Problem 39 - Applied Calculus

Question

With \(t\) in years since January 1,2010 , the population \(P\) of Slim Chance is predicted by $$ P=35,000(0.98)^{t} $$ At what rate will the population be changing on January \(1,2023 ?\)

Step-by-Step Solution

Answer

The population will be decreasing by about 554 people per year.

Step 1: Identify the Problem

We need to find the rate at which the population is changing on January 1, 2023. This involves finding the derivative of the population function with respect to time, which will give us the rate of change.

Step 2: Determine the Time Since 2010

Calculate the number of years since January 1, 2010, to January 1, 2023. This is 2023 - 2010 = 13 years.

Step 3: Write the Given Population Function

The population is given by the function: \[ P(t) = 35,000 imes (0.98)^t \] where \( t \) is the number of years since 2010.

Step 4: Differentiate the Population Function

To find the rate of change, we differentiate \( P(t) \) with respect to \( t \) using the chain rule. The derivative, \( \frac{dP}{dt} \), is given by: \[ \frac{dP}{dt} = 35,000 \times \ln(0.98) \times (0.98)^t \]

Step 5: Evaluate the Derivative at \( t = 13 \)

Substitute \( t = 13 \) into the derivative to find the rate of change at that specific time:\[ \frac{dP}{dt} = 35,000 \times \ln(0.98) \times (0.98)^{13} \]

Step 6: Calculate the Specific Rate of Change

Calculate the numerical value of the expression: - First, find \( \ln(0.98) \) which is approximately -0.0202.- Then calculate \( (0.98)^{13} \) which is approximately 0.7835.- Multiply these with 35,000: \[ \frac{dP}{dt} \approx 35,000 \times (-0.0202) \times 0.7835 \]\[ \frac{dP}{dt} \approx -554.12 \]

Step 7: Interpret the Result

The population is decreasing at a rate of approximately 554 people per year on January 1, 2023.

Key Concepts

Calculus Made Easy

Calculus is a branch of mathematics focusing on rates of change and accumulation. In this problem, we're using calculus to determine how quickly the population is changing over time. Calculus often deals with two main operations:

• Differentiation, which is used to find the rate of change of a quantity.
• Integration, which is used to accumulate quantities over a range.

In our problem, differentiation is our focus since we're interested in the instantaneous rate of change in the population of Slim Chance at a specific time. By applying calculus, specifically differentiation, we can predict the trends and behaviors of dynamic systems like populations over time.

Understanding Exponential Decay

Exponential decay is a process by which a quantity decreases at a rate proportional to its current value. This means that as time progresses, the decline gets slower, forming a decay curve.In this context, the population function of Slim Chance is described by an equation involving exponential decay: \( P = 35,000(0.98)^t \), where 0.98 indicates that the population is shrinking by 2% every year.Key points to understand exponential decay include:

• The base (0.98 in our case) is less than one, showing a decrease.
• Even though the decrease rate is constant (2% annually), the actual decrease in population figure is less over time because it is always relative to the current population.

Such models are essential for predicting how natural populations decline, tracking radioactive decay, or understanding depreciation in asset values.

Deciphering Derivatives

Derivatives provide a way to measure how a function changes. They express how one quantity changes with respect to another. In the context of our population problem, they tell us the rate at which the population is increasing or decreasing at any given time.When we take the derivative of the population function, we're looking for how \( P(t) = 35,000 (0.98)^t \) changes as \( t \) changes. The derivative \( \frac{dP}{dt} \) is computed using specific rules:

• The derivative of a constant times a function is the constant times the derivative of the function.
• We make use of the natural logarithm \( \ln \) when differentiating exponential functions.

So, our population's rate of change is found by: \( \frac{dP}{dt} = 35,000 \times \ln(0.98) \times (0.98)^t \), which calculates how fast (or slow) the population is dwindling.

Mastering the Chain Rule

The chain rule is an essential tool in calculus, particularly when dealing with composite functions. It allows us to differentiate complex expressions by breaking them down into simpler parts.In our exercise, the population function is a composite function involving an exponential part. To find its derivative, we apply the chain rule:- Identify the outer function and the inner function. Here, our outer function is \( P(t) = 35000 \times f(t) \) and the inner is \( f(t) = (0.98)^t \).- The derivative of the outer function \( P(t) \) concerning \( f(t) \) is straightforwardly \( 35000 \times f'(t) \).- The inner function \( f(t) = (0.98)^t \) involves an exponential. Differentiating this involves multiplication by \( \ln(0.98) \).By systematically using the chain rule, we successfully find the rate of change for more complicated functions, ensuring our calculations encompass the nuances of composite relationships.