Second Derivatives of Parametric Equations (College Board AP® Calculus BC)

Study Guide

Written by: Mark Curtis

Reviewed by: Dan Finlay

Updated on 24 January 2025

Second derivatives of parametric equations

What is the parametric second derivative?

The second derivative of parametric equations is an expression for $\frac{d^{2} y}{d x^{2}}$ but in terms of the parameter, $t$ , only
The formula for the parametric second derivative is

$\frac{d^{2} y}{d x^{2}} = \frac{\frac{d}{d t} (\frac{d y}{d x})}{\frac{d x}{d t}}$

In words, the formula says:
- differentiate the parametric first derivative, $\frac{d y}{d x}$ , in terms of $t$
- then divide it by $\frac{d x}{d t}$
  - Remember that the parametric first derivative is given by the formula $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$
The formula comes from $\frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (\frac{d y}{d x}) = \frac{d}{d t} (\frac{d y}{d x}) \times \frac{d t}{d x} = \frac{d}{d t} (\frac{d y}{d x}) \times \frac{1}{\frac{d x}{d t}}$

Examiner Tips and Tricks

It helps to note that the formulas for both the first and second parametric derivatives divide by $\frac{d x}{d t}$ .

When do I use the parametric second derivative?

The parametric second derivative can be used to classify local extrema
- See the study guide on 'Second Derivative Test for Local Extrema'
First, set the parametric first derivative equal to zero, $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = 0$ , and solve to find the critical $t$ -values, $t = t_{0}$
Then substitute the critical $t$ -values, $t = t_{0}$ . into the parametric second derivative $\frac{d^{2} y}{d x^{2}} = \frac{\frac{d}{d t} (\frac{d y}{d x})}{\frac{d x}{d t}}$ to determine the nature of the local extrema
- If $\frac{d^{2} y}{d x^{2}} > 0 at t = t_{0}$ then there is a local minimum at $t = t_{0}$
- If $\frac{d^{2} y}{d x^{2}} < 0 at t = t_{0}$ then there is a local maximum at $t = t_{0}$
- If $\frac{d^{2} y}{d x^{2}} = 0 at t = t_{0}$ then you need to investigate further (e.g. a sketch)

Examiner Tips and Tricks

You can find the sign of $\frac{d^{2} y}{d x^{2}}$ by substituting in the critical $t$ -value without necessarily needing to fully simplify the algebraic expression for $\frac{d^{2} y}{d x^{2}}$ .

Examiner Tips and Tricks

When calculating the parametric second derivative, be prepared to use the quotient rule.

Worked Example

A curve is given parametrically by

$\begin{array}{rcl} x & = & t^{2} + t \\ y & = & t^{3} - 3 t \end{array}$

(a) Find and simplify an expression for $\frac{d^{2} y}{d x^{2}}$ in terms of $t$ .

Start by finding the parametric first derivative, given by $\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}$

$\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}} = \frac{3 t^{2} - 3}{2 t + 1}$

The parametric second derivative is given by $\frac{d^{2} y}{d x^{2}} = \frac{\frac{d}{d t} (\frac{d y}{d x})}{\frac{d x}{d t}}$ where the numerator means differentiate $\frac{d y}{d x}$ from above, $\frac{3 t^{2} - 3}{2 t + 1}$ , with respect to $t$

This requires the quotient rule, $\frac{u^{'} v - u v^{'}}{v^{2}}$ , with $u = 3 t^{2} - 3$ and $v = 2 t + 1$

$\begin{array}{rcl} \frac{d}{d t} (\frac{d y}{d x}) & = & \frac{(6 t) (2 t + 1) - (3 t^{2} - 3) (2)}{{(2 t + 1)}^{2}} \\ = & \frac{12 t^{2} + 6 t - 6 t^{2} + 6}{{(2 t + 1)}^{2}} \\ = & \frac{6 t^{2} + 6 t + 6}{{(2 t + 1)}^{2}} \end{array}$

The formula, $\frac{d^{2} y}{d x^{2}} = \frac{\frac{d}{d t} (\frac{d y}{d x})}{\frac{d x}{d t}}$ , says to divide the expression above by $\frac{d x}{d t}$ and the question says to simplify the answer

$\begin{array}{rcl} \frac{d^{2} y}{d x^{2}} & = & \frac{\frac{6 t^{2} + 6 t + 6}{{(2 t + 1)}^{2}}}{2 t + 1} \\ = & \frac{6 t^{2} + 6 t + 6}{{(2 t + 1)}^{2}} \div (2 t + 1) \\ = & \frac{6 t^{2} + 6 t + 6}{{(2 t + 1)}^{2}} \div \frac{(2 t + 1)}{1} \\ = & \frac{6 t^{2} + 6 t + 6}{{(2 t + 1)}^{2}} \times \frac{1}{(2 t + 1)} \\ = & \frac{6 t^{2} + 6 t + 6}{{(2 t + 1)}^{3}} \end{array}$

You could also factorize out a 6 from the numerator (this is an optional step)

$\frac{d^{2} y}{d x^{2}} = \frac{6 (t^{2} + t + 1)}{{(2 t + 1)}^{3}}$

(b) Find the coordinates of any local extrema on the curve, and classify the nature of these extrema.

Local extrema occur when $\frac{d y}{d x} = 0$ , so solve $\frac{d y}{d x} = 0$

$\frac{3 t^{2} - 3}{2 t + 1} = 0$

The left-hand side will be zero when the numerator is equal to zero

$\begin{array}{rcl} 3 t^{2} - 3 & = & 0 \\ t^{2} & = & 1 \\ t & = & \pm 1 \end{array}$

To find the coordinates of the critical points, substitute $t = \pm 1$ into the parametric equations $\begin{array}{rcl} x & = & t^{2} + t \end{array}$ and $y = t^{3} - 3 t$

$(2, - 2)$ and $(0, 2)$

To determine the type of extrema, substitute $t = \pm 1$ into the parametric second derivative

$t = 1$ gives $\frac{d^{2} y}{d x^{2}} = \frac{6 (1^{2} + 1 + 1)}{{(2 \times 1 + 1)}^{3}} = \frac{2}{3} > 0$ , local minimum

$t = - 1$ gives $\frac{d^{2} y}{d x^{2}} = \frac{6 ({(- 1)}^{2} + (- 1) + 1)}{{(2 (- 1) + 1)}^{3}} = - 6 < 0$ , local maximum

Write out the answer, relating the correct set of coordinates to the correct extremum

$(2, - 2)$ is a local minimum

$(0, 2)$ is a local maximum

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Second Derivatives of Parametric Equations (College Board AP® Calculus BC)

Study Guide

Second derivatives of parametric equations

What is the parametric second derivative?

When do I use the parametric second derivative?

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

Unit 1: Limits & Continuity

Limits

Definition of a Limit

Infinite Limits & Limits at Infinity

Properties of Limits

Evaluating Limits Analytically

Evaluating Limits Numerically & Graphically

Squeeze Theorem & Trigonometric Limits

Summary of Limits

Selecting Procedures for Determining Limits

Continuity

Continuity

Removable Discontinuities

Non-removable Discontinuities

Intermediate Value Theorem

Unit 2: Differentiation Definition & Fundamental Properties

Definition of Differentiation

Average Rate of Change

Instantaneous Rate of Change

Derivatives & Tangents

Estimating the Derivative at a Point

Differentiability & Continuity

Fundamental Properties of Differentiation

Derivative Rules

Derivatives of Exponentials and Logarithms

Derivatives of Sine and Cosine Functions

The Product Rule

The Quotient Rule

Derivatives of Tangent and Reciprocal Trig Functions

Unit 3: Differentiation of Composite, Implicit & Inverse Functions

Differentiation of Composite & Inverse Functions

The Chain Rule

The Inverse Function Theorem

Derivatives of Inverse Trigonometric Functions

Higher-Order Derivatives

Summary of Differentiation

Selecting Procedures for Calculating Derivatives

Unit 4: Contextual Applications of Differentiation

Rates of Change & Related Rates

Meaning of a Derivative in Context

Motion in a Straight Line

Related Rates

Linearization

Approximating Values of a Function

L'Hospital's Rule

L'Hospital's Rule

Unit 5: Analytical Applications of Differentiation

Graphs of Functions & Their Derivatives

Mean Value Theorem

Extreme Value Theorem

Critical Points

Increasing & Decreasing Functions

First Derivative Test for Local Extrema

Candidates Test for Global Extrema

Concavity of Functions

Second Derivative Test for Local Extrema

Graphs of f, f' & f''

Optimization Problems

Behaviors of Implicit Relations

Second Derivatives of Implicit Functions

Critical Points of Implicit Relations

Unit 6: Integration & Accumulation of Change

Integration & Antiderivatives

Indefinite Integrals

Derivatives & Antiderivatives

Indefinite Integral Rules

Constant of Integration

Riemann Sums & Definite Integrals

Accumulation of Change

Riemann Sums

Trapezoidal sums

Accumulation Functions

Fundamental Theorem of Calculus