities, \(Q_1\) and \(Q_2\), have the same initial amount, let's call this initial amount \(Q_0\). Each grows over time according to their respective doubling times.

Step 2: Define the Exponential Growth Formulas

For exponential growth, the formula is \(Q(t) = Q_0 \cdot 2^{t/T}\), where \(T\) is the doubling time. Thus, for \(Q_1\), \(Q_1(t) = Q_0 \cdot 2^{t/1}\) and for \(Q_2\), \(Q_2(t) = Q_0 \cdot 2^{t/2}\).

Step 3: Set Up the Equation for the Condition

We need to find \(t\) such that \(Q_1(t) = 2 \cdot Q_2(t)\). Substitute the growth formulas: \(Q_0 \cdot 2^t = 2 \cdot (Q_0 \cdot 2^{t/2})\).

Step 4: Simplify the Equation

Cancel \(Q_0\) from each side: \(2^t = 2 \cdot 2^{t/2}\). This simplifies to \(2^t = 2^{1 + t/2}\).

Step 5: Solve for \(t\)

Equating the exponents gives \(t = 1 + \frac{t}{2}\). Solve for \(t\) by multiplying the entire equation by 2, resulting in \(2t = 2 + t\). Thus, \(t = 2\).

Key Concepts

Doubling Time

Doubling time is a key concept when dealing with exponential growth. It refers to the amount of time it takes for a quantity to double in size or value. In our problem, quantities \(Q_1\) and \(Q_2\) have doubling times of 1 year and 2 years respectively. This means:

• For \(Q_1\), the size doubles every year.
• For \(Q_2\), doubling happens every two years.

Understanding doubling time helps in gauging how quickly a quantity is growing. It provides a clear and intuitive measure of exponential growth, making it easier to compare different growth scenarios.

Exponential Functions

Exponential functions are crucial in modeling situations where growth rate keeps increasing over time. The general form of an exponential growth function is:\[ Q(t) = Q_0 \cdot 2^{t/T} \]Here:

• \(Q(t)\) represents the quantity at time \(t\).
• \(Q_0\) is the initial amount.
• \(T\) is the doubling time.

In our exercise, both \(Q_1\) and \(Q_2\) grow according to this formula but with different values for \(T\). For \(Q_1\), since \(T = 1\), its growth function becomes \(Q_1(t) = Q_0 \cdot 2^t\). Similarly, for \(Q_2\), with \(T = 2\), the growth function becomes \(Q_2(t) = Q_0 \cdot 2^{t/2}\). These functions provide the mathematical framework to describe and analyze how quickly each quantity is growing.

Solving Equations

Solving equations in this context involves finding the time \(t\) at which one quantity is a multiple of another. For our exercise, we want \(Q_1(t)\) to be twice \(Q_2(t)\). We set up the equation:\[ Q_0 \cdot 2^t = 2 \cdot (Q_0 \cdot 2^{t/2}) \]Start by cancelling \(Q_0\) from both sides:\[ 2^t = 2 \cdot 2^{t/2} \]This simplifies to:\[ 2^t = 2^{1 + t/2} \]Since the bases are the same, the exponents must be equal:\[ t = 1 + \frac{t}{2} \]Multiply everything by 2 to get:\[ 2t = 2 + t \]Subtract \(t\) from both sides, yielding:\[ t = 2 \]Therefore, it will take 2 years for \(Q_1\) to be twice the size of \(Q_2\). This process highlights the power of algebra in solving real-world problems involving exponential growth.