1.) Air pollution: According to the South Coast Air Quality Management District, the level of Nitrogen dioxide, a brown gas that impairs breathing, present in the atmosphere on a certain May day in down town Los Angeles is approximated by
A(t) = (0.03t^3)(t-7)^4 + 60.2
(0< = t < = 7)
where A(t) is measured in pollutant standard index (PSI) and t is measured in hours, with t = 0 corresponding to 7 a.m. At what time of day is the air pollution increasing, and at what time is it decreasing? /use the first derivative/
OR
GDP of a Developing Country: A developing country’s gross domestic product (GDP) from 2000 to 2008 is approximated by the function
G(t) = -0.2t^3 + 2.4t^2 + 60
(0< = t < = 8)
where G(t) is measured in billions of dollars and t = 0 corresponds to 2000. Show that the growth rate of the country’s GDP was maximal in 2004. /use the second derivative/
A is decreasing when A' < 0
The growth rate G' is maximal when its derivative (G") is zero.
Just as G has a max when G' = 0.
So, I just plug the numbers (1-7) or (1-8) in for x and make a graph?
To determine when air pollution is increasing or decreasing, we need to find the first derivative of the function A(t).
Step 1: Find the first derivative of A(t) by differentiating the function with respect to t.
A'(t) = 12t^2(t - 7)^4 - 4t^3(t - 7)^3
Step 2: To find when the air pollution is increasing, we need to identify the intervals where the derivative is positive (greater than zero).
Consider A'(t) > 0
12t^2(t - 7)^4 - 4t^3(t - 7)^3 > 0
Step 3: Solve the inequality to determine the intervals when the pollution is increasing.
Using the factored form of the equation, we can see that the terms (t - 7)^4 and t^2(t - 7)^3 are always positive or equal to zero.
So, we can ignore those terms and focus on the quadratic factors 12t^2 and -4t^3. We only need to find when these quadratic factors are positive.
12t^2 > 0
t^2 > 0
Since t^2 is always positive except at t = 0, we can ignore this factor.
-4t^3 > 0
t^3 < 0
The cubic factor t^3 is negative when t < 0 and positive when t > 0.
Hence, the intervals when the air pollution is increasing are when t > 0.
To determine when the air pollution is decreasing, we need to identify the intervals where the derivative is negative (less than zero).
Consider A'(t) < 0
12t^2(t - 7)^4 - 4t^3(t - 7)^3 < 0
Using the same logic, we can ignore the terms (t - 7)^4 and t^2(t - 7)^3 since they are always positive or equal to zero.
We focus on the quadratic factors.
12t^2 < 0
t^2 < 0
Since t^2 is always positive except at t = 0, we can ignore this factor.
-4t^3 < 0
t^3 > 0
The cubic factor t^3 is positive when t < 0 and negative when t > 0.
Hence, the intervals when the air pollution is decreasing are when t < 0.
In conclusion, the air pollution is increasing before 7 a.m. (t < 0) and decreasing after 7 a.m. (t > 0).
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To show that the growth rate of the country's GDP is maximal in 2004, we need to use the second derivative of the function G(t).
Step 1: Find the second derivative of G(t) by differentiating the function G'(t) with respect to t.
G''(t) = 2(-0.6t + 4.8)
Step 2: Set G''(t) equal to zero to find critical points.
-0.6t + 4.8 = 0
Step 3: Solve the equation to find the value of t that corresponds to the critical point.
-0.6t = -4.8
t = (-4.8)/(-0.6)
t = 8
Step 4: Determine the nature of the critical point using the second derivative test.
Since G''(t) = 2(-0.6t + 4.8), the coefficient of t in G''(t) is -0.6.
When t = 8, G''(t) = 2(-0.6(8) + 4.8) = -9.6 < 0.
Since the second derivative is negative at t = 8, it indicates a maximum point.
Hence, the growth rate of the country's GDP was maximal in 2004 (corresponding to t = 4).
Note: The years are represented by the values t = 0 corresponds to 2000, so t = 4 represents the year 2004.