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 \(\frac{dN}{dt} = -5N^2\). What will be the plot of \(\frac{N_0}{N}\) vs. \(t\) after 1 hour?
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The Correct Option is D
Solution and Explanation
Separate variables: \(\frac{dN}{N^2} = -5\,dt\).
Integrate: \(-\frac{1}{N} = -5t + C\), i.e. \(\frac{1}{N} = 5t + C'\).
Multiply by \(N_0\): \(\frac{N_0}{N} = 5N_0\,t + C''\).
This is a straight line in \(t\) with a positive slope \(5N_0\) and a positive intercept (the process is second order in \(N\), so the reciprocal of the population is linear in time).
So the plot of \(\frac{N_0}{N}\) vs \(t\) after 1 hour is a straight line rising with time, which is the graph in option (4).
Option (2) is a falling line, option (3) is a curve (not linear), and option (1) is a two-segment line.
Answer: (4) a straight line with positive slope and positive intercept
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