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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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**Understanding the Growth Rate of Protozoa**

**Objective**: Determine the population size of protozoa after 6 days given the initial conditions and growth rate.

**1. Given Data**:
- Initial Population (Day 0): 2 members
- Growth rate of protozoa: \( 0.8947 \text{ per day} \)

**2. Problem Statement**:
Find the population size after 6 days.

**3. Steps to Solve**:

**a. Differential Equation**:
\[ \frac{dp}{dt} = ?? \]
\[ \frac{dp}{dt} = 0.8947p \]

**b. Separation of Variables**:
Separate the variables p and t:
\[ \frac{1}{p} dp = 0.8947 dt \]

**c. Integration**:
Integrate both sides:
\[ \int \frac{1}{p} dp = \int 0.8947 dt \]

**d. Results of Integration**:
\[ \ln|p| = 0.8947t + C \]
Where C is the constant of integration.

**e. Solve for \( p \)**:
Rewriting the equation, we get:
\[ p = e^{0.8947t + C} \]
\[ p = e^C \cdot e^{0.8947t} \]

Let \( k = e^C \), then:
\[ p = k \cdot e^{0.8947t} \]

**f. Determine the Constant \( k \)**:
Using the initial condition \( p(0) = 2 \):
\[ 2 = k \cdot e^{0} \]
\[ k = 2 \]

Therefore, the population equation is:
\[ p(t) = 2 \cdot e^{0.8947t} \]

**4. Population Size After 6 Days**:
Plug in \( t = 6 \):
\[ p(6) = 2 \cdot e^{0.8947 \times 6} \]
\[ p(6) = 2 \cdot e^{5.3682} \]
\[ p(6) \approx 2 \cdot 213.6 \]
\[ p(6) \approx 427.2 \]

Thus, the population size after 6 days is approximately 427 protozoa.

---
Transcribed Image Text:**Understanding the Growth Rate of Protozoa** **Objective**: Determine the population size of protozoa after 6 days given the initial conditions and growth rate. **1. Given Data**: - Initial Population (Day 0): 2 members - Growth rate of protozoa: \( 0.8947 \text{ per day} \) **2. Problem Statement**: Find the population size after 6 days. **3. Steps to Solve**: **a. Differential Equation**: \[ \frac{dp}{dt} = ?? \] \[ \frac{dp}{dt} = 0.8947p \] **b. Separation of Variables**: Separate the variables p and t: \[ \frac{1}{p} dp = 0.8947 dt \] **c. Integration**: Integrate both sides: \[ \int \frac{1}{p} dp = \int 0.8947 dt \] **d. Results of Integration**: \[ \ln|p| = 0.8947t + C \] Where C is the constant of integration. **e. Solve for \( p \)**: Rewriting the equation, we get: \[ p = e^{0.8947t + C} \] \[ p = e^C \cdot e^{0.8947t} \] Let \( k = e^C \), then: \[ p = k \cdot e^{0.8947t} \] **f. Determine the Constant \( k \)**: Using the initial condition \( p(0) = 2 \): \[ 2 = k \cdot e^{0} \] \[ k = 2 \] Therefore, the population equation is: \[ p(t) = 2 \cdot e^{0.8947t} \] **4. Population Size After 6 Days**: Plug in \( t = 6 \): \[ p(6) = 2 \cdot e^{0.8947 \times 6} \] \[ p(6) = 2 \cdot e^{5.3682} \] \[ p(6) \approx 2 \cdot 213.6 \] \[ p(6) \approx 427.2 \] Thus, the population size after 6 days is approximately 427 protozoa. ---
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