The surface area of a balloon, being inflated, changes at a rate proportional to time t.
(i) If initially its radius is 1 unit and after 3 seconds it is 2 units, find the radius after time t.
(ii) If initially its radius is 3 units and after 2 seconds it is 5 units, find the radius after t seconds.

Answer

To solve the problem, we need to derive the relationship between the radius \( r \) of the balloon and time \( t \) based on the information provided. We will tackle both parts of the question step by step.

### Part (i)

1. **Understanding the relationship**:
The surface area \( S \) of a balloon is given by the formula:
\[ S = 4\pi r^2 \]
The rate of change of surface area with respect to time is given as:
\[ \frac{dS}{dt} = 8\pi r \frac{dr}{dt} \]
We are told that this rate is proportional to time \( t \):
\[ \frac{dS}{dt} = kt \]
where \( k \) is a constant.

2. **Setting up the equation**:
Equating the two expressions for \( \frac{dS}{dt} \):
\[ 8\pi r \frac{dr}{dt} = kt \]

3. **Separating variables**:
Rearranging gives:
\[ 8\pi r \, dr = k \, t \, dt \]

4. **Integrating both sides**:
Integrate both sides:
\[ \int 8\pi r \, dr = \int k \, t \, dt \]
This results in:
\[ 4\pi r^2 = \frac{k}{2} t^2 + C \]
where \( C \) is the constant of integration.

5. **Finding constants using initial conditions**:
We know that initially (at \( t = 0 \)), the radius \( r = 1 \):
\[ 4\pi(1^2) = \frac{k}{2}(0^2) + C \implies C = 4\pi \]

6. **Substituting back into the equation**:
Now substituting \( C \) back into our equation:
\[ 4\pi r^2 = \frac{k}{2} t^2 + 4\pi \]

7. **Using the second condition**:
After 3 seconds, the radius is 2 units:
\[ 4\pi(2^2) = \frac{k}{2}(3^2) + 4\pi \]
Simplifying gives:
\[ 16\pi = \frac{9k}{2} + 4\pi \]
\[ 12\pi = \frac{9k}{2} \implies k = \frac{24\pi}{9} = \frac{8\pi}{3} \]

8. **Final equation for radius**:
Substituting \( k \) back into the equation:
\[ 4\pi r^2 = \frac{8\pi}{3} \cdot \frac{t^2}{2} + 4\pi \]
Simplifying gives:
\[ 4\pi r^2 = \frac{4\pi}{3} t^2 + 4\pi \]
Dividing through by \( 4\pi \):
\[ r^2 = \frac{t^2}{3} + 1 \]
Therefore, the radius after time \( t \) is:
\[ r = \sqrt{\frac{t^2}{3} + 1} \]

### Part (ii)

1. **Setting up the equation**:
We start again from:
\[ 4\pi r^2 = \frac{k}{2} t^2 + C \]

2. **Using the initial condition**:
Initially (at \( t = 0 \)), the radius \( r = 3 \):
\[ 4\pi(3^2) = \frac{k}{2}(0^2) + C \implies C = 36\pi \]

3. **Substituting back into the equation**:
Thus, we have:
\[ 4\pi r^2 = \frac{k}{2} t^2 + 36\pi \]

4. **Using the second condition**:
After 2 seconds, the radius is 5 units:
\[ 4\pi(5^2) = \frac{k}{2}(2^2) + 36\pi \]
Simplifying gives:
\[ 100\pi = 2k + 36\pi \]
\[ 64\pi = 2k \implies k = 32\pi \]

5. **Final equation for radius**:
Substituting \( k \) back into the equation:
\[ 4\pi r^2 = 32\pi \cdot \frac{t^2}{2} + 36\pi \]
Simplifying gives:
\[ 4\pi r^2 = 16\pi t^2 + 36\pi \]
Dividing through by \( 4\pi \):
\[ r^2 = 4t^2 + 9 \]
Therefore, the radius after time \( t \) is:
\[ r = \sqrt{4t^2 + 9} \]