AP Calculus BC only

Unit 9: Parametric Equations, Polar Coordinates, and Vector-Valued Functions

Exam weighting: AB n/a · BC 10-15%

Unit 9 is BC-only and extends calculus to curves not written as y = f(x): parametric and vector-valued functions and polar coordinates. You find tangent slopes and second derivatives, arc length, planar motion (speed, displacement, distance), and areas of polar regions. It is 10-15% of the BC exam.

The core idea

Every earlier unit worked with curves written as y=f(x)y = f(x). Unit 9 (BC only) drops that restriction: a curve can be traced by a parameter, x=x(t)x = x(t) and y=y(t)y = y(t), packaged as a vector-valued function, or described by a distance and angle in polar form, r=f(θ)r = f(\theta). The calculus itself does not change. The chain rule still produces slopes and the definite integral still produces lengths and areas; you just apply them through the new variable. Every problem in the unit is one recognition question: which representation am I in, and which quantity (slope, length, motion, or area) does the problem want?

The unit runs in two halves. Topics 9.1 through 9.3 are parametric calculus: the tangent slope dydx\frac{dy}{dx} (9.1), the second derivative for concavity (9.2), and arc length (9.3). Topics 9.4 through 9.6 recast the same curve as a vector-valued function to handle motion in the plane, differentiating for velocity and acceleration (9.4), integrating to recover velocity and position (9.5), and combining both for speed, displacement, and total distance (9.6). Topics 9.7 through 9.9 switch to polar coordinates: differentiating a polar curve (9.7), then the area of a single polar region (9.8) and the area between two polar curves (9.9).

  • Tangent slope, parametric or polar (Topics 9.1, 9.7): divide, dydx=dy/dtdx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt} when dx/dt≠0dx/dt \neq 0. A horizontal tangent needs dy/dt=0dy/dt = 0; a vertical tangent needs dx/dt=0dx/dt = 0. In polar, first convert with x=rcos⁡θx = r\cos\theta and y=rsin⁡θy = r\sin\theta, then differentiate against θ\theta.
  • Second derivative or concavity (Topic 9.2): differentiate dydx\frac{dy}{dx} with respect to tt, then divide by dxdt\frac{dx}{dt} again. Do not divide the two second derivatives.
  • Arc length or total distance (Topics 9.3, 9.6): integrate speed, (dx/dt)2+(dy/dt)2\sqrt{(dx/dt)^2 + (dy/dt)^2}, over the tt-interval. A parametric curve's length and a particle's distance traveled use the same integrand.
  • Displacement versus position (Topics 9.5, 9.6): integrate the velocity vector component by component to get displacement, then add the starting position to get the final position.
  • Polar area (Topics 9.8, 9.9): use 12∫r2 dθ\frac{1}{2}\int r^2 \,d\theta for one curve and 12∫(router2−rinner2) dθ\frac{1}{2}\int (r_{\text{outer}}^2 - r_{\text{inner}}^2) \,d\theta between two.
dydx=dy/dtdx/dt(dx/dt≠0),d2ydx2=ddt ⁣(dydx)dx/dt\frac{dy}{dx} = \frac{dy/dt}{dx/dt} \quad (dx/dt \neq 0), \qquad \frac{d^2y}{dx^2} = \frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{dx/dt}

A vector-valued function bundles the two parametric equations into one position ⟨x(t),y(t)⟩\left\langle x(t), y(t) \right\rangle. Differentiating component by component gives the velocity vector ⟨dxdt,dydt⟩\left\langle \frac{dx}{dt}, \frac{dy}{dt} \right\rangle and differentiating again gives acceleration. Speed is the magnitude of velocity, (dxdt)2+(dydt)2\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}, a scalar, not a vector. Integration runs the process backward: integrate acceleration to recover velocity and velocity to recover position, using an initial condition to pin the constant on each component. Two integrals answer the classic motion questions: the definite integral of the velocity vector is displacement (net change in position, which can be zero even after a long trip), while the definite integral of speed is total distance traveled (never negative, zero only if the particle never moves).

A polar curve r=f(θ)r = f(\theta) gives distance rr from the origin as a function of the angle θ\theta. To differentiate it, convert to x=rcos⁡θ=f(θ)cos⁡θx = r\cos\theta = f(\theta)\cos\theta and y=rsin⁡θ=f(θ)sin⁡θy = r\sin\theta = f(\theta)\sin\theta, then apply the parametric slope rule with θ\theta as the parameter. Area works differently from rectangular area: instead of thin rectangles you sum thin circular sectors, which is where the factor of 12\frac{1}{2} and the r2r^2 come from. The whole challenge in Topics 9.8 and 9.9 is the limits of integration, not the formula. Find them by setting r=0r = 0 (where a petal starts and ends) or, for two curves, by setting the two rr values equal to locate intersection angles. Sketch the region first so you know which curve is outer.

A=12∫αβ[f(θ)]2 dθ,A=12∫αβ(router2−rinner2) dθA = \frac{1}{2}\int_{\alpha}^{\beta} \left[f(\theta)\right]^2 \,d\theta, \qquad A = \frac{1}{2}\int_{\alpha}^{\beta} \left(r_{\text{outer}}^2 - r_{\text{inner}}^2\right) \,d\theta

What the exam asks

Unit 9 is BC only and worth 10-15% of the BC exam, so it carries real weight. Expect a full free-response question built on one representation: a particle in planar motion asking for speed, displacement, and total distance (Topic 9.6), or a polar region asking for area between two curves (Topic 9.9), both usually in the calculator-allowed section where you set up the integral and evaluate numerically. Multiple-choice items test the parametric slope and second derivative (Topics 9.1 and 9.2) and arc length (Topic 9.3). The recurring trap across the unit is notation: dividing the wrong pair of derivatives for d2ydx2\frac{d^2y}{dx^2}, or confusing displacement with distance. Progress Check 9 has about 25 multiple-choice questions and 3 free-response questions.

Topics in this unit

Topic numbers and titles from the College Board Course and Exam Description.

  • 9.1Defining and Differentiating Parametric Equations
  • 9.2Second Derivatives of Parametric Equations
  • 9.3Finding Arc Lengths of Curves Given by Parametric Equations
  • 9.4Defining and Differentiating Vector-Valued Functions
  • 9.5Integrating Vector-Valued Functions
  • 9.6Solving Motion Problems Using Parametric and Vector-Valued Functions
  • 9.7Defining Polar Coordinates and Differentiating in Polar Form
  • 9.8Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve
  • 9.9Finding the Area of the Region Bounded by Two Polar Curves

How to study this unit

  • Topic 9.2: the parametric second derivative is $\frac{d^2y}{dx^2} = \frac{d/dt\left(dy/dx\right)}{dx/dt}$. Never compute $\frac{d^2y/dt^2}{d^2x/dt^2}$. Find $\frac{dy}{dx}$ first, differentiate that whole expression with respect to $t$, then divide by $\frac{dx}{dt}$ one more time.
  • Topics 9.3 and 9.6: arc length of a parametric curve and total distance traveled by a particle are the same integral, $\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\,dt$. Learn this integrand once and recognize both prompts as asking for it.
  • Topic 9.6: keep four quantities separate. Velocity is a vector, speed is its nonnegative magnitude, displacement is the integral of the velocity vector (a vector whose components can each be negative, and whose net change can be zero even after a long trip), and total distance is the integral of speed (never negative, zero only if the particle never moves). Final position equals initial position plus displacement.
  • Topics 9.8 and 9.9: the polar area formula is easy; the limits are the work. Sketch the curve, then find bounds by setting $r = 0$ for a single petal or by setting the two equations equal to find intersection angles, and always integrate $r_{\text{outer}}^2$ minus $r_{\text{inner}}^2$ between two curves.
  • Topics 9.1 and 9.7: for horizontal and vertical tangents, test the numerator and denominator of $\frac{dy}{dx}$ separately. Horizontal tangent when $\frac{dy}{dt} = 0$ and $\frac{dx}{dt} \neq 0$; vertical tangent when $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} \neq 0$. Convert polar to $x$ and $y$ first, then apply the same test.

Guides for this unit