Question

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.

Answer 1

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.