The concentration, C, of phenylbutazone, in micrograms per milliliter $(\mu \mathrm{g} / \mathrm{mL})$, in the plasma of a calf injected with this anti-inflammatory agent is given approximately by $\mathrm{C}(\mathrm{t})=42.52 e^{-0.01055 t}$, where $\mathrm{t}$ is the number of hours after the injection and $0 \leq t \leq 120$.
a) Given that the model is accurate for $0 \leq t \leq 120$, what is the initial dosage?
b) What is the average amount of phenylbutazone in the calf's body for the time between 5 and 120 hours?
Answer
So, the average amount of phenylbutazone in the calf's body for the time between 5 and 120 hours is approximately \(\boxed{23.36}\) micrograms per milliliter.
Step 1 ：The concentration, C, of phenylbutazone, in micrograms per milliliter $(\mu \mathrm{g} / \mathrm{mL})$, in the plasma of a calf injected with this anti-inflammatory agent is given approximately by $\mathrm{C}(\mathrm{t})=42.52 e^{-0.01055 t}$, where $\mathrm{t}$ is the number of hours after the injection and $0 \leq t \leq 120$.
Step 2 ：For part a), the initial dosage is the concentration at time t=0. We can find this by substitifying t=0 into the given equation.
Step 3 ：\(C(0) = 42.52 e^{-0.01055 \times 0} = 42.52\)
Step 4 ：So, the initial dosage is \(\boxed{42.52}\) micrograms per milliliter.
Step 5 ：For part b), the average amount of phenylbutazone in the calf's body for the time between 5 and 120 hours can be found by integrating the function from 5 to 120 and then dividing by the length of the interval (120-5). This is because the average value of a function on an interval [a, b] is given by (1/(b-a)) * integral from a to b of f(x) dx.
Step 6 ：\(\frac{1}{120-5} \int_{5}^{120} 42.52 e^{-0.01055 t} dt = 23.364006276623954\)
Step 7 ：So, the average amount of phenylbutazone in the calf's body for the time between 5 and 120 hours is approximately \(\boxed{23.36}\) micrograms per milliliter.
