3. ( 10 marks) Penicillin absorption The concentration of penicillin in the blood (in $$ \mu \mathrm{g} / \mathrm{mL} $$ ) varies with time ( $$ t $$ in hours) according to the relationship, $$ c(t)=5 e^{-0.2 t}-5 e^{-0.3 t} $$ a) What is the rate of absorption at time $$ t=1 $$ ? b) What is the rate of absorption at time $$ t=8 $$ ? c) When is the concentration of penicillin highest?

Question

UpStudy AI · Accepted Answer

$$
\textbf{Given:}\quad c(t)=5e^{-0.2t}-5e^{-0.3t}
$$

$$
\textbf{Step 1: Differentiate } c(t) \textbf{ to find the rate of absorption } c'(t).
$$

Differentiate term by term:

$$
\frac{d}{dt}\left[5e^{-0.2t}\right] = 5\left(-0.2e^{-0.2t}\right) = -e^{-0.2t}
$$
$$
\frac{d}{dt}\left[-5e^{-0.3t}\right] = -5\left(-0.3e^{-0.3t}\right) = 1.5e^{-0.3t}
$$

Thus, the derivative is:

$$
c'(t) = -e^{-0.2t} + 1.5e^{-0.3t}
$$

---

$$
\textbf{(a) Rate of absorption at } t=1:
$$

Substitute $$ t=1 $$ into $$ c'(t) $$:

$$
c'(1)= -e^{-0.2(1)} + 1.5e^{-0.3(1)} = -e^{-0.2} + 1.5e^{-0.3}
$$

Evaluating the exponentials (to 5 decimal places):

$$
e^{-0.2} \approx 0.81873, \quad e^{-0.3} \approx 0.74082
$$

Thus:

$$
c'(1) \approx -0.81873 + 1.5(0.74082) = -0.81873 + 1.11123 \approx 0.29250
$$

---

$$
\textbf{(b) Rate of absorption at } t=8:
$$

Substitute $$ t=8 $$ into $$ c'(t) $$:

$$
c'(8)= -e^{-0.2(8)} + 1.5e^{-0.3(8)} = -e^{-1.6} + 1.5e^{-2.4}
$$

Evaluating the exponentials:

$$
e^{-1.6} \approx 0.20190, \quad e^{-2.4} \approx 0.09072
$$

Thus:

$$
c'(8) \approx -0.20190 + 1.5(0.09072) = -0.20190 + 0.13608 \approx -0.06582
$$

---

$$
\textbf{(c) Time when the concentration is highest:}
$$

The maximum concentration occurs when $$ c'(t)=0 $$:

$$
-e^{-0.2t} + 1.5e^{-0.3t} = 0
$$

Solve for $$ t $$:

$$
1.5e^{-0.3t} = e^{-0.2t}
$$

Divide both sides by $$ e^{-0.3t} $$ (which is nonzero):

$$
1.5 = e^{-0.2t+0.3t} = e^{0.1t}
$$

Take natural logarithms:

$$
\ln(1.5)=0.1t \quad \Rightarrow \quad t = \frac{\ln(1.5)}{0.1}
$$

Evaluating:

$$
\ln(1.5) \approx 0.405465 \quad \Rightarrow \quad t \approx \frac{0.405465}{0.1} \approx 4.05465
$$

---

$$
\boxed{
\begin{aligned}
\text{(a)}\quad & c'(1) \approx 0.29250\ \mu\text{g/mL per hour,} \
\text{(b)}\quad & c'(8) \approx -0.06582\ \mu\text{g/mL per hour,} \
\text{(c)}\quad & \text{Maximum concentration at } t \approx 4.05\ \text{hours.}
\end{aligned}
}
$$
a) The rate of absorption at $$ t=1 $$ hour is approximately $$ 0.29250 \ \mu\text{g/mL per hour} $$. b) The rate of absorption at $$ t=8 $$ hours is approximately $$ -0.06582 \ \mu\text{g/mL per hour} $$. c) The concentration of penicillin is highest at approximately $$ 4.05 $$ hours.

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Calculus Questions