Question 1162793: The size P of a certain insect population at time t (in days) obeys the function P(t)=900e^0.03t
(a) Determine the number of insects at t=0 days.
(b) What is the growth rate of the insect population?
(c) What is the population after 10 days?
(d) When will the insect population reach 1440?
(e) When will the insect population double?
Found 2 solutions by Boreal, Theo:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! growth rate is 3% the 0.03 in the e^rt, where r is the rate of growth.
in 10 days the population is 900*e^0.3=1214.87 or 1215
reaches 1440 when 1440=900*e^0.03t
1.6=e^0.03 t
ln both sides
0.47=0.03t
t=15 2/3 days
doubles in 24 days using the rule of 72, but that is empirical
2=e^0.03t
ln2=0.03t
0.693=0.03t
t=23.1 days
Some use the rule of 70, which would give 23.3 days and is a better estimate.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! p(t) = 900 * e^(.03*t)
p(t) = number of insects at t days.
t = number of days.
at t = 0, the population is equal to 900 * e^(.03*0) = 900 * e^0 = 900 * 1 = 900.
at t = 10, the population is equal to 900 * e^(.03*10) = 900 * e^(.3) = 1214.872927.
since you can't have part of an ant, round the answer to the nearest integer = 1215.
the formula for the number of days when the population will be equal to 1414 is 1440 = 900 * e^(.03*t)
divide both sides of this equation by 900 to get:
1440/900 = e^(.03*t)
take the natural log of both sides of the equation to get ln(1440/900) = ln(e^(.03*t))
since ln(e^x) = x*ln(e), this becomes ln(1440/900) = .03*t * ln(e)
since ln(e) = 1, this becomes ln(1440/900) = .03*t
solve for t to get t = ln(1440/900) / .03 = 15.66678764 days.
to confirm this is correct, replace t with 15.0594361 in the original equation to get p(t) = 900 * e^(.03 * 15.66678764) = 1440.
the insect population will double when it becomes 1800.
the formula for the number of days when the population will be equal to 1800 is 1800 = 900 * e^(.03*t)
divide both sides of this equation by 900 to get:
1800/900 = e^(.03*t)
take the natural log of both sides of the equation to get ln(1800/900) = ln(e^(.03*t))
since ln(e^x) = x*ln(e), this becomes ln(1800/900) = .03*t * ln(e)
since ln(e) = 1, this becomes ln(1800/900) = .03*t
solve for t to get t = ln(1800/900) / .03 = 23.10490602 days.
to confirm this is correct, replace t with 23.10490602 in the original equation to get p(t) = 900 * e^(.03 * 23.10490602) = 1800.
to find the daily growth rate, do the following.
start with 900.
multiply it by e^(.03 * 1 to get 927.4090806.
divide that by 900 to get 1.030454534.
subtract 1 from that to get .030454534 = 3.0454534%.
that's the daily growth rate.
to confirm, use that in the discrete growth rate formula of f = p * ( 1 + r) ^ n
f is the future value
p is the present value
r is the growth rate per time period.
n is the number of time periods.
for example:
the population in 10 days will be equal to 900 * (1 + .030454534) ^ 10 = 1214.872927.
this is the same result we got before with p(t) = 900 * e^(.03 * 10)
p(t) means the same as f.
they both represent the future value of the insect population.
p(t) and f tell you what the future value is.
p tells you what the present value is.
the discrete growth rate formula of f = p * (1 + .030454534) ^ n will give you the same answer as the continuous growth rate formula of p(t) = p * e^(.03 * t)
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