A bacterial infection in an internal wound grows as N'(t) = N_0 exp(t), where the time t is in hours

Q: Why do we integrate from $t=1$ instead of $t=0$?

The antibiotic only reaches the wound and starts acting after 1 hour. Therefore, the differential equation $dN/dt = -5N^2$ is only valid for $t \ge 1$.

Q: What does $N'$ mean in the solution?

It's a notation used to represent the bacterial population exactly at $t=1$ hour, which acts as the initial population for the decay phase.

Q: Why is the slope of the graph positive when the population is decreasing?

The graph plots $N_0/N$ against $t$. As the population $N$ decreases, the denominator gets smaller, making the fraction $N_0/N$ larger. Hence, the slope is positive.

Q: How do we know the y-intercept is positive?

The equation is $N_0/N = 5N_0(t-1) + 1/e$. At $t=1$, the value is $1/e$, which is a positive number. The graph starts from this positive value, not from zero.

Q: Can we use the formula for first-order kinetics here?

No, the rate of decay is given as $-5N^2$, which means it follows second-order kinetics, not first-order. You must integrate the specific differential equation provided.

Question

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 ?

Sigma Percentile · Accepted Answer

### The Two Phases of Bacterial Growth

Imagine a bacterial infection as a ticking time bomb. For the first hour, the bacteria are having a party, multiplying exponentially without any resistance. The problem tells us that during this initial phase, the population grows according to the function $N(t) = N_0 \exp(t)$.

We need to find out exactly how many bacteria are present when the party gets crashed by the antibiotic. Since the antibiotic takes exactly 1 hour to reach the wound, we simply plug $t = 1$ into our growth equation.

This gives us the population at the 1-hour mark: $N(1) = N_0 e^1 = N_0 e$. This value is crucial because it becomes the starting point—the initial condition—for the next phase of our mathematical journey.

### The Antibiotic Strikes

Once the clock strikes 1 hour, the antibiotic arrives and starts destroying the bacteria. The rate of this destruction is given by the differential equation $\frac{dN}{dt} = -5N^2$.

Notice the negative sign? It beautifully captures the essence of decay. The $N^2$ term tells us that this is a second-order process; the bacteria are dying off at a rate proportional to the square of their current population.

To find out how the population $N$ changes over time $t$, we need to solve this differential equation. We do this by separating the variables—gathering all the $N$ terms on one side and the $t$ terms on the other.

This gives us $\frac{dN}{N^2} = -5 dt$.

Now, we integrate both sides. But we must be incredibly careful with our limits! The time $t$ starts from $1$ (not $0$), and the population $N$ starts from our previously calculated value, $N_0 e$.

So, our integral setup looks like this:
$\int_{N_0 e}^{N} N^{-2} dN = -5 \int_{1}^{t} dt$

### Solving the Integral

Let's execute the integration. The integral of $N^{-2}$ is simply $-\frac{1}{N}$.

Applying the upper and lower limits, the left side becomes:
$\left[ -\frac{1}{N} \right]_{N_0 e}^{N} = -\frac{1}{N} - \left( -\frac{1}{N_0 e} \right) = -\frac{1}{N} + \frac{1}{N_0 e}$

On the right side, the integral of $dt$ is just $t$. Applying the limits from $1$ to $t$, we get:
$-5 [t]_{1}^{t} = -5(t - 1)$

Equating both sides, we have our raw mathematical relationship:
$-\frac{1}{N} + \frac{1}{N_0 e} = -5(t - 1)$

### Unveiling the Graph

The question doesn't just ask for $N$; it specifically asks for the plot of $\frac{N_0}{N}$ versus $t$. This means we need to algebraically manipulate our equation to isolate $\frac{N_0}{N}$.

Let's multiply the entire equation by $N_0$ to clear the path:
$-\frac{N_0}{N} + \frac{1}{e} = -5N_0(t - 1)$

Now, let's move terms around to isolate $\frac{N_0}{N}$:
$\frac{N_0}{N} = 5N_0(t - 1) + \frac{1}{e}$

Let's expand this to see its true geometric form:
$\frac{N_0}{N} = (5N_0)t + \left( \frac{1}{e} - 5N_0 \right)$

Take a step back and look at this equation. It perfectly matches the equation of a straight line, $y = mx + c$!

Here, our y-axis variable is $\frac{N_0}{N}$, and our x-axis variable is $t$. The slope of this line, $m$, is $5N_0$. Since $N_0$ (the initial population) is a positive number, the slope is strictly positive.

Furthermore, let's check where this line starts. At $t = 1$, the value of $\frac{N_0}{N}$ is exactly $\frac{1}{e}$, which is a positive constant.

Therefore, the graph must be a straight line with a positive slope that starts from a positive value on the y-axis. Looking at our options, the first graph perfectly captures this linear, upward trajectory starting from a positive intercept.

Select Answer:

Visualized Solution

$N (t) at t = 1$

$Differential Equation$

$Separation of Variables$

$Integration$

$Algebraic Rearrangement$

$Equation of the Plot$

$Conclusion$

The Sigma Insight: Rate of Chemical Reaction

The Two Phases of Bacterial Growth

The Antibiotic Strikes

Solving the Integral

Unveiling the Graph

Similar Questions

For a reaction $\frac{1}{2} A ⟶ 2 B$ , rate of disappearance of 'A' is related to the rate of appearance of 'B' by the expression

Which of the following plots is(are) correct for the given reaction? ( $[P]_{0}$ is the initial concentration of P)

Rate constant, $k$ of the first order reaction when initial concentration $(C_{0})$ and concentration $(C_{t})$ at time $t$ , is given by equation $k t = lo g C_{0} - lo g C_{t}$ Graph is a straight line if we plot

For an elementary chemical reaction, $A_{2} k_{- 1} ⇌ k_{1} 2 A$ , the expression for $\frac{d [ A ]}{d t}$ is

For a first order reaction, $A ⟶ products$ the concentration of $A$ changes from $0.1 M$ to $0.025 M$ in $40 min$ . The rate of reaction when the concentration of $A$ is $0.01 M$ is

For the following reaction $2 X + Y k P$ the rate of reaction is $\frac{d [ P ]}{d t} = k [X]$ . Two moles of X are mixed with one mole of Y to make 1.0 L of solution. At 50 s, 0.5 mole of Y is left in the reaction mixture. The correct statement(s) about the reaction is(are) (Use: $ln 2 = 0.693$ )

For the reaction $2 A + 3 B + \frac{3}{2} C \to 3 P$ , which statement is correct?

For the reaction, $H_{2} + I_{2} ⟶ 2 H I$ , the differential rate law is

For a chemical reaction $A ⟶ B$ , it was found that concentration of $B$ is increased by $0.2 mol L^{- 1}$ in $30 min$ . The average rate of the reaction is $\dots \times 1 0^{- 1} mol L^{- 1} h^{- 1}$ . (Nearest integer)

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } N(t) at t=1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Differential Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Separation of Variables

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Integration

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Algebraic Rearrangement

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equation of the Plot

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion