Lesson 9: Parametric Calculus | Level 3 Calculus

Lesson at a Glance | He Tirohanga Whakamua

Do NowFinding tangent slope of a particle whose position is given by x = t² and y = t³10 min

First Parametric DerivativeCalculating

\frac{d y}{d x} = \frac{d y / d t}{d x / d t}

for trigonometric and polynomial parameter

t

15 min

Second Parametric DerivativeCalculating

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

for concavity15 min

Portfolio EntryWrite Section 9: Parametric Derivatives & Motion Analysis10 min

Exit DrillFind

\frac{d y}{d x}

t = \frac{π}{4}

for

x = 3 \cos t, y = 3 \sin t

5 min

Ngā Whāinga Ako | Learning Intentions

Students will know

First Parametric Derivative: $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ (where $\frac{d x}{d t} \neq 0$ ).
Second Parametric Derivative: $\frac{d^{2} y}{d x^{2}} = \frac{\frac{d}{d t} [\frac{d y}{d x}]}{\frac{d x}{d t}}$ .
Kinematic application: Rectilinear motion $s (t)$ , velocity $v (t) = s^{'} (t)$ , acceleration $a (t) = v^{'} (t) = s^{″} (t)$ .

Students will demonstrate

By calculating $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ for curves like $x = 2 t^{2}, y = t^{4} - 2 t$ .
By completing Section 9 of their Level 3 Calculus Differentiation Portfolio.

Do Now | Tīmatanga Whakaaro (10 min)

Parametric Motion Prompt:

"A rocket's horizontal position is $x = 5 t$ and vertical height is $y = 20 t - 5 t^{2}$ . How do you find the trajectory slope $\frac{d y}{d x}$ at time $t = 2 s$ without eliminating time $t$ ?"

Unpack: By taking time derivatives! $\frac{d x}{d t} = 5$ and $\frac{d y}{d t} = 20 - 10 t$ . Then $\frac{d y}{d x} = \frac{20 - 10 t}{5} = 4 - 2 t$ . At $t = 2 s$ , slope $\frac{d y}{d x} = 0$ (peak height)!

Parametric Derivative Worked Examples (15 min)

1. First Derivative Example

$x = 4 \cos t, y = 4 \sin t$
$\frac{d x}{d t} = - 4 \sin t, \frac{d y}{d t} = 4 \cos t$
$\frac{d y}{d x} = \frac{4 \cos t}{- 4 \sin t} = - \cot t$ .

2. Second Derivative Example

$\frac{d}{d t} [\frac{d y}{d x}] = \frac{d}{d t} [- \cot t] = \csc^{2} t$
$\frac{d^{2} y}{d x^{2}} = \frac{\csc^{2} t}{- 4 \sin t} = - \frac{1}{4 \sin^{3} t}$ .

📁 Calculus Differentiation Portfolio — Section 9: Parametric Calculus

Students open their Level 3 Calculus Portfolio and complete Section 9:

Section 9 Requirements:

1. Parametric Derivative Formula Proof: Write out both $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ formulas, emphasising why dividing $\frac{d^{2} y / d t^{2}}{d x / d t^{2}}$ is mathematically invalid.

2. Parametric Tangent Solver: Find the equation of the tangent line at $t = 2$ for the curve $x = 3 t^{2} - 1, y = t^{3} - 4 t$ .

3. Extension: Kinematic Acceleration Rationale: 1-paragraph explanation linking calculus derivatives $v (t) = s^{'} (t)$ and $a (t) = s^{″} (t)$ to Newton's 2nd Law $F = m s^{″} (t)$ .

Exit Verification | Ka Mutu Hoki (5 min)

Exit Check:

"My Section 9 calculates dy/dx = -cot(pi/4) = -1 for circle x = 3 cos t, y = 3 sin t at t = pi/4."

Teacher Planning & NCEA Alignment

NCEA Level 3 Calculus Alignment (6 Credits External):

Differentiation Methods: Differentiate parametric equations $x = f (t), y = g (t)$ , find second parametric derivatives $\frac{d^{2} y}{d x^{2}}$ , and solve kinematic motion problems.

Vocabulary: Parametric equation, parameter ( $t$ ), first parametric derivative ( $\frac{d y}{d x}$ ), second parametric derivative ( $\frac{d^{2} y}{d x^{2}}$ ), displacement ( $s$ ), velocity ( $v$ ), acceleration ( $a$ ).

Lesson 9: Parametric Differentiation & Motion Applications