Derivatives of parametric curves

Warm-up

Three quick questions from earlier lessons.

Discover

A point moves with $x = t^2$ , $y = t^3$ . At $t = 1$ , $x$ grows at 2 units per second and $y$ at 3 units per second. What is the slope of its path there?

$\tfrac{3}{2}$
$\tfrac{2}{3}$

Key idea

Divide the rates.

For

x = x(t)

,

y = y(t)

:

\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}

where

\dfrac{dx}{dt} \ne 0

. Horizontal tangent:

\dfrac{dy}{dt} = 0

,

\dfrac{dx}{dt} \ne 0

. Vertical tangent:

\dfrac{dx}{dt} = 0

,

\dfrac{dy}{dt} \ne 0

.

Worked example

$x = t^2 + 1$ , $y = t^3 - 3t$ . Slope at $t = 2$ .

\dfrac{dx}{dt} = 2t

,

\dfrac{dy}{dt} = 3t^2 - 3

Differentiate each.

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

Divide.

= \dfrac{9}{4}

t = 2.

Your turn, guided

Your turn: same curve, where is the tangent horizontal ( $t > 0$ )?

t = 1

3t² − 3 = 0, and dx/dt = 2 ≠ 0 there.

Practice ★

$x = t^2$ , $y = t^3 + 9t$ . Find $\dfrac{dy}{dx}$ at $t = 3$ .
$x = t^2$ , $y = t^3 + 4t$ . Find $\dfrac{dy}{dx}$ at $t = 1$ .
$x = t^2$ , $y = t^3 + 6t$ . Find $\dfrac{dy}{dx}$ at $t = 3$ .

Watch out!

Practice ★★

$x = 3\cos t$ , $y = 3\sin t$ . Find $\dfrac{dy}{dx}$ at $t = \dfrac{\pi}{ 3 }$ . Give an exact value.
$x = 8\cos t$ , $y = 8\sin t$ . Find $\dfrac{dy}{dx}$ at $t = \dfrac{\pi}{ 6 }$ . Give an exact value.

$x = t^2$ , $y = t^3 - 27t$ . For which $t > 0$ is the tangent horizontal?

Boss problem. $x = t^2 + 1$ , $y = 2t^3$ . Study the curve at $t = 1$ , the point $(2, 2)$ .

Exit ticket

Three questions to check you have got it.

$x = t^2$ , $y = t^3 + 9t$ . Find $\dfrac{dy}{dx}$ at $t = 3$ .
$x = t^2$ , $y = t^3 + 4t$ . Find $\dfrac{dy}{dx}$ at $t = 1$ .
$x = t^2$ , $y = t^3 + 6t$ . Find $\dfrac{dy}{dx}$ at $t = 3$ .

Summary card

Parametric slopes

$\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}$ .
Horizontal: $y' = 0$ . Vertical: $x' = 0$ .

Watch out: $y$ 's rate on top.

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