How Fast Does a Ripple Spread When a Stone Hits a Pond?

When a stone drops into a still pond, circular ripples spread out and the disturbed area grows every second. How fast it grows is a derivative — and the same idea tells a business whether its costs are rising, or a doctor how quickly a medicine level is falling.

This part covers derivatives as rates of change, increasing and decreasing functions, and applications to areas, volumes and costs.

**The derivative $\frac{dy}{dx}$ is the rate at which $y$ changes with $x$, and in related rates problems two quantities that both depend on time are linked by an equation that is differentiated with respect to time using the chain rule.

Rates through time.** If $y$ depends on $x$ and $x$ depends on time $t$:

$$\frac{dy}{dt}=\frac{dy}{dx}\cdot \frac{dx}{dt}$$

Worked example 1 — a spreading ripple. The radius grows at $3$ cm/s. How fast is the area growing when $r=10$ cm?

$$A=\pi r^2\Rightarrow \frac{dA}{dt}=2\pi r\frac{dr}{dt}=2\pi (10)(3)=60\pi {\text{ cm}}^2/\text{s}$$

Worked example 2 — a balloon. A spherical balloon's radius grows at $0.5$ cm/s. When $r=6$ cm:

$$V=\frac{4}{3}\pi r^3\Rightarrow \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}=4\pi (36)(0.5)=72\pi {\text{ cm}}^3/\text{s}$$

Method:

- Name the variables and note the rates given and asked for
- Write an equation linking the variables
- Differentiate with respect to $t$, and substitute values only after differentiating

An everyday example. Blowing air into a balloon at a steady rate, its radius grows quickly at first and then more slowly, because the same volume is spread over a larger surface.

The substance. Substituting numbers before differentiating turns variables into constants and wrongly gives a rate of zero.

**On an interval where $f^{\prime }(x)>0$ the function is strictly increasing, where $f^{\prime }(x)<0$ it is strictly decreasing, and where $f^{\prime }(x)=0$ throughout it is constant.

The test**, for $f$ continuous on $[a,b]$ and differentiable on $(a,b)$:

- $f^{\prime }(x)>0$ on $(a,b)$ — strictly increasing on $[a,b]$
- $f^{\prime }(x)<0$ on $(a,b)$ — strictly decreasing on $[a,b]$
- $f^{\prime }(x)=0$ on $(a,b)$ — constant on $[a,b]$

Worked example — find the intervals. $f(x)=x^3-3x^2-9x+5$.

$$f^{\prime }(x)=3x^2-6x-9=3(x-3)(x+1)$$

$f^{\prime }(x)=0$ at $x=-1$ and $x=3$, which split the line into three intervals:

- $(-\infty ,-1)$: $f^{\prime }(-2)=3(-5)(-1)=15>0$ — increasing
- $(-1,3)$: $f^{\prime }(0)=-9<0$ — decreasing
- $(3,\infty )$: $f^{\prime }(4)=3(1)(5)=15>0$ — increasing

Always increasing. $f(x)=e^x$ has $f^{\prime }(x)=e^x>0$ everywhere, so it is strictly increasing on $\mathbb{R}$.

An everyday example. A savings balance with regular deposits and no withdrawals is an increasing function of time — its rate of change never turns negative.

The substance. **$f^{\prime }(x)>0$ is sufficient but not necessary** — $x^3$ is strictly increasing on $\mathbb{R}$ even though its derivative is $0$ at $x=0$.

Writing a real quantity such as area, volume, cost or revenue as a function and studying the sign of its derivative shows when the quantity rises or falls, while the derivative's value shows how fast.

Worked example 1 — marginal cost. Producing $x$ units costs $C(x)=0.005x^3-0.02x^2+30x+5000$ rupees. The marginal cost is

$$C^{\prime }(x)=0.015x^2-0.04x+30$$

$$C^{\prime }(3)=0.015(9)-0.04(3)+30=0.135-0.12+30=30.015$$

so producing one more unit at that level costs about ₹30.02.

Worked example 2 — a melting cube. An ice cube's edge shrinks at $0.1$ cm/min. When the edge is $5$ cm:

$$S=6a^2\Rightarrow \frac{dS}{dt}=12a\frac{da}{dt}=12(5)(-0.1)=-6{\text{ cm}}^2/\text{min}$$

The negative sign shows the surface area is decreasing.

Worked example 3 — where revenue rises. $R(x)=40x-x^2$ gives $R^{\prime }(x)=40-2x$, positive for $x<20$, so revenue increases up to $20$ units and decreases after that.

An everyday example. A tea stall owner who finds each extra hour open brings in less extra money is watching a revenue function whose derivative is shrinking.

The substance. A negative rate is not an error — it simply means the quantity is decreasing.