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AP Calculus BC 9.2 Second Derivatives of Parametric Equations Study Notes - New Syllabus

AP Calculus BC 9.2 Second Derivatives of Parametric Equations Study Notes- New syllabus

AP Calculus BC 9.2 Second Derivatives of Parametric Equations Study Notes – AP Calculus BC-  per latest AP Calculus BC Syllabus.

LEARNING OBJECTIVE

  • Derivatives allow us to solve real-world problems involving rates of change.

Key Concepts: 

  • Second Derivatives of Parametric Equations

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Second Derivatives of Parametric Equations

Second Derivatives of Parametric Equations

In parametric equations, both \( x \) and \( y \) are expressed in terms of a parameter \( t \).

To find the slope of the curve, we use the chain rule:

$ \frac{dy}{dx} = \frac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}}, \quad \text{provided } \frac{dx}{dt} \neq 0 $

The second derivative \( \dfrac{d^{2}y}{dx^{2}} \) gives the rate of change of slope with respect to \( x \) and provides information about the curvature and concavity of the curve.

Formula for the Second Derivative:

From the chain rule for derivatives:

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

That is:

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

Step-by-Step Process:

  1. Find \( \dfrac{dy}{dt} \) and \( \dfrac{dx}{dt} \).
  2. Compute \( \dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} \).
  3. Differentiate \( \dfrac{dy}{dx} \) with respect to \( t \).
  4. Divide the result by \( \dfrac{dx}{dt} \) to get \( \dfrac{d^{2}y}{dx^{2}} \).

Key Notes:

  • The second derivative helps determine concavity:
    • If \( \dfrac{d^{2}y}{dx^{2}} > 0 \), the curve is concave upward.
    • If \( \dfrac{d^{2}y}{dx^{2}} < 0 \), the curve is concave downward.
  • The formula works in both 2D and higher dimensions, but here we focus on the 2D plane.
  • The denominator \( \dfrac{dx}{dt} \) must not be zero when computing \( \dfrac{dy}{dx} \) or \( \dfrac{d^{2}y}{dx^{2}} \).

Example

Given \( x = t^{2} \) and \( y = t^{3} \), find \( \dfrac{d^{2}y}{dx^{2}} \) when \( t = 2 \).

▶️ Answer/Explanation

Step 1 — First derivatives with respect to \( t \): $ \frac{dx}{dt} = 2t, \quad \frac{dy}{dt} = 3t^{2} $

First derivative \( \dfrac{dy}{dx} \): $ \frac{dy}{dx} = \frac{3t^{2}}{2t} = \frac{3t}{2} $

Differentiate \( \dfrac{dy}{dx} \) with respect to \( t \): $ \frac{d}{dt} \left( \frac{3t}{2} \right) = \frac{3}{2} $

Step 4 — Divide by \( \dfrac{dx}{dt} \): $ \frac{d^{2}y}{dx^{2}} = \frac{\frac{3}{2}}{2t} = \frac{3}{4t} $

At \( t = 2 \): $ \frac{d^{2}y}{dx^{2}} = \frac{3}{8} $

Example

Given \( x = \sin t \) and \( y = \cos t \), find \( \dfrac{d^{2}y}{dx^{2}} \) when \( t = \dfrac{\pi}{4} \).

▶️ Answer/Explanation

First derivatives with respect to \( t \): $ \frac{dx}{dt} = \cos t, \quad \frac{dy}{dt} = -\sin t $

First derivative \( \dfrac{dy}{dx} \): $ \frac{dy}{dx} = \frac{-\sin t}{\cos t} = -\tan t $

Differentiate \( \dfrac{dy}{dx} \) with respect to \( t \): $ \frac{d}{dt} \left( -\tan t \right) = -\sec^{2} t $

Step 4 — Divide by \( \dfrac{dx}{dt} \): $ \frac{d^{2}y}{dx^{2}} = \frac{-\sec^{2} t}{\cos t} = -\frac{\sec^{2} t}{\cos t} = -\sec^{3} t $

At \( t = \dfrac{\pi}{4} \): $ \frac{d^{2}y}{dx^{2}} = -\sec^{3} \left( \frac{\pi}{4} \right) = -\left( \sqrt{2} \right)^{3} = -2\sqrt{2} $

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