an initial population of 340 fish is introduced into a lake. this fish population grows according to a…

an initial population of 340 fish is introduced into a lake. this fish population grows according to a continuous exponential growth model. there are 544 fish in the lake after 18 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 = e^(t) (b) how many fish are there 22 years after the initial population is introduced? do not round any intermediate computations, and round your answer to the nearest whole number. fish

an initial population of 340 fish is introduced into a lake. this fish population grows according to a continuous exponential growth model. there are 544 fish in the lake after 18 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 = e^(t) (b) how many fish are there 22 years after the initial population is introduced? do not round any intermediate computations, and round your answer to the nearest whole number. fish

Answer

Explanation:

Step1: Identificar la fórmula general

La fórmula para el crecimiento exponencial continuo es $y = y_0e^{kt}$, donde $y_0$ es la población inicial, $k$ es la tasa de crecimiento y $t$ es el tiempo. Dado que $y_0 = 340$, la fórmula es $y=340e^{kt}$.

Step2: Encontrar la tasa de crecimiento $k$

Sustituimos $y = 544$, $y_0 = 340$ y $t = 18$ en la fórmula $y = y_0e^{kt}$. Obtenemos $544=340e^{18k}$. Dividimos ambos lados por 340: $\frac{544}{340}=e^{18k}$, es decir $ \frac{8}{5}=e^{18k}$. Aplicamos logaritmo natural a ambos lados: $\ln(\frac{8}{5})=\ln(e^{18k})$. Como $\ln(e^{18k}) = 18k$, entonces $k=\frac{\ln(\frac{8}{5})}{18}$.

Step3: Escribir la fórmula final para (a)

Sustituyendo $k=\frac{\ln(\frac{8}{5})}{18}$ en $y = 340e^{kt}$, la fórmula es $y = 340e^{\frac{\ln(\frac{8}{5})}{18}t}$.

Step4: Resolver (b)

Sustituimos $t = 22$ en la fórmula $y = 340e^{\frac{\ln(\frac{8}{5})}{18}t}$. Tenemos $y = 340e^{\frac{\ln(\frac{8}{5})}{18}\times22}$. Primero, $\frac{\ln(\frac{8}{5})}{18}\times22=\frac{11}{9}\ln(\frac{8}{5})$. Entonces $y = 340e^{\frac{11}{9}\ln(\frac{8}{5})}$. Como $a\ln(b)=\ln(b^{a})$, entonces $\frac{11}{9}\ln(\frac{8}{5})=\ln((\frac{8}{5})^{\frac{11}{9}})$ y $y = 340(\frac{8}{5})^{\frac{11}{9}}$. Calculando, $y\approx607$.

Answer:

(a) $y = 340e^{\frac{\ln(\frac{8}{5})}{18}t}$ (b) 607