Derivative as a Rate measure

If $y=f(x)$ is a given function, a change in $x$ from $x_1$ to $x_2$ is denoted $\delta x=x_2-x_1$, and the corresponding change in $y$ is $\delta y=f(x_2)-f(x_1)$. The ratio $\dfrac{\delta y}{\delta x}=\dfrac{f(x_2)-f(x_1)}{x_2-x_1}$ is called the average rate of change of $y$ with respect to $x$ over that interval; geometrically it is the slope of the secant line joining $P(x_1,f(x_1))$ and $Q(x_2,f(x_2))$ on the graph.

Letting $x_2$ approach $x_1$ (equivalently, letting $\delta x\to0$), the limit of this average rate of change is called the instantaneous rate of change of $y$ with respect to $x$ at $x=x_1$:

This limit is exactly the derivative $f'(x_1)$. So the derivative has two equivalent interpretations: it is the instantaneous rate of change of $y=f(x)$ with respect to $x$ at $x=a$, and it is also the slope of the tangent to $y=f(x)$ at $(a,f(a))$.

When several quantities are all changing with time (or with each other), and they are linked by a known geometric or physical formula, the chain rule lets us relate their rates: if $y$ depends on $x$ which in turn depends on time $t$, then $\dfrac{dy}{dt}=\dfrac{dy}{dx}\cdot\dfrac{dx}{dt}$. In practice: write down the formula connecting the changing quantities, differentiate both sides with respect to time, then substitute the given numerical rate(s) and the value(s) at the instant asked about.

Worked Example 1 — Expanding circular wave. A stone dropped into a lake creates a circular wave whose radius increases at $5$ cm/sec. Letting $R$ be the radius and $A=\pi R^2$ the enclosed area: $\dfrac{dA}{dt}=2\pi R\dfrac{dR}{dt}$. At $R=8$ cm, with $\dfrac{dR}{dt}=5$: $\dfrac{dA}{dt}=2\pi(8)(5)=80\pi$ cm²/sec — the area is increasing at $80\pi$ cm²/sec when the radius is $8$ cm.

Worked Example 2 — Spherical balloon, volume and surface area both asked. The volume of a spherical ball increases at $4\pi$ cc/sec; find the rates of change of the radius and surface area when the volume is $288\pi$ cc. With $V=\dfrac43\pi R^3$: differentiating gives $4\pi=\dfrac{4\pi}{3}(3R^2)\dfrac{dR}{dt}$, so $\dfrac{dR}{dt}=\dfrac{1}{R^2}$. Setting $\dfrac43\pi R^3=288\pi$ gives $R^3=216\Rightarrow R=6$, so $\dfrac{dR}{dt}=\dfrac{1}{36}$. For the surface area $S=4\pi R^2$: $\dfrac{dS}{dt}=8\pi R\dfrac{dR}{dt}$; at $R=6$, $\dfrac{dS}{dt}=8\pi(6)\left(\dfrac{1}{36}\right)=\dfrac{4\pi}{3}$ cm²/sec.

Worked Example 3 — Water filling a cylindrical vessel. Water is poured at $36$ m³/sec into a cylindrical vessel of fixed base radius $3$ m. With $V=\pi R^2H=9\pi H$ (since $R=3$ is fixed): $\dfrac{dV}{dt}=9\pi\dfrac{dH}{dt}$, so $\dfrac{dH}{dt}=\dfrac{1}{9\pi}\dfrac{dV}{dt}=\dfrac{36}{9\pi}=\dfrac{4}{\pi}$ meter/sec — the water level rises at $\dfrac4\pi$ m/sec. …