Problem 19 - Calculus for Biology and Medicine

Question

whose size at time \(t\) is denoted by \(N(t)\), grows according to $$ \frac{d N}{d t}=0.3 N \quad \text { with } N(0)=20 $$ Solve this differential equation, and find the size of the population at time \(t=5\).

Step-by-Step Solution

Answer

The population size at time \( t = 5 \) is approximately 90.

Step 1: Understanding the Differential Equation

We have the differential equation \( \frac{dN}{dt} = 0.3N \). This is a first-order linear differential equation that models exponential growth.

Step 2: Formulating the Solution Expression

For a differential equation of the form \( \frac{dN}{dt} = kN \), the general solution is given by \( N(t) = N(0) e^{kt} \). In this problem, the growth constant \( k \) is 0.3 and the initial population \( N(0) = 20 \).

Step 3: Applying Initial Conditions

Substituting the given values into the general solution formula, we have:\[ N(t) = 20 e^{0.3t} \]. This is the expression for the population at any time \( t \).

Step 4: Finding the Population at Time \( t=5 \)

Now we need to find \( N(5) \). Substitute \( t = 5 \) into the solution:\[ N(5) = 20 e^{0.3 \times 5} \].

Step 5: Calculating the Exponential Term

First, calculate the exponent by multiplying:\[ 0.3 \times 5 = 1.5 \]. Therefore, the expression becomes\[ N(5) = 20 e^{1.5} \].

Step 6: Evaluating the Expression

Calculate \( e^{1.5} \). This is approximately 4.4817. Thus,\[ N(5) \approx 20 \times 4.4817 \approx 89.634 \].

Step 7: Final Answer

The size of the population at time \( t = 5 \) is approximately 90.

Key Concepts

Exponential Growth

Understanding exponential growth is crucial when dealing with population dynamics and numerous scientific processes. At its core, exponential growth describes a process where a quantity increases at a rate proportional to its current size. This means, the larger the quantity, the faster it grows.In the context of our exercise, the population size is defined by a differential equation that reflects exponential growth:

• This equation is \( \frac{dN}{dt} = 0.3N \).
• The '0.3' in the equation is known as the growth rate or growth constant.
• Exponential growth happens because the change in population size, \( \frac{dN}{dt} \), is proportional to the current population size, \( N \).

By solving this kind of equation, we can describe how the population evolves over time. Exponential growth often results in rapid and dramatic increases, which can deeply impact ecosystems, businesses, and more. Thus, understanding its underlying principles is very valuable.

First-order Linear Differential Equations

First-order linear differential equations are a fundamental part of understanding many natural and engineered systems. These equations involve rates of change that are proportional to the current state of the variable in question.The simplest form, as shown in the exercise, is \( \frac{dN}{dt} = kN \), where \( k \) is a constant. Here's a breakdown of its components:

• The equation represents a relationship between a function and its derivative.
• These equations are called 'first-order' because they involve the first derivative of the function.
• 'Linear' indicates that both terms are of the first degree.

For our population growth problem, solving the first-order linear differential equation provided us with an exponential solution: \( N(t) = N(0) e^{kt} \). This highlights how first-order linear differential equations often model situations where the rate of change of a quantity is directly proportional to its size, leading to exponential behaviors.

Initial Value Problems

An initial value problem (IVP) is a type of differential equation that, in addition to providing the equation itself, includes initial conditions that must be satisfied by the solution. This term refers to knowing the state of a system at a specific starting point, which helps in finding a particular solution to the differential equation.In the example given, the initial condition is \( N(0) = 20 \). This implies:

• The population at time \( t=0 \) is 20.
• This condition allows us to solve for the specific solution rather than a family of solutions.

With this initial condition, we applied it to the general solution \( N(t) = N(0) e^{kt} \) and substituted the known values to obtain \( N(t) = 20 e^{0.3t} \). IVPs are common in real-world applications because they allow for tailored solutions that account for the initial setup or condition of the system being studied.