# Second derivatives of parametric equations

> AP Calculus BC · CED Unit 9
> Source: https://www.owlsprep.com/study/ap-calculus-bc-u9-second-derivatives-of-parametric-equations/

This module covers derivation of the second derivative formula for parametric curves, step-by-step computation of $\frac{d^2y}{dx^2}$, concavity analysis, and application to planar particle motion problems, a commonly tested topic on AP Calculus BC.

**Prerequisites:** [How to compute first derivatives of parametric functions](https://www.owlsprep.com/study/ap-calculus-bc-u9-first-derivatives-parametric/); Chain rule and basic differentiation rules; Definition of concavity for Cartesian curves

## Learning objectives

- Derive and recall the formula for the second derivative of a parametric curve
- Calculate $\frac{d^2y}{dx^2}$ for parametric equations at a given point
- Analyze concavity and find inflection points of parametric curves
- Distinguish between path concavity and particle acceleration for planar motion

## Core Definition and Formula Derivation

For a parametric curve defined by $x = x(t)$ and $y = y(t)$, the second derivative of $y$ with respect to $x$ is the rate of change of the first derivative $\frac{dy}{dx}$ as $x$ changes, not as the parameter $t$ changes. This topic extends concavity analysis from Cartesian functions to parametric curves, which can describe closed curves, non-functions, and particle motion trajectories.

**Second Derivative of a Parametric Curve** — The derivative of the first derivative $\frac{dy}{dx}$ with respect to $x$, for a parametric curve $x=x(t), y=y(t)$ where $x'(t) \neq 0$.

*Notation:* $\frac{d^2y}{dx^2}$

**Derivation:** Derive the general formula for $\frac{d^2y}{dx^2}$

*Starting from:* We know the first derivative $\frac{dy}{dx} = \frac{y'(t)}{x'(t)}$ for $x'(t) \neq 0$.

1. The second derivative is defined as $\frac{d}{dx}\left(\frac{dy}{dx}\right)$. Apply the chain rule to rewrite this derivative with respect to $t$:

   $$\frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx}$$
2. We know from the inverse derivative relationship that $\frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{x'(t)}$, so:

   $$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left( \frac{y'(t)}{x'(t)} \right)}{x'(t)}$$
3. Apply the quotient rule to the numerator: $\frac{d}{dt}\left(\frac{y'}{x'}\right) = \frac{y''x' - y'x''}{(x')^2}$. Substitute back to get the simplified formula:

   $$\frac{d^2y}{dx^2} = \frac{y''(t)x'(t) - y'(t)x''(t)}{[x'(t)]^3}$$

*Conclusion:* This is the simplified general formula for the second derivative of a parametric curve, valid for all $x'(t) \neq 0$.

**Worked example:** Find $\frac{d^2y}{dx^2}$ for the parametric curve $x(t) = t^2 + 2t$, $y(t) = t^3 - 3t$ at $t = 1$.

1. Compute first and second derivatives of $x$ and $y$ with respect to $t$, then evaluate at $t=1$:

   $$x'(t) = 2t + 2 \implies x'(1) = 4 \\ y'(t) = 3t^2 - 3 \implies y'(1) = 0 \\ x''(t) = 2 \implies x''(1) = 2 \\ y''(t) = 6t \implies y''(1) = 6$$
2. Substitute values into the simplified second derivative formula:

   $$\frac{d^2y}{dx^2} = \frac{y''x' - y'x''}{(x')^3} = \frac{(6)(4) - (0)(2)}{4^3} = \frac{24}{64} = \frac{3}{8}$$
3. Final result at $t=1$ is $\frac{3}{8}$.

> **Exam tip:** If you only need the second derivative at a specific value of $t$, compute $x', y', x'', y''$ at that $t$ first before plugging into the formula. This avoids messy algebraic simplification, saving time on multiple-choice questions.

## Analyzing Concavity of Parametric Curves

The most common AP exam application of parametric second derivatives is analyzing where a parametric curve is concave up or down, just like with Cartesian curves. The same concavity rules apply, but $\frac{d^2y}{dx^2}$ is a function of $t$, so we solve sign inequalities in terms of $t$ and map back to coordinates if required.

Inflection point (concavity change) candidates occur where $\frac{d^2y}{dx^2} = 0$ *or* $\frac{d^2y}{dx^2}$ is undefined (which happens when $x'(t) = 0$, since that makes the denominator zero). We test intervals between these candidates to confirm a concavity change, just like with Cartesian curves.

**Worked example:** For the parametric curve $x(t) = t^2$, $y(t) = e^t$, $t > 0$, determine if the curve is concave up or concave down at $x = 4$.

1. First find the value of $t$ corresponding to $x=4$: $t^2 = 4 \implies t=2$ (since $t>0$).
2. Compute and evaluate derivatives at $t=2$:

   $$x'(t) = 2t \implies x'(2) = 4 \\ y'(t) = e^t \implies y'(2) = e^2 \\ x''(t) = 2 \implies x''(2) = 2 \\ y''(t) = e^t \implies y''(2) = e^2$$
3. Calculate $\frac{d^2y}{dx^2}$ and check its sign:

   $$\frac{d^2y}{dx^2} = \frac{(e^2)(4) - (e^2)(2)}{4^3} = \frac{2e^2}{64} = \frac{e^2}{32} \approx 0.23 > 0$$
4. Conclusion: the curve is concave up at $x=4$.

> **Exam tip:** Always confirm what value of $t$ corresponds to the requested $x$ or $y$ before evaluating the second derivative. Exam questions regularly hide this step to test understanding that parametric derivatives are functions of $t$, not $x$.

## Second Derivatives for Planar Particle Motion

When a particle moves in the $xy$-plane with position $(x(t), y(t))$ at time $t$, $\frac{dy}{dx}$ gives the slope of the particle's trajectory (the path it follows through the plane), and $\frac{d^2y}{dx^2}$ gives the concavity of that path. This is *not the same* as the particle's acceleration vector, which is $\langle x''(t), y''(t) \rangle$, a vector describing how the particle's velocity changes with time. AP exam questions regularly test this distinction.

**Worked example:** A particle moves in the $xy$-plane with position at time $t$ given by $x(t) = 2\cos t$, $y(t) = \sin t$, $0 < t < \pi$. What is the concavity of the particle's path at $t = \frac{\pi}{4}$?

1. Compute first and second derivatives of $x(t)$ and $y(t)$:

   $$x'(t) = -2\sin t, \quad y'(t) = \cos t \\ x''(t) = -2\cos t, \quad y''(t) = -\sin t$$
2. Evaluate all derivatives at $t = \frac{\pi}{4}$:

   $$x' = -\sqrt{2}, \quad y' = \frac{\sqrt{2}}{2} \\ x'' = -\sqrt{2}, \quad y'' = -\frac{\sqrt{2}}{2}$$
3. Substitute into the second derivative formula and simplify:

   $$\frac{d^2y}{dx^2} = \frac{y''x' - y'x''}{(x')^3} = \frac{\left(-\frac{\sqrt{2}}{2}\right)\left(-\sqrt{2}\right) - \left(\frac{\sqrt{2}}{2}\right)\left(-\sqrt{2}\right)}{(-\sqrt{2})^3} = \frac{1 + 1}{-2\sqrt{2}} = -\frac{1}{\sqrt{2}} < 0$$
4. Conclusion: the particle's path is concave down at $t = \frac{\pi}{4}$. Note that incorrectly using $y''(t)$ would give the same sign here by coincidence, but this approach fails for most problems.

**Check your understanding**

Test your understanding of the key distinction here:

1. A question asks for the acceleration of a particle moving along a parametric trajectory. Which of the following should you calculate?

   - $\frac{d^2y}{dx^2}$ at the given time $t$
   - $y''(t)$ only
   - The vector $\langle x''(t), y''(t) \rangle$
   - $\frac{y''(t)}{x''(t)}$

   *Answer:* The vector $\langle x''(t), y''(t) \rangle$

   *Why:* Correct! Acceleration of the particle is a vector quantity that depends only on time $t$, not on the concavity of the path.

> **Exam tip:** Always read the question carefully: if it asks for "concavity of the path" or "slope of the trajectory", use $\frac{d^2y}{dx^2}$. If it asks for "acceleration of the particle", give the vector $\langle x''(t), y''(t) \rangle$.

## AP-Style Practice Problems

**Check your understanding**

Test your calculation skills with this multiple-choice question:

1. For the parametric curve $x(t) = t^2 + 1$, $y(t) = t^3 + t$, what is $\frac{d^2y}{dx^2}$ at $t=1$?

   - $\frac{1}{2}$
   - $1$
   - $2$
   - $\frac{1}{4}$

   *Answer:* $\frac{1}{2}$

   *Why:* Correct! The calculation: $x'(1)=2, y'(1)=4, x''(1)=2, y''(1)=6$, so $\frac{(6)(2)-(4)(2)}{2^3} = \frac{4}{8} = \frac{1}{2}$. Incorrect options match common mistakes: B is $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ without dividing by $x'$, C is the wrong ratio $\frac{y''}{x''}$, D is the first derivative $\frac{dy}{dx}$ at $t=1$.

**Worked example:** A small projectile is launched from the edge of a cliff, with position at time $t$ seconds given by $x(t) = 25t$ (horizontal distance in meters), $y(t) = -4.9t^2 + 40t + 120$ (vertical height in meters). Find $\frac{d^2y}{dx^2}$ for the trajectory at $t=2$, and interpret the result.

1. Compute derivatives of $x(t)$ and $y(t)$ with respect to $t$:

   $$x'(t) = 25, \quad y'(t) = -9.8t + 40 \\ x''(t) = 0, \quad y''(t) = -9.8$$
2. Substitute into the second derivative formula:

   $$\frac{d^2y}{dx^2} = \frac{y''x' - y'x''}{(x')^3} = \frac{(-9.8)(25) - 0}{25^3} = \frac{-245}{15625} = -0.01568$$
3. Interpretation: The second derivative is negative for all $t$ in the domain, so the projectile's trajectory is always concave down, curving downward as horizontal distance from launch increases. This matches the expected shape of a projectile under constant gravitational acceleration.

## Common pitfalls

- **Wrong:** Computing $\frac{d^2y}{dx^2}$ as $\frac{y''(t)}{x''(t)}$
  - Why it fails: Students incorrectly extend the first derivative ratio $\frac{dy}{dx} = \frac{y'(t)}{x'(t)}$ to second derivatives, assuming the same pattern holds.
  - Correct: Always use the full formula $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{x'(t)}$, never just the ratio of the second derivatives of $x$ and $y$.
- **Wrong:** Forgetting to divide by $x'(t)$ after differentiating $\frac{dy}{dx}$ with respect to $t$
  - Why it fails: Students remember to differentiate the first derivative, but stop after differentiating with respect to $t$, forgetting we need the derivative with respect to $x$.
  - Correct: After calculating $\frac{d}{dt}\left(\frac{dy}{dx}\right)$, always explicitly divide by $x'(t)$ to convert to derivative with respect to $x$.
- **Wrong:** Writing $[y'(t)]^3$ instead of $[x'(t)]^3$ in the denominator of the simplified formula
  - Why it fails: Students mix up terms when memorizing the formula without understanding its derivation.
  - Correct: If you forget the simplified formula, re-derive it quickly by differentiating $\frac{y'}{x'}$ with respect to $t$, then divide by $x'$ — this takes only 30 seconds and avoids denominator errors.
- **Wrong:** Using $y''(t)$ to determine concavity of a particle's trajectory
  - Why it fails: Students confuse the $y$-component of the particle's acceleration ($y''(t)$) with the second derivative of $y$ with respect to $x$ along the path.
  - Correct: Explicitly match the question's request: use $\frac{d^2y}{dx^2}$ for path concavity, and $\langle x''(t), y''(t) \rangle$ for particle acceleration.
- **Wrong:** Only checking numerator zeros when finding concavity critical points, ignoring undefined points
  - Why it fails: Students forget that $\frac{d^2y}{dx^2}$ is undefined when $x'(t) = 0$, which can also be a point of concavity change.
  - Correct: When analyzing concavity, collect both points where the numerator is zero ($\frac{d^2y}{dx^2} = 0$) and points where $x'(t) = 0$ ($\frac{d^2y}{dx^2}$ undefined) as critical points.

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| First derivative of parametric curve | $\frac{dy}{dx} = \frac{y'(t)}{x'(t)}, \quad x'(t) \neq 0$ | Precursor to second derivative; gives slope of the tangent at parameter $t$ |
| Second derivative definition | $\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$ | Derivative of the first derivative *with respect to $x$*, not $t$ |
| Second derivative (unsimplified) | $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{x'(t)}$ | Easier to remember than the simplified form; works for all $x'(t) \neq 0$ |
| Second derivative (simplified) | $\frac{d^2y}{dx^2} = \frac{y''(t)x'(t) - y'(t)x''(t)}{[x'(t)]^3}$ | General formula for computation; requires $x'(t) \neq 0$ |
| Concavity test | Concave up: $\frac{d^2y}{dx^2} > 0$<br>Concave down: $\frac{d^2y}{dx^2} < 0$ | Same rule as Cartesian curves; uses $\frac{d^2y}{dx^2}$, not $y''(t)$ |
| Inflection point candidates | Occur where $\frac{d^2y}{dx^2} = 0$ *or* $\frac{d^2y}{dx^2}$ is undefined | $\frac{d^2y}{dx^2}$ is undefined when $x'(t) = 0$; always check these points |
| Trajectory concavity (particle motion) | Use $\frac{d^2y}{dx^2}$ formula | Not equal to $y''(t)$, which is the y-component of the particle's acceleration |
| Particle acceleration vector | $\vec{a}(t) = \langle x''(t), y''(t) \rangle$ | For motion problems, this is the acceleration of the particle, not the path second derivative |

## What's next

Mastering second derivatives of parametric equations is a critical prerequisite for upcoming topics in AP Calculus BC Unit 9, including derivatives of vector-valued functions and arc length of parametric curves. The core skill of converting derivatives with respect to the parameter $t$ to derivatives with respect to $x$ builds the chain rule intuition needed for all further work with parametric and polar curves. This topic also clarifies the key distinction between particle acceleration and concavity of the particle's trajectory, a common testing point in AP free-response questions. Without mastering the second derivative formula and its derivation, upcoming topics will be much more challenging.

- [Arc Length of a Parametric Curve](https://www.owlsprep.com/study/ap-calculus-bc-u9-arc-length-of-a-parametric/)
- [Defining and differentiating vector-valued functions](https://www.owlsprep.com/study/ap-calculus-bc-u9-defining-and-differentiating-vector-valued/)
- [Integrating vector-valued functions](https://www.owlsprep.com/study/ap-calculus-bc-u9-integrating-vector-valued-functions/)

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