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The carbon monoxide level in a city is predicted to be \(0.02 x^{3 / 2}+1\) ppm (parts per million), where \(x\) is the population in thousands. In \(t\) years the population of the city is predicted to be \(x(t)=12+2 t\) thousand people. Therefore, in \(t\) years the carbon monoxide level will be $$ P(t)=0.02(12+2 t)^{3 / 2}+1 \quad \overline{\mathrm{ppm}} $$ Find \(P^{\prime}(2)\), the rate at which carbon monoxide pollution will be increasing in 2 years.

Short Answer

Expert verified
The rate of increase of carbon monoxide in 2 years is 0.24 ppm/year.

Step by step solution

01

Substitute Population Function into P(t) Function

Given the carbon monoxide level function \( P(t) = 0.02(12 + 2t)^{3/2} + 1 \), substitute the population function \( x(t) = 12 + 2t \) into the carbon monoxide function. This results in:\[ P(t) = 0.02(x(t))^{3/2} + 1 = 0.02(12 + 2t)^{3/2} + 1 \].This substitution indicates how \( P(t) \) varies with time \( t \) as the population changes.
02

Differentiate P(t) with Respect to t

To find the rate at which carbon monoxide pollution increases, differentiate \( P(t) \) with respect to \( t \). Recall that by the chain rule:\[ \frac{d}{dt}(a(u(t))^{b}) = ab(u(t))^{b-1} \cdot \frac{du(t)}{dt} \].Here, \( a = 0.02 \), \( u(t) = 12 + 2t \), and \( b = \frac{3}{2} \). Differentiate:\[ P'(t) = 0.02 \times \frac{3}{2} \times (12 + 2t)^{\frac{1}{2}} \times 2 \].
03

Simplify the Derivative

Simplify the expression obtained by differentiated \( P(t) \):\[ P'(t) = 0.02 \times \frac{3}{2} \times 2 \times (12 + 2t)^{\frac{1}{2}} \].This further simplifies to:\[ P'(t) = 0.06 (12 + 2t)^{\frac{1}{2}} \].
04

Evaluate Derivative at t = 2

Substitute \( t = 2 \) into the derivative \( P'(t) = 0.06 (12 + 2t)^{\frac{1}{2}} \):\[ P'(2) = 0.06 (12 + 4)^{\frac{1}{2}} = 0.06 \times \sqrt{16} \].Since \( \sqrt{16} = 4 \), the expression simplifies to:\[ P'(2) = 0.06 \times 4 = 0.24 \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differentiation
Differentiation is a fundamental concept in calculus. It involves finding the rate at which a quantity changes with respect to another quantity. When you differentiate a function, you determine how rapidly the output of the function is changing at a specific point. This concept is key in many scientific fields because it provides insights into the behavior of mathematical models.

The main tool in differentiation is the derivative. The derivative of a function measures the slope of the tangent line to the curve at any particular point. The process of differentiation can be performed using several rules, one of the most important being the chain rule. This rule is especially handy for functions composed of functions, like in our carbon monoxide problem.

Utilizing the chain rule, it's possible to differentiate complex functions step-by-step. This ensures accuracy when assessing how changes in input (like time) affect the output (like pollution levels). Understanding differentiation allows us to compute the rate of change of pollution levels over time, providing actionable data for environmental studies.
Carbon Monoxide Level
Carbon monoxide (CO) is a colorless, odorless gas that can be harmful when inhaled in large amounts. It is important to monitor CO levels in urban areas due to its adverse effects on human health and the environment. In mathematical terms, we express the concentration of CO as a function of other related factors, such as population density or industrial activities.

In our example, the carbon monoxide level is given as a function of the population, expressed by the formula \( P(t) = 0.02 (12 + 2t)^{3/2} + 1 \). This indicates that the CO levels depend on how many people live in the city. As the population increases, the activity and traffic likely do too, causing CO levels to rise. Understanding this relationship helps city planners and policymakers develop strategies to manage air quality effectively.

Analyzing CO levels with the help of mathematical models enables us to predict future pollution trends. This information is essential for designing policies aimed at reducing emissions and ultimately protecting public health.
Population Growth Function
Population growth functions are mathematical expressions that predict how populations change over time. These functions are crucial for urban planning, resource allocation, and environmental management because they provide insights into future demographic trends.

In the exercise, the population is modeled by the function \( x(t) = 12 + 2t \), where \( t \) represents years into the future. This specific linear function suggests a steady growth rate, meaning the population increases by 2000 people each year. By substituting this function into pollution models, we can predict how changes in population affect environmental conditions, like CO levels.

Understanding how to work with population growth functions is vital for solving complex problems related to urbanization and pollution. It allows us to connect various environmental variables and make informed decisions about future strategies to maintain sustainable and healthy city environments.

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Most popular questions from this chapter

The windchill index (revised in 2001 ) for a temperature of 32 degrees Fahrenheit and wind speed \(x\) miles per hour is \(W(x)=55.628-22.07 x^{0.16}\). a. Graph the windchill index on a graphing calculator using the window \([0,50]\) by \([0,40]\). Then find the windchill index for wind speeds of \(x=15\) and \(x=30\) mph. b. Notice from your graph that the windchill index has first derivative negative and second derivative positive. What does this mean about how successive 1-mph increases in wind speed affect the windchill index? c. Verify your answer to part (b) by defining \(1 / 2\) to be the derivative of \(y_{1}\) (using NDERIV), evaluating it at \(x=15\) and \(x=30\), and interpreting your answers.

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The strength of a patient's reaction to a dose of \(x\) milligrams of a certain drug is \(R(x)=4 x \sqrt{11+0.5 x}\) for \(0 \leq x \leq 140\). The derivative \(R^{\prime}(x)\) is called the sensitivity to the drug. Find \(R^{\prime}(50)\), the sensitivity to a dose of \(50 \mathrm{mg}\).

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