Problem 6 - Calculus with Concepts in Calculus

Question

Suppose the population \(f(t)\) of a given species grows exponentially, so that \(f(t)=f(0) e^{k t}\) for some positive constant \(k\) a. Show that the population doubles during any time interval of duration \((\ln 2) / k\). Thus \((\ln 2) / k\) is the doubling time \(d\). b. Show that \(f(t)=f(0) 2^{t / d}\).

Step-by-Step Solution

Answer

Doubling time is \((\ln 2)/k\); population is \(f(0) 2^{t/d}\).

Step 1: Understand the Problem

We are given an exponential population growth model, and our goal is to find the doubling time in terms of a logarithm and then express the population in an alternative form.

Step 1: Find Doubling Time

We need to determine the time it takes for the population to double, starting with the equation \(f(t) = f(0) e^{kt}\). If \(f(t)\) doubles, we have \(f(t) = 2f(0)\). Substituting into the exponential form, we get:\[2f(0) = f(0) e^{kt}\]Divide both sides by \(f(0)\), which gives:\[2 = e^{kt}\]Taking the natural logarithm of both sides, we obtain:\[\ln 2 = kt\]Solving for \(t\), we find:\[t = \frac{\ln 2}{k}\]Therefore, the doubling time \(d = \frac{\ln 2}{k}\).

Step 2: Express Population with Doubling Time

Given that \(d = \frac{\ln 2}{k}\), we want to rewrite the initial equation to show the population as a function of \(t/d\). Start with:\[f(t) = f(0) e^{kt}\]Express \(e^{kt}\) using the doubling time \(d = \frac{\ln 2}{k}\). We know that \(kd = \ln 2\), so \(k = \frac{\ln 2}{d}\). Substitute this into the exponent:\[f(t) = f(0) e^{\left(\frac{\ln 2}{d} \right)t}\]We can rewrite this expression as:\[f(t) = f(0) \left(e^{\ln 2} \right)^{t/d}\]Since \(e^{\ln 2} = 2\), this simplifies to:\[f(t) = f(0) 2^{t/d}\]

Step 3: Confirm Simplified Population Equation

Verify that the alternative expression accurately depicts doubling behavior. Each increment of \(d\) in \(t\) results in the population multiplying by \(2\). This confirms:\[f(t) = f(0) 2^{t/d}\] demonstrates the same doubling as the original expression.

Key Concepts

Doubling Time

When we talk about doubling time in exponential growth, we're referring to how long it takes for a population to double in size. This is a crucial concept because it helps us understand the speed of growth. In any system that grows exponentially, like populations or investments, doubling time gives a clear picture of the pace at which growth happens.To calculate the doubling time (\(d\)), we use the formula:\[d = \frac{\ln 2}{k}\]- \(\ln 2\) is the natural logarithm of 2, which approximates to 0.693.- \(k\) is the growth rate constant.This formula shows that doubling time (\(d\)) is directly connected to the natural logarithm and the growth rate. If \(k\) is large, meaning the population grows quickly, the doubling time will be shorter. Conversely, a smaller \(k\) leads to a longer doubling time. Understanding this relationship helps in predicting how fast populations increase.

Natural Logarithm

The natural logarithm is a mathematical function denoted as \(\ln\). It is the inverse operation of exponentiation with the base of the natural constant, \(e\), roughly equal to 2.718. When solving exponential growth problems, the natural logarithm often pops up because it is crucial in unraveling values from an exponential equation. In this context:- If we have an equation such as \(e^x = a\), the value of \(x\) can be determined by \(x = \ln a\).- In solving for doubling time, we take the natural logarithm of both sides of an equation involving \(e^{kt}\), allowing us to easily isolate and solve for time \(t\).By using the natural logarithm, we turn multiplication into addition, which simplifies calculations immensely, making complex growth problems manageable.

Population Growth Model

The population growth model in this context refers to a way of expressing how a population increases over time exponentially. The core equation for this is given by:\[f(t) = f(0) e^{kt}\]- \(f(t)\) is the population at time \(t\).- \(f(0)\) is the initial population size.- \(k\) is the growth rate constant, and more significant values indicate faster growth.As populations double, we can express this process with a simpler model using the doubling time:\[f(t) = f(0) 2^{t/d}\]- \(d\) is the doubling time, calculated using the relationship with the natural logarithm as explained before.This model highlights how every period of doubling time, the population multiplies by 2. It's intuitive and straightforward for understanding exponential growth. Furthermore, shifting from using \(e\) to \(2\) when looking at population doubling provides an alternative perspective that aligns with how we naturally think about growth, making predictions and comparisons easier to grasp.