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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Question
![**Differential Equations Problem: Solving for H(6)**
**Problem Statement:**
A rate of change of \( H \) is given by
\[ \frac{dH}{dt} = k(5 + H) \]
where:
- When \( t = 0 \), \( H = 4 \).
- When \( t = 3 \), \( H = 10 \).
**Objective:**
Solve for \( H(6) \).
**Multiple Choice Options:**
- \( \circ \) 9.52
- \( \circ \) 10.24
- \( \circ \) 15.765
- \( \circ \) 19.989
**Solution Steps:**
1. **First, identify the form of a differential equation:**
The differential equation given is a first-order linear ordinary differential equation of the form:
\[
\frac{dH}{dt} = k(5 + H)
\]
2. **Determine the constant \( k \):**
Use the initial conditions provided:
\[
H(0) = 4 \quad \text{and} \quad H(3) = 10
\]
3. **Solve the differential equation:**
\[
\frac{dH}{dt} = k(5 + H)
\]
One method to solve this is separation of variables. Integrate both sides:
\[
\int \frac{1}{5+H} \, dH = k \int \, dt
\]
After integrating:
\[
\ln|5+H| = kt + C
\]
Exponentiate both sides:
\[
5 + H = Ce^{kt}
\]
Use the initial condition \( H(0) = 4 \):
\[
5 + 4 = Ce^{k \cdot 0} \implies 9 = C
\]
Hence,
\[
5 + H = 9e^{kt}
\]
or
\[
H = 9e^{kt} - 5
\]
4. **Determine \( k \) using \( H(3) = 10 \):**
\[
10 = 9e^{3k](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0b1d198d-95a1-4d0b-9915-fffff0f3c2fd%2F19fe84f2-112e-45ea-9cb8-baeb58484055%2Fcuuxjhq_processed.png&w=3840&q=75)
Transcribed Image Text:**Differential Equations Problem: Solving for H(6)**
**Problem Statement:**
A rate of change of \( H \) is given by
\[ \frac{dH}{dt} = k(5 + H) \]
where:
- When \( t = 0 \), \( H = 4 \).
- When \( t = 3 \), \( H = 10 \).
**Objective:**
Solve for \( H(6) \).
**Multiple Choice Options:**
- \( \circ \) 9.52
- \( \circ \) 10.24
- \( \circ \) 15.765
- \( \circ \) 19.989
**Solution Steps:**
1. **First, identify the form of a differential equation:**
The differential equation given is a first-order linear ordinary differential equation of the form:
\[
\frac{dH}{dt} = k(5 + H)
\]
2. **Determine the constant \( k \):**
Use the initial conditions provided:
\[
H(0) = 4 \quad \text{and} \quad H(3) = 10
\]
3. **Solve the differential equation:**
\[
\frac{dH}{dt} = k(5 + H)
\]
One method to solve this is separation of variables. Integrate both sides:
\[
\int \frac{1}{5+H} \, dH = k \int \, dt
\]
After integrating:
\[
\ln|5+H| = kt + C
\]
Exponentiate both sides:
\[
5 + H = Ce^{kt}
\]
Use the initial condition \( H(0) = 4 \):
\[
5 + 4 = Ce^{k \cdot 0} \implies 9 = C
\]
Hence,
\[
5 + H = 9e^{kt}
\]
or
\[
H = 9e^{kt} - 5
\]
4. **Determine \( k \) using \( H(3) = 10 \):**
\[
10 = 9e^{3k
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