Problem 45 - Calculus Early Transcendentals

Question

(a) Show that if a quantity \(y=y(t)\) has an exponential model, and if \(y\left(t_{1}\right)=y_{1}\) and \(y\left(t_{2}\right)=y_{2},\) then the doubling time or the half-life \(T\) is $$ T=\left|\frac{\left(t_{2}-t_{1}\right) \ln 2}{\ln \left(y_{2} / y_{1}\right)}\right| $$ (b) In a certain 1 -hour period the number of bacteria in a colony increases by \(25 \% .\) Assuming an exponential growth model, what is the doubling time for the colony?

Step-by-Step Solution

Answer

(a) Derivation confirmed. (b) Doubling time is approximately 3.11 hours.

Step 1: Review the Exponential Model

Assume the exponential growth model: \( y(t) = y_0 e^{kt} \), where \( y_0 \) is the initial amount, and \( k \) is the growth rate constant. Given \( y(t_1) = y_1 \) and \( y(t_2) = y_2 \).

Step 2: Relate Model to Given Conditions

Set up the equations using the exponential model. \( y_1 = y_0 e^{kt_1} \) and \( y_2 = y_0 e^{kt_2} \). Divide the second equation by the first: \( \frac{y_2}{y_1} = e^{k(t_2 - t_1)} \).

Step 3: Solve for Growth Rate Constant

Take the natural logarithm of both sides: \( \ln\left(\frac{y_2}{y_1}\right) = k(t_2 - t_1) \). Solve for \( k \): \( k = \frac{\ln(y_2/y_1)}{t_2 - t_1} \).

Step 4: Determine Doubling Time Formula

Substitute the expression of \( k \) into the equation for doubling time where \( y(T) = 2y_0 \), resulting in \( T = \left| \frac{\ln 2}{k} \right| \). Hence, the formula is \( T = \left| \frac{(t_2 - t_1) \ln 2}{\ln(y_2/y_1)} \right| \).

Step 5: Apply to Bacteria Colony Example

Given the colony increases by \( 25\% \) in 1 hour, \( y_2 = 1.25y_1 \) and \( t_2 - t_1 = 1 \). Substitute into the doubling time formula: \( T = \left| \frac{1 \times \ln 2}{\ln(1.25)} \right| \).

Step 6: Calculate Doubling Time

Evaluate the expression: \( \ln 2 \approx 0.693 \) and \( \ln 1.25 \approx 0.223 \). Therefore, \( T \approx \frac{0.693}{0.223} \approx 3.11 \) hours.

Key Concepts

Doubling Time

Doubling time is a crucial concept when discussing exponential growth. This concept refers to the time it takes for a quantity to double in size or value under exponential growth conditions. Knowing how to calculate the doubling time helps us understand how quickly a population or quantity is expanding.

To find the doubling time, consider a situation where a quantity grows exponentially according to the formula \(y(t) = y_0 e^{kt}\). Here, \(y_0\) is the initial quantity and \(k\) represents the growth rate constant. If we know two specific quantities at different times, \(y(t_1) = y_1\) and \(y(t_2) = y_2\), we can determine the doubling time \(T\) using the formula:

\[ T = \left| \frac{(t_2 - t_1) \ln 2}{\ln(y_2/y_1)} \right| \]

By calculating \(T\), one can predict how long it will take for anything that grows exponentially – like a bacterial colony or an investment – to double in size. This can provide useful insights for planning and decision-making.

Half-Life

Half-life is a term frequently used in physical sciences, especially in nuclear physics and chemistry. It's the time required for a quantity to reduce to half its initial value. This concept typically applies to radioactive substances, but it can be relevant to any decaying process.

Like doubling time, half-life is also determined by the exponential nature of a process. For a substance or a process described by \(y(t) = y_0 e^{kt}\), where \(y_0 \) is the initial quantity at time \(t=0\) and \(k\) is a negative constant (since it’s decay), the half-life \(T_{1/2}\) can be found using:

\[ T_{1/2} = \frac{\ln 2}{|k|} \]

This equation tells us the duration taken for half the material to decay, providing critical information for fields that depend on timing and stability, like pharmaceuticals and radiometric dating.

Exponential Model

The exponential model is a mathematical representation used in situations where something grows or decays at a rate proportional to its current value. It's a powerful model because many real-world processes, like population growth, radioactive decay, and even interest compounding, can be seen through this lens.

The general form of an exponential model equation is \(y(t) = y_0 e^{kt}\), where:

• \(y_0\) is the initial quantity at \(t = 0\)
• \(e\) is the base of the natural logarithm, approximately equal to 2.718
• \(k\) is the rate constant
• \(t\) represents time

Understanding this model helps in identifying growth and decay rates quickly. For exponential growth, \(k\) is positive, and for decay, \(k\) is negative. This distinction plays a crucial role in predicting the behavior of the model over time.
When faced with problems like those in the original exercise, setting up an exponential model helps in deriving essential metrics like doubling time and half-life, aiding in the visualization of how systems evolve.