Question

the mass of a substance, which follows a continuous exponential growth model, is being studied in a lab. the doubling time for this substance was observed to be 11 minutes. there were 45.6 mg of the substance present at the beginning of the study. (a) let t be the time (in minutes) since the beginning of the study, and let y be the amount of the substance 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 much will be present in 12 minutes? do not round any intermediate computations, and round your answer to the nearest tenth. mg

Explanation:

Step1: Recall exponential - growth formula

The general form of 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. Given that $y_0 = 45.6$ mg. When $t = 11$ (doubling - time), $y = 2y_0=91.2$ mg.

Step2: Find the growth constant $k$

Substitute $y_0 = 45.6$, $y = 91.2$, and $t = 11$ into $y = y_0e^{kt}$:
$91.2=45.6e^{11k}$. Divide both sides by $45.6$: $\frac{91.2}{45.6}=e^{11k}$, so $2 = e^{11k}$. Take the natural logarithm of both sides: $\ln(2)=\ln(e^{11k})$. Since $\ln(e^{x}) = x$, we have $\ln(2)=11k$. Then $k=\frac{\ln(2)}{11}$.

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

Substitute $y_0 = 45.6$ and $k=\frac{\ln(2)}{11}$ into $y = y_0e^{kt}$: $y = 45.6e^{\frac{\ln(2)}{11}t}$.

Step4: Find the amount at $t = 12$

Substitute $t = 12$ into $y = 45.6e^{\frac{\ln(2)}{11}t}$:
$y = 45.6e^{\frac{\ln(2)}{11}\times12}$. First, $\frac{\ln(2)}{11}\times12=\frac{12\ln(2)}{11}$. Since $a\ln(b)=\ln(b^{a})$, $\frac{12\ln(2)}{11}=\ln(2^{\frac{12}{11}})$. Then $y = 45.6e^{\ln(2^{\frac{12}{11}})}$. Since $e^{\ln(x)} = x$, $y = 45.6\times2^{\frac{12}{11}}$.
$2^{\frac{12}{11}}=2^{1+\frac{1}{11}}=2\times2^{\frac{1}{11}}$. Calculate $2^{\frac{1}{11}}\approx1.065$. Then $2^{\frac{12}{11}}\approx2\times1.065 = 2.13$. So $y = 45.6\times2.13=97.128\approx97.1$ mg.

Answer:

(a) $y = 45.6e^{\frac{\ln(2)}{11}t}$
(b) $97.1$ mg