Applications of Derivatives

Rate of Change of Quantities

Understand

If \(y=f(x)\), then \(\frac{dy}{dx}\) (or \(f\,'(x_0)\) at a point) is the rate of change of \(y\) with respect to \(x\). When both quantities depend on time \(t\), the chain rule links their rates:

\[\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}\]

Example. The radius of a circle grows at \(0.5\) cm/s. How fast is the area growing when \(r=4\) cm? \(A=\pi r^2\Rightarrow\frac{dA}{dt}=2\pi r\frac{dr}{dt}=2\pi(4)(0.5)=4\pi\) cm²/s.

Strategy for related rates

  1. Draw a diagram; name the variables and write the given rates with signs (decreasing ⇒ negative).
  2. Write an equation linking the variables (area, volume, Pythagoras, similar triangles).
  3. Differentiate with respect to t.
  4. Substitute the values at the required instant – only after differentiating.

Economics. If \(C(x)\) is the total cost of producing \(x\) units, the marginal cost is \(C\,'(x)\), the approximate cost of one more unit. Likewise marginal revenue is \(R\,'(x)\).

Key Concepts

  • Rate of change = derivative.
  • Related rates: differentiate the linking equation with respect to t.
  • Substitute values only after differentiating.
  • Marginal cost = C′(x); marginal revenue = R′(x).

Formula Bank

Related rates

\[\frac{dy}{dt}=\frac{dy}{dx}\cdot\frac{dx}{dt}\]

Common Mistakes

Substituting too early

Differentiate \(V=\frac43\pi r^3\) first, then put \(r=5\). If you substitute first, \(V\) becomes a constant with derivative 0.

Sign of a decreasing rate

"Decreasing at 2 cm/s" means \(\frac{dx}{dt}=-2\).

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Topic Summary

Write the relation, differentiate with respect to time, then substitute the instant’s values. Units of a rate: (units of y) per (unit of t).

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