<PRE><B>Question 87056</B>
A) What was the initial amount of bacteria in the population?
Initial amount would be determined after 0 hours ... t = 0
{{{P(t) = 1270/(1 + 27.22e^(-0.348t))}}}
{{{P(0) = 1270/(1 + 27.22e^(0))}}}
{{{P(0) = 1270/(1 + 27.22)}}}
{{{P(0) = 1270/28.22}}}
Initial amount would be about 45
B)After how many hours is the population of bacteria 1000? Round to the nearest hour.
{{{P(t) = 1270/(1 + 27.22e^(-0.348t))}}}
{{{1000 = 1270/(1 + 27.22e^(-0.348t))}}}
{{{1 + 27.22e^(-0.348t) = 1270/1000}}}
{{{1 + 27.22e^(-0.348t) = 1.27}}}
{{{27.22e^(-0.348t) = 0.27}}}
{{{e^(-0.348t) = 0.27/27.22}}}
{{{log(e,e^(-0.348t)) = log(e,0.27/27.22)}}}
{{{-0.348t = log(e,0.27/27.22)}}}
{{{t = -log(e,0.27/27.22)/0.348}}}
C) What is the limitig size of P(t), the poluation of bacterium?
{{{P(t) = 1270/(1 + 27.22e^(-0.348t))}}}
Firstly, we would have to map out 1 + 27.22e^(-0.348t). Since time is positive, we must see how 1 + 27.22e^(-0.348t) reacts as {{{t}}} increases positively.
Red: 1 + 27.22/e^(0.348t)
Green: 1270/(1 + 27.22e^(-0.348t))
{{{graph(300,800,-20,50,-2,1400,1 + 27.22/2.718^(0.348x),1270/(1 + 27.22/(2.718)^(0.348x)),1270)}}}
1270 is being divided by an exponentially smaller number every hour
as 1 + 27.22/e^(0.348t) approaches infinity for time ... it equals 1
1270 / 1 is the limit</PRE>