A ballon , spherical in shape is inflated and its surface area is increasing with a constant rate , initially the radius is 3 unit , after 5 sec radius is 7 units then the radius after 9 second is

Options

A. 9
B. 7
C. 5
D. 3

Answer

To solve the problem, we need to find the radius of a spherical balloon after 9 seconds, given that its surface area is increasing at a constant rate. We start with the information provided:

1. **Initial Radius**: \( r_0 = 3 \) units at \( t = 0 \) seconds.
2. **Radius after 5 seconds**: \( r(5) = 7 \) units.

### Step-by-Step Solution:

**Step 1: Surface Area Formula**
The surface area \( A \) of a sphere is given by the formula:
\[ A = 4\pi r^2 \]

**Step 2: Differentiate the Surface Area**
To find the rate of change of the surface area with respect to time, we differentiate \( A \) with respect to \( t \):
\[ \frac{dA}{dt} = \frac{d}{dt}(4\pi r^2) = 8\pi r \frac{dr}{dt} \]

**Step 3: Set Up the Constant Rate of Change**
Let \( k \) be the constant rate of change of the surface area:
\[ \frac{dA}{dt} = k \]
Thus, we have:
\[ 8\pi r \frac{dr}{dt} = k \]

**Step 4: Find the Value of \( k \)**
We know the radius at \( t = 5 \) seconds is \( r(5) = 7 \) units. We can find \( k \) using the initial conditions. First, we need to find \( k \) at \( t = 0 \) and \( t = 5 \).

At \( t = 0 \):
\[ A(0) = 4\pi (3^2) = 36\pi \]

At \( t = 5 \):
\[ A(5) = 4\pi (7^2) = 196\pi \]

The change in surface area from \( t = 0 \) to \( t = 5 \) is:
\[ \Delta A = A(5) - A(0) = 196\pi - 36\pi = 160\pi \]

The time interval is \( 5 \) seconds, so the rate \( k \) is:
\[ k = \frac{\Delta A}{\Delta t} = \frac{160\pi}{5} = 32\pi \]

**Step 5: Write the Equation for Surface Area**
Now we can express the surface area as a function of time:
\[ A(t) = kt + A(0) = 32\pi t + 36\pi \]

**Step 6: Relate Surface Area to Radius**
Since \( A(t) = 4\pi r^2 \), we can set the two expressions for \( A \) equal:
\[ 4\pi r^2 = 32\pi t + 36\pi \]

**Step 7: Solve for \( r \)**
Dividing through by \( 4\pi \):
\[ r^2 = 8t + 9 \]
Taking the square root:
\[ r = \sqrt{8t + 9} \]

**Step 8: Find the Radius at \( t = 9 \) seconds**
Substituting \( t = 9 \):
\[ r(9) = \sqrt{8(9) + 9} = \sqrt{72 + 9} = \sqrt{81} = 9 \]

### Final Answer:
The radius of the balloon after 9 seconds is \( 9 \) units.