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?
- Differentiate dy/dx with respect to t, then divide by dx/dt
- Differentiate dy/dx with respect to t, and stop there
- Square the first derivative dy/dx, then divide by t
- 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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