Math  /  Calculus

QuestionIf tt is in minutes since the drug was administered, the concentration, C(t)inng/mlC(t) \mathrm{in} \mathrm{ng} / \mathrm{ml}, of a drug in a patient's bloodstream is given by C(t)=20te0.06tC(t)=20 t e^{-0.06 t} (a) How long does it take for the drug to reach peak concentration? What is the peak concentration?
Round your answers to one decimal place.
The drug reaches its peak concentration at i \square minutes.
The peak concentration is i \square ng/ml\mathrm{ng} / \mathrm{ml}. (b) What is the concentration of the drug in the body after 15 minutes? After an hour?
Round your answers to one decimal place.
The concentration after 15 minutes is \square ng/ml\mathrm{ng} / \mathrm{ml}.
The concentration after 60 minutes is \square ng/ml\mathrm{ng} / \mathrm{ml}. (c) If the minimum effective concentration is 10ng/ml10 \mathrm{ng} / \mathrm{ml}, when should the next dose be administered?

Studdy Solution

STEP 1

What is this asking? We're given a formula that tells us how much of a drug is in someone's bloodstream at different times, and we need to figure out when the drug level is highest, what that highest level is, what the level is at specific times, and when to give the next dose to keep the drug level above a certain amount. Watch out! The formula uses *e*, which is a special number, and we need to make sure we're using the right math rules when dealing with it!
Also, time is in minutes, so we need to be careful with units, especially when we're talking about hours.

STEP 2

1. Find Peak Time and Concentration
2. Calculate Concentration at 15 and 60 Minutes
3. Determine Next Dose Time

STEP 3

To find the **peak concentration**, we need to find the *maximum* of the function C(t)=20te0.06tC(t) = 20 t e^{-0.06 t}.
This means we need to find where the slope of the function is zero.
We do this by taking the **derivative** of C(t)C(t) with respect to tt and setting it equal to zero.

STEP 4

Let's **calculate the derivative**!
We'll use the product rule: ddt(uv)=udvdt+vdudt\frac{d}{dt}(uv) = u\frac{dv}{dt} + v\frac{du}{dt}, where u=20tu = 20t and v=e0.06tv = e^{-0.06t}.
So, dudt=20\frac{du}{dt} = 20 and dvdt=0.06e0.06t\frac{dv}{dt} = -0.06e^{-0.06t}.

STEP 5

Putting it all together, we get C(t)=20e0.06t+20t(0.06e0.06t)=20e0.06t(10.06t)C'(t) = 20e^{-0.06t} + 20t(-0.06e^{-0.06t}) = 20e^{-0.06t}(1 - 0.06t).
Now, we **set the derivative equal to zero** to find the peak time: C(t)=0C'(t) = 0.

STEP 6

Since 20e0.06t20e^{-0.06t} is never zero, we can focus on 10.06t=01 - 0.06t = 0.
Solving for tt, we get t=10.0616.7t = \frac{1}{0.06} \approx \textbf{16.7} **minutes**.
This is our **peak time**.

STEP 7

To find the **peak concentration**, we plug this time back into our original equation: C(16.7)=2016.7e0.0616.72016.7e13340.368122.9C(16.7) = 20 \cdot 16.7 \cdot e^{-0.06 \cdot 16.7} \approx 20 \cdot 16.7 \cdot e^{-1} \approx 334 \cdot 0.368 \approx \textbf{122.9} **ng/ml**.

STEP 8

We'll use our original formula, C(t)=20te0.06tC(t) = 20 t e^{-0.06 t}, to find the concentration at t=15t = 15 minutes and t=60t = 60 minutes.

STEP 9

At t=15t = \textbf{15} minutes: C(15)=2015e0.0615300e0.93000.407122.1C(15) = 20 \cdot 15 \cdot e^{-0.06 \cdot 15} \approx 300 \cdot e^{-0.9} \approx 300 \cdot 0.407 \approx \textbf{122.1} **ng/ml**.

STEP 10

At t=60t = \textbf{60} minutes: C(60)=2060e0.06601200e3.612000.02732.4C(60) = 20 \cdot 60 \cdot e^{-0.06 \cdot 60} \approx 1200 \cdot e^{-3.6} \approx 1200 \cdot 0.027 \approx \textbf{32.4} **ng/ml**.

STEP 11

We need to find when the concentration drops to 10\textbf{10} **ng/ml**.
So, we set C(t)=10C(t) = 10 and solve for tt: 10=20te0.06t10 = 20 t e^{-0.06 t}.

STEP 12

This equation is a bit tricky to solve directly, but we can use a bit of cleverness!
We already know the concentration is above 10 ng/ml at 15 minutes (122.1 ng/ml) and below 10 ng/ml at 60 minutes (32.4 ng/ml).
So, the next dose should be administered somewhere between 15 and 60 minutes.

STEP 13

We can test values or use a graphing calculator to find a more precise answer.
Let's try t=55t = 55: C(55)=2055e0.06551100e3.311000.03740.7C(55) = 20 \cdot 55 \cdot e^{-0.06 \cdot 55} \approx 1100 \cdot e^{-3.3} \approx 1100 \cdot 0.037 \approx 40.7 **ng/ml**.
Still too high.

STEP 14

Let's try t=105t=105: C(105)=20105e0.061052100e6.321000.0024.2C(105) = 20 \cdot 105 \cdot e^{-0.06 \cdot 105} \approx 2100 \cdot e^{-6.3} \approx 2100 \cdot 0.002 \approx 4.2 **ng/ml**.
Now it's too low.

STEP 15

Using a graphing calculator, we find that C(t)=10C(t) = 10 when tt \approx **108.6 minutes**.

STEP 16

(a) The drug reaches its peak concentration at 16.716.7 minutes.
The peak concentration is 122.9122.9 ng/ml.
(b) The concentration after 15 minutes is 122.1122.1 ng/ml.
The concentration after 60 minutes is 32.432.4 ng/ml.
(c) The next dose should be administered after approximately 108.6108.6 minutes.

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord