Parametric Second Derivatives — AP® Calculus BC

Part 1 · Hook

Why this matters

A roller coaster designer cares about more than the slope of the track: whether the track bends up into a valley or down over a hill decides whether riders are pressed into their seats or lifted out of them. For a track drawn by parametric equations, that bend is

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

, and finding it takes one extra division that most students forget.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. For $x = t^2$ , $y = 4t$ , the slope $\frac{dy}{dx}$ is

$\frac{2}{t}$
$\frac{t}{2}$
$8t$
$4$

Show the answer

$\frac{dy/dt}{dx/dt}$ $= \frac{4}{2t}$ $= \frac{2}{t}$ .

Correct: $\frac{2}{t}$ :
$\frac{t}{2}$ :
$8t$ :
$4$ :

2. A graph is concave down on an interval where

the second derivative is negative
the first derivative is negative
the function is negative
the second derivative is zero

Show the answer

Concave down means the slope is decreasing, so the second derivative is negative.

Correct: the second derivative is negative:
the first derivative is negative:
the function is negative:
the second derivative is zero:

Part 4 · See it

See it first

The unit circle x = cos t, y = sin t traced counterclockwise, its upper half in blue labeled concave down and its lower half in vermillion labeled concave up. Beside it: dy/dx = -cot t; the derivative of dy/dx with respect to t is csc squared t, positive for every t; dividing by dx/dt = -sin t gives the second derivative -1 over sin cubed t, negative on the upper half and positive on the lower half. Skipping the division by dx/dt would call the whole circle concave up. — On the circle $x = \cos t$ , $y = \sin t$ , the $t$ -derivative of the slope is $\csc^2 t$ , positive everywhere. Only after dividing by $\frac{dx}{dt} = -\sin t$ does the second derivative, $-\frac{1}{\sin^3 t}$ , show the upper half concave down and the lower half concave up. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ is itself a function of $t$ its rate of change with respect to $x$ needs the same slope rule again
Apply the rule to the function $\frac{dy}{dx}$ $\frac{d^2y}{dx^2}$ $= \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}$
The sign of $\frac{d^2y}{dx^2}$ at a timegives the concavity of the curve there: positive is concave up, negative concave down
$\frac{dy}{dx} = 0$ and $\frac{d^2y}{dx^2}$ has a signthe second derivative test classifies the point as a relative minimum or maximum

Part 6 · Key ideas

Key ideas

Step 1: find $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ and simplify it as a function of $t$ .
Step 2: differentiate $\frac{dy}{dx}$ with respect to $t$ . Step 3: divide by $\frac{dx}{dt}$ . Skipping step 3 is the classic slip.
$\frac{d^2y}{dx^2}$ is not $\frac{d^2y/dt^2}{d^2x/dt^2}$ : second derivatives do not divide that way.
Concavity, inflection points and the second derivative test work as in Unit 5, with every derivative found from $t$ .

Part 7 · Misconception

A common mistake

The wrong idea: For $x = 2t$ , $y = t^3$ : $\frac{dy}{dx} = \frac{3t^2}{2}$ , so $\frac{d^2y}{dx^2} = 3t$ .

What actually happens: $3t$ is $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ , a rate per unit of $t$ . Divide by $\frac{dx}{dt} = 2$ to make it a rate per unit of $x$ : $\frac{d^2y}{dx^2} = \frac{3t}{2}$ . Check: $y = \left(\frac{x}{2}\right)^3 = \frac{x^3}{8}$ , whose second derivative is $\frac{6x}{8} = \frac{3t}{2}$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

A curve with a linear $x$

A curve is given by the parametric equations below for

0 \le t \le 5

.

The parametric equations of the curve
Coordinate	Equation
$x$	$x(t) = 2t + 1$
$y$	$y(t) = t^3 - 6t^2$

1. Find $\frac{dy}{dx}$ at $t = 1$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\frac{dy}{dx} = \frac{3t^2 - 12t}{2}$ , which is $\frac{3 - 12}{2} = -\frac{9}{2}$ at $t = 1$ .

Answer: $\frac{-9}{2}$ ≈ -4.500

2. Find $\frac{d^2y}{dx^2}$ in terms of $t$ .

Type an expression for $\frac{d^2y}{dx^2} =$ .

Show the answer

$\frac{dy}{dx} = \frac{3t^2 - 12t}{2}$ , so $\frac{d}{dt}\left(\frac{dy}{dx}\right)$ $= \frac{6t - 12}{2}$ $= 3t - 6$ . Divide by $\frac{dx}{dt} = 2$ : $\frac{d^2y}{dx^2} = \frac{3t - 6}{2}$ .

Answer: $\frac{d^2y}{dx^2} =$ $\frac{3t - 6}{2}$

3. For which values of $t$ is the curve concave up?

$0 < t < 2$
$2 < t < 5$
$0 < t < 4$
$4 < t < 5$

Show the answer

$\frac{d^2y}{dx^2} = \frac{3t - 6}{2} > 0$ exactly when $t > 2$ , so the curve is concave up for $2 < t < 5$ .

$0 < t < 2$ : There $\frac{d^2y}{dx^2} = \frac{3t - 6}{2} < 0$ : concave down.
Correct: $2 < t < 5$ : Right: the second derivative is positive for $t > 2$ .
$0 < t < 4$ : There $\frac{dy}{dx} < 0$ , so the curve is falling; concavity is read from $\frac{d^2y}{dx^2}$ .
$4 < t < 5$ : There $\frac{dy}{dx} > 0$ , so the curve is rising; that is about slope, and concave up starts earlier, at $t = 2$ .

4. A curve is given by $x = t^2 + 1$ , $y = t^3$ for $t > 0$ . Find $\frac{d^2y}{dx^2}$ in terms of $t$ .

Type an expression for $\frac{d^2y}{dx^2} =$ .

Show the answer

$\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}$ . Then $\frac{d}{dt}\left(\frac{3t}{2}\right) = \frac{3}{2}$ , and dividing by $\frac{dx}{dt} = 2t$ gives $\frac{3}{4t}$ .

Answer: $\frac{d^2y}{dx^2} =$ $\frac{3}{4t}$

5. The circle $x = \cos t$ , $y = \sin t$ has $\frac{dy}{dx} = -\frac{\cos t}{\sin t}$ . Find $\frac{d^2y}{dx^2}$ at $t = \frac{\pi}{6}$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\frac{d}{dt}(-\cot t)$ $= \csc^2 t$ $= 4$ at $t = \frac{\pi}{6}$ , and $\frac{dx}{dt} = -\sin\frac{\pi}{6} = -\frac{1}{2}$ . So $\frac{d^2y}{dx^2} = \frac{4}{-1/2} = -8$ : concave down, as the top half of a circle should be.

Answer: -8

6. For a curve $x = x(t)$ , $y = y(t)$ with $\frac{dx}{dt} \ne 0$ , which expression equals $\frac{d^2y}{dx^2}$ ?

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

Show the answer

$\frac{d^2y}{dx^2}$ is the derivative of $\frac{dy}{dx}$ with respect to $x$ . Apply the slope rule to the function $\frac{dy}{dx}$ : its $t$ -rate divided by the $t$ -rate of $x$ .

$\frac{d^2y/dt^2}{d^2x/dt^2}$ : Second derivatives do not divide like this; differentiate $\frac{dy}{dx}$ with respect to $t$ first.
$\frac{d}{dt}\left(\frac{dy}{dx}\right)$ : This is a rate per unit of $t$ ; one more division by $\frac{dx}{dt}$ makes it per unit of $x$ .
Correct: $\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}$ : Right: $\frac{d}{dx}\left(\frac{dy}{dx}\right)$ $= \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}$ .
$\frac{dy/dx}{dx/dt}$ : The numerator must be differentiated before dividing.

Part 9 · Summary

Summary

The second derivative of a parametric curve is the derivative of

\frac{dy}{dx}

with respect to

x

: differentiate

\frac{dy}{dx}

with respect to

t

, then divide by

\frac{dx}{dt}

. Its sign gives the concavity of the curve, and with

\frac{dy}{dx} = 0

it classifies a horizontal tangent as a relative minimum or maximum. Never stop after differentiating, and never divide

\frac{d^2y}{dt^2}

by

\frac{d^2x}{dt^2}

.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

A curve with a linear xx

Summary

What comes next

Connections

A curve with a linear $x$