Expanding Ripple: How Fast Does the Area Grow?

A pebble drops into a still pond. The ripple spreads outward as a perfect circle, with the radius growing at a constant rate. The area, however, does not grow at a constant rate — it grows faster as the ripple gets bigger, because there's more circumference to push outward.

This is the most elegant related-rates problem in calculus: a single chain rule, two derivatives, and you're done.

The Physics

For a circle, the area as a function of radius is:

$A = \pi r^{2}$

If both $A$ and $r$ depend on time, differentiate both sides with respect to $t$:

$\dfrac{dA}{dt} = \dfrac{dA}{dr}\cdot\dfrac{dr}{dt} = 2\pi r\,\dfrac{dr}{dt}$

This is the chain rule in action — and it tells us the area grows at a rate proportional to the current radius times the rate of radius growth.

Step-by-Step Solution

Given:

• $\dfrac{dr}{dt} = 2\;\text{m/s}$ (constant outward expansion)
• Snapshot: $r = 10\;\text{m}$

Find: $\dfrac{dA}{dt}$ at that instant.

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Step 1 — Write down the area formula.

$A(t) = \pi\,[r(t)]^{2}$

Both $A$ and $r$ are functions of time, but the formula relating them at any instant is just $A = \pi r^{2}$.

Step 2 — Differentiate with respect to time using the chain rule.

The "outer function" is $\pi u^{2}$ where $u = r(t)$. The derivative of $\pi u^{2}$ with respect to $u$ is $2\pi u$. Then multiply by $\dfrac{du}{dt} = \dfrac{dr}{dt}$:

$\dfrac{dA}{dt} = 2\pi r\,\dfrac{dr}{dt}$

Step 3 — Substitute the given values at the snapshot.

Plug $r = 10$ and $\dfrac{dr}{dt} = 2$:

$\dfrac{dA}{dt} = 2\pi (10)(2) = 40\pi$

Step 4 — Convert to a decimal.

$\dfrac{dA}{dt} = 40\pi \approx 40 \times 3.14159 \approx 125.66\;\text{m}^{2}/\text{s}$

Step 5 — Compare to a smaller radius.

How does this compare to the rate when $r$ was just 1 m?

$\dfrac{dA}{dt}\bigg|_{r=1} = 2\pi(1)(2) = 4\pi \approx 12.57\;\text{m}^{2}/\text{s}$

So at $r = 10$ the area is growing 10 times faster than at $r = 1$, even though the radius is growing at the same constant rate. The area rate is proportional to the circumference ($2\pi r$) — which makes sense: a larger ripple has more boundary to push outward.

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Answer: When the radius reaches $10\;\text{m}$, the area is growing at the rate

$\boxed{\;\dfrac{dA}{dt} = 40\pi \approx 125.66\;\text{m}^{2}/\text{s}\;}$

The rate scales linearly with $r$ — every additional meter of radius adds another $4\pi \approx 12.57\;\text{m}^{2}/\text{s}$ to the rate.

Try It

• The animation runs forward in time — watch the radius grow at a constant 2 m/s while the area accelerates.
• The HUD shows the current radius and current $dA/dt$ in real time.
• Adjust the radial speed widget — the area rate scales linearly.
• The counter in the corner accumulates the total area swept.