4. Uninhibited Growth At any time $$ t $$ in hours, the rate of increase in the area, in millimeters squared $$ \left(\mathrm{mm}^{2}\right) $$, of a culture of bacteria is twice the area $$ A $$ of the culture.

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To solve the problem, we need to establish a relationship between the rate of increase in the area of the bacteria culture and the area itself.

### Step 1: Set up the equation

We are given that the rate of increase in the area $$ \frac{dA}{dt} $$ is twice the area $$ A $$. This can be expressed mathematically as:

$$
\frac{dA}{dt} = 2A
$$

### Step 2: Solve the differential equation

This is a first-order linear differential equation. We can solve it using the method of separation of variables.

1. Rearranging the equation gives us:

$$
\frac{dA}{A} = 2 dt
$$

2. Now, we integrate both sides:

$$
\int \frac{dA}{A} = \int 2 dt
$$

### Step 3: Perform the integration

The left side integrates to $$ \ln |A| $$ and the right side integrates to $$ 2t + C $$, where $$ C $$ is the constant of integration:

$$
\ln |A| = 2t + C
$$

### Step 4: Exponentiate both sides

To solve for $$ A $$, we exponentiate both sides:

$$
|A| = e^{2t + C} = e^{2t} \cdot e^{C}
$$

Let $$ k = e^{C} $$, which is a positive constant. Thus, we can write:

$$
A = ke^{2t}
$$

### Step 5: Conclusion

The solution to the differential equation shows that the area $$ A $$ of the bacteria culture at any time $$ t $$ is given by:

$$
A(t) = ke^{2t}
$$

where $$ k $$ is a constant that can be determined if an initial condition is provided (e.g., the area at $$ t = 0 $$).

This equation describes the uninhibited growth of the bacteria culture, indicating that the area increases exponentially over time.
The area $$ A $$ of the bacteria culture grows exponentially over time and is given by $$ A(t) = ke^{2t} $$, where $$ k $$ is a constant.

4. Uninhibited Growth At any time \( t \) in hours, the rate of increase in the area, in millimeters squared \( \left(\mathrm{mm}^{2}\right) \), of a culture of bacteria is twice the area \( A \) of the culture.

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