Question

pulation 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=?

Answer 1

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:

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

Answer 2

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