Question 1157354
The size of certain insect population is given by

{{{P(t) =2000 e^(0.01t)}}}, where {{{t}}} is measured in days.

a) How many insects were present initially?

 the initial amount is {{{P(0)=2000 }}}

b) what is the growth rate?

The growth rate is {{{0.01}}}

c) At what time will the population double?

 double amount is {{{P(t) =4000}}}

{{{4000 =2000 e^(0.01t)}}}

{{{4000/2000 =e^(0.01t)}}}

{{{e^(0.01t)=2}}}......take natural log of both sides

{{{ln(e^(0.01t))=ln(2)}}}

{{{(0.01t)ln(e)=ln(2)}}}

{{{0.01t=ln(2)/ln(e)}}}.........{{{ln(e)=1}}}

{{{0.01t=0.69314718}}}

{{{t=0.69314718/0.01}}}

{{{t}}}≈{{{69.315 }}}days


d) give a differential equation satisfied by {{{p(t)}}}?

take the constant out : {{{(a* f )}}}'={{{a}}}* {{{f}}}'

{{{P}}}'{{{(t) =2000 *(d/dt)(e^(0.01t))}}}

apply the chain rule:{{{(df(u)/dt)=(df/du)*(du/dx)}}}

{{{f=e^u}}}, {{{u=0.01t}}}

{{{P}}}'{{{(t) =2000 *(d/du)(e^u)*(d/dt)(0.01t)}}}

{{{(d/du)(e^u)=e^u}}}

{{{(d/dt)(0.01t)=0.01}}}

substitute {{{u}}} back


{{{P}}}'{{{(t)=2000*e^(0.01t)*0.01}}}


{{{P}}}'{{{(t) = 20*e^(0.01t)}}}


d) how fast the bacteria is growing when it reaches {{{10000}}}?

{{{10000 =2000*e^(0.01t)}}}

{{{10000 /2000 =e^(0.01t)}}}

{{{5 =e^(0.01t)}}}

{{{ln(e^(0.01t))=ln(5)}}}

{{{(0.01t)ln(e)=ln(5)}}}

{{{0.01t=ln(5)}}}

{{{t=ln(5)/0.01}}}

{{{t}}}≈{{{160.944 }}}days