A culture of bacteria in a particular dish has an initial population of 400 cells grows at a rate of
cells/day.
a. Find the population at any time .
b. What is the population after 17 days?
a. Find a formula for the number of cells, , after t days.

(Round any numbers in exponents to five decimal places. Round all other numbers to the nearest tenth.)

Ask by Ford Pritchard. in the United States

Mar 20,2025

Question

A culture of bacteria in a particular dish has an initial population of 400 cells grows at a rate of
cells/day.
a. Find the population at any time .
b. What is the population after 17 days?
a. Find a formula for the number of cells, , after t days.

(Round any numbers in exponents to five decimal places. Round all other numbers to the nearest tenth.)

Ask by Ford Pritchard. in the United States

Mar 20,2025

UpStudy AI · Accepted Answer

To solve the problem, we will follow these steps:

### Part a: Find the population $$ N(t) $$

1. **Given the rate of growth**: 
   $$
   N'(t) = 50 e^{0.29863 t}
   $$

2. **To find $$ N(t) $$**, we need to integrate $$ N'(t) $$:
   $$
   N(t) = \int N'(t) \, dt = \int 50 e^{0.29863 t} \, dt
   $$

3. **Perform the integration**:
   - The integral of $$ e^{kt} $$ is $$ \frac{1}{k} e^{kt} + C $$, where $$ k $$ is a constant.

4. **Calculate the integral**:
   $$
   N(t) = 50 \cdot \frac{1}{0.29863} e^{0.29863 t} + C
   $$

5. **Substituting the initial condition**: 
   - At $$ t = 0 $$, $$ N(0) = 400 $$:
   $$
   N(0) = 50 \cdot \frac{1}{0.29863} e^{0} + C = 400
   $$

Now, let's calculate the integral and find the constant $$ C $$.

### Part b: Find the population after 17 days

1. **Substitute $$ t = 17 $$ into the formula for $$ N(t) $$**.

Let's perform the calculations for both parts.

First, we will calculate the integral and find $$ C $$. Then, we will use that to find $$ N(17) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$e^{0.29863\times 17}$$
- step1: Multiply the numbers:
  $$e^{5.07671}$$
- step2: Convert the expressions:
  $$e^{\frac{507671}{100000}}$$
Calculate or simplify the expression $$ 50 / 0.29863 $$.
Calculate the value by following steps:
- step0: Calculate:
  $$\frac{50}{0.29863}$$
- step1: Convert the expressions:
  $$\frac{50}{\frac{29863}{100000}}$$
- step2: Multiply by the reciprocal:
  $$50\times \frac{100000}{29863}$$
- step3: Multiply:
  $$\frac{50\times 100000}{29863}$$
- step4: Multiply:
  $$\frac{5000000}{29863}$$
### Part a: Find the population $$ N(t) $$

1. **Calculate the integral**:
   $$
   \frac{50}{0.29863} \approx 167.431269
   $$
   Thus, we have:
   $$
   N(t) = 167.431269 e^{0.29863 t} + C
   $$

2. **Using the initial condition $$ N(0) = 400 $$**:
   $$
   400 = 167.431269 e^{0} + C
   $$
   $$
   C = 400 - 167.431269 \approx 232.568731
   $$

3. **Final formula for $$ N(t) $$**:
   $$
   N(t) = 167.431269 e^{0.29863 t} + 232.568731
   $$

### Part b: Find the population after 17 days

1. **Calculate $$ e^{0.29863 \cdot 17} $$**:
   $$
   e^{0.29863 \cdot 17} \approx 160.245978
   $$

2. **Substituting $$ t = 17 $$ into the formula**:
   $$
   N(17) = 167.431269 \cdot 160.245978 + 232.568731
   $$

Let's calculate $$ N(17) $$.
Calculate the value by following steps:
- step0: Calculate:
  $$167.431269\times 160.245978+232.568731$$
- step1: Multiply the numbers:
  $$\frac{13415093724343041}{500000000000}+232.568731$$
- step2: Convert the expressions:
  $$\frac{13415093724343041}{500000000000}+\frac{232568731}{1000000}$$
- step3: Reduce fractions to a common denominator:
  $$\frac{13415093724343041}{500000000000}+\frac{232568731\times 500000}{1000000\times 500000}$$
- step4: Multiply the numbers:
  $$\frac{13415093724343041}{500000000000}+\frac{232568731\times 500000}{500000000000}$$
- step5: Transform the expression:
  $$\frac{13415093724343041+232568731\times 500000}{500000000000}$$
- step6: Multiply the numbers:
  $$\frac{13415093724343041+116284365500000}{500000000000}$$
- step7: Add the numbers:
  $$\frac{13531378089843041}{500000000000}$$
### Final Results

1. **Formula for the number of cells $$ N(t) $$**:
   $$
   N(t) = 167.43127 e^{0.29863 t} + 232.56873
   $$

2. **Population after 17 days**:
   $$
   N(17) \approx 27062.8
   $$

Thus, the population after 17 days is approximately $$ 27062.8 $$ cells.
The population $$ N(t) $$ at any time $$ t $$ is: $$ N(t) = 167.43127 e^{0.29863 t} + 232.56873 $$ After 17 days, the population is approximately 27,062.8 cells.

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Solución

Part a: Find the population

Part b: Find the population after 17 days

Part a: Find the population

Part b: Find the population after 17 days

Final Results

The Deep Dive

a. Finding the Population

b. Finding the Population After 17 Days

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