Parametric Derivatives

You’ll be able to

Compute dy/dx for a parametrically defined curve
Find the second derivative d²y/dx² of a parametric curve
Locate horizontal and vertical tangents from the parametric derivatives

Curves traced by a parameter

A parametric curve gives x and y each as functions of a parameter t: x = x(t), y = y(t). As t advances, the point (x(t), y(t)) traces a path — which may loop or cross itself, something a plain y = f(x) cannot do. To do calculus we need the slope dy/dx of that path, but x and y are both tied to t, so we relate their t-derivatives rather than differentiating y with respect to x directly.

Parametric first derivative

dy/dx = (dy/dt) / (dx/dt), provided dx/dt ≠ 0

The slope of the curve is the ratio of the two t-derivatives — a direct consequence of the chain rule.

The second derivative

The second derivative d²y/dx² measures concavity, but you cannot just divide second t-derivatives. Instead, take the t-derivative of the first derivative dy/dx (itself a function of t) and divide by dx/dt again: d²y/dx² = (d/dt[dy/dx]) / (dx/dt). The recurring principle is "differentiate with respect to t, then divide by dx/dt." Skipping that final division by dx/dt is the most common error in parametric second derivatives.

Horizontal and vertical tangents

Tangent lines reveal themselves through the two t-derivatives. A horizontal tangent occurs where dy/dt = 0 (and dx/dt ≠ 0) — the numerator of the slope vanishes. A vertical tangent occurs where dx/dt = 0 (and dy/dt ≠ 0) — the denominator vanishes, sending the slope to infinity. If both derivatives are zero at once, the point needs closer analysis (it may be a cusp).

Worked example

A curve is given by x = t² and y = t³ − 3t. Find dy/dx and locate any horizontal tangents.

1.Differentiate each with respect to t: dx/dt = 2t and dy/dt = 3t² − 3.
2.Form the slope: dy/dx = (3t² − 3)/(2t).
3.Horizontal tangents need dy/dt = 0: 3t² − 3 = 0 gives t = ±1 (and dx/dt = ±2 ≠ 0 there).
4.So horizontal tangents occur at t = 1 and t = −1.

Answer: dy/dx = (3t² − 3)/(2t), with horizontal tangents at t = ±1, where the numerator (dy/dt) is zero but the denominator is not.

Watch out

For d²y/dx², divide by dx/dt — not by (dx/dt)² and not by d²x/dt². The rule is: differentiate dy/dx with respect to t, then divide once more by dx/dt. Getting the denominator wrong is the signature parametric mistake.

Checkpoint

For x = t² and y = 4t, what is dy/dx?

Checkpoint

A parametric curve has a vertical tangent line where:

On the exam

Keep the pattern straight: dy/dt = 0 → horizontal tangent, dx/dt = 0 → vertical tangent. Because the slope is a fraction (dy/dt)/(dx/dt), the numerator controls "flat" and the denominator controls "steep/vertical."

Answer the 2 checkpoints as you read.