Let A(t)=15(0.8)^0.2t (in milligrams) be the amount of drug in a patients body at time where t is the number of hours after the drug was given to the patient. Find the instantaneous rate if change of the amount of drug in the patients body 6 hours after drug was
Let A(t)=15(0.8)^0.2t (in milligrams) be the amount of drug in a patients body at time where t is the number of hours after the drug was given to the patient. Find the instantaneous rate if change of the amount of drug in the patients body 6 hours after drug was
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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![### Exponential and Logarithmic Calculus Problem
**Problem Statement:**
Given the function \( A(t) = 15(0.8)^{0.2t} \) (in milligrams), which represents the amount of drug in a patient's body at time \( t \), where \( t \) is the number of hours after the drug was administered:
- Find the instantaneous rate of change of the amount of drug in the patient's body 6 hours after the drug was given.
- Provide the answer in appropriate units.
**Solution Approach:**
To find the instantaneous rate of change (which is the derivative) of the function \( A(t) \) at \( t = 6 \) hours, follow these steps:
1. **Function Given:**
\[
A(t) = 15(0.8)^{0.2t}
\]
2. **Differentiate \( A(t) \):**
- Use the chain rule and the properties of exponential functions to differentiate.
Let’s denote \( A(t) \) in terms of a simpler variable:
\[
A(t) = 15 \cdot e^{0.2t \ln(0.8)}
\]
Now, differentiate \( A(t) \):
\[
\frac{dA(t)}{dt} = 15 \cdot \left(0.2 \ln(0.8) \cdot e^{0.2t \ln(0.8)}\right)
\]
Therefore,
\[
\frac{dA(t)}{dt} = 15 \cdot 0.2 \ln(0.8) \cdot (0.8)^{0.2t}
\]
3. **Evaluate the Derivative at \( t = 6 \):**
\[
\frac{dA}{dt} \Bigg|_{t=6} = 15 \cdot 0.2 \ln(0.8) \cdot (0.8)^{1.2}
\]
4. **Simplify the Expression:**
- \( \ln(0.8) \) is a constant.
- \( (0.8)^{1.2} \) can be calculated for the specific value.
\(\ln(0.8) \approx](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F77922221-9e32-493b-ab71-d15b7753ec9e%2Ffd2f7a46-c8e2-408d-a2c1-66fae14fe3a5%2Fs1hmygg_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Exponential and Logarithmic Calculus Problem
**Problem Statement:**
Given the function \( A(t) = 15(0.8)^{0.2t} \) (in milligrams), which represents the amount of drug in a patient's body at time \( t \), where \( t \) is the number of hours after the drug was administered:
- Find the instantaneous rate of change of the amount of drug in the patient's body 6 hours after the drug was given.
- Provide the answer in appropriate units.
**Solution Approach:**
To find the instantaneous rate of change (which is the derivative) of the function \( A(t) \) at \( t = 6 \) hours, follow these steps:
1. **Function Given:**
\[
A(t) = 15(0.8)^{0.2t}
\]
2. **Differentiate \( A(t) \):**
- Use the chain rule and the properties of exponential functions to differentiate.
Let’s denote \( A(t) \) in terms of a simpler variable:
\[
A(t) = 15 \cdot e^{0.2t \ln(0.8)}
\]
Now, differentiate \( A(t) \):
\[
\frac{dA(t)}{dt} = 15 \cdot \left(0.2 \ln(0.8) \cdot e^{0.2t \ln(0.8)}\right)
\]
Therefore,
\[
\frac{dA(t)}{dt} = 15 \cdot 0.2 \ln(0.8) \cdot (0.8)^{0.2t}
\]
3. **Evaluate the Derivative at \( t = 6 \):**
\[
\frac{dA}{dt} \Bigg|_{t=6} = 15 \cdot 0.2 \ln(0.8) \cdot (0.8)^{1.2}
\]
4. **Simplify the Expression:**
- \( \ln(0.8) \) is a constant.
- \( (0.8)^{1.2} \) can be calculated for the specific value.
\(\ln(0.8) \approx
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