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The problem asks to compute certain quantities for a growing sphere. Let's break it down:

- The volume of the sphere is increasing at a constant rate of \( 10 \, \text{m}^3/\text{second} \).
- \( r(t) \), \( V(t) \), and \( S(t) \) represent the radius, volume, and surface area of the sphere as functions of time \( t \).
- \( r(1) = 2 \) means that at \( t = 1 \), the radius of the sphere is 2 meters.

The general formulas for a sphere are:
- Volume: \( V(t) = \frac{4}{3} \pi r(t)^3 \)
- Surface Area: \( S(t) = 4 \pi r(t)^2 \)

### Step 1: Relating the volume rate to the radius
We are given that the volume is increasing at a rate of \( \frac{dV}{dt} = 10 \, \text{m}^3/\text{second} \). 

From the formula for the volume:
\[
V(t) = \frac{4}{3} \pi r(t)^3
\]
Differentiate both sides with respect to \( t \) to find \( \frac{dV}{dt} \):
\[
\frac{dV}{dt} = 4 \pi r(t)^2 \frac{dr}{dt}
\]
We know \( \frac{dV}{dt} = 10 \), so:
\[
10 = 4 \pi r(t)^2 \frac{dr}{dt}
\]
This equation relates the rate of change of the radius \( \frac{dr}{dt} \) to the radius \( r(t) \) and the given rate of volume change.

### Step 2: Solve for \( \frac{dr}{dt} \) at \( t = 1 \)
We are given that \( r(1) = 2 \). Plug this into the equation:
\[
10 = 4 \pi (2)^2 \frac{dr}{dt}
\]
Simplify:
\[
10 = 16 \pi \frac{dr}{dt}
\]
\[
\frac{dr}{dt} = \frac{10}{16 \pi} = \frac{5}{8 \pi}
\]

### Final Answer:
At \( t = 1 \), the rate of change of the radius is:
\[
\frac{dr}{dt} = \frac{5}{8 \pi} \, \text{meters/second}
\]

Would you like further details or help with the next part of the problem?

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Here are five related questions you might find interesting:
1. How is the surface area of the sphere changing over time?
2. How would the problem change if the rate of volume increase were different?
3. What would the radius of the sphere be after 2 seconds, assuming constant volume rate?
4. How can we find the surface area at \( t = 1 \)?
5. How does the volume change relate to the surface area change?

**Tip**: When working with rates of change in geometry, always differentiate carefully and identify how variables are linked through geometric relationships!

A sphere is growing, its volume is increasing at a constant rate of 10 m³ per second. Let r(t), V(t), and S(t) be the radius, volume, and surface area of the sphere at time t. If r(1) = 2, compute: 

This problem involves calculating the rate of change of a sphere's radius as its volume increases at a constant rate of 10 m³/second. Learn how to apply related rates using differentiation formulas for volume and surface area of a sphere, suitable for students in grades 11-12.

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