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AP Calculus BC — Unit 9.2

Second Derivatives of Parametric Equations

Key Concepts Packet + Worked Examples + Interactive MCQ Practice (36 Questions)

Unit 9.2 Key Concepts Packet
1) The Second Derivative for Parametric Curves
Core Formulas
First derivative: \( \displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} \) (when \( \frac{dx}{dt}\ne0 \))
Second derivative: \( \displaystyle \frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}} \) (also requires \( \frac{dx}{dt}\ne0 \))
  • The expression \( \frac{d}{dt}\left(\frac{dy}{dx}\right) \) is computed by differentiating your \(dy/dx\) (as a function of \(t\)) with respect to \(t\).
  • You must divide by \(dx/dt\) again to convert “per \(t\)” into “per \(x\).”
2) Reliable Step-by-Step Process (AP-Friendly)
The 3-step workflow
  1. Compute \(dx/dt\) and \(dy/dt\).
  2. Compute \(dy/dx=(dy/dt)/(dx/dt)\) and simplify.
  3. Differentiate \(dy/dx\) with respect to \(t\), then divide by \(dx/dt\): \( \displaystyle d^2y/dx^2=\big(d/dt(dy/dx)\big)/(dx/dt) \).
  • Always simplify \(dy/dx\) first; it makes the second derivative dramatically easier.
  • If the question asks “at \(t=t_0\),” plug in at the end (unless simplification suggests otherwise).
3) Concavity + Inflection Points in Parametric Form
Concavity
\( \frac{d^2y}{dx^2}>0 \) → concave up    \( \frac{d^2y}{dx^2}<0 \) → concave down
  • Inflection points occur where concavity changes (typically where \(d^2y/dx^2=0\) or undefined) and the curve is continuous through the point.
  • Parametric curves can have special points (cusps/loops) where \(dy/dx\) or \(d^2y/dx^2\) is undefined—handle those with care.
4) When the Formula Breaks (and What to Do)
The key warning
If \(dx/dt=0\) at a parameter value, then both \(dy/dx\) and \(d^2y/dx^2\) may be undefined using the standard formulas.
  • If needed, eliminate \(t\) (when possible) and analyze in Cartesian form.
  • Or analyze the behavior near the parameter value (limits/sign charts) to determine concavity changes.
5) Calculator Expectations + Common AP Traps
  • Do not confuse \(d^2y/dt^2\) with \(d^2y/dx^2\). The AP question almost always wants the latter.
  • When solving for inflection points, confirm a sign change in \(d^2y/dx^2\).
  • If \(dx/dt=0\), the standard second-derivative formula is not valid at that parameter value.
Worked Examples + Notes

Example 1 — Compute \( \left.\frac{d^2y}{dx^2}\right|_{t=1} \)

Given \(x=t^2\) and \(y=t^3\), find \( \left.\frac{d^2y}{dx^2}\right|_{t=1} \).

Step 1: \(dx/dt=2t\), \(dy/dt=3t^2\).
Step 2: \(dy/dx=\frac{3t^2}{2t}=\frac{3t}{2}\) (for \(t\ne0\)).
Step 3: \( \frac{d}{dt}(dy/dx)=\frac{d}{dt}\left(\frac{3t}{2}\right)=\frac{3}{2} \).
Step 4: \( d^2y/dx^2=\frac{(d/dt)(dy/dx)}{dx/dt}=\frac{\frac{3}{2}}{2t}=\frac{3}{4t} \).
Step 5: At \(t=1\), \(d^2y/dx^2=\frac{3}{4}\).

Example 2 — Concavity at a parameter value

For \(x=t\), \(y=\sin t\), determine concavity at \(t=\frac{\pi}{2}\).

Step 1: Since \(x=t\), we have \(dy/dx=dy/dt=\cos t\).
Step 2: \(d^2y/dx^2=d/dx(\cos t)=d/dt(\cos t)=-\sin t\) (because \(dx/dt=1\)).
Step 3: At \(t=\frac{\pi}{2}\), \(d^2y/dx^2=-1<0\) so the curve is concave down.

Example 3 — Second derivative with trig param

For \(x=\cos t\), \(y=\sin t\), compute \(d^2y/dx^2\).

Step 1: \(dx/dt=-\sin t\), \(dy/dt=\cos t\).
Step 2: \(dy/dx=\frac{\cos t}{-\sin t}=-\cot t\).
Step 3: \(d/dt(dy/dx)=d/dt(-\cot t)=\csc^2 t\).
Step 4: \(d^2y/dx^2=\frac{\csc^2 t}{-\sin t}=-\csc^3 t\).

Example 4 — Inflection point in param form (easy case)

For \(x=t\), \(y=t^3\), identify the inflection point.

Step 1: Since \(x=t\), we have \(y=x^3\).
Step 2: \(y''=6x\), which changes sign at \(x=0\).
Conclusion: Inflection point is \((0,0)\).
Note: If you ever see raw backslashes instead of formatted math, it means MathJax did not typeset that section—this page forces typesetting on load and when solutions open.
Unit 9.2 Multiple-Choice Practice (36 Questions)
Answer Key