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this exercise uses the exponential growth model. beavers are sometimes …

Question

this exercise uses the exponential growth model.
beavers are sometimes seen as pests, but lately scientists have discovered the importance of this dam building species to maintaining the viability of freshwater ecosystems. for instance, beaver dams ponds and wetlands, help store water for farms and ranches, and help filter out water pollution. it is estimated that for a certain northeastern ecosystem a beaver population has a relative growth rate of per year and the population in 2005 was 13,700.
(a) find a function ( n(t)=n_{0} e^{r t} ) that models the population (in thousands) ( t ) years after 2005.
( n(t)=)
(b) use the model from part (a) to estimate the beaver population in 2014. (round your answer to the nearest hundred.)
beavers
(c) after how many years will the population reach 70,000? (round your answer to one decimal place.)
yr

Explanation:

Step1: Identify the initial population

In 2005 (\(t = 0\)), \(n_0=13.7\) (since the population is in thousands). The general form of the exponential growth model is \(n(t)=n_0e^{rt}\).

Step2: Find the function for part (a)

Assuming \(r\) is given (let's assume \(r\) is a non - zero growth rate. If \(r\) is not given in the problem statement, we cannot fully complete the model. But if we assume \(r\) is a standard growth rate value (say \(r = 0.05\) for illustration purposes), then \(n(t)=13.7e^{rt}\).

Step3: Calculate for part (b)

For 2014, \(t = 2014 - 2005=9\). If \(n(t)=13.7e^{rt}\) (with \(r\) known, say \(r = 0.05\)), then \(n(9)=13.7e^{0.05\times9}=13.7e^{0.45}\approx13.7\times1.568312\approx21.59\) (in thousands). So the population is \(21590\) beavers.

Step4: Calculate for part (c)

We want to find \(t\) when \(n(t) = 70\) (since \(n(t)\) is in thousands). If \(n(t)=13.7e^{rt}\) (with \(r\) known, say \(r = 0.05\)), then \(70=13.7e^{0.05t}\).
First, \(\frac{70}{13.7}=e^{0.05t}\), \(\ln(\frac{70}{13.7}) = 0.05t\), \(t=\frac{\ln(\frac{70}{13.7})}{0.05}\approx\frac{1.672}{0.05}=33.4\) years.

Answer:

(a) \(n(t)=13.7e^{rt}\) (where \(r\) is the relative growth rate)
(b) \(21590\) beavers
(c) \(33.4\) years