	A sphere magically grows in size every second. Its radius increases at a rate of 
3 x 2^t mm/s, where t is time in seconds. The initial radius of the ball is 1mm. 


To solve this problem, we will explore the dynamics of the sphere's radius and calculate some key details.

The radius of the sphere increases at a rate of \( 3 \times 2^t \) mm/s, where \( t \) is the time in seconds. The initial radius of the sphere is 1 mm. Our task is to analyze the situation and find the expression for the radius \( r(t) \) as a function of time.

### Step 1: Define the rate of change of the radius.
The rate of change of the radius is given as:
\[
\frac{dr}{dt} = 3 \times 2^t
\]
This is a first-order differential equation, where the derivative of the radius with respect to time is provided. 

### Step 2: Solve the differential equation.
To find \( r(t) \), we need to integrate the rate of change \( \frac{dr}{dt} \).

\[
r(t) = \int 3 \times 2^t \, dt
\]

The integral of \( 2^t \) is \( \frac{2^t}{\ln 2} \), so:

\[
r(t) = 3 \times \frac{2^t}{\ln 2} + C
\]

where \( C \) is the constant of integration.

### Step 3: Apply the initial condition.
The initial radius of the sphere is 1 mm at \( t = 0 \). So we substitute \( r(0) = 1 \) to solve for \( C \):

\[
1 = 3 \times \frac{2^0}{\ln 2} + C
\]
\[
1 = 3 \times \frac{1}{\ln 2} + C
\]
\[
C = 1 - \frac{3}{\ln 2}
\]

Thus, the full expression for the radius as a function of time is:

\[
r(t) = 3 \times \frac{2^t}{\ln 2} + \left(1 - \frac{3}{\ln 2}\right)
\]

This gives us the radius of the sphere at any time \( t \).

Would you like to explore the volume of the sphere, the total surface area, or any other related properties? Let me know if you have any questions.

### 5 Related Questions:
1. What is the volume of the sphere as a function of time \( t \)?
2. How does the surface area of the sphere change over time?
3. Can you calculate the radius at \( t = 5 \) seconds?
4. How would you describe the behavior of the radius for very large values of \( t \)?
5. How does the growth rate of the sphere’s radius compare to its surface area?

### Tip:
When working with exponential growth, remember that small changes in time can lead to large differences in the size of the object due to the rapid increase.

A sphere magically grows in size every second. Its radius increases at a rate of 3 x 2^t mm/s, where t is time in seconds. The initial radius of the ball is 1mm.

Explore how the radius of a sphere grows at a rate of 3 x 2^t mm/s. Solve a differential equation to find the radius as a function of time, using calculus concepts. Suitable for advanced high school and college-level students.

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