Learn the method, study an example, then explain your reasoning in an exam-style problem. Work on paper before opening the solution and compare each step with the marking guidance.
Related rates and small increments
Relate changing quantities through a derivative, keeping signs, units and approximation status.
Notes
Before you start: differentiate powers, use geometric formulas and convert units consistently.
If y=f(x) and x changes smoothly with time t, the chain rule gives
\[\frac{dy}{dt}=\frac{dy}{dx}\frac{dx}{dt}.\]
Evaluate the geometric derivative at the specified instant. A negative rate means a quantity is decreasing; it is not an instruction to discard a solution. If a volume has units cm³ and time has units s, its rate has units cm³/s, while a radius rate has units cm/s.
For a small change Δx, the linear estimate is \(\Delta y\approx f^{\prime}(x)\Delta x\). Use the derivative at the initial x. The exact change is f(x+Δx)−f(x); the two need not agree. Similarly, over a short time step, \(x(t+\Delta t)\approx x(t)+\dot x(t)\Delta t\). This is a local approximation, not permission to assume the rate stays constant indefinitely.
Recall later
If y=x², then Δy=2xΔx+(Δx)² exactly. The first-order estimate omits (Δx)².
Common mistakes
Confusing a geometric derivative with a time derivative.
Dropping a negative rate or using inconsistent length/volume units.
Writing an approximate small-increment relation as an exact finite change.
Three worked examples
Example 1
A sphere has V=4πr³/3. At r=2 cm, dr/dt=0.1 cm/s. Find dV/dt.
Show the reasoning
\(dV/dr=4\pi r^2\), so \(dV/dt=4\pi(2^2)(0.1)=1.6\pi\) cm³/s. Both the sign and the volume-per-time units are retained.
Example 2
An area y=3x² cm² depends on a length x cm. At x=5, dx/dt=−0.4 cm/s. Find dy/dt.
Show the reasoning
\(dy/dx=6x\), so \(dy/dt=6(5)(-0.4)=-12\) cm²/s. The area is decreasing at 12 cm²/s.
Example 3
Use differentiation to estimate the change in x² when x increases from 4 to 4.03; compare with the exact change.
Show the reasoning
Δx=0.03 and f′(4)=8, so Δy≈8(0.03)=0.24. The exact change is 4.03²−4²=0.2409. The linear estimate is low by 0.0009, the omitted square (0.03)².
