Given that air is being pumped into a spherical balloon so that its volume increases at a rate of $\frac{dV}{dt} = 100 \, \text{cm}^3/\text{s}$, we need to determine how fast the radius of the balloon is increasing when the diameter is 50 cm.

First, let's recall the formula for the volume $V$ of a sphere in terms of its radius $r$:

$V = \frac{4}{3} \pi r^3$

To find the rate at which the radius is increasing, we differentiate 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:

$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$

We are given $\frac{dV}{dt} = 100 \, \text{cm}^3/\text{s}$ and need to find $\frac{dr}{dt}$ when the diameter is 50 cm. The radius $r$ is half of the diameter, so:

$r = \frac{50}{2} = 25 \, \text{cm}$

Now, substitute $r = 25 \, \text{cm}$ and $\frac{dV}{dt} = 100 \, \text{cm}^3/\text{s}$ into the differentiated volume formula:

$100 = 4 \pi (25)^2 \frac{dr}{dt}$

Simplify the equation:

$100 = 4 \pi (625) \frac{dr}{dt}$

$100 = 2500 \pi \frac{dr}{dt}$

Solve for $\frac{dr}{dt}$:

$\frac{dr}{dt} = \frac{100}{2500 \pi}$

$\frac{dr}{dt} = \frac{1}{25 \pi} \, \text{cm/s}$

Therefore, the radius of the balloon is increasing at a rate of $\frac{1}{25 \pi} \, \text{cm/s}$ when the diameter is 50 cm.

Would you like more details or have any questions?

Here are 5 more questions you might want to ask:

1. How do you calculate the surface area of a sphere?
2. What is the relationship between the volume and radius of a sphere?
3. How do you differentiate implicit functions?
4. Can you explain the chain rule in calculus?
5. How do you solve related rates problems in general?

Tip: When solving related rates problems, always start by identifying what rates are given and what rates you need to find. Use the relationships between the variables to set up the necessary equations before differentiating.