Problem 1 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

(a) Suppose a population grows at a rate of \(5 \%\) per year: \(P=P_{0}(1.05)^{t}\). i. Express this in the form \(P=P_{0} e^{r t}\). ii. Compute \(\frac{d P}{d t}\). iii. Find the proportionality constant \(k\) so that \(\frac{d P}{d t}=k P\). (b) Suppose a population grows according to \(P=P_{0} e^{0.05 t}\) i. Find the proportionality constant \(k\) so that \(\frac{d P}{d t}=k P\). ii. By what percent does the population grow each year? Look back over this problem and think about it. Do your answers make sense to you?

Step-by-Step Solution

Answer

The constant rate \(r\) is given by \(\ln(1.05)\). The derivative \(\frac{d P}{d t}\) is \(\ln(1.05)P\) and the constant of proportionality \(k\) for this equation is \(\ln(1.05)\). For the equation \(P=P_{0} e^{0.05 t}\), \(k=0.05\). The population grows annually about \(5.13%\).

Step 1: Converting the given formula to exponential function

Given \(P=P_{0}(1.05)^{t}\), we can express this in the form \(P=P_{0} e^{r t}\) by taking the natural logarithm of both sides. As we know, \(e^{ \ln a}=a\). Therefore, we rewrite the equation as \(P=P_{0} e^{ \ln (1.05) t}\), so \(r= \ln(1.05)\). Now \(P=P_{0} e^{ \ln (1.05) t}\). Hence, \(r= \ln(1.05)\).

Step 2: Computing the derivative of P wrt t

Using the formula of exponential function \(f(x)=ae^{bx}\), the derivative \(f'(x)= bae^{bx}\), we get \(\frac{d P}{d t}= \ln(1.05) P_{0} e^{ \ln (1.05) t}= \ln(1.05)P\).

Step 3: Finding the proportionality constant k

From the derivative \(\frac{d P}{d t}= \ln(1.05)P\), we can observe that the proportionality constant \(k= \ln(1.05)\).

Step 4: Finding proportionality constant for \(P=P_{0} e^{0.05 t}\)

The population growth according to \(P=P_{0} e^{0.05 t}\). The proportionality constant \(k=0.05\) for this equation.

Step 5: Calculating annual growth rate

To identify the annual growth rate we rewrite the generated exponent and take the inverse of the natural logarithm. The rate can be given by \(e^{k} - 1\), which equates to \(e^{0.05} -1 approximately 0.0513 = 5.13%\).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus, dealing with how a function changes as its input changes. When we differentiate, we find the derivative, which is essentially the rate of change of a function.
In the context of exponential growth, as used in the exercise, differentiation helps us understand how quickly a population is growing over time. For an exponential function such as \(P = P_{0} e^{r t}\), where \(P\) is a function of time \(t\), the derivative \(\frac{dP}{dt}\) indicates how fast the population \(P\) grows as time goes by.

• To compute this derivative, we apply the rule for differentiating exponential functions: if \(f(x) = ae^{bx}\), then \(f'(x) = abe^{bx}\).
• This results in \(\frac{dP}{dt} = rP_{0} e^{r t} = rP\), highlighting that the rate of change of \(P\) is directly proportional to its current size.

This relationship between functions and their rates of change is key in understanding both differential calculus and dynamic systems like population growth.

Population Growth

Population growth is a model that describes how the number of individuals in a population increases over time. In natural contexts, populations often grow exponentially, which implies that as time progresses, the population multiplies by a consistent factor.
In our exercise, the population grows following the function \(P = P_{0} e^{0.05 t}\), indicating exponential growth at a steady rate of \(5\%\) annually. This is what we call an exponential growth model, where \(P_{0}\) is the initial population at time \(t = 0\).

• Such a model assumes that resources are unlimited and that the growth rate is static, an idealization often used for theoretical purposes.
• Real-world growth usually encounters limits, such as scarcity of resources or environmental pressure, causing growth rates to vary.

Understanding exponential growth helps us predict and analyze trends in populations, which is crucial for planning and decision-making in areas like law, economics, and environmental science.

Proportionality Constant

A proportionality constant \(k\) in the context of exponential growth is a value that describes the fixed ratio of change between two proportional quantities, illustrating how one quantity increments continuously with respect to another.
For the population growth model \(P = P_{0}e^{r t}\), the derivative \(\frac{dP}{dt} = kP\) showcases this constant. Here \(k\) is the growth rate derived from the formula, and it indicates that the rate of change of the population \(P\) is proportional to its current size.

• In the exercise, the constant \(k = \ln(1.05)\) for the first part implies a logarithmic transformation of the growth factor, while \(k = 0.05\) represents the straightforward rate in the given exponential expression.

This constant is significant because it gives precise insight into the dynamics of growth, being pivotal to formulating realistic models in population studies and other fields like epidemiology and finance.

Natural Logarithm

The natural logarithm, denoted as \(\ln\), is the logarithm to the base \(e\), where \(e\) is an irrational constant approximately equal to \(2.71828\). In the field of exponential growth and calculus, the natural logarithm plays a crucial role by simplifying the manipulation of exponential functions.
When converting an expression like \(P = P_{0}(1.05)^{t}\) to \(P = P_{0} e^{rt}\), the natural logarithm is used to express the growth factor \(1.05\) with base \(e\).

• By utilizing \(\ln(a)\), we can rewrite any positive number as an exponent of \(e\), which is essential in calculus due to its smooth and continuous nature.
• This conversion simplifies differentiation and integration, allowing for more manageable calculations of rates of change.

Understanding \(\ln\) is beneficial in translating practical exponential growth problems into mathematically tractable forms, aiding in richer insights and accurate predictions in varied applications.