AP Calculus BC

Parametric and Polar Calculus: Derivatives and Area

Q: How do you find the second derivative of a parametric equation?

Find \frac{dy}{dx} first, then differentiate that expression with respect to t and divide by \frac{dx}{dt}. In symbols, \frac{d^2y}{dx^2}=\dfrac{\frac{d}{dt}(dy/dx)}{dx/dt}. Dividing by \frac{d^2x}{dt^2} is the classic wrong turn.

Q: How do you find horizontal tangents on a polar curve?

Convert to y=r\sin\theta, then set \frac{dy}{d\theta}=0 and check that \frac{dx}{d\theta}\neq 0 at that angle. Do not set \frac{dr}{d\theta}=0; that finds where the radius stops changing, not where the tangent is horizontal.

Which formula to use depends on what you are given and what is asked: divide dy/dt by dx/dt for a parametric slope, integrate the speed sqrt((dx/dt)^2+(dy/dt)^2) for arc length, or take one-half the integral of r^2 dtheta for polar area. The skill is reading the prompt to pick which one applies.

Recognize which Unit 9 tool the problem wants

Unit 9 is BC only and worth 11 to 12 percent of the exam. It bundles three curve systems that run on one engine, the chain rule. Parametric equations give $x$ and $y$ separately as functions of a parameter $t$ . Polar equations give a radius $r$ as a function of an angle $\theta$ . Vector-valued functions repackage the same parametric information as a position vector. Because the three setups look different on the page, the first move on any Unit 9 problem is not computation. It is matching what you are given, and what is asked, to the right formula.

Read every prompt for two signals: what you are handed, and what it wants. If you have $x(t)$ and $y(t)$ and the question asks for a slope, reach for the parametric derivative. If that same curve is asked for the length of its path, reach for the arc length integral instead. A polar equation $r=f(\theta)$ paired with the word area almost always means the one-half r-squared integral, while a request for a tangent line to a polar curve means you rewrite it in $x$ and $y$ first. The table below is the decision map for the entire unit.

The payoff for slowing down at this step is large. A student who writes the polar area integral when the question asked for a slope has lost the points before touching any algebra. Method selection, not execution, is where most of the exam's Unit 9 difficulty actually lives, and it is the one habit that carries across all nine topics.

What you are given	What is asked	Formula to reach for
$x(t)$ and $y(t)$	Slope of the tangent	$\frac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$ (9.1)
$x(t)$ and $y(t)$	Concavity	$\dfrac{d^2y}{dx^2}$ (9.2)
$x(t)$ and $y(t)$	Length of the path	$\int_a^b\sqrt{(dx/dt)^2+(dy/dt)^2}\,dt$ (9.3)
Velocity vector, time interval	Total distance	$\int \text{speed}\,dt$ (9.6)
$r=f(\theta)$	Slope of the tangent	Rewrite as $x,y$ , then $\frac{dy}{dx}$ (9.7)
$r=f(\theta)$	Area swept once	$\frac{1}{2}\int r^2\,d\theta$ (9.8)
Two polar curves	Area between them	$\frac{1}{2}\int (r_{\text{outer}}^2-r_{\text{inner}}^2)\,d\theta$ (9.9)

Parametric derivatives: slope first, then concavity

For a curve defined by $x(t)$ and $y(t)$ , the slope $\frac{dy}{dx}$ is the ratio of the two rates of change (Topic 9.1). Divide the vertical rate by the horizontal rate.

\frac{dy}{dx}=\frac{dy/dt}{dx/dt},\qquad \frac{dx}{dt}\neq 0

The formula holds wherever $\frac{dx}{dt}$ is not zero. Where $\frac{dx}{dt}=0$ but $\frac{dy}{dt}\neq 0$ , the tangent line is vertical and the slope is undefined, which is often exactly what a free-response part is testing. Where both rates are zero at once, the point may be a cusp, and you have to look closer.

The second derivative (Topic 9.2) costs students more points than any other step in the unit, because the reflex is to differentiate $\frac{dy}{dx}$ with respect to $x$ . You cannot, since everything is written in $t$ . Differentiate the slope with respect to $t$ , then divide by $\frac{dx}{dt}$ one more time.

\frac{d^2y}{dx^2}=\frac{\dfrac{d}{dt}\!\left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}}

The denominator trap

The bottom of the second-derivative formula is $\frac{dx}{dt}$ , never $\frac{d^2x}{dt^2}$ . Once you have $\frac{d^2y}{dx^2}$ , its sign reports concavity exactly as it does for an ordinary function: positive means concave up, negative means concave down.

Arc length and motion in the plane

Arc length (Topic 9.3) measures the distance traveled along a curve as $t$ runs across an interval $[a,b]$ . Square the two rates, add them, take the root, and integrate.

L=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt

The integrand is a speed, so a well-designed no-calculator problem almost always hides a perfect square you can pull out from under the radical. When the radical refuses to simplify, the problem is calculator-active: set the integral up correctly and evaluate it numerically. Getting the setup right earns the setup points even if the arithmetic runs long.

Topics 9.4 through 9.6 restate all of this in vector language. A particle's position is $\langle x(t), y(t)\rangle$ . Differentiate component by component for the velocity vector $\left\langle \frac{dx}{dt}, \frac{dy}{dt}\right\rangle$ , and differentiate again for acceleration. Speed is the magnitude of velocity, which is the same radical sitting inside the arc length integral.

\text{speed}=\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2},\qquad \text{distance}=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt

That link is worth committing to memory. Total distance traveled is the integral of speed, and it equals the arc length of the path. Displacement is a different quantity: it is the integral of the velocity vector taken component by component, and it can be far smaller than distance whenever the particle reverses direction. Mixing up the two is a reliable way to lose an otherwise easy free-response point.

Polar derivatives and area

A polar curve $r=f(\theta)$ is secretly parametric, with $\theta$ as the parameter, through $x=r\cos\theta$ and $y=r\sin\theta$ (Topic 9.7). To find the slope of a tangent line, differentiate both with the product rule and divide, just as in the parametric case.

\frac{dy}{dx}=\frac{\dfrac{dr}{d\theta}\sin\theta+r\cos\theta}{\dfrac{dr}{d\theta}\cos\theta-r\sin\theta}

Horizontal tangents occur where the numerator $\frac{dy}{d\theta}=0$ while the denominator is nonzero; vertical tangents occur where the denominator $\frac{dx}{d\theta}=0$ while the numerator $\frac{dy}{d\theta}\neq 0$ . Setting $\frac{dr}{d\theta}=0$ and calling the result a horizontal tangent is a common and costly mistake, because $r$ is only one piece of $y=r\sin\theta$ .

Polar area (Topic 9.8) sums thin circular sectors instead of rectangles, which is why the formula carries a factor of one-half and squares the radius.

A=\frac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta=\frac{1}{2}\int_{\alpha}^{\beta}\bigl[f(\theta)\bigr]^2\,d\theta

The limits $\alpha$ and $\beta$ are the angles that sweep the region exactly once. For a region between two polar curves (Topic 9.9), subtract the inner radius squared from the outer radius squared inside a single integral.

A=\frac{1}{2}\int_{\alpha}^{\beta}\left(r_{\text{outer}}^2-r_{\text{inner}}^2\right)\,d\theta

The difficulty here is entirely in the setup. Find where the curves meet by solving $f(\theta)=g(\theta)$ , decide which curve is the outer boundary on each interval, and pick limits that trace the region a single time. Sketching both curves before you integrate is not optional; it is how you catch a region that has to be split into two integrals, and how you notice a curve that passes through the pole (the origin).

Symmetry is your best ally for choosing limits. Many exam curves (cardioids, rose petals, limacons) are symmetric about an axis, so you can integrate over half the region and double the result, which keeps the arithmetic short and lowers the chance of a bookkeeping slip. For a rose like $r=\cos(3\theta)$ , one petal runs between two consecutive angles where $r=0$ , and multiplying a single petal's area by the number of petals is far safer than trying to sweep the whole flower in one integral.

Where BC students lose points

Most Unit 9 errors are setup errors, not algebra errors. These are the ones graders see most often, and each one is avoidable by reading the prompt before reaching for a formula. Slow down for one deliberate beat to name the technique, then execute; that single habit recovers more Unit 9 points than any amount of extra computational speed.

Dividing by $\frac{d^2x}{dt^2}$ instead of $\frac{dx}{dt}$ when computing the parametric second derivative.
Dropping the factor of one-half, or forgetting to square the radius, in the polar area formula.
Trying to integrate $\cos^2\theta$ or $\sin^2\theta$ directly instead of using the power-reduction identity first.
Using $\frac{dr}{d\theta}=0$ to locate horizontal tangents on a polar curve instead of $\frac{dy}{d\theta}=0$ .
Choosing polar limits that trace a region twice, which doubles the area; sketch first and confirm that $\theta$ sweeps the region once.
Confusing displacement (the integral of the velocity vector) with total distance (the integral of speed).

Worked examples

Worked example

Parametric slope and concavity at a point

A curve is defined by $x=t^2$ and $y=t^3-3t$ . Find $\frac{dy}{dx}$ and $\frac{d^2y}{dx^2}$ at $t=2$ , and write the equation of the tangent line there.

Differentiate each equation with respect to $t$ : $\frac{dx}{dt}=2t$ and $\frac{dy}{dt}=3t^2-3$ .
Form the slope as the ratio of these rates (Topic 9.1): $\frac{dy}{dx}=\frac{3t^2-3}{2t}$ , valid where $\frac{dx}{dt}=2t\neq 0$ .
Evaluate at $t=2$ : $\frac{dy}{dx}=\frac{3(4)-3}{2(2)}=\frac{9}{4}$ .
For the second derivative (Topic 9.2), first rewrite the slope as $\frac{dy}{dx}=\frac{3}{2}\left(t-t^{-1}\right)$ , then differentiate with respect to $t$ : $\frac{d}{dt}\!\left(\frac{dy}{dx}\right)=\frac{3}{2}\left(1+t^{-2}\right)$ .
Divide that by $\frac{dx}{dt}=2t$ , not by $\frac{d^2x}{dt^2}$ : $\frac{d^2y}{dx^2}=\frac{\frac{3}{2}\left(1+t^{-2}\right)}{2t}=\frac{3\left(t^2+1\right)}{4t^3}$ .
At $t=2$ : $\frac{d^2y}{dx^2}=\frac{3(4+1)}{4(8)}=\frac{15}{32}>0$ , so the curve is concave up there.
Find the point of tangency: $x=2^2=4$ and $y=2^3-3(2)=2$ . The tangent line is $y-2=\frac{9}{4}(x-4)$ .

dy/dx = 9/4 and d^2y/dx^2 = 15/32 at t = 2. The tangent line is y - 2 = (9/4)(x - 4), and the curve is concave up there.

Worked example

Arc length of a parametric curve

Find the length of the curve $x=\frac{1}{3}t^3$ , $y=\frac{1}{2}t^2$ for $0\le t\le 2$ .

Read the cue: a parametric curve with a $t$ -interval and the word length points to the arc length integral (Topic 9.3), not a slope or an area formula.
Differentiate: $\frac{dx}{dt}=t^2$ and $\frac{dy}{dt}=t$ .
Build the integrand $\sqrt{(dx/dt)^2+(dy/dt)^2}=\sqrt{t^4+t^2}$ .
Factor $t^2$ out of the radical. Since $t\ge 0$ on this interval, $\sqrt{t^2\left(t^2+1\right)}=t\sqrt{t^2+1}$ .
Set up the definite integral: $L=\int_0^2 t\sqrt{t^2+1}\,dt$ .
Substitute $u=t^2+1$ , so $du=2t\,dt$ and $t\,dt=\frac{1}{2}\,du$ . When $t=0$ , $u=1$ ; when $t=2$ , $u=5$ .
Integrate: $L=\frac{1}{2}\int_1^5 \sqrt{u}\,du=\frac{1}{2}\cdot\frac{2}{3}u^{3/2}\Big|_1^5=\frac{1}{3}\left(5^{3/2}-1\right)$ .
Simplify $5^{3/2}=5\sqrt{5}$ : $L=\frac{1}{3}\left(5\sqrt{5}-1\right)\approx 3.39$ .

L = (1/3)(5*sqrt(5) - 1), which is about 3.39.

Worked example

Area enclosed by a cardioid

Find the area enclosed by the polar curve $r=1+\cos\theta$ .

A single polar curve traced once, plus the word area, points to $A=\frac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta$ (Topic 9.8). The cardioid is traced exactly once as $\theta$ runs from $0$ to $2\pi$ .
Set up the integral: $A=\frac{1}{2}\int_0^{2\pi}(1+\cos\theta)^2\,d\theta$ .
Expand the square: $(1+\cos\theta)^2=1+2\cos\theta+\cos^2\theta$ .
Replace $\cos^2\theta$ with the power-reduction identity $\cos^2\theta=\frac{1+\cos 2\theta}{2}$ , the step students skip most often: $A=\frac{1}{2}\int_0^{2\pi}\left(1+2\cos\theta+\frac{1+\cos 2\theta}{2}\right)\,d\theta$ .
Combine the constants, $1+\frac{1}{2}=\frac{3}{2}$ : $A=\frac{1}{2}\int_0^{2\pi}\left(\frac{3}{2}+2\cos\theta+\frac{1}{2}\cos 2\theta\right)\,d\theta$ .
Integrate term by term. Over a full period, $\int_0^{2\pi} 2\cos\theta\,d\theta=0$ and $\int_0^{2\pi}\frac{1}{2}\cos 2\theta\,d\theta=0$ , while $\int_0^{2\pi}\frac{3}{2}\,d\theta=3\pi$ .
So $A=\frac{1}{2}(3\pi)=\frac{3\pi}{2}$ .

A = 3*pi/2.

Frequently asked questions

Is parametric and polar calculus on the AP Calculus AB exam?

No. All of Unit 9 is BC only. AB students are not tested on parametric derivatives, polar area, or vector-valued functions. On the BC exam the unit carries 11 to 12 percent of the score.

How do you find the second derivative of a parametric equation?

Find $\frac{dy}{dx}$ first, then differentiate that expression with respect to $t$ and divide by $\frac{dx}{dt}$ . In symbols, $\frac{d^2y}{dx^2}=\dfrac{\frac{d}{dt}(dy/dx)}{dx/dt}$ . Dividing by $\frac{d^2x}{dt^2}$ is the classic wrong turn.

Why is there a one-half in the polar area formula?

Because polar area adds up thin circular sectors, not rectangles. A sector of radius $r$ and angle $d\theta$ has area $\frac{1}{2}r^2\,d\theta$ , and integrating those sectors gives $A=\frac{1}{2}\int_{\alpha}^{\beta} r^2\,d\theta$ .

How do you find horizontal tangents on a polar curve?

Convert to $y=r\sin\theta$ , then set $\frac{dy}{d\theta}=0$ and check that $\frac{dx}{d\theta}\neq 0$ at that angle. Do not set $\frac{dr}{d\theta}=0$ ; that finds where the radius stops changing, not where the tangent is horizontal.