Second Derivatives and Concavity of Parametric Curves

You’ll be able to

Compute d²y/dx² for a parametric curve using the correct two-step quotient
Determine concavity of a parametric curve at a given parameter value
Locate horizontal and vertical tangents on a parametric curve

The second derivative is not the second derivatives divided

For a parametric curve, dy/dx = (dy/dt)/(dx/dt). The tempting extension — d²y/dx² = (d²y/dt²)/(d²x/dt²) — is false, and it is the most-punished error in Unit 9. The correct rule differentiates the first derivative expression with respect to t and then divides by dx/dt again: d²y/dx² = [d/dt(dy/dx)] / (dx/dt). The extra division by dx/dt is what converts a t-rate back into an x-rate, and it never goes away.

Why the extra division appears

d²y/dx² means "the rate at which the slope dy/dx changes per unit x." You can only differentiate with respect to t, which gives the change per unit t. Converting that to a per-unit-x rate is one more application of the chain rule: (d/dt)(dy/dx) · (dt/dx), and dt/dx is 1/(dx/dt). Seeing it as a chain rule rather than a memorized formula makes the structure hard to forget.

Horizontal and vertical tangents

A horizontal tangent occurs where dy/dt = 0 while dx/dt ≠ 0 — the curve is momentarily moving purely sideways. A vertical tangent occurs where dx/dt = 0 while dy/dt ≠ 0. When both vanish at the same t the point is a cusp or a stop, and the tangent must be investigated with a limit rather than declared. Concavity then follows from the sign of d²y/dx²: positive is concave up, negative is concave down, exactly as for a function of x.

Parametric derivatives

dy/dx = (dy/dt)/(dx/dt) · d²y/dx² = [d/dt (dy/dx)] / (dx/dt)

Differentiate the slope expression with respect to t, then divide by dx/dt once more. Never divide the second t-derivatives.

Worked example

A curve is given by x = t² + 1 and y = t³ − 3t. Find d²y/dx² at t = 2 and state the concavity there.

1.dx/dt = 2t and dy/dt = 3t² − 3, so dy/dx = (3t² − 3)/(2t).
2.Differentiate dy/dx with respect to t using the quotient rule: d/dt[(3t² − 3)/(2t)] = [(6t)(2t) − (3t² − 3)(2)]/(2t)² = (12t² − 6t² + 6)/(4t²) = (6t² + 6)/(4t²).
3.Divide by dx/dt = 2t: d²y/dx² = [(6t² + 6)/(4t²)] / (2t) = (6t² + 6)/(8t³).
4.At t = 2: (6·4 + 6)/(8·8) = 30/64 = 15/32 ≈ 0.469.
5.Since d²y/dx² > 0 at t = 2, the curve is concave up there.

Answer: d²y/dx² = 15/32 at t = 2, so the curve is concave up. The final division by dx/dt = 2t is the step that distinguishes the right answer from the common wrong one.

Watch out

If your parametric second derivative came out as a ratio of two second derivatives, it is wrong. A quick check: the correct expression always has dx/dt appearing in the denominator an extra time.

Checkpoint

For a parametric curve, d²y/dx² equals which expression?

Checkpoint

At t = 3, a curve has dx/dt = 0 and dy/dt = 5. What does the curve have at that point?

On the exam

Parametric second-derivative questions are usually multiple choice and usually include the ratio-of-second-derivatives distractor. Compute dy/dx first, write it down, then differentiate that written expression — the discipline prevents the shortcut.

Answer the 2 checkpoints as you read.