A balloon which always remains spherical, is being inflated by pumping in 900 cubic centimetres of gas per second. Find the rate at which the radius of the balloon is increasing when the radius is 15 cm.

Answer

To solve the problem step by step, we need to find the rate at which the radius of the balloon is increasing when the radius is 15 cm, given that the volume of gas being pumped into the balloon is 900 cubic centimeters per second.

### Step 1: Understand the relationship between volume and radius
The volume \( V \) of a sphere is given by the formula:
\[ V = \frac{4}{3} \pi r^3 \]
where \( r \) is the radius of the sphere.

### Step 2: Differentiate the volume with respect to time
To find the rate of change of the radius with respect to time, we differentiate both sides of the volume formula with respect to time \( t \):
\[ \frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3} \pi r^3\right) \]
Using the chain rule, we have:
\[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \]

### Step 3: Substitute the known values
We know from the problem that:
\[ \frac{dV}{dt} = 900 \, \text{cm}^3/\text{s} \]
and we need to find \( \frac{dr}{dt} \) when \( r = 15 \, \text{cm} \).

Substituting these values into the differentiated equation:
\[ 900 = 4 \pi (15^2) \frac{dr}{dt} \]

### Step 4: Simplify the equation
Calculating \( 15^2 \):
\[ 15^2 = 225 \]
Thus, the equation becomes:
\[ 900 = 4 \pi (225) \frac{dr}{dt} \]
\[ 900 = 900 \pi \frac{dr}{dt} \]

### Step 5: Solve for \( \frac{dr}{dt} \)
To isolate \( \frac{dr}{dt} \), divide both sides by \( 900 \pi \):
\[ \frac{dr}{dt} = \frac{900}{900 \pi} = \frac{1}{\pi} \]

### Step 6: Substitute the value of \( \pi \)
Using \( \pi \approx \frac{22}{7} \):
\[ \frac{dr}{dt} = \frac{1}{\frac{22}{7}} = \frac{7}{22} \, \text{cm/s} \]

### Final Answer
Thus, the rate at which the radius of the balloon is increasing when the radius is 15 cm is:
\[ \frac{dr}{dt} = \frac{7}{22} \, \text{cm/s} \]

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