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Chapter 4: Problem 77

The rate of change of the level of ozone, an invisible gas that is an irritant and impairs breathing, present in the atmosphere on a certain May day in the city of Riverside is given by $$ R(t)=3.2922 t^{2}-0.366 t^{3} \quad 0

Short Answer

Expert verified
The ozone level function A(t) at any time t can be found by integrating the given rate function and using the initial condition. The final function is: $$ A(t) = \frac{3.2922}{3}t^3 - \frac{0.366}{4}t^4 + 34 $$

Step by step solution

01

Integrate the rate function with respect to t

To find the ozone level function A(t), we need to integrate the given rate function, R(t). So, integrate R(t) with respect to t: $$ \int(3.2922t^2 - 0.366t^3) dt $$
02

Find the anti-derivative of the given function

Now, find the anti-derivative of the function inside the integral: $$ \int(3.2922t^2 - 0.366t^3) dt = (3.2922\int t^2 dt) - (0.366 \int t^3 dt)\\ = (3.2922 \times \frac{1}{3}t^3) - (0.366 \times \frac{1}{4}t^4) + C $$ So, we have: $$ A(t) = \frac{3.2922}{3}t^3 - \frac{0.366}{4}t^4 + C $$
03

Determine the constant of integration using the initial condition

We are given that at t=0 (7 A.M.), the ozone level A(t) is 34. We can use this initial condition to solve for the constant C. $$ A(0) = 34 = \frac{3.2922}{3}(0)^3 - \frac{0.366}{4}(0)^4 + C $$ Since both terms including t become zero, we get: $$ C = 34 $$
04

Write the ozone level function A(t)

Now that we have the constant of integration C, we can write the function A(t) representing the ozone level at any time t: $$ A(t) = \frac{3.2922}{3}t^3 - \frac{0.366}{4}t^4 + 34 $$ Now, we have the ozone level function A(t) for any time t (measured in hours).

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Key Concepts

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

Integrating Rate Functions
When faced with a problem involving the rate of change, such as the level of ozone in the atmosphere over time, we turn to calculus to handle the situation more precisely. Integrating rate functions is a crucial concept in calculus, allowing us to find the accumulated quantity, such as pollutants, that changes at a known rate over time.

Integrating a rate function essentially involves finding the total amount of change from a starting point to an endpoint. In the given problem, the rate function for the level of ozone, identified as R(t), is expressed as a polynomial in terms of time t. To determine the cumulative level of ozone at any given time, we integrate the function R(t).
  • The integral of a rate function gives us a new function which represents the total amount of the quantity at any time.
  • The integration process reveals the accumulated change from the initial state (the initial amount of ozone in our case).
  • The resulting function after integration is expressed as an anti-derivative, which contains a constant of integration due to the indefinite nature of the integral.
By integrating, we move from understanding the rate of change to understanding the total quantity that has changed up to a certain time, which is the essence of solving problems involving rates in calculus.
Anti-derivative
Understanding the anti-derivative is fundamental to solving integration problems. An anti-derivative, also known as an indefinite integral, is essentially the reverse of differentiation. If a function represents the rate of change, then its anti-derivative represents the original quantity before it was changed by that rate.

To find the anti-derivative of a function, we use the rules of integration which are based on the rules of differentiation but applied in reverse. For example, the power rule for integration states that the anti-derivative of t^n is t^(n+1)/(n+1), as long as n is not equal to -1. In our ozone level problem, applying this rule helps us move from the rate of change R(t) to the ozone level function A(t).

When to use the anti-derivative:

  • When you need to find the original quantity from its rate of change.
  • When solving indefinite integrals to find a general form of anti-derivatives.
The anti-derivative is not complete without adding a constant of integration, usually denoted as C, because differentiation of a constant is zero. This reflects the fact that there could be many functions with the same rate of change, differing only by a constant amount.
Initial Conditions in Calculus
Initial conditions play a central role when dealing with integration problems. They are used to determine the specific instance of an anti-derivative function, effectively finding the constant of integration, C, which makes the solution unique to a given situation.

For problems such as the one provided, the initial condition is the known value of the function at a specific point in time, which in our case is the ozone level at 7 A.M. This piece of information is essential because it allows us to solve for the constant C and turn our general solution into a specific one that applies to our scenario.
  • Initial conditions allow us to tailor the general anti-derivative to the particular situation by giving us a specific value to work with.
  • The constant C can be found by substituting the initial condition into the anti-derivative.
In the exercise related to the ozone level, we have the initial condition A(0) = 34. When we plug t = 0 into our general anti-derivative A(t), we find that the constant of integration C is indeed 34. This correlates the mathematical solution to the real-world problem, providing us with a complete function that describes the ozone level at any time during the day.

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Most popular questions from this chapter

To test air purifiers, engineers ran a purifier in a smoke-filled \(10-\mathrm{ft} \times 20\) -ft room. While conducting a test for a certain brand of air purifier, it was determined that the amount of smoke in the room was decreasing at the rate of \(R(t)\) percent of the (original) amount of smoke per minute, \(t\) min after the start of the test, where \(R\) is given by $$\begin{array}{r} R(t)=0.00032 t^{4}-0.01872 t^{3}+0.3948 t^{2}-3.83 t+17.63 \\ 0 \leq t \leq 20 \end{array}$$ How much smoke was left in the room 5 min after the start of the test? How much smoke was left in the room \(10 \mathrm{~min}\) after the start of the test?

Evaluate \(\int_{-1}^{1} \frac{2 x^{5}+x^{4}-3 x^{3}+2 x^{2}+8 x+1}{x^{2}+1} d x\)

Determine whether the statement is true or false. If it is true, explain why it is true. If it is false, explain why or give an example to show why it is false. $$ \begin{array}{l} \int\left[\int f(x) d x\right] d x=G(x)+C_{1} x+C_{2} \\ \text { where } G^{\prime}=F \text { and } F^{\prime}=f \end{array} $$

a. Show that if \(f\) is a continuous function, then $$\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x$$ and give a geometric interpretation of this result. b. Use the result of part (a) to prove that $$\int_{0}^{\pi} \frac{\sin 2 k x}{\sin x} d x=0$$ where \(k\) is an integer. c. Plot the graph of $$f(x)=\frac{\sin 2 k x}{\sin x}$$ for \(k=1,2,3\), and 4 . Do these graphs support the result of part (b)? d. Prove that the graph of $$f(x)=\frac{\sin 2 k x}{\sin x}$$ on \([0, \pi]\) is antisymmetric with respect to the line \(x=\pi / 2\) by showing that \(f\left(x+\frac{\pi}{2}\right)=-f\left(x-\frac{\pi}{2}\right)\) for \(0 \leq x \leq \frac{\pi}{2}\), and use this result to explain part (b).

Life Expectancy of a Female Suppose that in a certain country the life expectancy at birth of a female is changing at the rate of $$ g^{\prime}(t)=\frac{5.45218}{(1+1.09 t)^{0.9}} $$ years per year. Here, \(t\) is measured in years, with \(t=0\) corresponding to the beginning of 1900 . Find an expression \(g(t)\) giving the life expectancy at birth (in years) of a female in that country if the life expectancy at the beginning of 1900 is \(50.02\) years. What is the life expectancy at birth of a female born at the beginning of 2000 in that country?
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