A bacterial infection in an internal wound grows as `N'(t) = N_(0) exp (t)`, where the time t is in hours.A dose of antibiotic, taken orally, needs 1 hour to reach the wound. Once it reaches there, the bacterial population goes down as `(dN)/(dt)= -5 N^(2)`. What will be the plot of `N_(0)/N` vs. t after 1 hour?

Options

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Explanation

To solve the problem, we need to analyze the growth of the bacterial population and the effect of the antibiotic over time. Here’s a step-by-step breakdown of the solution:

### Step 1: Determine the bacterial growth before the antibiotic reaches the wound
The bacterial growth is given by the equation:
\[ N(t) = N_0 e^t \]
where \( N_0 \) is the initial bacterial population and \( t \) is the time in hours.

### Step 2: Calculate the bacterial population at \( t = 1 \) hour
At \( t = 1 \):
\[ N(1) = N_0 e^1 = N_0 e \]

### Step 3: Analyze the effect of the antibiotic after 1 hour
Once the antibiotic reaches the wound (after 1 hour), the bacterial population decreases according to the differential equation:
\[ \frac{dN}{dt} = -5N^2 \]

### Step 4: Rearrange the differential equation
Rearranging gives:
\[ \frac{dN}{N^2} = -5 dt \]

### Step 5: Integrate the equation
Integrate both sides. The left side integrates to:
\[ \int \frac{dN}{N^2} = -\frac{1}{N} \]
The right side integrates to:
\[ \int -5 dt = -5t + C \]
Thus, we have:
\[ -\frac{1}{N} = -5t + C \]

### Step 6: Solve for the constant \( C \)
At \( t = 1 \), we know:
\[ N(1) = N_0 e \]
Substituting this into the equation gives:
\[ -\frac{1}{N_0 e} = -5(1) + C \]
So,
\[ C = -\frac{1}{N_0 e} + 5 \]

### Step 7: Substitute \( C \) back into the equation
Now substituting \( C \) back into the integrated equation:
\[ -\frac{1}{N} = -5t - \frac{1}{N_0 e} + 5 \]

### Step 8: Rearranging to find \( N \)
Rearranging gives:
\[ \frac{1}{N} = 5t - 5 + \frac{1}{N_0 e} \]
Thus,
\[ N = \frac{1}{5t - 5 + \frac{1}{N_0 e}} \]

### Step 9: Finding \( \frac{N_0}{N} \)
Taking the reciprocal:
\[ \frac{N_0}{N} = N_0 \left(5t - 5 + \frac{1}{N_0 e}\right) \]
This simplifies to:
\[ \frac{N_0}{N} = 5N_0t - 5N_0 + \frac{1}{e} \]

### Step 10: Plotting \( \frac{N_0}{N} \) vs. \( t \)
The equation \( \frac{N_0}{N} = 5N_0t - 5N_0 + \frac{1}{e} \) represents a linear function in the form \( y = mx + c \), where:
- \( m = 5N_0 \)
- \( c = -5N_0 + \frac{1}{e} \)

This indicates that the plot of \( \frac{N_0}{N} \) vs. \( t \) will be a straight line with a positive slope.

### Conclusion
The plot of \( \frac{N_0}{N} \) vs. \( t \) after 1 hour will be a straight line with a positive slope.

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