9.2 Second Derivatives of Parametric Equations

The second derivative of a parametric curve tells you about concavity, and you find it by taking the derivative of $dy/dx$ with respect to $t$, then dividing by $dx/dt$. The key formula is $\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$.

Why This Matters for the AP Calculus Exam

Second derivatives of parametric curves show up only on the AP Calculus BC exam, inside the unit on parametric, polar, and vector-valued functions. Once you can find dy/dx from 9.1, this topic extends that skill so you can analyze concavity and curve shape, not just slope. On the exam you may be asked to set up the formula, compute the value at a specific t, or use the sign of $\frac{d^2y}{dx^2}$ to decide whether a curve is concave up or down. Clear notation matters here because it is easy to confuse derivatives with respect to t and derivatives with respect to x.

Key Takeaways

• Use $\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$ every time. Differentiate dy/dx with respect to t, then divide by dx/dt.
• Find dy/dx first, since you cannot get the second derivative without it.
• Differentiating dy/dx usually needs the quotient rule, because dy/dx is a ratio in terms of t.
• The sign of $\frac{d^2y}{dx^2}$ tells you concavity: positive means concave up, negative means concave down.
• Watch where dx/dt = 0, since the second derivative is undefined there.
• Always divide by dx/dt, never by $\frac{d^2x}{dt^2}$.

Finding Second Derivatives of Parametric Equations

For a parametric curve, the second derivative is written as:

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

To build it, apply the chain rule. Since dy/dx is itself a function of t, you differentiate it with respect to t and then divide by dx/dt:

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

Breaking Down the Formula

Finding a second derivative always means taking the derivative of the first derivative. So you first need dy/dx, the parametric derivative from 9.1.

Once you have dy/dx, differentiate it with respect to the parameter t. Because dy/dx is a ratio of $\frac{dy}{dt}$ to $\frac{dx}{dt}$, this step usually calls for the quotient rule. The quotient rule says the derivative of a ratio equals the derivative of the top times the bottom, minus the top times the derivative of the bottom, all over the bottom squared.

You do not need to memorize the full chain-rule derivation. Focus on the working formula:

Second Derivative of a Parametric Function: Given a parametric curve defined by $x(t)$ and $y(t)$, the second derivative is:

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

This gets more natural with practice, so work through the examples below.

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Practice Finding Second Derivatives of Parametric Functions

Three worked examples with full steps.

Example 1

Show that the cycloid defined by $x(t) = 2(t - \sin(t))$ and $y(t) = 2(1 - \cos(t))$ is concave down on $t \in (0, 2)$.

Remember: the second derivative gives you information about concavity.

$$x(t)=2t-2\sin(t),\quad y(t)=2-2\cos(t)$$

Finding $dx/dt$ and $dy/dt$:

$$\frac{dx}{dt}=2-2\cos(t),\quad \frac{dy}{dt}=2\sin(t)$$

Combine them to get dy/dx, the first derivative:

$$\frac{dy}{dx}=\frac{2\sin(t)}{2(1-\cos(t))}=\frac{\sin(t)}{1-\cos(t)}$$

To find the second derivative, differentiate dy/dx with respect to t, then divide by dx/dt. Using the quotient rule, the numerator becomes:

$$\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{\cos(t)(1-\cos(t))-(\sin(t))(\sin(t))}{(1-\cos(t))^2}$$

Divide by dx/dt and simplify:

$$\frac{d^2y}{dx^2}=\frac{\frac{\cos(t)(1-\cos(t))-\sin^2(t)}{(1-\cos(t))^2}}{2(1-\cos(t))}=\frac{\cos(t)-\cos^2(t)-\sin^2(t)}{2(1-\cos(t))^3}$$

Using the identity $\sin^2(t)+\cos^2(t)=1$:

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

Since this is always negative, the cycloid is always concave down.

Graph displaying the cycloid function, repeatedly concave down. Courtesy of Desmos.

Example 2

Consider the parametric equations $x = t^3 - 3t$ and $y = t^2 + 2t - 5$. Find the second derivative of $y$ with respect to $x$.

$$x(t)=t^3-3t,\quad y(t)=t^2+2t-5$$

Finding $dx/dt$ and $dy/dt$:

$$\frac{dx}{dt}=3t^2-3,\quad \frac{dy}{dt}=2t+2$$

Combine to get dy/dx:

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

Differentiate dy/dx with respect to t:

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

Divide by dx/dt and simplify:

$$\frac{d^2y}{dx^2}=\frac{\frac{-2}{3(t-1)^2}}{3(t^2-1)}=\frac{-2}{9(t-1)(t^2-1)}$$

Example 3

A car moves along a curved road represented by the parametric equations $x = t^3$ and $y = t^4$. Find the value of the second derivative of $y$ with respect to $x$.

Finding $dx/dt$ and $dy/dt$:

$$\frac{dx}{dt}=3t^2,\quad \frac{dy}{dt}=4t^3$$

Combine to get dy/dx:

$$\frac{dy}{dx}=\frac{4t^3}{3t^2}=\frac{4}{3}t$$

Differentiate dy/dx with respect to t:

$$\frac{d}{dt}\left(\frac{dy}{dx}\right)=\frac{4}{3}$$

Divide by dx/dt and simplify:

$$\frac{d^2y}{dx^2}=\frac{4/3}{3t^2}=\frac{4}{9t^2}$$

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How to Use This on the AP Calculus Exam

Problem Solving

• Get dy/dx first, then differentiate that with respect to t. Skipping straight to second derivatives leads to errors.
• After differentiating dy/dx, divide by dx/dt, not by anything else.
• If the question asks about concavity, find the sign of $\frac{d^2y}{dx^2}$ at the given t value or interval. Positive is concave up, negative is concave down.
• Simplify dy/dx before differentiating when you can. Cleaner expressions make the quotient rule easier and reduce arithmetic slips.

Common Trap

• The single biggest error is computing $\frac{d^2y}{dx^2}$ as $\frac{d^2y/dt^2}{d^2x/dt^2}$. That is not how the formula works. You must divide the t-derivative of dy/dx by dx/dt.
• Keep careful track of which variable you are differentiating with respect to. Mixing up t and x is easy when expressions get long.

Common Misconceptions

• The second derivative is not the second derivative of y over the second derivative of x. $\frac{d^2y}{dx^2}$ does not equal $\frac{d^2y/dt^2}{d^2x/dt^2}$. You differentiate dy/dx with respect to t, then divide by dx/dt.
• You divide by dx/dt, not by the derivative of the denominator. The final division uses dx/dt, the same first derivative you used in 9.1.
• dy/dx is still in terms of t. Since dy/dx is a function of t, you differentiate it with respect to t, which is why the quotient rule usually appears.
• A negative second derivative means concave down, not decreasing. Concavity and increasing/decreasing are different ideas. The second derivative describes concavity.
• The second derivative can be undefined where dx/dt = 0. At points where dx/dt = 0, the formula breaks down, so check those points separately.

Related AP Calculus Guides

• 9.1 Defining and Differentiating Parametric Equations
• 9.4 Defining and Differentiating Vector-Valued Functions
• Unit 9 Overview: Parametric Equations, Polar Coordinates, and Vector-Valued Functions
• 9.3 Finding Arc Lengths of Curves Given by Parametric Equations
• 9.7 Defining Polar Coordinates and Differentiating in Polar Form
• 9.8 Find the Area of a Polar Region or the Area Bounded by a Single Polar Curve