Aim for FiveCalculus BC

Unit 9 · Topic 9.2

BC only

9.2 Second Derivatives of Parametric Equations

BC only. The second derivative of a parametric curve is d²y/dx² = [d/dt (dy/dx)] ÷ (dx/dt). You differentiate the slope with respect to t, then divide by dx/dt again, and the result tells you concavity.

Key terms

  • second derivative
  • d²y/dx²
  • concavity
  • chain rule

The formula

This whole unit is BC only. d²y/dx² means the derivative of dy/dx with respect to x. But your expression for dy/dx is in terms of t, so use the same trick as before: differentiate with respect to t, then divide by dx/dt.

d²y/dx² = [d/dt (dy/dx)] / (dx/dt).

It's the chain rule again: d/dt (dy/dx) = [d/dx (dy/dx)] · (dx/dt), so dividing by dx/dt leaves d/dx (dy/dx).

The same caution applies as for the first derivative: the formula only makes sense where dx/dt ≠ 0. At a point with a vertical tangent, d²y/dx² isn't defined.

The steps

  • Find dy/dx = (dy/dt)/(dx/dt) and simplify it as much as you can.
  • Differentiate that expression with respect to t (quotient rule if needed).
  • Divide the result by dx/dt.
  • Plug in the t-value if you need a number.

What the sign means

As with ordinary functions, d²y/dx² > 0 means the curve is concave up at that point and d²y/dx² < 0 means it's concave down. Notice that the sign of dx/dt matters. Where dx/dt < 0, d²y/dx² has the opposite sign from d/dt (dy/dx). So always compute the full formula rather than guessing from d/dt (dy/dx) alone.

You can also find where a parametric curve changes concavity: set the numerator of d²y/dx² equal to zero, and watch for t-values where dx/dt = 0, since the expression can change sign there too.

Why the extra division matters

d/dt (dy/dx) tells you how fast the slope changes per unit of time. Concavity is about how fast the slope changes per unit of x. If the point is moving horizontally very fast (dx/dt large), the slope changes slowly per unit of x even if it changes quickly per unit of time. Dividing by dx/dt converts from “per t” to “per x.” Skipping that division is the most common mistake on this topic.

Simplify first

Before you differentiate dy/dx with respect to t, simplify it. In the example below, (3t² − 3)/(2t) becomes (3/2)(t − 1/t), which you can differentiate without the quotient rule. Fewer steps means fewer chances for an algebra slip, and the AP exam usually asks for a number at a particular t, so you can plug in early once the derivative is done.

Worked examples

Try each one yourself first, then open the solution.

  1. Example 1

    Concavity at two points

    For x = t² − 1 and y = t³ − 3t, find d²y/dx² and determine the concavity at t = 1 and t = −1.

    Show the solution
    1. Step 1: dy/dx = (3t² − 3)/(2t) = (3/2)(t − 1/t).
    2. Step 2: d/dt (dy/dx) = (3/2)(1 + 1/t²).
    3. Step 3: Divide by dx/dt = 2t: d²y/dx² = (3/2)(1 + 1/t²)/(2t) = 3(t² + 1)/(4t³).
    4. Step 4: At t = 1: 3(2)/4 = 3/2 > 0, concave up.
    5. Step 5: At t = −1: 3(2)/(−4) = −3/2 < 0, concave down.

    Answer: d²y/dx² = 3(t² + 1)/(4t³); concave up at t = 1, concave down at t = −1.

  2. Example 2

    Trap: forgetting the last division

    For x = eᵗ and y = t², find d²y/dx², and say for which t the curve is concave up.

    Show the solution
    1. Step 1: dy/dx = 2t/eᵗ = 2te^(−t).
    2. Step 2: d/dt (dy/dx) = 2e^(−t) − 2te^(−t) = 2e^(−t)(1 − t).
    3. Step 3: Stopping here is the classic mistake.
    4. Step 4: Divide by dx/dt = eᵗ: d²y/dx² = 2e^(−t)(1 − t)/eᵗ = 2e^(−2t)(1 − t).
    5. Step 5: e^(−2t) > 0 always, so d²y/dx² > 0 exactly when 1 − t > 0, that is, t < 1.

    Answer: d²y/dx² = 2e^(−2t)(1 − t); concave up for t < 1.

Common mistakes

  • Stopping at d/dt (dy/dx) without dividing by dx/dt.
  • Computing (d²y/dt²)/(d²x/dt²). That's not the second derivative.
  • Differentiating the unsimplified quotient and making an algebra slip. Simplify dy/dx first.
  • Reading concavity from dy/dx's sign instead of from d²y/dx².

On the exam

  • Multiple-choice questions often list d/dt (dy/dx) as a wrong answer choice. Make sure you divided by dx/dt.
  • If a free-response question asks whether a curve is concave up at a point, compute d²y/dx² there and state its sign as the reason.

Connected topics

Videos

  • Calculus BC – 9.2 Second Derivatives of Parametric Equations

    The AlgebrosWatch on YouTube (opens in a new tab)

  • Second derivatives (parametric functions) | Advanced derivatives | AP Calculus BC | Khan Academy

    Khan AcademyWatch on YouTube (opens in a new tab)

  • Second Derivative of Parametric Equations with Example

    turksvidsWatch on YouTube (opens in a new tab)

  • Parametric Curves - Finding Second Derivatives

    Patrick JWatch on YouTube (opens in a new tab)

  • Second Derivative of a Parametric Curve (KristaKingMath)

    Krista KingWatch on YouTube (opens in a new tab)

  • Second Derivatives of Parametric Equations With Concavity

    The Organic Chemistry TutorWatch on YouTube (opens in a new tab)

Check yourself

4 questions on 9.2 Second Derivatives of Parametric Equations. Pick an answer to see if you got it, and why.

Question 1 of 4

A curve is defined by x = t² + 1 and y = t³. What is the value of d²y/dx² at the point where t = 2?

Question 2 of 4

A curve is defined by x = t² + t and y = t² − t. For which values of t is the curve concave up?

Question 3 of 4

A curve is defined by x = ln t and y = t² for t > 0. What is the value of d²y/dx² at the point where t = 3?

Question 4 of 4

A curve is defined by x = 2t − 1 and y = t³ − 3t². What is the value of d²y/dx² at the point where t = 2?

0 of 4 answered