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AP Calculus BC — Unit 9.1

Defining and Differentiating Parametric Equations

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

Unit 9.1 Key Concepts Packet
1) What Are Parametric Equations?
Definition
A parametric curve is given by \( x=x(t) \), \( y=y(t) \), where \(t\) is the parameter.
  • The parameter \(t\) controls both \(x\) and \(y\) simultaneously.
  • Parametric form preserves direction/orientation as \(t\) increases.
  • Eliminating \(t\) can lose orientation and may introduce domain restrictions.
2) Converting Parametric ↔ Cartesian
Strategy
Solve one equation for \(t\) (if possible) and substitute into the other.
  • Example: If \(x=1+t\), then \(t=x-1\).
  • Watch for restrictions (e.g., if \(x=t^2\) then \(x\ge 0\)).
  • A curve can be traced more than once as \(t\) varies.
3) Derivative for Parametric Curves: \( \frac{dy}{dx} \)
Core Formula
\( \displaystyle \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}} \)   (when \( \frac{dx}{dt}\ne 0 \))
  • If \(dx/dt=0\) and \(dy/dt\ne 0\), the tangent is vertical.
  • If \(dy/dt=0\) and \(dx/dt\ne 0\), the tangent is horizontal.
  • If both are zero, \(dy/dx\) is indeterminate; you must analyze further.
4) Tangent Lines + Horizontal/Vertical Tangents
Tangent line at \(t=t_0\)
1) Compute point \( (x(t_0),y(t_0)) \)
2) Compute slope \( \left.\frac{dy}{dx}\right|_{t=t_0} \)
3) Use point-slope form: \( y-y_0=m(x-x_0) \)
  • Horizontal tangent: solve \(dy/dt=0\) and check \(dx/dt\ne 0\).
  • Vertical tangent: solve \(dx/dt=0\) and check \(dy/dt\ne 0\).
5) Velocity and Speed Basics
Velocity + Speed
Velocity: \( \langle x'(t),\,y'(t)\rangle \)
Speed: \( \displaystyle \sqrt{(x'(t))^2+(y'(t))^2} \)
  • Sign of \(x'(t)\) indicates left/right motion; sign of \(y'(t)\) indicates down/up motion.
  • Speed is always nonnegative (a magnitude).
6) Calculator Expectations + Common AP Traps
  • If asked for a tangent line, report the point and the slope (or the full equation).
  • When solving for horizontal/vertical tangents, always include the “check” condition.
  • If both \(dx/dt\) and \(dy/dt\) are zero, do not guess—analyze by eliminating \(t\) or using limits.
Worked Examples + Notes

Example 1 — Find \( \left.\frac{dy}{dx}\right|_{t=1} \)

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

Step 1: \( \frac{dx}{dt}=2t \), \( \frac{dy}{dt}=3t^2 \).
Step 2: \( \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{3t^2}{2t}=\frac{3t}{2} \) (for \(t\ne0\)).
Step 3: At \(t=1\), \( \frac{dy}{dx}=\frac{3}{2} \).

Example 2 — Equation of a tangent line at \(t=1\)

Using the same curve, find the tangent line at \(t=1\).

Step 1: Point: \(x(1)=0\), \(y(1)=3\) so \((0,3)\).
Step 2: Slope: \(m=\frac{3}{2}\).
Step 3: Tangent line: \(y-3=\frac{3}{2}(x-0)\).

Example 3 — Horizontal & vertical tangents

Let \(x=t^2+1\), \(y=t^3-3t\). Find where tangents are horizontal or vertical.

Step 1: \(dx/dt=2t\), \(dy/dt=3t^2-3=3(t^2-1)\).
Horizontal: \(dy/dt=0\Rightarrow t=\pm1\), and \(dx/dt\ne0\) at \(t=\pm1\).
Vertical: \(dx/dt=0\Rightarrow t=0\), and \(dy/dt=-3\ne0\).

Example 4 — Eliminate the parameter

If \(x=1+t\) and \(y=1-t^2\), write a Cartesian equation.

Step 1: \(t=x-1\).
Step 2: Substitute: \(y=1-(x-1)^2\).
Step 3: Domain note: this curve is traced as \(t\) varies, but the Cartesian equation is still valid.
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Unit 9.1 Multiple-Choice Practice (36 Questions)
Answer Key