Problem 40
Question
It is projected that \(t\) months from now, the population of a certain town will be \(P(t)=3 t+5 t^{3 / 2}+6,000 .\) At what percentage rate will the population be changing with respect to time 4 months from now?
Step-by-Step Solution
Verified Answer
The population is changing at approximately 0.2974% per month at the 4-month mark.
1Step 1 - Differentiate the population function
The population function is given by \[P(t) = 3t + 5t^{3/2} + 6,000.\]To find the rate of change of the population with respect to time, differentiate the function with respect to \(t\).
2Step 2 - Apply the power rule
Using the power rule for differentiation \( \frac{d}{dt}[t^n] = nt^{n-1}, \)we differentiate each term individually:\[P'(t) = \frac{d}{dt}[3t] + \frac{d}{dt}[5t^{3/2}] + \frac{d}{dt}[6,000].\]
3Step 3 - Differentiate each term
Differentiate each term using the power rule:1. \( \frac{d}{dt}[3t] = 3 2. \)\( \frac{d}{dt}[5t^{3/2}] = 5 \cdot \frac{3}{2}t^{1/2} = \frac{15}{2}t^{1/2} \)3. \( \frac{d}{dt}[6,000] = 0 \)So,\[P'(t) = 3 + \frac{15}{2}t^{1/2}.\]
4Step 4 - Evaluate the derivative at \(t=4\)
Substitute \(t = 4\) into the derivative to find the rate of change at \(t = 4\): \[P'(4) = 3 + \frac{15}{2}\cdot 4^{1/2}\]Since \(4^{1/2} = 2\): \[P'(4) = 3 + \frac{15}{2}\cdot 2 = 3 + 15 = 18.\]
5Step 5 - Convert rate of change to a percentage
To express the rate of change as a percentage, consider the rate of change per month. Here, \(P'(4)=18\) people per month. The percentage rate is given as \[\left(\frac{P'(4)}{P(4)}\right) \cdot 100\%\]First, find \(P(4)\): \[P(4) = 3\cdot 4 + 5\cdot 4^{3/2} + 6,000\]\[= 12 + 5 \cdot 8 + 6,000\]\[= 12 + 40 + 6,000 = 6,052.\]Now, calculate the percentage: \[\left(\frac{18}{6,052}\right) \cdot 100\% \approx 0.2974\%.\]
Key Concepts
DifferentiationRate of ChangePower Rule
Differentiation
Differentiation is a fundamental concept in calculus. It is the process of finding the derivative of a function. The derivative represents how a function changes as its input changes, which is known as the rate of change. In the context of population growth, differentiation helps us determine how the population size changes over time.
For the function given in the exercise, we need to differentiate the population function \( P(t) = 3t + 5t^{3/2} + 6,000 \) with respect to time \( t \). This process allows us to find the rate at which the population is changing at any given moment.
Differentiation involves applying certain rules, such as the power rule, which we will discuss later. The result of differentiation is a new function, called the derivative, which gives us the rate of change for the original function at different points.
For the function given in the exercise, we need to differentiate the population function \( P(t) = 3t + 5t^{3/2} + 6,000 \) with respect to time \( t \). This process allows us to find the rate at which the population is changing at any given moment.
Differentiation involves applying certain rules, such as the power rule, which we will discuss later. The result of differentiation is a new function, called the derivative, which gives us the rate of change for the original function at different points.
Rate of Change
The rate of change measures how one quantity changes in relation to another. In the context of population growth, it tells us how fast the population is increasing or decreasing over time. This is extremely important for understanding trends and making predictions.
In the exercise, we found the rate of change of the population with respect to time by calculating the derivative of the population function \( P(t) = 3t + 5t^{3/2} + 6,000 \). After differentiating, we get \( P'(t) = 3 + \frac{15}{2}t^{1/2} \).
To determine the rate of change at a specific time, say 4 months from now, we substitute \( t = 4 \) into the derivative function. This gives us \( P'(4) = 3 + 15 = 18 \).
This means the population is growing by 18 people per month four months from now.
In the exercise, we found the rate of change of the population with respect to time by calculating the derivative of the population function \( P(t) = 3t + 5t^{3/2} + 6,000 \). After differentiating, we get \( P'(t) = 3 + \frac{15}{2}t^{1/2} \).
To determine the rate of change at a specific time, say 4 months from now, we substitute \( t = 4 \) into the derivative function. This gives us \( P'(4) = 3 + 15 = 18 \).
This means the population is growing by 18 people per month four months from now.
Power Rule
The power rule is a quick way to find the derivative of functions where the variable is raised to a power. The rule states: \[ \frac{d}{dt}[t^n] = nt^{n-1} \]. This is highly useful for differentiating polynomial functions.
In the exercise, we applied the power rule to differentiate each term of the population function. Let's break down the steps:
In the exercise, we applied the power rule to differentiate each term of the population function. Let's break down the steps:
- For \( 3t \), we get \( 3 \).
- For \( 5t^{3/2} \), we get \( 5 \cdot \frac{3}{2} t^{(3/2)-1} = \frac{15}{2} t^{1/2} \).
- For the constant term 6,000, its derivative is 0 since constants do not change.
Other exercises in this chapter
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