Problem 11 - Calculus for Biology and Medicine

Question

In Problems give a formula for \(N(t), t=0,1,2, \ldots\), on the basis of the information provided. Suppose \(N_{t}=20 \cdot 4^{t}, t=0,1,2, \ldots\), and one unit of time corresponds to 3 hours. Determine the amount of time it takes the population to double in size.

Step-by-Step Solution

Answer

The population doubles in 1.5 hours.

Step 1: Understanding the Problem

We are given a formula for the population growth: \(N_t = 20 \cdot 4^t\). One unit of time \(t\) corresponds to 3 hours. We need to find the time it takes for the population to double.

Step 2: Set Up the Equation

To find the time for the population to double, we start by setting \(N_{t+1} = 2N_t\). Thus, the equation becomes:\[20 \cdot 4^{t+1} = 2 \cdot 20 \cdot 4^t\]

Step 3: Simplify the Equation

Divide both sides of the equation by \(20 \cdot 4^t\):\[4 = 2\]This tells us that \(4^{t+1-t} = 2\), so \(4^1 = 2 \cdot 4^0\), simplifies to \(4 = 2\cdot 1\), which seems inconsistent with common math operations.

Step 4: Correct Physical Interpretation

We interpret this as needing the next exponent of 4 to be a double multiple of 2 relative to previous, thus needed formulaic inaccuracies arise when possible typographical error is not addressed right.

Step 5: Use Logarithms for Clarity

To be precise, set \(k\) such that \(20 \cdot 4^{k} = 40 \cdot 4^t\) directly means \(4^k = 40 / 40\) equal to clear \(k = t+ 1.15\) and scientifically same invariance was checked for accuracy.

Step 6: Find the Doubling Time in terms of Target

Determine when \(4^t = 2\) using logarithms:\[t \cdot \log(4) = \log(2)\] which gives \[t = \frac{\log(2)}{\log(4)} = \frac{1}{2}\] cells. Thus one unit time is equivalent to half unit time transcription rate determined accordingly.

Step 7: Convert Time Units

Since each unit \(t\) corresponds to 3 hours, \(\frac{1}{2}\) unit of \(t\) corresponds to \(\frac{1}{2} \cdot 3 = 1.5\) hours.

Key Concepts

Exponential Growth

Exponential growth is a fundamental concept in understanding how populations or quantities increase over time. In simple terms, it's a growth pattern where the rate of growth is proportional to the current value, leading to a rapid increase. This type of growth forms the basis for the equation given in the exercise:

• \(N_t = 20 \cdot 4^t\)

The key here is the base of the exponent, which is 4 in this case. Each increase by one unit in time means the population multiplies by 4.
The formula indicates that every step forward in time results in the population being 4 times what it was at the previous step. It's crucial to remember that exponential growth can lead to very large numbers quite quickly due to this compounding effect.
Think of it like this: if you start with a certain population size, after one time unit, it's quadrupled, and after two units, it's 16 times the original size. This emphasizes the dramatic effect of exponential growth, especially over extended periods.

Doubling Time

Doubling time is another important concept that helps measure how quickly a growing quantity doubles. It's particularly useful in the context of exponential growth. In our problem, we want to find out how much time it takes for the population to double its size, given the formula

• \(N_t = 20 \cdot 4^t\)

To find the doubling time, you essentially identify how long it takes for the population to become twice as big. Mathematically, this is done by setting up an equation where the new population size is double the original.
The solution to finding the doubling time involves finding when \(4^t = 2\).
This is solved using logarithms, a powerful mathematical tool that provides clarity when working with exponential functions. In this case, the doubling time in terms of \(t\) is \( \frac{1}{2} \), meaning each population doubling happens every half a unit time period. Since each unit of \(t\) is equivalent to 3 hours, the population doubles in 1.5 hours.

Logarithms

Logarithms are essential in solving equations involving exponential growth, especially when dealing with problems like finding the doubling time. The logarithm essentially "undoes" an exponential function, making it manageable to work with.

• When you take the logarithm of both sides of an exponential equation, you can solve for the exponent efficiently.

In our exercise, to calculate the value of time \(t\) needed for the population to double, we use the equation

• \(t \cdot \log(4) = \log(2)\).

This step converts our exponential problem into a simple division:

• \(t = \frac{\log(2)}{\log(4)}\).

By applying the properties of logarithms, this yields \(t = \frac{1}{2}\).
Thus, logarithms not only make the process of finding unknown exponents easier but also allow us to express and understand scalability in quantitative terms, which is exactly what was needed to determine the doubling time in this scenario.