The rate of growth of bacteria in a culture is proportional to the number of bacteria present and th

Q: Why can't we just use simple interest formulas like $A = P(1+rt)$? Why do we need calculus?

That is a fantastic question! Simple interest assumes the growth rate applies only to the initial amount. However, in biological systems, the bacteria present at any given moment are all reproducing. This means the 'offspring' also start reproducing, leading to continuous compounding. The differential equation $\frac{dN}{dt} = \lambda N$ captures this continuous, self-reinforcing nature of growth perfectly.

Q: Why does the $\ln(\frac{6}{5})$ term appear in the final time expression? It looks so messy!

It looks messy, but it is actually the 'DNA' of your growth rate. When you solve $1.2 = e^{2\lambda}$, you get $2\lambda = \ln(1.2)$, or $\ln(\frac{6}{5})$. This logarithmic term is essentially the 'speedometer' of your bacterial culture. It is designed to cancel out later when you substitute it into the equation for $t$, leaving you with a clean, elegant integer.

Q: Wait, why do we reject the negative root or constant? Is there ever a case where $\lambda$ is negative?

In this specific problem, we are dealing with growth, so $\lambda$ must be positive. However, you are absolutely right to be cautious! If the problem described radioactive decay or a dying population, $\lambda$ would indeed be negative. Always check the physical context of the problem—if the population is increasing, your growth constant must be positive.

Question

The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is 1000 at initial time $t = 0$. The number of bacteria is increased by 20% in 2 hours. If the population of bacteria is 2000 after $\frac{k}{\log_e(\frac{6}{5})}$ hours, then $(\log_e(\frac{k}{2}))^2$ is equal to

Options:

(A) 4

(B) 2

(C) 16

(D) 8

Sigma Percentile · Accepted Answer

### The Elegance of Exponential Growth

Imagine you are standing in a quiet laboratory, peering into a petri dish. You are not just looking at a collection of cells; you are witnessing a mathematical phenomenon.

At time $t = 0$, you have exactly $1000$ bacteria. These bacteria are alive, and they are multiplying. This is the world of population dynamics.

### The Differential Equation: The Heart of the Problem

The problem gives us a vital clue: the rate of growth is proportional to the number of bacteria present. In the language of calculus, this is expressed as:

$$\frac{dN}{dt} \propto N$$

When we remove the proportionality sign, we introduce a constant, $\lambda$, which represents the growth rate. Thus, we arrive at our governing equation:

$$\frac{dN}{dt} = \lambda N$$

This equation is the heartbeat of the system. It tells us that the more bacteria there are, the faster the population grows.

### Separating the Variables

To solve this, we use the method of separation of variables. We group all the $N$ terms on one side and all the $t$ terms on the other:

$$\frac{dN}{N} = \lambda dt$$

Now, we integrate both sides:

$$\int \frac{dN}{N} = \int \lambda dt$$

This integration is a classic. The integral of $\frac{1}{N}$ is $\ln N$, and the integral of $\lambda$ with respect to $t$ is $\lambda t + C$. Exponentiating both sides, we find $N(t) = e^{\lambda t + C}$, which simplifies to:

$$N(t) = A e^{\lambda t}$$

Here, $A = e^C$ is our constant of integration. This is the master key to understanding the population at any time $t$.

### Applying the Initial Conditions

We know that at $t = 0$, the population $N(0) = 1000$. Substituting this into our equation:

$$1000 = A e^{\lambda(0)}$$

This immediately tells us that $A = 1000$. Our specific equation is now:

$$N(t) = 1000 e^{\lambda t}$$

### The Climax: Finding the Growth Constant

The problem states that after $2$ hours, the population increases by $20\%$. Since $20\%$ of $1000$ is $200$, the new population is $1200$. Plugging this into our model:

$$1200 = 1000 e^{2\lambda}$$

Dividing by $1000$, we get $1.2 = e^{2\lambda}$, or $\frac{6}{5} = e^{2\lambda}$. Taking the natural log of both sides, we find:

$$2\lambda = \ln\left(\frac{6}{5}\right) \Rightarrow \lambda = \frac{1}{2} \ln\left(\frac{6}{5}\right)$$

### The Elegant Cancellation

We want to find the time $t$ when the population reaches $2000$. Setting $N(t) = 2000$:

$$2000 = 1000 e^{\lambda t} \Rightarrow 2 = e^{\lambda t} \Rightarrow \ln 2 = \lambda t$$

We are given that this time $t$ is expressed as $\frac{k}{\ln(\frac{6}{5})}$. Substituting our value of $\lambda$ into the equation $\ln 2 = \lambda t$:

$$\ln 2 = \left[ \frac{1}{2} \ln\left(\frac{6}{5}\right) \right] \cdot \left[ \frac{k}{\ln(\frac{6}{5})} \right]$$

The term $\ln(\frac{6}{5})$ cancels out perfectly. We are left with $\ln 2 = \frac{k}{2}$, which means **$k = 2 \ln 2$**.

### The Final Victory

The question asks us to evaluate the square of the expression involving $k$. Substituting $k = 2 \ln 2$ into our final expression:

$$\left( \frac{2 \ln 2}{\ln 2} \right)^2 = 2^2 = 4$$

The complexity of the exponential growth, the differential equations, and the logarithmic constants all collapse into a single, beautiful integer: **$4$**.

Select Answer:

Visualized Solution

Visualizing Bacterial Growth

The Differential Equation

Separating Variables

General Solution Form

Applying Initial Conditions

The $20%$ Increase

Solving for $λ$

Reaching $2000$ Bacteria

Substituting $t$ and $λ$

Finding the value of $k$

Final Calculation

The Sigma Insight: Variable Separable Method

The Elegance of Exponential Growth

The Differential Equation

Separating the Variables

Applying the Initial Conditions

The Climax

The Elegant Cancellation

The Final Victory

Similar Questions

The population P = P(t) at time 't' of a certain species follows the differential equation $\frac{d P}{d t} = 0.5 P - 450$ . If $P (0) = 850$ , then the time at which population becomes zero is :

The population $p (t)$ at time $t$ of a certain mouse species satisfies the differential equation $\frac{d p ( t )}{d t} = 0.5 p (t) - 450$ . If $p (0) = 850$ , then the time at which the population becomes zero is :

Let the population of rabbits surviving at time $t$ be governed by the differential equation $\frac{d p ( t )}{d t} = \frac{1}{2} p (t) - 200$ . If $p (0) = 100$ , then $p (t)$ equals:

Let a smooth curve $y = f (x)$ be such that the slope of the tangent at any point $(x, y)$ on it is directly proportional to $\frac{- y}{x}$ . If the curve passes through the point $(1, 2)$ and $(8, 1)$ , then $∣ y (\frac{1}{8}) ∣$ is equal to

Let $y = y (x)$ be the solution of the differential equation $d y = e^{α x + y} d x; α \in N$ . If $y (lo g_{e} 2) = lo g_{e} 2$ and $y (0) = lo g_{e} (1/2)$ , then the value of $α$ is equal to

Let $x = x (t)$ and $y = y (t)$ be solutions of the differential equations $\frac{d x}{d t} + a x = 0$ and $\frac{d y}{d t} + b y = 0$ respectively, $a, b \in R$ . Given that $x (0) = 2; y (0) = 1$ and $3 y (1) = 2 x (1)$ , the value of $t$ , for which $x (t) = y (t)$ , is :

If $y = y (x)$ is the solution of the differential equation, $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to

If $\frac{d y}{d x} = \frac{2 ^{x + y} - 2 ^{x}}{2 ^{y}}, y (0) = 1$ , then $y (1)$ is equal to :

At present, a firm is manufacturing 2000 items. It is estimated that the rate of change of production $P$ w.r.t. additional number of workers $x$ is given by $\frac{d P}{d x} = 100 - 12 x$ . If the firm employs 25 more workers, then the new level of production of items is

Let a curve $y = f (x)$ pass through the points $(0, 5)$ and $(lo g_{e} 2, k)$ . If the curve satisfies the differential equation $2 (3 + y) e^{2 x} d x - (7 + e^{2 x}) d y = 0$ , then k is equal to

Select Answer:

Visualized Solution

Visualizing Bacterial Growth

The Differential Equation

Separating Variables

General Solution Form

Applying Initial Conditions

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 20% Increase

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

Reaching .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2000 Bacteria

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

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

Final Calculation

The Sigma Insight: Variable Separable Method

The Elegance of Exponential Growth

The Differential Equation

Separating the Variables

Applying the Initial Conditions

The Climax

The Elegant Cancellation

The Final Victory

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the population of rabbits surviving at time t be governed by the differential equation dtdp(t)​=21​p(t)−200. If p(0)=100, then p(t) equals:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let y=y(x) be the solution of the differential equation dy=eαx+ydx;α∈N. If y(loge​2)=loge​2 and y(0)=loge​(1/2), then the value of α is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=y(x) is the solution of the differential equation, ey(dxdy​−1)=ex such that y(0)=0, then y(1) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If dxdy​=2y2x+y−2x​,y(0)=1, then y(1) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a curve y=f(x) pass through the points (0,5) and (loge​2,k). If the curve satisfies the differential equation 2(3+y)e2xdx−(7+e2x)dy=0, then k is equal to

The $20%$ Increase

Solving for $λ$

Reaching $2000$ Bacteria

Substituting $t$ and $λ$

Finding the value of $k$

Let the population of rabbits surviving at time $t$ be governed by the differential equation $\frac{d p ( t )}{d t} = \frac{1}{2} p (t) - 200$ . If $p (0) = 100$ , then $p (t)$ equals:

Let $y = y (x)$ be the solution of the differential equation $d y = e^{α x + y} d x; α \in N$ . If $y (lo g_{e} 2) = lo g_{e} 2$ and $y (0) = lo g_{e} (1/2)$ , then the value of $α$ is equal to

If $y = y (x)$ is the solution of the differential equation, $e^{y} (\frac{d y}{d x} - 1) = e^{x}$ such that $y (0) = 0$ , then $y (1)$ is equal to

If $\frac{d y}{d x} = \frac{2 ^{x + y} - 2 ^{x}}{2 ^{y}}, y (0) = 1$ , then $y (1)$ is equal to :

Let a curve $y = f (x)$ pass through the points $(0, 5)$ and $(lo g_{e} 2, k)$ . If the curve satisfies the differential equation $2 (3 + y) e^{2 x} d x - (7 + e^{2 x}) d y = 0$ , then k is equal to