A spherical balloon is inflated so that its volume increases at a rate of 100 cubic centimeters per second. How fast is the radius increasing when the radius is 10 cm? - Parker Core Knowledge

Understanding the Context

Why This Inflation Rate Matters Now

The steady flow of 100 cubic centimeters per second isn’t just a classroom example—it reflects real-world pressures shaping product development, safety standards, and creative engineering. In an era where automation and precision matter, understanding volume expansion rates helps designers ensure integrity, efficiency, and performance.

This steady inflow sparks curiosity because it’s relatable: imagine inflating balloons for events, balloons used in science kits, or even inflatable toys and air accessories. When volume increases predictably, it enables better planning—whether in packaging, air storage, or even medical inflation systems requiring controlled growth. This kind of predictable behavior is key to innovation across industries.

Key Insights

How Volume Growth Translates to Radius Expansion

At the heart of the question lies a simple yet profound physics principle: volume and surface geometry are deeply connected in a sphere. When volume increases, the balloon expands, and the radius grows at a regular rate determined by the derivative relationship.

Starting from the formula for the volume of a sphere, V = (4/3)πr³, differentiating with respect to time shows how volume rate (dV/dt) links directly to radius growth (dr/dt):

dV/dt = 4πr² × dr/dt

Final Thoughts

With dV/dt = 100 cm³/s, and r = 10 cm, solve for dr/dt.

Plugging in:
100 = 4π(10)² × dr/dt → 100 = 400π × dr/dt
dr/dt = 100 / (400π) = 1 / (4π) cm/s ≈ 0.0796 cm/s

This means the radius is increasing by roughly 0.08 centimeters per second—slow