Solutions — Related Rates of Change — Grade 12 Specialist Maths

A sphere’s volume is increasing at 10 cm³/s. Find dr/dt when r = 5 cm. Fluency

Step 1 — Identify the relationship: Volume of a sphere: V = (4/3)πr³.

Step 2 — Differentiate with respect to time t (chain rule):

dV/dt = 4πr² × dr/dt

Step 3 — Substitute known values: dV/dt = 10 cm³/s, r = 5 cm.

10 = 4π(5)² × dr/dt = 4π(25) × dr/dt = 100π × dr/dt

dr/dt = 10 / (100π) = 1/(10π) cm/s ≈ 0.0318 cm/s

Check dimensions: [cm³/s] = [cm²] × [cm/s] ✓

Sliding ladder: a 5 m ladder leans against a wall. The base slides out at 0.5 m/s. Find the rate at which the top slides down when the base is 3 m from the wall. Fluency

Step 1 — Draw and label: Let x = distance of foot from wall (m), y = height of top on wall (m). Both x and y are functions of t.

Step 2 — Geometric relationship: Pythagorean theorem: x² + y² = 5² = 25.

Step 3 — Differentiate with respect to t:

2x(dx/dt) + 2y(dy/dt) = 0 ⇒ x(dx/dt) + y(dy/dt) = 0

Step 4 — Find y when x = 3: y = √(25 − 9) = √16 = 4 m.

Step 5 — Substitute x = 3, y = 4, dx/dt = 0.5:

3(0.5) + 4(dy/dt) = 0 ⇒ 1.5 + 4(dy/dt) = 0

dy/dt = −1.5/4 = −3/8 = −0.375 m/s

The negative sign means the top is sliding down. The top slides down at 0.375 m/s.

A conical tank (height 10 cm, radius 4 cm) fills with water at 2 cm³/s. Find dh/dt when h = 6 cm. Fluency

Step 1 — Similar triangles: The cone has H = 10, R = 4, so r/h = 4/10 = 2/5, giving r = 2h/5.

Step 2 — Volume in terms of h only:

V = (1/3)πr²h = (1/3)π(2h/5)²h = (1/3)π(4h²/25)h = 4πh³/75

Step 3 — Differentiate with respect to t:

dV/dt = (4π/75) × 3h² × dh/dt = (4πh²/25)(dh/dt)

Step 4 — Substitute dV/dt = 2, h = 6:

2 = (4π × 36 / 25)(dh/dt) = (144π/25)(dh/dt)

dh/dt = 2 × 25 / (144π) = 50/(144π) = 25/(72π) ≈ 0.110 cm/s

Given x² + y² = 25 and dx/dt = 3, find dy/dt when x = 4. Fluency

Step 1 — Differentiate x² + y² = 25 with respect to t:

2x(dx/dt) + 2y(dy/dt) = 0

Step 2 — Find y when x = 4: y = √(25 − 16) = √9 = 3 (taking positive root).

Step 3 — Substitute x = 4, y = 3, dx/dt = 3:

2(4)(3) + 2(3)(dy/dt) = 0

24 + 6(dy/dt) = 0 ⇒ dy/dt = −24/6 = −4

y is decreasing at 4 units per time unit when x = 4.

A kite flies at constant height 60 m; the wind carries it horizontally at 4 m/s. Find the rate at which the string length increases when 100 m of string is out. Understanding

Step 1 — Variables: Let x = horizontal distance from person (m), h = 60 m (constant), L = length of string (m).

Step 2 — Relationship: L² = x² + 60² = x² + 3600.

Step 3 — Find x when L = 100: 10000 = x² + 3600, so x = √6400 = 80 m.

Step 4 — Differentiate L² = x² + 3600 with respect to t:

2L(dL/dt) = 2x(dx/dt)

Step 5 — Substitute L = 100, x = 80, dx/dt = 4:

2(100)(dL/dt) = 2(80)(4) = 640

dL/dt = 640/200 = 3.2 m/s

Two ships leave a port simultaneously — Ship A travels north at 15 km/h, Ship B travels east at 20 km/h. How fast is the distance between them increasing after 2 hours? Understanding

Step 1 — After 2 hours: Ship A is at (0, 30) km north; Ship B is at (40, 0) km east.

Let y = distance north = 15t, x = distance east = 20t, D = distance apart.

Step 2 — Relationship: D² = x² + y².

Step 3 — At t = 2: x = 40, y = 30, D = √(1600 + 900) = √2500 = 50 km.

Step 4 — Differentiate: 2D(dD/dt) = 2x(dx/dt) + 2y(dy/dt).

Step 5 — Substitute D = 50, x = 40, y = 30, dx/dt = 20, dy/dt = 15:

2(50)(dD/dt) = 2(40)(20) + 2(30)(15) = 1600 + 900 = 2500

dD/dt = 2500/100 = 25 km/h

y = x³ − 3x. If x increases at 2 units/s, find the rate of change of y when x = 2. Understanding

Step 1 — Differentiate y = x³ − 3x with respect to t using the chain rule:

dy/dt = (dy/dx)(dx/dt) = (3x² − 3)(dx/dt)

Step 2 — Substitute x = 2 and dx/dt = 2:

dy/dt = (3(4) − 3)(2) = (12 − 3)(2) = 9 × 2 = 18 units/s

y is increasing at 18 units per second when x = 2.

A rectangle has constant area 100 cm². The length increases at 3 cm/s. Find the rate of width decrease when the length is 20 cm. Understanding

Step 1 — Relationship: Area A = lw = 100 (constant).

Step 2 — Find w when l = 20: w = 100/20 = 5 cm.

Step 3 — Differentiate lw = 100 with respect to t (product rule):

l(dw/dt) + w(dl/dt) = 0

Step 4 — Substitute l = 20, w = 5, dl/dt = 3:

20(dw/dt) + 5(3) = 0 ⇒ 20(dw/dt) = −15

dw/dt = −3/4 = −0.75 cm/s

The width is decreasing at 0.75 cm/s. The negative sign confirms the decrease.

Water drains from a hemispherical bowl of radius R at rate Q = k√h. Set up the differential equation for dh/dt given V = πh²(R − h/3). Problem Solving

Step 1 — Given: V = πh²(R − h/3) = πRh² − πh³/3.

Step 2 — Differentiate V with respect to h:

dV/dh = 2πRh − πh² = πh(2R − h)

Step 3 — Apply the chain rule: dV/dt = (dV/dh)(dh/dt)

πh(2R − h)(dh/dt) = dV/dt

Step 4 — Apply the drain condition: dV/dt = −k√h (negative: volume decreasing).

πh(2R − h)(dh/dt) = −k√h

Divide both sides by h (h > 0):

π(2R − h)(dh/dt) = −k/√h = −kh^−1/2

dh/dt = −k / [π√h(2R − h)]

This separable DE shows that the rate of depth decrease depends on both the current depth h and the bowl geometry R.

A rocket at height h(t) is observed from 1000 m away. The angle of elevation θ satisfies tan θ = h/1000. If h increases at 200 m/s, find dθ/dt when θ = 30°. Problem Solving

Step 1 — Relationship: tan θ = h/1000 (horizontal distance is constant at 1000 m).

Step 2 — Differentiate both sides with respect to t:

sec²θ × dθ/dt = (1/1000)(dh/dt)

Step 3 — Evaluate sec²θ at θ = 30°:

cos(30°) = √3/2, so sec(30°) = 2/√3, and sec²(30°) = 4/3.

Step 4 — Substitute sec²θ = 4/3, dh/dt = 200:

(4/3)(dθ/dt) = 200/1000 = 0.2 = 1/5

dθ/dt = (1/5) × (3/4) = 3/20

dθ/dt = 3/20 = 0.15 rad/s ≈ 8.59°/s

Related Rates of Change — Full Worked Solutions