9 min readPathway sheet2-mark and 3-mark questions; the second derivative version is a common 3-mark question; often a case-based question on a moving object

Derivatives of functions in parametric form

When x = f(t) and y = g(t), find dx/dt and dy/dt separately; then dy/dx = (dy/dt) ÷ (dx/dt), provided dx/dt ≠ 0.

Parametric formx = f(t) and y = g(t): both are given in terms of a third variable t, the parameter

First derivativedy/dx = (dy/dt) ÷ (dx/dt), provided dx/dt ≠ 0

Second derivatived²y/dx² = [d/dt (dy/dx)] ÷ (dx/dt)

Common slipd²y/dx² is not (d²y/dt²) ÷ (d²x/dt²)

What parametric form is

Sometimes x and y are both given through a third variable, called a parameter (often t or θ). For example, x = 4 cos t, y = 3 sin t traces an ellipse as t changes. Instead of removing t, differentiate x and y separately with respect to t and divide: dy/dx = (dy/dt)/(dx/dt).

Method

1
Find dx/dt.
2
Find dy/dt.
3★
Divide: dy/dx = (dy/dt) ÷ (dx/dt), with dx/dt ≠ 0.
4
Simplify, keeping the answer in terms of t.
5
If a value of t is given, put it in at the end.

★ key step

Example 1: an ellipse

Problem

If x = 4 cos t and y = 3 sin t, find dy/dx at t = π/4.

Working

dx/dt = −4 sin t and dy/dt = 3 cos t
dy/dx = (3 cos t)/(−4 sin t) = −(3/4) cot t
At t = π/4, cot(π/4) = 1.

Answer

dy/dx = −3/4 at t = π/4

Example 2: with eᵗ

Problem

If x = eᵗ cos t and y = eᵗ sin t, find dy/dx.

Working

Product rule: dx/dt = eᵗ cos t − eᵗ sin t = eᵗ(cos t − sin t)
dy/dt = eᵗ sin t + eᵗ cos t = eᵗ(sin t + cos t)
dy/dx = eᵗ(sin t + cos t)/[eᵗ(cos t − sin t)]
eᵗ cancels.

Answer

dy/dx = (sin t + cos t)/(cos t − sin t)

Example 3: second derivative

Problem

If x = sec θ and y = tan θ, find d²y/dx².

Working

dx/dθ = sec θ tan θ and dy/dθ = sec²θ
dy/dx = sec²θ/(sec θ tan θ) = sec θ/tan θ = cosec θ
d/dθ (dy/dx) = d/dθ (cosec θ) = −cosec θ cot θ
d²y/dx² = (−cosec θ cot θ) ÷ (sec θ tan θ)
= −(cos θ/sin²θ) × (cos²θ/sin θ) = −cos³θ/sin³θ

Answer

d²y/dx² = −cot³θ

In real life

The path of a cricket ball or a javelin is easiest to describe in parametric form: x and y both depend on time t. Then dy/dx gives the direction of motion at any moment.

Exam angle

2 marks: “If x = t + 1/t and y = t − 1/t, find dy/dx.” dx/dt = 1 − 1/t² and dy/dt = 1 + 1/t². dy/dx = (1 + 1/t²)/(1 − 1/t²) = (t² + 1)/(t² − 1).

3 marks: “If x = 2(θ + sin θ) and y = 2(1 − cos θ), find dy/dx at θ = π/2.” dx/dθ = 2(1 + cos θ) and dy/dθ = 2 sin θ. dy/dx = sin θ/(1 + cos θ) = tan(θ/2). At θ = π/2, dy/dx = tan(π/4) = 1.

3 marks: “If x = t² and y = t³, find d²y/dx².” dy/dx = 3t²/(2t) = 3t/2. d²y/dx² = [d/dt (3t/2)] ÷ (dx/dt) = (3/2) ÷ 2t = 3/(4t).

Remember it

dy over dt, divided by dx over dt.