To solve this problem, let's break it down step-by-step:

Given Information:

• The rate at which the radius $r$ of the sphere is increasing is $\frac{dr}{dt} = 5 \ \text{mm/s}$.
• The diameter of the sphere is $100 \ \text{mm}$, which means the radius $r = \frac{100}{2} = 50 \ \text{mm}$.
• We are asked to find the rate of change of the volume $V$ of the sphere when $r = 50 \ \text{mm}$.

Formula for the Volume of a Sphere:

The volume $V$ of a sphere is given by: $V = \frac{4}{3} \pi r^3$

Differentiate with Respect to Time $t$:

To find how fast the volume is increasing, we differentiate both sides with respect to $t$: $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$

Substitute the Known Values:

Now, substitute $r = 50 \ \text{mm}$ and $\frac{dr}{dt} = 5 \ \text{mm/s}$ into the differentiated formula: $\frac{dV}{dt} = 4 \pi (50)^2 \times 5$

Calculating each part:

1. $50^2 = 2500$
2. $4 \pi \times 2500 = 10000 \pi$
3. $10000 \pi \times 5 = 50000 \pi$

So, $\frac{dV}{dt} = 50000 \pi \ \text{mm}^3/\text{s}$

Approximating the Value:

Using $\pi \approx 3.14159$: $\frac{dV}{dt} \approx 50000 \times 3.14159 = 157079.63 \ \text{mm}^3/\text{s}$

Final Answer:

The rate at which the volume is increasing is approximately: $\frac{dV}{dt} \approx 157079.63 \ \text{mm}^3/\text{s}$

Would you like further explanation on any step?

Here are some follow-up questions for practice:

1. What would be the rate of volume increase if the diameter were 200 mm instead?
2. How would the rate of volume increase change if $\frac{dr}{dt}$ were 10 mm/s?
3. How does the volume change relate to the surface area of the sphere as it expands?
4. What is the significance of using the chain rule in related rates problems?
5. How can you apply related rates to other geometric shapes, like cylinders or cones?

Tip: In related rates problems, always start by writing down what rates are given and what is being asked. Then, find a formula that relates the quantities and differentiate with respect to time.