ate of change of S is directly proportional to S(x), where S(0) = 10 and S(4) = 40 k, the proportionality constant. k=3.689 k=2.303 k=1.386 k=0.347
ate of change of S is directly proportional to S(x), where S(0) = 10 and S(4) = 40 k, the proportionality constant. k=3.689 k=2.303 k=1.386 k=0.347
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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![### 6.01 Semester Test: Calculus
#### Question:
The rate of change of S is directly proportional to S(x), where S(0) = 10 and S(4) = 40.
Find \( k \), the proportionality constant.
#### Options:
- \( k = 3.689 \)
- \( k = 2.303 \)
- \( k = 1.386 \)
- \( k = 0.347 \)
---
To answer this question, you need to use the concept of direct proportionality in calculus. Given that the rate of change of \( S \) is directly proportional to \( S(x) \), and knowing the values of \( S \) at specific points, you can calculate the proportionality constant \( k \).
### How to Approach the Problem:
1. **Understand the relationship**: Since the rate of change of \( S \) is directly proportional to \( S(x) \), we have:
\[
\frac{dS}{dx} = kS
\]
2. **Given values**: \( S(0) = 10 \) and \( S(4) = 40 \).
3. **Use the exponential growth model**: The solution to the differential equation is:
\[
S(x) = S(0) \cdot e^{kx}
\]
4. **Plug in the given values**:
When \( x = 0 \):
\[
S(0) = 10
\]
When \( x = 4 \):
\[
S(4) = 10 \cdot e^{4k} = 40
\]
5. **Solve for \( k \)**:
\[
40 = 10 \cdot e^{4k} \implies e^{4k} = 4 \implies 4k = \ln{4} \implies k = \frac{\ln{4}}{4}
\]
Calculating the value:
\[
\ln{4} \approx 1.386 \implies k \approx \frac{1.386}{4} \approx 0.347
\]
Therefore, the correct choice for the value of \( k \) is:
**\( k = 0.347 \)**.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1a92b920-e019-446e-b580-35fb69c82c25%2F6e429af4-a3c1-4477-9089-6be5cc818dfa%2Ff6ok8pw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### 6.01 Semester Test: Calculus
#### Question:
The rate of change of S is directly proportional to S(x), where S(0) = 10 and S(4) = 40.
Find \( k \), the proportionality constant.
#### Options:
- \( k = 3.689 \)
- \( k = 2.303 \)
- \( k = 1.386 \)
- \( k = 0.347 \)
---
To answer this question, you need to use the concept of direct proportionality in calculus. Given that the rate of change of \( S \) is directly proportional to \( S(x) \), and knowing the values of \( S \) at specific points, you can calculate the proportionality constant \( k \).
### How to Approach the Problem:
1. **Understand the relationship**: Since the rate of change of \( S \) is directly proportional to \( S(x) \), we have:
\[
\frac{dS}{dx} = kS
\]
2. **Given values**: \( S(0) = 10 \) and \( S(4) = 40 \).
3. **Use the exponential growth model**: The solution to the differential equation is:
\[
S(x) = S(0) \cdot e^{kx}
\]
4. **Plug in the given values**:
When \( x = 0 \):
\[
S(0) = 10
\]
When \( x = 4 \):
\[
S(4) = 10 \cdot e^{4k} = 40
\]
5. **Solve for \( k \)**:
\[
40 = 10 \cdot e^{4k} \implies e^{4k} = 4 \implies 4k = \ln{4} \implies k = \frac{\ln{4}}{4}
\]
Calculating the value:
\[
\ln{4} \approx 1.386 \implies k \approx \frac{1.386}{4} \approx 0.347
\]
Therefore, the correct choice for the value of \( k \) is:
**\( k = 0.347 \)**.
Transcribed Image Text:**6.01 Semester Test: Calculus**
---
Bob's grade in his science class is related to the number of hours he studies. The rate of change of his grade G(t) is proportional to (100 - G), where t is measured in hours and G(t) is his grade after t hours of study.
- The highest grade is a 100.
- G(0) = 40, meaning he could get a 40 without any studying.
- After 20 hours Bob's grade would be a 70.
**Question:**
How many hours would Bob need to study to get an A, 90 or better?
- Find your answer to the nearest hour.
**Options:**
- 20
- 43
- 52
- 75
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