Question

the area covered by a certain population of bacteria increases according to a continuous exponential growth model. suppose that a sample culture has an initial area of 55.6 mm² and an observed doubling - time of 9 hours. (a) let t be the time (in hours) passed, and let y be the area of the sample 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) what will the area of the sample be in 14 hours? do not round any intermediate computations, and round your answer to the nearest tenth. mm²

Explanation:

Step1: Recall exponential - growth formula

The general formula for continuous exponential growth is $y = y_0e^{kt}$, where $y_0$ is the initial amount, $k$ is the growth constant, and $t$ is the time. We know that $y_0=55.6$. When $t = 9$ (doubling - time), $y = 2y_0$. Substitute these values into the formula: $2y_0=y_0e^{9k}$.

Step2: Solve for $k$

Since $y_0
eq0$ (because $y_0 = 55.6$), we can divide both sides of the equation $2y_0=y_0e^{9k}$ by $y_0$. We get $2=e^{9k}$. Take the natural logarithm of both sides: $\ln(2)=\ln(e^{9k})$. Using the property $\ln(e^x)=x$, we have $\ln(2) = 9k$. Then $k=\frac{\ln(2)}{9}$.

Step3: Write the formula relating $y$ to $t$

Substitute $y_0 = 55.6$ and $k=\frac{\ln(2)}{9}$ into the formula $y = y_0e^{kt}$. So $y = 55.6e^{\frac{\ln(2)}{9}t}$.

Step4: Find the area at $t = 14$ hours

Substitute $t = 14$ into the formula $y = 55.6e^{\frac{\ln(2)}{9}t}$. First, $\frac{\ln(2)}{9}\times14=\frac{14\ln(2)}{9}$. Then $y = 55.6e^{\frac{14\ln(2)}{9}}$. Since $a\ln(b)=\ln(b^a)$, $\frac{14\ln(2)}{9}=\ln(2^{\frac{14}{9}})$. So $y = 55.6\times2^{\frac{14}{9}}$. Calculate $2^{\frac{14}{9}}\approx3.0379$. Then $y = 55.6\times3.0379\approx168.9$.

Answer:

(a) $y = 55.6e^{\frac{\ln(2)}{9}t}$
(b) $168.9$