Second derivatives of parametric equations — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Derivation of the second derivative formula for parametric curves, step-by-step computation of $\frac{d^{2}y}{d{x}^2}$, concavity analysis, and application to planar particle motion problems.

You should already know: How to compute first derivatives of parametric functions, the chain rule for differentiation, the definition of concavity for Cartesian curves.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Second derivatives of parametric equations?

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 is a core topic in AP Calculus BC Unit 9, which accounts for roughly 11-12% of the total exam weight per the College Board CED. Second derivatives of parametric equations appear in both multiple-choice (MCQ) and free-response (FRQ) sections of the exam; they are commonly paired with concavity questions, planar particle motion problems, or general curve analysis, making up 1-3 points on a typical FRQ and 1-2 MCQ questions per exam. Unlike the second derivative of a Cartesian function $y(x)$, the parametric second derivative requires an extra chain rule step to account for differentiation with respect to $x$, not $t$. This topic extends concavity analysis (learned for Cartesian functions) to parametric curves, which can describe closed curves, non-functions, and particle motion trajectories.

2. Derivation and Formula for the Parametric Second Derivative

We start with a parametric curve $x=x(t)$, $y=y(t)$, where both functions are differentiable and $\frac{dx}{dt}\ne 0$. We already know the first derivative of $y$ with respect to $x$ is: $$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{y^{\prime }(t)}{x^{\prime }(t)}$$ The second derivative $\frac{d^{2}y}{d{x}^2}$ is defined as the derivative of $\frac{dy}{dx}$ with respect to $x$, not with respect to $t$. To find this derivative, we apply the chain rule: $\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot \frac{dt}{dx}$. From the inverse derivative rule, $\frac{dt}{dx}=\frac{1}{\frac{dx}{dt}}=\frac{1}{x^{\prime }(t)}$.

Combining these gives the unsimplified formula: $$\frac{d^{2}y}{d{x}^2}=\frac{\frac{d}{dt}\left(\frac{y^{\prime }(t)}{x^{\prime }(t)}\right)}{x^{\prime }(t)}$$ Applying the quotient rule to the numerator gives the simplified general formula: $$\frac{d^{2}y}{d{x}^2}=\frac{y^{\prime \prime }(t)x^{\prime }(t)-y^{\prime }(t)x^{\prime \prime }(t)}{[x^{\prime }(t){]}^3}$$ The extra denominator of $x^{\prime }(t)$ (and the final cube) comes directly from the chain rule conversion from derivative with respect to $t$ to derivative with respect to $x$, the step most students forget.

Worked Example

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

1. Compute first derivatives of $x$ and $y$ with respect to $t$: $x^{\prime }(t)=2t+2$, $y^{\prime }(t)=3t^2-3$. At $t=1$, $x^{\prime }(1)=4$, $y^{\prime }(1)=0$.
2. Compute second derivatives of $x$ and $y$ with respect to $t$: $x^{\prime \prime }(t)=2$, $y^{\prime \prime }(t)=6t$. At $t=1$, $x^{\prime \prime }(1)=2$, $y^{\prime \prime }(1)=6$.
3. Substitute into the simplified second derivative formula: $\frac{d^{2}y}{d{x}^2}=\frac{y^{\prime \prime }{x}^{\prime }-y^{\prime }{x}^{\prime \prime }}{(x^{\prime }{)}^3}$.
4. Plug in $t=1$ values: numerator $=(6)(4)-(0)(2)=24$, denominator $=4^3=64$, so $\frac{d^{2}y}{d{x}^2}=\frac{24}{64}=\frac{3}{8}$.

Exam tip: If you only need the second derivative at a specific value of $t$, compute $x^{\prime },y^{\prime },x^{\prime \prime },y^{\prime \prime }$ at that $t$ first before plugging into the formula. This avoids messy algebraic simplification of the general $\frac{d^{2}y}{d{x}^2}$, saving valuable time on MCQ.

3. 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 rule applies: a curve is concave up when $\frac{d^{2}y}{d{x}^2}>0$ and concave down when $\frac{d^{2}y}{d{x}^2}<0$, regardless of the parameter $t$. The only difference is that $\frac{d^{2}y}{d{x}^2}$ is a function of $t$, so we solve sign inequalities in terms of $t$, then map back to $x/y$ coordinates if required.

Inflection point (concavity change) candidates occur where $\frac{d^{2}y}{d{x}^2}=0$ or $\frac{d^{2}y}{d{x}^2}$ is undefined (which happens when $\frac{dx}{dt}=0$, 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 $x=4$: $x(t)=t^2=4$, so $t=2$ (since $t>0$).
2. Compute derivatives: $x^{\prime }(t)=2t$, $y^{\prime }(t)=e^t$, $x^{\prime \prime }(t)=2$, $y^{\prime \prime }(t)=e^t$.
3. Evaluate derivatives at $t=2$: $x^{\prime }(2)=4$, $y^{\prime }(2)=e^2$, $x^{\prime \prime }(2)=2$, $y^{\prime \prime }(2)=e^2$.
4. Calculate $\frac{d^{2}y}{d{x}^2}$: $\frac{(e^2)(4)-(e^2)(2)}{4^3}=\frac{2e^2}{64}=\frac{e^2}{32}\approx 0.23>0$.
5. 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 if you remember that parametric derivatives are functions of $t$, not $x$.

4. 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 $\frac{d^{2}y}{d{x}^2}$ gives the concavity of that path. This is not the same as the particle's acceleration vector, which is $⟨x^{\prime \prime }(t),y^{\prime \prime }(t)⟩$, a vector describing how the particle's velocity changes with time. AP exam questions regularly test this distinction, so it is critical to separate the two concepts. If asked for concavity of the particle's path, always compute $\frac{d^{2}y}{d{x}^2}$, not just use $y^{\prime \prime }(t)$.

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 derivatives: $x^{\prime }(t)=-2\sin t$, $y^{\prime }(t)=\cos t$, $x^{\prime \prime }(t)=-2\cos t$, $y^{\prime \prime }(t)=-\sin t$.
2. Evaluate at $t=\frac{\pi }{4}$: $x^{\prime }=-\sqrt{2}$, $y^{\prime }=\frac{\sqrt{2}}{2}$, $x^{\prime \prime }=-\sqrt{2}$, $y^{\prime \prime }=-\frac{\sqrt{2}}{2}$.
3. Substitute into the second derivative formula: $\frac{d^{2}y}{d{x}^2}=\frac{y^{\prime \prime }{x}^{\prime }-y^{\prime }{x}^{\prime \prime }}{(x^{\prime }{)}^3}=\frac{\left(-\frac{\sqrt{2}}{2}\right)\left(-\sqrt{2}\right)-\left(\frac{\sqrt{2}}{2}\right)\left(-\sqrt{2}\right)}{(-\sqrt{2}{)}^3}=\frac{(1)+(1)}{-2\sqrt{2}}=-\frac{1}{\sqrt{2}}<0$.
4. Conclusion: The particle's path is concave down at $t=\frac{\pi }{4}$. If we had incorrectly used $y^{\prime \prime }(t)=-\frac{\sqrt{2}}{2}<0$, we would have gotten the correct sign by coincidence here, but this approach fails for most problems.

Exam tip: Always read the question carefully: if it asks for "concavity of the path" or "slope of the trajectory", use $\frac{d^{2}y}{d{x}^2}$. If it asks for "acceleration of the particle", give the vector $⟨x^{\prime \prime }(t),y^{\prime \prime }(t)⟩$.

5. Common Pitfalls (and how to avoid them)

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

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

For the parametric curve $x(t)=t^2+1$, $y(t)=t^3+t$, what is $\frac{d^{2}y}{d{x}^2}$ at $t=1$? A) $\frac{1}{2}$ B) $1$ C) $2$ D) $\frac{1}{4}$

Worked Solution: First, calculate the first and second derivatives of $x(t)$ and $y(t)$ with respect to $t$: $x^{\prime }(t)=2t$, $y^{\prime }(t)=3t^2+1$, $x^{\prime \prime }(t)=2$, $y^{\prime \prime }(t)=6t$. Evaluating at $t=1$ gives $x^{\prime }(1)=2$, $y^{\prime }(1)=4$, $x^{\prime \prime }(1)=2$, $y^{\prime \prime }(1)=6$. Substitute into the parametric second derivative formula: $\frac{d^{2}y}{d{x}^2}=\frac{y^{\prime \prime }{x}^{\prime }-y^{\prime }{x}^{\prime \prime }}{(x^{\prime }{)}^3}=\frac{(6)(2)-(4)(2)}{2^3}=\frac{12-8}{8}=\frac{4}{8}=\frac{1}{2}$. The incorrect options correspond to common mistakes: B is $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ without dividing by $x^{\prime }$, C is the ratio $\frac{y^{\prime \prime }}{x^{\prime \prime }}$, D is the first derivative $\frac{dy}{dx}$ at $t=1$. The correct answer is A.

Question 2 (Free Response)

Consider the parametric curve $C$ given by $x(t)=\cos t$, $y(t)=\sin 2t$, for $0\le t\le \pi$. (a) Find $\frac{d^{2}y}{d{x}^2}$ in terms of $t$. (b) Determine all values of $t$ in $0<t<\pi$ for which $C$ is concave up. (c) Find the coordinates of all inflection points on $C$ for $0\le t\le \pi$.

Worked Solution: (a) First, compute derivatives: $x^{\prime }(t)=-\sin t$, $y^{\prime }(t)=2\cos 2t$. The first derivative is $\frac{dy}{dx}=-\frac{2\cos 2t}{\sin t}$. Differentiate $\frac{dy}{dx}$ with respect to $t$ using the quotient rule: $\frac{d}{dt}\left(\frac{dy}{dx}\right)=-\frac{2(-2\sin 2t\sin t-\cos 2t\cos t)}{{\sin }^{2}t}=\frac{2\cos t(2{\sin }^{2}t+1)}{{\sin }^{2}t}$. Divide by $x^{\prime }(t)=-\sin t$ to get $\frac{d^{2}y}{d{x}^2}=-\frac{2\cos t(2{\sin }^{2}t+1)}{{\sin }^{3}t}$. (b) For $0<t<\pi$, $\sin t>0$, so ${\sin }^{3}t>0$, and $2{\sin }^{2}t+1>0$ always. The sign of $\frac{d^{2}y}{d{x}^2}$ depends only on $-\cos t$. $\frac{d^{2}y}{d{x}^2}>0$ when $-\cos t>0⟹\cos t<0$, which occurs for $\frac{\pi }{2}<t<\pi$. So $C$ is concave up on $\left(\frac{\pi }{2},\pi \right)$. (c) Inflection points occur where concavity changes. $\frac{d^{2}y}{d{x}^2}=0$ only when $\cos t=0⟹t=\frac{\pi }{2}$ in the interval. $\frac{d^{2}y}{d{x}^2}$ is undefined at the endpoints $t=0$ and $t=\pi$. Concavity changes from down (for $0<t<\frac{\pi }{2}$) to up (for $\frac{\pi }{2}<t<\pi$) at $t=\frac{\pi }{2}$. The coordinates are $x\left(\frac{\pi }{2}\right)=0$, $y\left(\frac{\pi }{2}\right)=\sin \pi =0$, so the only inflection point is $\boxed{(0,0)}$.

Question 3 (Application / Real-World Style)

A small projectile is launched from the edge of a cliff, and its position at time $t$ seconds after launch is given by $x(t)=25t$ (horizontal distance in meters), $y(t)=-4.9t^2+40t+120$ (vertical height in meters above the ground below the cliff), for $0\le t\le 8.5$ (when it hits the ground). Find the value of $\frac{d^{2}y}{d{x}^2}$ for the trajectory at $t=2$, and interpret your result in context.

Worked Solution: First, compute derivatives: $x^{\prime }(t)=25$, $y^{\prime }(t)=-9.8t+40$, $x^{\prime \prime }(t)=0$, $y^{\prime \prime }(t)=-9.8$. Substitute into the second derivative formula: $$\frac{d^{2}y}{d{x}^2}=\frac{y^{\prime \prime }{x}^{\prime }-y^{\prime }{x}^{\prime \prime }}{(x^{\prime }{)}^3}=\frac{(-9.8)(25)-0}{25^3}=\frac{-245}{15625}=-0.01568$$ The second derivative is negative for all $t$ in the interval. In context, this means the projectile's trajectory through the air is always concave down, curving downward as horizontal distance from the launch point increases, which matches the expected shape of a projectile under constant gravitational acceleration.

7. Quick Reference Cheatsheet

8. What's Next

Mastering second derivatives of parametric equations is a critical prerequisite for upcoming topics in Unit 9, including derivatives of vector-valued functions and arc length of parametric curves. The core skill of this topic — converting derivatives with respect to the parameter $t$ to derivatives with respect to $x$ or $y$ — 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 (a vector quantity based on $t$) and concavity of the particle's trajectory (a path property based on $x$), which is a common point of testing in AP FRQs. Next, you will apply similar chain rule reasoning to find derivatives of polar curves and to compute arc length of parametric curves; without mastering the second derivative formula and its derivation, these topics will be much more challenging.

First derivatives of parametric equations Derivatives of vector-valued functions Derivatives of polar curves Arc length of parametric curves