Problem 58 - Calculus for Biology and Medicine

Question

Assume that a population size at time \(t\) is \(N(t)\) and that $$ N(t)=40 \cdot 2^{t}, \quad t \geq 0 $$ (a) Find the population size at time \(t=0\). (b) Show that $$ N(t)=40 e^{t \ln 2}, \quad t \geq 0 $$

Step-by-Step Solution

Answer

Population at \( t=0 \) is 40; \( N(t) = 40 e^{t \ln 2} \) is correct.

Step 1: Calculate Population at t=0

Substitute \( t=0 \) into the given population function \( N(t) = 40 \cdot 2^t \). We have \( N(0) = 40 \cdot 2^0 \). Since \( 2^0 = 1 \), the equation simplifies to \( N(0) = 40 \cdot 1 = 40 \). Therefore, the population size at \( t=0 \) is 40.

Step 2: Use Exponentiation Properties

Recall that \( a^b = e^{b \ln a} \) for any positive number \( a \). We can rewrite the function \( N(t) = 40 \cdot 2^t \) using this property. So, it becomes \( N(t) = 40 \cdot e^{t \ln 2} \).

Step 3: Verify the Condition

To ensure the rewritten function matches our original condition, substitute the expression \( 2^t = e^{t \ln 2} \) back into \( N(t) = 40 e^{t \ln 2} \). We see that both forms are equivalent ways of expressing exponential growth in terms of \( e \) and \( 2 \), confirming that \( N(t) = 40 e^{t \ln 2} \) is valid for \( t \geq 0 \).

Key Concepts

Understanding Population Dynamics

Population dynamics is a branch of biology that studies the changes in population size and composition over time. It helps us comprehend how populations grow, shrink, and maintain themselves. One common model for analyzing population growth is exponential growth, which assumes that the population grows at a constant relative rate. This means that as the population increases, it grows even faster — a hallmark of exponential growth.

- **Exponential Growth:** In our given exercise, the population size at any time \( t \) is represented by the function \( N(t) = 40 \cdot 2^t \). This illustrates an exponential growth pattern as every increase in time leads to a multiplication of the population size by a factor of 2.- **Initial Population:** At \( t = 0 \), the population is simply the multiplier at the start of the function, which in this case is 40. As you see, this was confirmed in Step 1 of the solution, reinforcing the concept of initial population size when we set \( t = 0 \) in our equation.
Understanding population dynamics through simple equations like this helps us model and predict patterns, which can be crucial for fields like ecology, resource management, and public health.

Exponential Functions Explained

Exponential functions are mathematical expressions that model growth or decay processes. These functions are characterized by the formula \( a \cdot b^t \), where \( a \) is the initial amount and \( b \) is the base of the exponential, dictating its rate of growth or decay.

- **Growth Factor:** In our exercise, \( 2^t \) is the exponential part of the function, with 2 being the base—telling us how the population doubles as time progresses. This is a key component of what makes an exponential function powerful, as each increment in time exponentially influences the result.- **Natural Base \( e \):** Sometimes, it's useful to rewrite exponential functions using the natural number \( e \) approximately equal to 2.718. In step 2 of the solution, we used the natural base representation equation \( e^{b \ln a} \), allowing us to interchangeably express \( 2^t \) as \( e^{t \ln 2} \). This is particularly applicable in continuous growth models found in natural sciences.
Using exponential functions allows mathematicians and scientists to predict the behavior of dynamic systems, whether they be in physics, biology, or economics, particularly when dealing with growth or spread.

The Properties of Exponents

Mastery of the properties of exponents is key to working with exponential functions effectively. These properties simplify complex equations, making calculations more approachable and understandable.

**Key Properties to Remember:**

• Zero Exponent Rule: Any non-zero number raised to the power of 0 is 1. In our exercise, we used this when calculating \( 2^0 = 1 \) to find the initial population.
• Power of a Power: \( (a^m)^n = a^{m\cdot n} \). This helps when working with combined exponential terms.
• Product of Powers: \( a^m \cdot a^n = a^{m+n} \). It is useful when multiplying similar bases, maintaining order with exponents.
• Exponentiation of Logarithms: Utilized in step 2 of the solution as \( b^t = e^{t \ln b} \). This is extremely useful in rewriting expressions involving varying exponents.

The properties of exponents are foundational in mathematics and are essential in both theoretical and applied scenarios—particularly when dealing with exponential growth models, making calculations and transformations more feasible.