The birth rate in a certain city is described by the following function The city
ID: 3372237 • Letter: T
Question
The birth rate in a certain city is described by the following function
The city's death rate is given by
Here, t is measured in years, and t = 0 corresponds to the start of the year 1990.
The birth and death rates are measured in thousands of births or deaths per year.
At the start of 1990, the population of the city is 225 thousand.
Enter all the following answers correct to two decimal places.
Number of births = thousand.
Number of deaths = thousand.
Population : thousand.
(Enter your dates as years, not the number of years since 1990.)
Increasing from until .
Over what interval is the population decreasing?
Decreasing from until .
Area :
Explanation / Answer
Calculate the total number of births between the start of 1990 and the end of 1999:
integrate (b(t) = 5.64 - .03t^2) from 0 - 10
= 46,400 births were given
Calculate the total number of deaths over the same period
integrate (d(t) = 5 + .01t^2) from 0 - 10
=53,333 deaths
What is the population of the city at the start of year 2000?
Merely adding the births and subtracting the deaths.
200,000 + 46,400 - 53.333 = 193,067
Considering just the period from the start of 1990 to the start of 2000, over what interval is the population increasing?
As given by the birth and deathrates the birthrates are falling and the deathrate is increasing. the birthrate at t = 0 is higher than the deathrate at t = 0, 5.64 > 5.
Thus the population is increasing from t = 0 --> b(t) = d(t)
5.64 - .03t^2 = 5 + .01t^2
=> t = 4
Thus, the population is increating from t = 0 --> t = 4, and decreasing ever after.
The interval in which the population is increasing would be january 1, 1990 --> december 31, 1993.
The population is decreasing from january 1, 1994 --> ever after
Calculate the area between the curves y = b ( t ) and y = d ( t ) for t between 0 and 10.
from t=0 --> t=4 he area between the two curves is 1.71
from t = 4 --> t=10 the area is 8,64;
= the total area is 10.35
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