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AP Calculus BC · Unit 9: Parametric, Polar & Vector Functions

9.2 Second Derivatives of Parametric Equations

Finding the second derivative

For parametric equations x(t) and y(t), the first derivative is dy/dx = (dy/dt)/(dx/dt). To get the second derivative, don't differentiate dy/dx directly with respect to t — instead compute d/dt(dy/dx) and divide by dx/dt again: d²y/dx² = [d/dt(dy/dx)] / (dx/dt).

Common mistake to avoid

A frequent error is writing d²y/dx² = (d²y/dt²)/(d²x/dt²). This is wrong. You must treat dy/dx as a new function of t, differentiate it with respect to t, then divide by dx/dt — not divide the raw second derivatives of x and y.

Using the second derivative

Once you have d²y/dx², you can analyze concavity of the parametric curve. If d²y/dx² > 0, the curve is concave up; if < 0, concave down. Setting d²y/dx² = 0 (where dx/dt ≠ 0) helps locate points of inflection, just like in standard calculus.

Key terms

Formula for d²y/dx² (parametric)
d²y/dx² = [d/dt(dy/dx)] / (dx/dt)
Common mistake
Writing d²y/dx² = (d²y/dt²)/(d²x/dt²) — this is incorrect; must differentiate dy/dx first.
Steps to find second derivative
1) Find dy/dx as a function of t. 2) Differentiate that with respect to t. 3) Divide the result by dx/dt.
Concavity from d²y/dx²
Positive → concave up; Negative → concave down (same interpretation as in standard calculus).
Finding points of inflection parametrically
Set d²y/dx² = 0 (where dx/dt ≠ 0) and solve for t to locate potential inflection points.

Try an AP-style question

How is d²y/dx² found for a parametric curve?

  1. Differentiate dy/dx with respect to t, then divide by dx/dt
  2. Differentiate dy/dx with respect to t, and stop there
  3. Square the first derivative dy/dx, then divide by t
  4. Divide d²y/dt² by d²x/dt², the two second derivatives
Show the answer

A. d²y/dx² = [d/dt(dy/dx)] / (dx/dt).

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