Lesson 5: Implicit & Related Rates

Lesson at a Glance | He Tirohanga Whakamua

Do NowHow to find tangent slope on a circle x² + y² = 25 where y cannot be isolated easily10 min

Implicit DifferentiationDifferentiating term-by-term with respect to x:

\frac{d}{d x} [y^{2}] = 2 y \frac{d y}{d x}

15 min

Related Rates ProblemsConnecting

\frac{d V}{d t} = \frac{d V}{d r} \cdot \frac{d r}{d t}

for expanding spheres & conical tanks15 min

Portfolio EntryWrite Section 5: Implicit Curves & Related Rates Kinetics10 min

Exit DrillFind

\frac{d r}{d t}

for a spherical balloon expanding at

\frac{d V}{d t} = 20 {cm}^{3} / s

when

r = 5 cm

5 min

Ngā Whāinga Ako | Learning Intentions

Students will know

Implicit Differentiation: Differentiate every term with respect to $x$ , applying the chain rule whenever differentiating $y$ terms ( $\frac{d}{d x} [g (y)] = g^{'} (y) \frac{d y}{d x}$ ).
Related Rates of Change: Use the chain rule with respect to time $t$ :
$\frac{d A}{d t} = \frac{d A}{d r} \cdot \frac{d r}{d t}$
How to solve geometric related rates scenarios (spherical volume $V = \frac{4}{3} π r^{3}$ , conical tank $V = \frac{1}{3} π r^{2} h$ ).

Students will demonstrate

By finding $\frac{d y}{d x}$ for implicitly defined curves like $x^{2} + 3 x y + y^{2} = 19$ .
By completing Section 5 of their Level 3 Calculus Differentiation Portfolio.

Do Now | Tīmatanga Whakaaro (10 min)

Expanding Spherical Balloon Prompt:

"Air is pumped into a spherical balloon at a constant rate of $\frac{d V}{d t} = 100 {cm}^{3} / s$ . As the balloon gets larger, does the radius increase faster, slower, or at the exact same rate?"

Unpack: Slower! $V = \frac{4}{3} π r^{3} \to \frac{d V}{d r} = 4 π r^{2}$ . Since $\frac{d V}{d t} = \frac{d V}{d r} \cdot \frac{d r}{d t}$ , we get $\frac{d r}{d t} = \frac{100}{4 π r^{2}}$ . As $r$ grows, $r^{2}$ is in the denominator, so the radius expands much slower at larger sizes!

Implicit & Related Rates Worked Examples (15 min)

1. Implicit Curve Example ( $x^{2} + y^{2} = 25$ )

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [y^{2}] = \frac{d}{d x} [25]$
$2 x + 2 y \frac{d y}{d x} = 0 \to \frac{d y}{d x} = - \frac{x}{y}$ .

2. Related Rates Conical Tank Example

Conical tank $r = \frac{h}{2} \to V = \frac{1}{3} π {(\frac{h}{2})}^{2} h = \frac{π}{12} h^{3}$
$\frac{d V}{d h} = \frac{π}{4} h^{2} \to \frac{d h}{d t} = \frac{d V / d t}{\frac{π}{4} h^{2}}$ .

📁 Calculus Differentiation Portfolio — Section 5: Implicit & Related Rates

Students open their Level 3 Calculus Portfolio and complete Section 5:

Section 5 Requirements:

1. Implicit Product Derivative Proof: Find $\frac{d y}{d x}$ for $x^{3} + x y^{2} - y^{3} = 7$ using product and chain rules on $x y^{2}$ .

2. Sliding Ladder Related Rates Solver: A $5 m$ ladder leans against a wall. The foot slides away at $0.8 m/s$ . Calculate how fast the top slides down when the foot is $3 m$ from the wall ( $x^{2} + y^{2} = 25$ ).

3. Similar-Triangle Rationale: 1-paragraph explanation showing how similar triangles reduce 2 variables $(r, h)$ to 1 variable before taking time derivatives.

Kaiako answer and checking guide

Scope: Mathematical checking support for the shipped tasks. This is not an NZQA marking schedule or an assessment-evidence requirement.

Exit drill: dr/dt = 20/[4π(5)²] = 1/(5π) ≈ 0.063662 cm/s.
Portfolio 1: Differentiation gives 3x² + y² + 2xyy′ − 3y²y′ = 0, so dy/dx = (3x² + y²)/(3y² − 2xy).
Portfolio 2: When x = 3 m, y = 4 m. From x dx/dt + y dy/dt = 0, dy/dt = −0.6 m/s, or 0.6 m/s downward.
Portfolio 3: If similar triangles give a fixed ratio r = kh, then V = (πk²/3)h³ and dV/dt = πk²h²dh/dt. The ratio must be established before differentiating.

Exit Verification | Ka Mutu Hoki (5 min)

Exit Check:

"My Section 5 calculates radius rate dr/dt = 0.0637 cm/s for a 5cm balloon and solves sliding ladder rates via x^2 + y^2 = 25."

Teacher Planning & NCEA Alignment

NCEA Level 3 Calculus Alignment (6 Credits External):

Differentiation Methods: Apply implicit differentiation and solve time-dependent related rates of change problems.

Vocabulary: Implicit differentiation, related rates, time derivative ( $\frac{d}{d t}$ ), chain rule, similar triangles, volume rates.

Lesson 5: Implicit Differentiation & Related Rates of Change