The population of a certain species of fish has a relative growth rate of 1.1% per year. It is estimated that the population in
2010 was 12 million. a. Find an exponential model n(t)= no e^rt for the population t years after 2010. b. Estimate the fish population in the year 2015. c. After how many years will the fish population reach 14 million? d. Sketch a graph of the fish population.
1 answer:
Answer:
(a) n(t) = P(0)*e^(0.010939940t)
(b) 12,674,681 (nearest unit)
(c) 14 years (nearest year)
Step-by-step explanation:
rate = 1.1% / year = 1.011
(a)
P(0) = 12,000,000 = population in 2010
In compound interest format, after t years
P(t) = P(0)* (1.011)^t
Given format = P(0)* e^(rt)
therefore
e^(rt) = 1.011^t use law of exponents
(e^r)^t = 1.011^t
e^r = 1.011
r = log_e(1.011) = 0.010939940 (to 9 decimal places)
required formula is
n(t) = P(0)*e^(0.010939940t)
(b)
in 2015,
P(0)=12000000, n = 5 (years after 2010)
n(5) = 12000000*e^( 0.010939940 * 5 ) = 12,674,680.6 = 12,674,681 (nearest unit)
(c)
to reach 14 million, we equate
n(t) = 14,000,000
12,000,000 *e^(0.010939940*t) = 14,000,000
e^(0.010939940*t) = 14000000/12000000 = 7/6
take log on both sides
0.010939940*t = log(7/6)
t = log(7/6) / 0.010939940 = 14.091 years = 14 years to the nearest year.
See graph attached. Y-axis is in millions, x-axis is in years.
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