How do I find the second derivative of a parametric curve?

Differentiate dy/dx with respect to t, then divide by dx/dt. The extra division is the step most often forgotten.

What's the formula for polar area?

One half the integral of r squared with respect to theta, over the angles that trace the region exactly once.

What's the difference between speed and velocity for parametric motion?

Velocity is the vector of x' and y'. Speed is the size of that vector, the square root of (x')² + (y')².

Parametric, polar, and vector calculus for AP Calculus BC

Updated November 9, 2026

Unit 9 is BC-only, and it comes down to a short list of formulas used carefully. Curves are described by a parameter or by an angle, and the calculus adapts to fit.

This guide works through the formulas you need for parametric, vector, and polar questions, and how to avoid the traps in each.

Parametric derivatives

Slope: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ , where $\frac{dx}{dt} \ne 0$ .
Second derivative: $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}$ . Don't forget to divide by $dx/dt$ again.

Slope and concavity

A curve is given by

x = t^2 - 2t

and

y = t^3 - 3t

. Find

\frac{dy}{dx}

and

\frac{d^2y}{dx^2}

t = 2

$\frac{dy}{dx} = \frac{3t^2 - 3}{2t - 2} = \frac{3(t + 1)}{2}$ for $t \ne 1$ . At $t = 2$ that's $\frac{9}{2}$ .

The derivative of $\frac{3(t + 1)}{2}$ with respect to $t$ is $\frac{3}{2}$ . Dividing by $\frac{dx}{dt} = 2$ at $t = 2$ gives $\frac{3}{4}$ .

Answer: $\frac{dy}{dx} = \frac{9}{2}$ and $\frac{d^2y}{dx^2} = \frac{3}{4}$

Motion: speed and distance

For motion in the plane, position is $\langle x(t), y(t) \rangle$ , and velocity is $\langle x'(t), y'(t) \rangle$ . Acceleration is the derivative of velocity.

Speed: $\sqrt{(x')^2 + (y')^2}$ , the size of the velocity vector.
Distance traveled from $t = a$ to $t = b$ : $\int_a^b \sqrt{(x')^2 + (y')^2}\,dt$ , which is also the arc length.
Position at a later time: starting position plus the integral of velocity, done for each coordinate separately.

Speed at an instant

Using

x = t^2 - 2t

and

y = t^3 - 3t

, find the speed at

t = 2

$x'(2) = 2$ and $y'(2) = 9$ , so the speed is $\sqrt{2^2 + 9^2} = \sqrt{85}$ .

Answer: $\sqrt{85} \approx 9.220$

Distance with a calculator

A particle moves with

x = 3\cos t + t

and

y = 2\sin t

. Find the distance it travels on

0 \le t \le \pi

$x' = 1 - 3\sin t$ and $y' = 2\cos t$ . Write $\int_0^{\pi} \sqrt{(1 - 3\sin t)^2 + (2\cos t)^2}\,dt$ , then evaluate in Desmos using $x$ as the variable. It gives about $6.01786$ .

Answer: $6.018$

Polar slope

A polar curve $r = f(\theta)$ becomes parametric if you use $x = r\cos\theta$ and $y = r\sin\theta$ . Then $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$ , with the product rule on each.

Tangent slope

Find the slope of the tangent line to

r = 2 + 2\cos\theta

\theta = \frac{\pi}{2}

Here $r = 2$ and $r' = -2\sin\theta = -2$ at $\theta = \frac{\pi}{2}$ .

$\frac{dy}{d\theta} = r'\sin\theta + r\cos\theta = -2 + 0 = -2$ , and $\frac{dx}{d\theta} = r'\cos\theta - r\sin\theta = 0 - 2 = -2$ .

Answer: $1$

Polar area

The area swept out from $\theta = \alpha$ to $\theta = \beta$ is $\frac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta$ . The hard part is the limits, so figure out which angles trace the region exactly once.

A full loop

Find the area enclosed by

r = 2 + 2\cos\theta

The curve traces once on $0 \le \theta \le 2\pi$ . Using $\cos^2\theta = \frac{1 + \cos 2\theta}{2}$ , $\frac{1}{2}\int_0^{2\pi} (2 + 2\cos\theta)^2\,d\theta = 6\pi$ .

Answer: $6\pi$

An inner loop

Find the area of the inner loop of

r = 1 + 2\cos\theta

The inner loop is traced while $r \le 0$ . Graph $r$ against $\theta$ like an ordinary function to see where. Setting $1 + 2\cos\theta = 0$ gives $\theta = \frac{2\pi}{3}$ and $\theta = \frac{4\pi}{3}$ .

r = 1 + 2\cos\theta

Plotting $r$ against $\theta$ shows the angles of the inner loop.

The area is $\frac{1}{2}\int_{2\pi/3}^{4\pi/3} (1 + 2\cos\theta)^2\,d\theta = \pi - \frac{3\sqrt{3}}{2}$ .

Answer: $\pi - \frac{3\sqrt{3}}{2} \approx 0.544$

Between two curves

Find the area inside

r = 6\cos\theta

and outside

r = 2 + 2\cos\theta

Set them equal: $6\cos\theta = 2 + 2\cos\theta$ gives $\cos\theta = \frac{1}{2}$ , so $\theta = \pm\frac{\pi}{3}$ .

Subtract inside the integral, outer minus inner, each squared separately: $\frac{1}{2}\int_{-\pi/3}^{\pi/3} \left[(6\cos\theta)^2 - (2 + 2\cos\theta)^2\right] d\theta = 4\pi$ .

Answer: $4\pi$

Graphing these in Desmos

On the calculator parts, Desmos graphs both types directly, which helps you see which angles trace a region. The Desmos guide shows how to enter them.

Keep going

Practice on exam-style questions

Intersect Prep has 50 multiple-choice questions for Unit 9, with a full solution for each one. Create a free account to try a full lesson with the calculator. One payment of $49 unlocks the course for 6 months.

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Questions

How do I find the second derivative of a parametric curve?: Differentiate dy/dx with respect to t, then divide by dx/dt. The extra division is the step most often forgotten.
What's the formula for polar area?: One half the integral of r squared with respect to theta, over the angles that trace the region exactly once.
What's the difference between speed and velocity for parametric motion?: Velocity is the vector of x' and y'. Speed is the size of that vector, the square root of (x')² + (y')².