The radius of spherical balloon is increasing at the rate of 5 cm per second. At what rate is the surface of the balloon increasing, when the radius is 10 cm?

Answer

To solve the problem, we need to find the rate at which the surface area of a spherical balloon is increasing when the radius is 10 cm. We are given that the radius is increasing at a rate of 5 cm per second.

### Step-by-Step Solution:

1. **Identify the formula for the surface area of a sphere:**
The surface area \( A \) of a sphere is given by the formula:
\[ A = 4\pi r^2 \]

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

3. **Substitute the known values:**
We know:
- \( r = 10 \) cm (the radius at the moment we are interested in)
- \( \frac{dr}{dt} = 5 \) cm/s (the rate at which the radius is increasing)

Now, substituting these values into the differentiated equation:
\[ \frac{dA}{dt} = 8\pi (10) (5) \]

4. **Calculate the rate of change of the surface area:**
\[ \frac{dA}{dt} = 8\pi \cdot 10 \cdot 5 = 400\pi \text{ cm}^2/\text{s} \]

5. **Convert to a numerical value (optional):**
If we want to express this in numerical terms, we can use \( \pi \approx 3.14 \):
\[ \frac{dA}{dt} \approx 400 \cdot 3.14 = 1256 \text{ cm}^2/\text{s} \]

### Final Answer:
The rate at which the surface area of the balloon is increasing when the radius is 10 cm is \( 400\pi \) cm²/s or approximately \( 1256 \) cm²/s.