A bacteria gives birth to two new bacterias in each second and the life span of each bacteria is 5 seconds. The process of the reproduction is continuous untill the death of the bacteria . Initially there is one newly born bacteria at time t=0 , then find the total number of live bacteria just after 10 seconds :

Options

A. a. `(3^(10))/(2)`
B. b. `3^5 (3^5 -1)`
C. c. `3^(10)-2^(10)`
D. d. `3^(10)-2^(5)`

Answer

To solve the problem of how many live bacteria there are just after 10 seconds, we can break it down step by step.

### Step 1: Understand the reproduction and lifespan of the bacteria
- Initially, at time \( t = 0 \), there is 1 bacteria.
- Each bacteria gives birth to 2 new bacteria every second.
- The lifespan of each bacteria is 5 seconds.

### Step 2: Calculate the number of bacteria at each second
- **At \( t = 0 \)**:
- Total bacteria = 1 (initial)

- **At \( t = 1 \)**:
- The initial bacteria gives birth to 2 new bacteria.
- Total bacteria = 1 (initial) + 2 (newborns) = 3

- **At \( t = 2 \)**:
- The initial bacteria gives birth to 2 more, and the 2 new bacteria from \( t = 1 \) each also give birth to 2 new bacteria.
- Total bacteria = 1 (initial) + 2 (from \( t = 1 \)) + 2 (from \( t = 1 \)) + 2 (newborns from the initial) = 7

- **At \( t = 3 \)**:
- The initial bacteria dies (after 5 seconds), but the 2 from \( t = 1 \) and their offspring continue to reproduce.
- Total bacteria = 0 (initial) + 2 (from \( t = 1 \)) + 4 (from \( t = 2 \)) + 4 (newborns from \( t = 2 \)) = 10

- **At \( t = 4 \)**:
- The 2 bacteria from \( t = 1 \) die, but the bacteria from \( t = 2 \) and their offspring continue to reproduce.
- Total bacteria = 0 (from \( t = 1 \)) + 4 (from \( t = 2 \)) + 4 (newborns from \( t = 3 \)) + 4 (newborns from \( t = 3 \)) = 12

- **At \( t = 5 \)**:
- The bacteria from \( t = 2 \) die, but the bacteria from \( t = 3 \) and their offspring continue to reproduce.
- Total bacteria = 0 (from \( t = 1 \)) + 0 (from \( t = 2 \)) + 4 (from \( t = 3 \)) + 4 (newborns from \( t = 4 \)) + 4 (newborns from \( t = 4 \)) = 12

- **At \( t = 6 \)**:
- The bacteria from \( t = 3 \) die, but the bacteria from \( t = 4 \) and their offspring continue to reproduce.
- Total bacteria = 0 (from \( t = 1 \)) + 0 (from \( t = 2 \)) + 0 (from \( t = 3 \)) + 4 (from \( t = 4 \)) + 4 (newborns from \( t = 5 \)) + 4 (newborns from \( t = 5 \)) = 8

- **At \( t = 7 \)**:
- The bacteria from \( t = 4 \) die, but the bacteria from \( t = 5 \) and their offspring continue to reproduce.
- Total bacteria = 0 (from \( t = 1 \)) + 0 (from \( t = 2 \)) + 0 (from \( t = 3 \)) + 0 (from \( t = 4 \)) + 4 (from \( t = 5 \)) + 4 (newborns from \( t = 6 \)) + 4 (newborns from \( t = 6 \)) = 8

- **At \( t = 8 \)**:
- The bacteria from \( t = 5 \) die.
- Total bacteria = 0 (from \( t = 1 \)) + 0 (from \( t = 2 \)) + 0 (from \( t = 3 \)) + 0 (from \( t = 4 \)) + 0 (from \( t = 5 \)) + 4 (from \( t = 6 \)) + 4 (newborns from \( t = 7 \)) + 4 (newborns from \( t = 7 \)) = 8

- **At \( t = 9 \)**:
- The bacteria from \( t = 6 \) die.
- Total bacteria = 0 (from \( t = 1 \)) + 0 (from \( t = 2 \)) + 0 (from \( t = 3 \)) + 0 (from \( t = 4 \)) + 0 (from \( t = 5 \)) + 0 (from \( t = 6 \)) + 4 (from \( t = 7 \)) + 4 (newborns from \( t = 8 \)) + 4 (newborns from \( t = 8 \)) = 8

- **At \( t = 10 \)**:
- The bacteria from \( t = 7 \) die.
- Total bacteria = 0 (from \( t = 1 \)) + 0 (from \( t = 2 \)) + 0 (from \( t = 3 \)) + 0 (from \( t = 4 \)) + 0 (from \( t = 5 \)) + 0 (from \( t = 6 \)) + 0 (from \( t = 7 \)) + 4 (from \( t = 8 \)) + 4 (newborns from \( t = 9 \)) + 4 (newborns from \( t = 9 \)) = 8

### Final Count
After calculating the total number of live bacteria just after 10 seconds, we find that there are **8 live bacteria**.