Mathematics, 14.09.2019 08:30 neariah24

The population of a mining city grows at a rate proportional to that population, in two years the population has doubled and a year later there were 10,000 inhabitants.
what was the initial population?

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Answer from: heroicblad

The initial population was approximatedly 3535 inhabitants.

Step-by-step explanation:

The population of the city can be given by the following differential equation.

$\frac{dP}{dt} = Pr$ ,

In which r is the rate of growth of the population.

We can solve this diffential equation by the variable separation method.

$\frac{dP}{dt} = Pr$

$\frac{dP}{P} = r dt$

Integrating both sides:

ln P = rt + c

Since ln and the exponential are inverse operations, to write P in function of t, we apply ln to both sides.

$e^{ln P} = e^{rt + C}$

$P(t) = Ce^{rt}$

C is the initial population, so:

$P(t) = P(0)e^{rt}$

Now, we apply the problem's statements to first find the growth rate and then the initial population.

The problem states that:

In two years the population has doubled:

P(2) = 2P(0)

$P(t) = P(0)e^{rt}$

$2P(0) = P(0)e^{2r}$

$2 = e^{2r}$

To isolate r, we apply ln both sides

$e^{2r} = 2$

$ln e^{2r} = ln 2$

2r = 0.69

$r = \frac{0.69}{2}$

r = 0.3466

$P(t) = P(0)e^{0.3466t}$

In two years the population has doubled and a year later there were 10,000 inhabitants.

P(3) = 10,000

$P(t) = P(0)e^{0.3466t}$

$10,000= P(0)e^{0.3466*3}$

$P(0) = \frac{10,000}{e^{1.04}}$

P(0) = 3534.55

The initial population was approximatedly 3535 inhabitants.

Answer from: Quest

part a

answer: a reasonable domain is $0 \le n \le 10$ where n is a real number. so n can be between 0 and 10. both endpoints are included.

work shown:

n is the number of days after the experiment starts. the smallest n can be is n = 0 which means that 0 days have gone by, and we're at the start. to find out how large n should be, then replace f(n) with 16.13 and solve for n. use logarithms to isolate the exponent.

f(n) = 12*(1.03)^n

16.13 = 12*(1.03)^n

16.13/12 = (1.03)^n

1.34416666666667 = (1.03)^n

(1.03)^n = 1.34416666666667

log[ (1.03)^n ] = log[ 1.34416666666667 ]

n*log[ 1.03 ] = log[ 1.34416666666667 ]

n = log[ 1.34416666666667 ]/log[ 1.03 ]

n = 10.0062999823929

this rounds to n = 10 which is fairly close but not 100% perfect. so this is the largest n can be.

note: if you plug n = 10 into f(n), you'll get roughly 16.127 which rounds to 16.13 (this comes up again in part c)

=====================================

part b

answer: the y intercept is 12. it represents the starting height of the plant in cm.

work shown:

plug n = 0 into the f(n) function. simplify

f(n) = 12*(1.03)^n

f(0) = 12*(1.03)^0

f(0) = 12*(1)

f(0) = 12

on day n = 0, aka the starting point, the height f(n) is 12 cm

=====================================

part c

answer: the average rate of change is approximately 0.43061036458991 (round however you need to). this represents the average growth rate from day n = 3 to day n = 10. so the plant grew roughly 0.43 cm per day during this timespan, assuming you round to 2 decimal places.

work shown:

compute f(3)

f(n) = 12*(1.03)^n

f(3) = 12*(1.03)^3

f(3) = 13.112724 < we'll use this later

compute f(10)

f(n) = 12*(1.03)^n

f(10) = 12*(1.03)^10

f(10) = 16.1269965521294 < we'll use this later

now use the formula below with a = 3 and b = 10

aroc = average rate of change

aroc = [ f(b) - f(a) ]/[ b - a ]

aroc = [ f(10) - f(3) ]/[ 10 - 3 ]

aroc = (16.1269965521294 - 13.112724)/(10 - 3)

aroc = 3.0142725521294/7

aroc = 0.43061036458991

round this however you need to

note: the plant grew approximately 3.01 cm over 7 days, so roughly 0.43 cm per day is the average growth rate (if you were to round to 2 decimal places).

Answer from: Quest

answer: what in the world is this

step-by-step explanation: