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

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Chapter1: Functions And Models
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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:
**\( k = 0.347 \)**.
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: **\( k = 0.347 \)**.
**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
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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