Parametric forms and second derivatives

Sometimes x and y are both given in terms of a third variable t, called a parameter: x = f(t), y = g(t). This is the parametric form.

Do not try to remove t. Instead:

dy/dx = (dy/dt) ÷ (dx/dt), as long as dx/dt ≠ 0.

Example: x = a cos θ, y = b sin θ. dx/dθ = –a sin θ, dy/dθ = b cos θ. So dy/dx = –(b/a) cot θ. The answer stays in terms of the parameter.

The second derivative d²y/dx² (also written y″ or y₂) is the derivative of dy/dx. Example: y = e^(2x) gives y′ = 2e^(2x) and y″ = 4e^(2x), so y″ – 4y = 0.

Questions often say 'show that y″ + ... = 0'. Differentiate twice, then put the original y back in. Example: y = sin 3x gives y″ = –9 sin 3x = –9y, so y″ + 9y = 0.

A common trap in parametric second derivatives: dy/dx is in terms of t, so differentiate it with respect to t and then divide by dx/dt again. Example: x = t², y = t³. dy/dx = 3t²/2t = 3t/2. Then d²y/dx² = (3/2) ÷ (2t) = 3/(4t), not just 3/2.

Tip: if dy/dx has a square root or fraction, first square or cross-multiply to clear it, then differentiate again. The working becomes much shorter.