Problem 63 - Single Variable Calculus

Question

Population Growth in Clark County Clark County in Nevada, which is dominated by greater Las Vegas, is one of the fastest-growing metropolitan areas in the United States. The population of the county from 1970 through 2000 is approximated by the function $$ \begin{aligned} P(t)=44,560 t^{3}-89,394 t^{2}+234,633 t+273,288 & \\ 0 & \leq t \leq 3 \end{aligned} $$ where \(t\) is measured in decades, with \(t=0\) corresponding to the beginning of 1970 . a. Show that the population of Clark County was always increasing over the time period in question. b. Show that the population of Clark County was increasing at the slowest pace some time around the middle of August 1976 . Source: U.S. Census Bureau.

Step-by-Step Solution

Answer

To show that the population of Clark County was always increasing during the given time period, we found the first derivative of the function \( P'(t) = 133,680t^2 - 178,788t + 234,633 \), which indicates the rate of change of the population. Since this quadratic expression is always positive for \( 0 \leq t \leq 3 \), the population was always increasing. To find the time when population growth was at its slowest pace, we computed the second derivative \( P''(t) = 267,360t - 178,788 \) and set it equal to 0. Solving for t, we got \( t \approx 0.6685 \) decades. Converting this to years and months, we found that the population growth rate was slowest around the middle of August 1976.

Step 1: Finding the first derivative of P(t)#

To show that the population of Clark County was always increasing over the given time period, we will find the first derivative of the function, which represents the rate of change of the population with respect to time. If the first derivative is positive, then the population is increasing. We differentiate P(t) with respect to t: \( P'(t) = \frac{d}{dt} (44,560t^3 - 89,394t^2 + 234,633t + 273,288) \) Apply the power rule for each term: \( P'(t) = 3(44,560) t^2 - 2(89,394)t + 234,633 \) \( P'(t) = 133,680t^2 - 178,788t + 234,633 \) #

Step 2: Checking if P'(t) is positive for the given time period #

Now we want to show that P'(t) is positive for all t in the interval [0, 3]. To do this, let's find the points where P'(t) equals 0 (which are the critical points) and check the signs between those points. We have the following quadratic equation: \( 133,680t^2 - 178,788t + 234,633 = 0 \) We can find the roots using the quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where a = 133,680, b = -178,788, and c = 234,633. However, we notice it's a quadratic with positive leading coefficient so it will have a parabolic shape which is upwards facing. The discriminant \(b^2 - 4ac\) is negative, which means the quadratic equation has no real roots and P'(t) never equals 0 in the given interval. Since the coefficient of the quadratic term is positive, the quadratic expression will always be positive for \( 0 \leq t \leq 3 \), proving that the population was always increasing during the given time period. #

Step 3: Finding the second derivative of P(t) #

To determine the time when the population growth was at its slowest pace, we need to find the inflection points in the original population function. For that, we need to compute the second derivative of P(t), set it to 0, and solve for t. Differentiate P'(t) with respect to t: \( P''(t) = \frac{d}{dt} (133,680t^2 - 178,788t + 234,633) \) Apply the power rule for each term: \( P''(t) = 2(133,680)t - 178,788 \) #

Step 4: Finding the inflection point #

Now, we will set P''(t) equal to 0 and solve for t to find the inflection point when the population growth was at its slowest pace: \( 0 = 267,360t - 178,788 \) \( t = \frac{178,788}{267,360} \approx 0.6685 \) (measured in decades, where t = 0 corresponds to the beginning of 1970) Now, let's convert 0.6685 decades to years and months to find the corresponding time: 0.6685 decades = 6.685 years 6 years is equal to 6 * 12 = 72 months 0.685 years = 8.22 months (approximately) So, the population growth rate was slowest around the middle of August 1976 (1970 + 6 years and 8 months).

Key Concepts

First Derivative

In population growth, understanding how quickly the population is changing can be unlocked through the first derivative of a function. The first derivative, denoted as \( P'(t) \), represents the rate of change of the population over time. If \( P'(t) \) is positive, it tells us the population is increasing.

For Clark County's population function \( P(t) = 44,560t^3 - 89,394t^2 + 234,633t + 273,288 \), taking the derivative involves applying the power rule to each term:

• Differentiate \( 44,560t^3 \) to get \( 133,680t^2 \).
• Differentiate \( -89,394t^2 \) to get \( -178,788t \).
• Differentiate \( 234,633t \) to get \( 234,633 \).

So, \( P'(t) = 133,680t^2 - 178,788t + 234,633 \). This quadratic polynomial tells us about the increasing nature of the population over time.

Quadratic Equation

The quadratic equation is a key player in determining the behavior of \( P'(t) \). The expression \( 133,680t^2 - 178,788t + 234,633 = 0 \) is a quadratic equation. Let's delve deeper into its structure:

The solutions to this equation help determine where growth changes pace. However, in this case:

• The discriminant \( b^2 - 4ac \) is negative, meaning no real roots exist.
• This indicates \( P'(t) \) does not cross zero and, given the leading positive coefficient, \( P'(t) \) remains positive.

This means that during the interval \( 0 \leq t \leq 3 \), the population was always increasing. The absence of real roots assures us no decrease occurs within this time frame.

Second Derivative

To understand how the rate of population growth changes, we compute the second derivative. This gives us insight into when the growth is slowing down or speeding up. Calculating the second derivative, \( P''(t) \), involves differentiating \( P'(t) = 133,680t^2 - 178,788t + 234,633 \):

• Differentiate \( 133,680t^2 \) to get \( 267,360t \).
• The constant term differentiates to \( 0 \).
• This results in \( P''(t) = 267,360t - 178,788 \).

The sign and value of \( P''(t) \) help explore when the rate of growth shifts. In particular, solving \( P''(t) = 0 \) identifies inflection points.

Inflection Point

The inflection point is crucial for determining when the population growth rate was at its slowest pace. An inflection point occurs where \( P''(t) = 0 \), meaning the acceleration of growth changes. Solving \( 267,360t - 178,788 = 0 \) finds this critical moment:

• \( t = \frac{178,788}{267,360} \approx 0.6685 \) decades.
• Converted to years, this is approximately 6.685 years after 1970.
• Roughly, this equals the middle of August 1976.

This point signals where the growth rate was slowing down, providing a key insight into demographic changes during this period.