Rate of Change — Volume of Sphere When Radius Increases

Question

The radius of a sphere is increasing at the rate of 0.2 cm/s. At what rate is the volume of the sphere increasing when the radius is 15 cm?

Solution — Step by Step

Why This Works

When we differentiate $V = \frac{4}{3}\pi r^3$ with respect to $t$, the chain rule forces us to multiply by $\frac{dr}{dt}$. We’re not asking “how does $V$ change as $r$ changes?” — we’re asking “how does $V$ change as time passes?” That extra $\frac{dr}{dt}$ converts the spatial relationship into a time-based one.

The result $\frac{dV}{dt} = 4\pi r^2 \cdot \frac{dr}{dt}$ has a beautiful interpretation: the surface area of the sphere acts as the multiplier. Think of it this way — a thin shell of thickness $dr$ added to a sphere of radius $r$ has volume $\approx 4\pi r^2 \cdot dr$. That’s exactly what we’re computing when $r$ grows by $dr$ in time $dt$.

This formula appears in NCERT Chapter 6 and has shown up in CBSE board papers almost every alternate year since 2018. At large $r$, even a slow $\frac{dr}{dt}$ produces a huge $\frac{dV}{dt}$ — because the surface area is large.

Alternative Method

Alternatively, express $V$ in terms of $t$ symbolically. If $r(t) = r_0 + 0.2t$, then $V(t) = \frac{4}{3}\pi(r_0 + 0.2t)^3$. Differentiating: $\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3(r_0 + 0.2t)^2 \cdot 0.2$, which at $r = 15$ gives the same $180\pi$.

This approach is more verbose but makes the time-dependence explicit — useful if you want to verify your chain rule work.

Common Mistake