Dabour Hub EGEgyptian Curriculum 2026/2027
Parametric differentiation

Parametric differentiation

Point 3 of 5 — Implicit and Parametric Differentiation and Connected Rates of ChangeQuestion bank: 0 questions (0 with pictures)

03 Parametric differentiation

If x = f(t) and y = g(t), then by the chain rule

  • dy/dx = (dy/dt) ÷ (dx/dt), provided dx/dt ≠ 0.
  • d²y/dx² = [d/dt (dy/dx)] ÷ (dx/dt) — a frequent error is to divide d²y/dt² by d²x/dt², which is wrong.

Worked example 1 — a circle

x = 5 cos t, y = 5 sin t. dx/dt = −5 sin t, dy/dt = 5 cos t ⟹ dy/dx = −cot t. At t = π/3: dy/dx = −cot(π/3) = −1/√3 at the point (2.5, 4.33).

Worked example 2 — a parabola

x = t², y = t³ − 3t. dy/dx = (3t² − 3)/(2t). At t = 1 the slope is 0 (horizontal tangent at (1, −2)); the tangent is y = −2.

  • d/dt(dy/dx) = d/dt[(3/2)(t − t⁻¹)] = (3/2)(1 + t⁻²); at t = 1 this is 3, and dx/dt = 2 ⟹ d²y/dx² = 3/2 > 0, so (1, −2) is a local minimum.

Motion reading

With t as time, dx/dt and dy/dt are the velocity components, the speed is √[(dx/dt)² + (dy/dt)²], and the slope dy/dx is the direction of motion.

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