In 2012, the growth of rate of the south african population was -0,412% Determine the population of South Africa in 2011 if the population was 48, 810, 427 in 2012

Guest Jul 29, 2015

#1**+10 **

Rate of growth is dN/dt = -0.412/100*N which can be solved to get N_{t} = N_{12}*e^{-0.412/100*t }where N_{12} = 48810427. (I've assumed the -0.412% means -0.412% per year)

So, to find N_{11} (the number in 2011) we set t = -1 (i.e. it's 1 year earlier) to get:

$${\mathtt{N11}} = {\mathtt{48\,810\,427}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{0.412}}}{{\mathtt{100}}}}{\mathtt{\,\times\,}}\left(-{\mathtt{1}}\right)\right)} \Rightarrow {\mathtt{N11}} = {\mathtt{49\,011\,940.792\: \!604\: \!870\: \!237\: \!79}}$$

or N_{11} ≈ 49011941

.

Alan Jul 29, 2015

#1**+10 **

Best Answer

Rate of growth is dN/dt = -0.412/100*N which can be solved to get N_{t} = N_{12}*e^{-0.412/100*t }where N_{12} = 48810427. (I've assumed the -0.412% means -0.412% per year)

So, to find N_{11} (the number in 2011) we set t = -1 (i.e. it's 1 year earlier) to get:

$${\mathtt{N11}} = {\mathtt{48\,810\,427}}{\mathtt{\,\times\,}}{{\mathtt{e}}}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{0.412}}}{{\mathtt{100}}}}{\mathtt{\,\times\,}}\left(-{\mathtt{1}}\right)\right)} \Rightarrow {\mathtt{N11}} = {\mathtt{49\,011\,940.792\: \!604\: \!870\: \!237\: \!79}}$$

or N_{11} ≈ 49011941

.

Alan Jul 29, 2015