AP Calculus BC Unit 9 Notes: Learning Parametric Curves from First Principles

Defining and Differentiating Parametric Equations

What a parametric equation is (and why you’d use one)

A parametric curve describes a path in the plane by giving both coordinates as functions of a third variable called a parameter (usually $t$). Instead of defining $y$ directly as a function of $x$ (like $y = f(x)$), you define

$x = x(t)$

$y = y(t)$

As $t$ changes, the point $\big(x(t), y(t)\big)$ moves—so you can think of the curve as a “motion picture” of a point traveling through the plane.

This matters because many curves are awkward or impossible to describe cleanly as $y$ in terms of $x$. Parametric equations let you:

• Model motion naturally (position depends on time).
• Describe curves that fail the vertical line test (loops, sideways curves).
• Keep track of direction and time along the curve, not just the shape.

A key mindset shift: with parametrics, $t$ is not just a trick; it carries information about where you are on the curve and how fast you’re moving.

Interpreting the curve: shape, direction, and time

When you graph a parametric curve, you are really graphing the set of points

$\{(x(t), y(t)) : t \text{ in some interval}\}$

Two important interpretive ideas:

1. Direction (orientation): As $t$ increases, the point traces the curve in a specific direction. Two different parameterizations can trace the same geometric curve but in opposite directions.

2. Speed along the curve: If $t$ is time, then $\frac{dx}{dt}$ and $\frac{dy}{dt}$ describe how the coordinates change per unit time. Even if the curve’s shape is the same, different parameterizations can move along it at different speeds.

Eliminating the parameter (when you can, and what you lose)

Sometimes you can rewrite the parametric equations to remove $t$ and get a relationship between $x$ and $y$.

Example idea (not a full worked problem yet): if $x = \cos t$ and $y = \sin t$, then eliminating $t$ gives $x^2 + y^2 = 1$.

This is useful for recognizing the shape. But you lose information about:

• which point corresponds to a specific $t$,
• the direction of travel,
• how quickly the point moves.

So on AP questions, eliminating $t$ can help identify a curve, but you often still need derivatives with respect to $t$ to answer calculus questions.

Differentiating parametric equations: finding $\frac{dy}{dx}$

To find slope on a parametric curve, you want $\frac{dy}{dx}$. But you are not given $y$ as a function of $x$; you are given both as functions of $t$. The key idea is the Chain Rule:

If $y = y(t)$ and $x = x(t)$, then

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

as long as $\frac{dx}{dt} \ne 0$.

Why this formula makes sense

Over a tiny change $\Delta t$, you get approximate changes

$\Delta y \approx \frac{dy}{dt}\Delta t$

$\Delta x \approx \frac{dx}{dt}\Delta t$

So the slope $\frac{\Delta y}{\Delta x}$ becomes approximately

$\frac{\frac{dy}{dt}\Delta t}{\frac{dx}{dt}\Delta t} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

The $\Delta t$ cancels, which is exactly the intuition: slope depends on the _relative_ rates of change of $y$ and $x$.

Tangent lines and normals

Once you compute $\frac{dy}{dx}$ at a specific parameter value $t = a$, you can write the tangent line in point-slope form:

• Point on curve: $\big(x(a), y(a)\big)$
• Slope: $m = \left.\frac{dy}{dx}\right|_{t=a}$

Equation:

$y - y(a) = m\big(x - x(a)\big)$

A normal line is perpendicular to the tangent line, so its slope is

$m_{\text{normal}} = -\frac{1}{m}$

provided $m \ne 0$.

Horizontal and vertical tangents (common AP targets)

Because $\frac{dy}{dx} = \frac{(dy/dt)}{(dx/dt)}$:

• Horizontal tangent: $\frac{dy}{dx} = 0$ happens when

$\frac{dy}{dt} = 0$

and $\frac{dx}{dt} \ne 0$.

• Vertical tangent: slope is undefined when

$\frac{dx}{dt} = 0$

and $\frac{dy}{dt} \ne 0$.

A subtle but important caution: if both $\frac{dx}{dt} = 0$ and $\frac{dy}{dt} = 0$, then the point may be a cusp, corner, or a moment where the motion “stops”—and you cannot conclude “vertical tangent” from $\frac{dx}{dt} = 0$ unless you also check $\frac{dy}{dt}$.

Worked Example 1: slope and tangent line

Let

$x = t^2 - 1$

$y = t^3 + t$

Find $\frac{dy}{dx}$ and the tangent line at $t = 1$.

Step 1: Differentiate each with respect to $t$.

$\frac{dx}{dt} = 2t$

$\frac{dy}{dt} = 3t^2 + 1$

Step 2: Use the parametric derivative formula.

$\frac{dy}{dx} = \frac{3t^2 + 1}{2t}$

Step 3: Evaluate at $t = 1$.

Point:

$x(1) = 1^2 - 1 = 0$

$y(1) = 1^3 + 1 = 2$

Slope:

$\left.\frac{dy}{dx}\right|_{t=1} = \frac{3(1)^2 + 1}{2(1)} = \frac{4}{2} = 2$

Step 4: Write the tangent line.

$y - 2 = 2(x - 0)$

So

$y = 2x + 2$

Worked Example 2: horizontal and vertical tangents

Let

$x = t^2 + t$

$y = t^3 - 3t$

Find parameter values where the curve has horizontal or vertical tangents.

Step 1: Compute derivatives.

$\frac{dx}{dt} = 2t + 1$

$\frac{dy}{dt} = 3t^2 - 3 = 3(t^2 - 1)$

Horizontal tangents: set $\frac{dy}{dt} = 0$ and require $\frac{dx}{dt} \ne 0$.

$3(t^2 - 1) = 0$

$t^2 = 1$

$t = 1, -1$

Check $\frac{dx}{dt}$:

$\frac{dx}{dt}(1) = 3 \ne 0$

$\frac{dx}{dt}(-1) = -1 \ne 0$

So both $t = 1$ and $t = -1$ give horizontal tangents.

Vertical tangents: set $\frac{dx}{dt} = 0$ and require $\frac{dy}{dt} \ne 0$.

$2t + 1 = 0$

$t = -\frac{1}{2}$

Check $\frac{dy}{dt}$:

$\frac{dy}{dt}\left(-\frac{1}{2}\right) = 3\left(\frac{1}{4} - 1\right) = 3\left(-\frac{3}{4}\right) = -\frac{9}{4} \ne 0$

So $t = -\frac{1}{2}$ gives a vertical tangent.

Exam Focus

• Typical question patterns
  • “Given $x(t)$ and $y(t)$, find $\frac{dy}{dx}$ at $t = a$ and write the tangent line equation.”
  • “Find all points where the curve has a horizontal/vertical tangent on a given $t$-interval.”
  • “Eliminate the parameter to identify the curve, then use calculus in parametric form to answer slope/velocity questions.”
• Common mistakes
  • Forgetting that $\frac{dy}{dx}$ is a _ratio_ and writing $\frac{dy}{dx} = \frac{dy}{dt}$.
  • Declaring “vertical tangent” when $\frac{dx}{dt} = 0$ without checking $\frac{dy}{dt} \ne 0$.
  • Plugging in $t$ too early (before simplifying) and missing cancellations or sign errors.

Second Derivatives of Parametric Equations

What the second derivative means in parametric form

The second derivative $\frac{d^2y}{dx^2}$ measures how the slope $\frac{dy}{dx}$ changes as $x$ changes. Geometrically, it tells you the curve’s **concavity** (concave up vs concave down) when you view $y$ as changing with $x$.

In parametric problems, you still care about concavity for the same reasons as in regular functions:

• identifying concavity changes (inflection behavior),
• understanding curve shape beyond just tangent lines,
• modeling motion (it also connects to acceleration when combined with vector ideas).

But since $y$ is not explicitly given as $y(x)$, you must compute $\frac{d^2y}{dx^2}$ carefully using the chain rule again.

How to compute $\frac{d^2y}{dx^2}$

Start with

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$

This is a function of $t$. To find how it changes with $x$, you can think in two stages:

1. Differentiate the slope with respect to $t$ to get $\frac{d}{dt}\left(\frac{dy}{dx}\right)$.
2. Convert “per unit $t$” to “per unit $x$” by dividing by $\frac{dx}{dt}$.

That gives the parametric second derivative formula:

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

This formula is easy to misuse if you forget what it’s saying: you’re doing the chain rule with $x$ as an intermediate variable.

Why dividing by $\frac{dx}{dt}$ appears again

If $\frac{dy}{dx}$ is a function of $t$, then

$\frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$

This mirrors the first-derivative logic: “rate of change with respect to $x$” equals “rate of change with respect to $t$” divided by “rate of change of $x$ with respect to $t$.”

Interpreting concavity and points to be careful about

• If $\frac{d^2y}{dx^2} > 0$, the curve is concave up at that parameter value.
• If $\frac{d^2y}{dx^2} < 0$, the curve is concave down.

However, concavity is about how $y$ bends as $x$ increases. If the curve doubles back (not one-to-one in $x$), concavity descriptions can still be made locally, but you must be careful interpreting “up” and “down” in a global way.

Also, note the second-derivative formula requires dividing by $\frac{dx}{dt}$. If $\frac{dx}{dt} = 0$ at a point (a vertical tangent or cusp possibility), $\frac{d^2y}{dx^2}$ may be undefined there, even if the curve exists.

Worked Example 1: computing $\frac{d^2y}{dx^2}$ and concavity

Let

$x = t^2 + 1$

$y = t^3 - 3t$

Find $\frac{d^2y}{dx^2}$ and determine concavity at $t = 1$.

Step 1: First derivatives with respect to $t$.

$\frac{dx}{dt} = 2t$

$\frac{dy}{dt} = 3t^2 - 3$

Step 2: Compute $\frac{dy}{dx}$.

$\frac{dy}{dx} = \frac{3t^2 - 3}{2t}$

You can simplify to reduce algebra errors:

$\frac{dy}{dx} = \frac{3(t^2 - 1)}{2t}$

Step 3: Differentiate $\frac{dy}{dx}$ with respect to $t$.

Using quotient rule on $\frac{3t^2 - 3}{2t}$ is straightforward:

Let $u = 3t^2 - 3$ and $v = 2t$.

$u' = 6t$

$v' = 2$

Then

$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{u'v - uv'}{v^2} = \frac{(6t)(2t) - (3t^2 - 3)(2)}{(2t)^2}$

Simplify:

$\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{12t^2 - 6t^2 + 6}{4t^2} = \frac{6t^2 + 6}{4t^2} = \frac{3(t^2 + 1)}{2t^2}$

Step 4: Divide by $\frac{dx}{dt}$ to get $\frac{d^2y}{dx^2}$.

$\frac{d^2y}{dx^2} = \frac{\frac{3(t^2 + 1)}{2t^2}}{2t} = \frac{3(t^2 + 1)}{4t^3}$

Step 5: Evaluate at $t = 1$.

$\left.\frac{d^2y}{dx^2}\right|_{t=1} = \frac{3(2)}{4} = \frac{3}{2}$

Since $\frac{3}{2} > 0$, the curve is concave up at $t = 1$.

Worked Example 2: noticing where the second derivative can fail

Let

$x = t^3$

$y = t^2$

Compute $\frac{dy}{dx}$ and discuss what happens at $t = 0$.

Step 1: First derivatives.

$\frac{dx}{dt} = 3t^2$

$\frac{dy}{dt} = 2t$

Step 2: First derivative $\frac{dy}{dx}$.

$\frac{dy}{dx} = \frac{2t}{3t^2} = \frac{2}{3t}$

At $t = 0$, this is undefined. That doesn’t mean the curve disappears—rather, it signals a vertical tangent or cusp-type behavior. If you eliminate the parameter for insight:

From $x = t^3$, you get $t = x^{1/3}$, so

$y = t^2 = x^{2/3}$

The graph $y = x^{2/3}$ has a cusp at the origin with a vertical tangent behavior, matching the undefined slope.

This example shows why you must treat points where $\frac{dx}{dt} = 0$ with extra care—both for $\frac{dy}{dx}$ and for $\frac{d^2y}{dx^2}$.

Exam Focus

• Typical question patterns
  • “Given parametric equations, find $\frac{d^2y}{dx^2}$ at $t = a$ and determine concavity.”
  • “Find where the curve changes concavity on a parameter interval (often by analyzing the sign of $\frac{d^2y}{dx^2}$).”
  • “At a time/value where $\frac{dx}{dt} = 0$, interpret what happens to slope/concavity.”
• Common mistakes
  • Computing $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ and stopping, forgetting to divide by $\frac{dx}{dt}$.
  • Plugging into the second-derivative formula at a value where $\frac{dx}{dt} = 0$ without discussing that $\frac{d^2y}{dx^2}$ may be undefined.
  • Algebra/quotient-rule errors inside $\frac{d}{dt}\left(\frac{dy}{dx}\right)$; simplifying $\frac{dy}{dx}$ first often reduces mistakes.

Arc Lengths of Parametric Curves

What arc length is and why parametrics make it natural

Arc length is the distance traveled along a curve between two points. For straight lines, distance is easy. For curves, you approximate by lots of tiny straight segments and take a limit (an integral).

Parametric equations are especially well-suited for arc length because they already describe a moving point. If $t$ is time, arc length over $t \in [a,b]$ is literally “total distance traveled” during that time interval.

Deriving the arc length formula (from the distance formula)

Over a tiny change $dt$, the point moves from $\big(x(t), y(t)\big)$ to approximately $\big(x(t)+dx, y(t)+dy\big)$. The tiny distance traveled is approximately the hypotenuse of a right triangle:

$ds \approx \sqrt{(dx)^2 + (dy)^2}$

Divide by $dt$ to convert to rates:

$\frac{ds}{dt} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}$

This quantity $\frac{ds}{dt}$ is the **speed** (magnitude of velocity) if $t$ represents time.

Integrating speed gives total distance (arc length):

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

When the formula works nicely (and when it doesn’t)

This integral is guaranteed to represent arc length for a smooth curve, but it may not be easy to compute exactly. On AP problems, the functions are often chosen so that the square root simplifies (frequently using a Pythagorean identity or a perfect square).

Be aware of two common “gotchas”:

• Parameter interval matters. The arc length depends on $t \in [a,b]$; the same geometric curve traced multiple times would accumulate extra length.
• Direction does not matter for length. Reversing direction changes signs of derivatives, but the squares make the integrand nonnegative.

Notation connection: arc length and speed

It helps to recognize these equivalent interpretations:

This is a powerful bridge: arc length is a geometry concept, but in parametric form it becomes a motion concept.

Worked Example 1: arc length with simplification

Let

$x = 3\cos t$

$y = 3\sin t$

Find the arc length from $t = 0$ to $t = \pi$.

Step 1: Compute derivatives.

$\frac{dx}{dt} = -3\sin t$

$\frac{dy}{dt} = 3\cos t$

Step 2: Build the integrand.

$\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \sqrt{9\sin^2 t + 9\cos^2 t}$

Use $\sin^2 t + \cos^2 t = 1$:

$\sqrt{9(1)} = 3$

Step 3: Integrate over the parameter interval.

$L = \int_0^\pi 3\,dt = 3\pi$

Interpretation: the curve is a circle of radius 3 traced from angle 0 to $\pi$, which is a semicircle. Its arc length should be half the circumference:

$\frac{1}{2}(2\pi r) = \pi r = 3\pi$

So the parametric arc length matches the geometry, which is a good reasonableness check.

Worked Example 2: arc length of a polynomial parameterization

Let

$x = t^2$

$y = \frac{2}{3}t^3$

Find arc length from $t = 0$ to $t = 1$.

Step 1: Differentiate.

$\frac{dx}{dt} = 2t$

$\frac{dy}{dt} = 2t^2$

Step 2: Form the speed.

$\sqrt{\left(2t\right)^2 + \left(2t^2\right)^2} = \sqrt{4t^2 + 4t^4}$

Factor inside:

$\sqrt{4t^2(1 + t^2)} = 2|t|\sqrt{1+t^2}$

On $[0,1]$, $|t| = t$, so the integrand is

$2t\sqrt{1+t^2}$

Step 3: Integrate.

$L = \int_0^1 2t\sqrt{1+t^2}\,dt$

Let $u = 1+t^2$, so $du = 2t\,dt$. When $t=0$, $u=1$. When $t=1$, $u=2$.

$L = \int_1^2 \sqrt{u}\,du = \int_1^2 u^{1/2}\,du$

$L = \left.\frac{2}{3}u^{3/2}\right|_1^2 = \frac{2}{3}\left(2^{3/2} - 1\right)$

You can also write $2^{3/2}$ as $2\sqrt{2}$:

$L = \frac{2}{3}\left(2\sqrt{2} - 1\right)$

A common conceptual trap: arc length vs straight-line distance

Arc length is not the distance between endpoints. Even if you find the endpoints $\big(x(a),y(a)\big)$ and $\big(x(b),y(b)\big)$, using the distance formula gives the chord length, not the arc length.

If an AP question asks for “distance traveled” by a particle with position $\big(x(t),y(t)\big)$, that is arc length, so you should think immediately of

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

not

$\sqrt{(x(b)-x(a))^2 + (y(b)-y(a))^2}$

Exam Focus

• Typical question patterns
  • “Find the length of the curve for $t$ from $a$ to $b$ (often chosen so the radical simplifies).”
  • “A particle moves with position $\big(x(t),y(t)\big)$; find total distance traveled on $[a,b]$.”
  • “Compute arc length and interpret it geometrically (circle, cycloid-like segments, etc.).”
• Common mistakes
  • Forgetting to square both derivatives inside the square root, or forgetting the square root entirely.
  • Dropping absolute values incorrectly when simplifying (for example, replacing $\sqrt{t^2}$ with $t$ on an interval that includes negatives).
  • Computing endpoint distance instead of integrating speed (confusing displacement with distance traveled).