Question

an initial population of 280 fish is introduced into a lake. this fish population grows according to a continuous exponential growth model. there are 588 fish in the lake after 15 years. (a) let t be the time (in years) since the initial population is introduced, and let y be the number of fish at time t. write a formula relating y to t. use exact expressions to fill in the missing parts of the formula. do not use approximations. y = (b) how many fish are there 16 years after the initial population is introduced? do not round any intermediate computations, and round your answer to the nearest whole number. fish

Explanation:

Step1: Recall exponential - growth formula

The continuous exponential - growth model is given by $y = y_0e^{rt}$, where $y_0$ is the initial amount, $r$ is the growth rate, and $t$ is the time. We know that $y_0 = 280$ (initial population of fish), when $t = 15$, $y=588$. Substitute these values into the formula: $588 = 280e^{15r}$.

Step2: Solve for $r$

First, divide both sides of the equation $588 = 280e^{15r}$ by 280: $\frac{588}{280}=e^{15r}$. Simplify $\frac{588}{280}=\frac{21}{10}$, so $\frac{21}{10}=e^{15r}$. Then take the natural - logarithm of both sides: $\ln(\frac{21}{10})=\ln(e^{15r})$. Since $\ln(e^{x}) = x$, we have $\ln(\frac{21}{10}) = 15r$. Solve for $r$: $r=\frac{\ln(\frac{21}{10})}{15}$.

Step3: Write the formula for $y$ in terms of $t$

Substitute $y_0 = 280$ and $r=\frac{\ln(\frac{21}{10})}{15}$ into the formula $y = y_0e^{rt}$, we get $y = 280e^{\frac{\ln(\frac{21}{10})}{15}t}$.

Step4: Find the number of fish at $t = 16$

Substitute $t = 16$ into the formula $y = 280e^{\frac{\ln(\frac{21}{10})}{15}t}$. First, calculate the exponent: $\frac{\ln(\frac{21}{10})}{15}\times16=\frac{16}{15}\ln(\frac{21}{10})$. Then $y = 280e^{\frac{16}{15}\ln(\frac{21}{10})}$. Since $a\ln(b)=\ln(b^{a})$, we have $y = 280e^{\ln((\frac{21}{10})^{\frac{16}{15}})}$. And since $e^{\ln(x)} = x$, $y = 280\times(\frac{21}{10})^{\frac{16}{15}}$.
$(\frac{21}{10})^{\frac{16}{15}}=\sqrt[15]{(\frac{21}{10})^{16}}=\frac{21^{\frac{16}{15}}}{10^{\frac{16}{15}}}$. Calculate $21^{\frac{16}{15}}\approx21^{1.067}\approx23.57$ and $10^{\frac{16}{15}}\approx10^{1.067}\approx11.61$. Then $(\frac{21}{10})^{\frac{16}{15}}\approx\frac{23.57}{11.61}\approx2.03$. So $y = 280\times2.03 = 568.4\approx568$.

Answer:

(a) $y = 280e^{\frac{\ln(\frac{21}{10})}{15}t}$
(b) 568