Topic 9.2 Notes – Second Derivatives of Parametric Equations

Second Derivatives of Parametric Equations

You’re given:

$$x = x(t), \quad y = y(t)$$

Both coordinates depend on $t$ (often time).

From 9.1, you already know:

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}, \quad \text{as long as } \frac{dx}{dt} \ne 0$$

That gives the slope of the tangent line in the $xy$-plane.

Now we want the second derivative:

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

Since everything is still written in terms of $t$, we apply the chain rule and get the formula you need to memorize:

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

That’s the entire skill for this topic.

What the Formula Means

Think about what’s happening:

• First derivative: slope $\frac{dy}{dx}$
• Second derivative: how that slope changes as $x$ changes

But we don’t have $y$ as a function of $x$. We only have everything in terms of $t$. So we:

1. Differenti​ate the slope with respect to $t$
2. Convert that rate into a rate with respect to $x$ by dividing by $dx/dt$

It’s chain rule logic:

$$\frac{d}{dx} = \frac{1}{dx/dt}\frac{d}{dt}$$

Step-by-Step Process

Every problem follows this same structure.

1. Find $dx/dt$ and $dy/dt$
2. Compute the first derivative

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

   Simplify here if possible.

3. Differentiate $dy/dx$ with respect to $t$

   Use product or quotient rule if needed.

4. Divide by $dx/dt$

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

If you forget the final division step, you haven’t finished.

Quick Example

Let

$$x = t^2 + 1, \quad y = t^3 - t$$

Step 1: First derivatives

$$\frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 3t^2 - 1$$

Step 2: First derivative

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

Step 3: Differentiate with respect to $t$
Use quotient rule:

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

Step 4: Divide by $dx/dt = 2t$

$$\frac{d^2y}{dx^2} = \frac{\frac{6t^2 + 2}{4t^2}}{2t} = \frac{6t^2 + 2}{8t^3}$$

Done.

On an FRQ, every one of those steps needs to be visible.

Concavity of Parametric Curves

The second derivative still controls concavity in the $xy$-plane.

• $\frac{d^2y}{dx^2} > 0$ → concave up
• $\frac{d^2y}{dx^2} < 0$ → concave down

Even though you’re calculating in terms of $t$, the interpretation is about the graph in $x$ and $y$.

Here’s a quick visual reminder of what concave up and concave down look like in the $xy$-plane:

Concave up and concave down (with tangent lines)

When slope increases as $x$ increases, the graph cups upward. When slope decreases as $x$ increases, it bends downward.

Important Conditions

When $dx/dt = 0$

The formula only works when $dx/dt \ne 0$.

If $dx/dt = 0$:

• $dy/dx$ is undefined
• You likely have a vertical tangent
• The second derivative formula also breaks there

AP questions sometimes sneak this in and ask about behavior at that value of $t$. Always check before plugging in.

Common Errors

• Stopping too early. Many students compute $\frac{d}{dt}(dy/dx)$ and forget to divide by $dx/dt$.
• Finding $d^2y/dt^2$ instead. That is not the same thing.
• Not simplifying $dy/dx$ first. Algebra gets messy fast.
• Plugging in a $t$-value before finishing the formula. Keep everything symbolic until the end.

On multiple choice, wrong answers often come from skipping the final division step.