Math / Calculus

QuestionQuestion Population Growth It is estimated that the population of a certain town changes at the rate of $2+t^{4 / 5}$ people per month. If the current population is 20,000 , what will the population be in 10 months?
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Population in 10 months = $\square$ FEEDBACK MORE INSTRUCTION SUBMIT

Studdy Solution

STEP 1

1. The rate of population change is given by the function $\frac{dP}{dt} = 2 + t^{4/5}$ .
2. The initial population at $t = 0$ is $20,000$ .
3. We need to find the population at $t = 10$ months.

STEP 2

1. Set up the integral to find the population function.
2. Integrate the rate of change function.
3. Solve for the constant of integration using the initial condition.
4. Calculate the population at $t = 10$ months.

STEP 3

Set up the integral to find the population function:
$P(t) = \int (2 + t^{4/5}) \, dt$

STEP 4

Integrate the rate of change function:
$P(t) = \int (2 + t^{4/5}) \, dt$ $= \int 2 \, dt + \int t^{4/5} \, dt$
$= 2t + \frac{t^{9/5}}{9/5} + C$ $= 2t + \frac{5}{9}t^{9/5} + C$

STEP 5

Solve for the constant of integration using the initial condition $P(0) = 20000$ :
$P(0) = 2(0) + \frac{5}{9}(0)^{9/5} + C = 20000$ $C = 20000$
Thus, the population function is:
$P(t) = 2t + \frac{5}{9}t^{9/5} + 20000$

STEP 6

Calculate the population at $t = 10$ months:
$P(10) = 2(10) + \frac{5}{9}(10)^{9/5} + 20000$
Calculate each term:
$= 20 + \frac{5}{9}(10)^{9/5} + 20000$
Calculate $(10)^{9/5}$ :
$(10)^{9/5} \approx 25.1189$
Substitute back:
$= 20 + \frac{5}{9} \times 25.1189 + 20000$ $= 20 + 13.9544 + 20000$ $\approx 20033.9544$
Therefore, the population in 10 months is approximately:
$\boxed{20034}$

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