Web Analytics
Answered

Assuming that Switzerland's population is growing exponentially at a continuous rate of 0.19 percent a year and that the 1988 population was 6.9 million, write an expression for the population as a function of time in years. (Let t=0 in 1988.)


P=?


hint: logistic function

Assuming that Switzerland's population is growing exponentially at a continuous rate of 0.19 percent a year and that the 1988 population was 6.9 million, write an expression for the population as a function of time in years. (Let t=0 in 1988.)


P=?

Verified Answer :

MissPhiladelphia

First, the formula would be like this dP / dt= KP.
Secondly, since dP is given which is 6.9. Substitute it from the equation.

P(0) = 6.9 x 10^6
integral 1 divided by pdp which will look like this,

integral 1/ PdP = integral kdt

the answer would be .0019 or
k=.0019

Answer:

\(P(t) = 6900000 e^{0.0019t}\)

Step-by-step explanation:

Equation for an exponential growth:

\(P(t) = P_{0} e^{kt}\).............(1)

Where \(P_{0}\) = The initial population of Switzerland at time t = 0

At t = 0 in 1988, the population was \(P_{0} = 6.9 million\)

k = 0.19% = 0.19/100 = 0.0019

The population as a function of time becomes:

\(P(t) = 6900000 e^{0.0019t}\)

Related Questions