Second derivatives of parametric curves

Warm-up

Three quick questions from earlier lessons.

Discover

For $x = t^2$ , $y = t^3$ ( $t > 0$ ), $\dfrac{dy}{dx} = \dfrac{3t}{2}$ . Is $\dfrac{d^2y}{dx^2}$ just $\dfrac{d}{dt}\left(\dfrac{3t}{2}\right) = \dfrac{3}{2}$ ?

No, it must be divided by $\tfrac{dx}{dt}$
Yes

Key idea

Differentiate the slope with respect to $t$ , then convert to $x$ .

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

. Its sign gives the concavity of the path.

Worked example

$x = t^2$ , $y = t^3 - 3t$ , $t > 0$ .

\dfrac{dy}{dx} = \dfrac{3t^2 - 3}{2t} = \tfrac{3}{2}t - \tfrac{3}{2t}

First derivative.

\tfrac{3}{2} + \tfrac{3}{2t^2}

d/dt of it.

\dfrac{d^2y}{dx^2} = \dfrac{3}{4t} + \dfrac{3}{4t^3} > 0

: concave up

Divide by 2t.

Your turn, guided

Your turn: $x = t$ , $y = t^3$ . (Here $x = t$ , so it is the ordinary second derivative.)

\dfrac{d^2y}{dx^2} = 6t

d/dt(3t²) ÷ 1.

Practice ★

$x = t^2$ , $y = 8t^3$ for $t > 0$ . Find $\dfrac{d^2y}{dx^2}$ in terms of $t$ .
$x = t^2$ , $y = 1t^3$ for $t > 0$ . Find $\dfrac{d^2y}{dx^2}$ in terms of $t$ .

$x = t^2$ , $y = 9t^3$ . Find $\dfrac{d^2y}{dx^2}$ at $t = 3$ .
$x = t^2$ , $y = 7t^3$ . Find $\dfrac{d^2y}{dx^2}$ at $t = 3$ .

Watch out!

Practice ★★

$x = e^t$ , $y = e^{-3t}$ . Find $\dfrac{d^2y}{dx^2}$ in terms of $t$ .
$x = e^t$ , $y = e^{-4t}$ . Find $\dfrac{d^2y}{dx^2}$ in terms of $t$ .

Exit ticket

Three questions to check you have got it.

$x = t^2$ , $y = 8t^3$ for $t > 0$ . Find $\dfrac{d^2y}{dx^2}$ in terms of $t$ .
$x = t^2$ , $y = 1t^3$ for $t > 0$ . Find $\dfrac{d^2y}{dx^2}$ in terms of $t$ .
$x = t^2$ , $y = 5t^3$ for $t > 0$ . Find $\dfrac{d^2y}{dx^2}$ in terms of $t$ .

Summary card

Second derivative

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

Watch out: the division by $dx/dt$ is needed again.

About this lesson