NCEA Level 3Related rates of changeAS 91578

Related rates of change: spheres, cones and the chain rule

What this lesson covers

  • When two quantities change with time, the chain rule links their rates.
  • Write every rate you know and the one you want before differentiating anything.
Key idea

Rates that are linked

  • Radius and volume both change with time.
  • Know one rate, find the other.
  • The chain rule makes the link.
Key idea

Write it down before you differentiate

  • The rate you are given.
  • The rate you want.
  • A formula linking the two quantities.
Key idea

The chain rule links the rates

  • dV/dt = dV/dr × dr/dt.
  • The dr parts cancel like fractions.
  • Want dr/dt? Divide by dV/dr.
Worked example

A ripple: dr/dt = 3 cm/s

  • Given dr/dt, want dA/dt.
  • Chain: dA/dr × dr/dt.
  1. given: dr/dt = 3 cm/s
  2. want: dA/dt when r = 10
  3. A = πr², dA/dr = 2πrthe link
  4. dA/dt = dA/dr × dr/dtchain rule
  5. = 2π(10) × 3r = 10, dr/dt = 3
  6. = 60π ≈ 188 cm²/s
Worked example

A balloon: dV/dt = 100 cm³/s

  • Given dV/dt, want dr/dt.
  • Divide by dV/dr.
  1. given: dV/dt = 100 cm³/s
  2. want: dr/dt when r = 5
  3. V = (4/3)πr³the link
  4. dV/dr = 4πr²(4/3) × 3 = 4
  5. r = 5: dV/dr = 4π(25) = 100π
  6. dV/dt = dV/dr × dr/dtchain rule
  7. 100 = 100π × dr/dtsubstitute
  8. dr/dt = 100/100π = 1/πdivide by 100π
  9. ≈ 0.318 cm/s
Worked example

Then the surface area

  • dr/dt = 1/π from before.
  • Chain: dS/dr × dr/dt.
  1. S = 4πr², dS/dr = 8πrthe link
  2. r = 5: dS/dr = 8π(5) = 40π
  3. dS/dt = dS/dr × dr/dtchain rule
  4. = 40π × 1/πdr/dt from before
  5. = 40 cm²/s
Worked example

Substituting too early

  • r is 5 at one instant only.
  • Differentiate first.
  1. V = (4/3)π(5)³ = 500π/3r = 5 too early
  2. dV/dt = 0a constant
  3. dV/dr = 4πr²differentiate first
  4. r = 5: dV/dr = 100πthen substitute
Worked example

A cone: two variables, one link

  • Radius 4 cm, height 12 cm, point down.
  • Water depth h, surface radius r.
  • Similar triangles: r = h/3.
A cone: two variables, one link
Worked example

Filling the cone: dV/dt = 8 cm³/s

  • Substitute r = h/3 into V.
  • One variable, then d/dh.
  1. given: dV/dt = 8 cm³/s
  2. want: dh/dt when h = 6
  3. V = (1/3)πr²hcone volume
  4. r = h/3similar triangles
  5. V = (1/3)π(h/3)²hsubstitute r = h/3
  6. = (1/3)π(h²/9)h(h/3)² = h²/9
  7. = πh³/27one variable
Worked example

Filling the cone: dh/dt at h = 6

  • Chain: dV/dh × dh/dt.
  • Then substitute h = 6.
  1. V = πh³/27
  2. dV/dh = 3πh²/27 = πh²/9differentiate
  3. h = 6: dV/dh = 36π/9 = 4π
  4. dV/dt = dV/dh × dh/dtchain rule
  5. 8 = 4π × dh/dtsubstitute
  6. dh/dt = 8/4π = 2/πdivide by 4π
  7. ≈ 0.637 cm/s
Worked example

Negative means falling

  • Leaking: dV/dt is negative.
  • Say it in words.
  1. dV/dt = −100 cm³/sleaking
  2. r = 5: dV/dr = 100πas before
  3. −100 = 100π × dr/dtchain rule
  4. dr/dt = −1/π ≈ −0.318 cm/s
  5. the radius falls at 0.318 cm/s
Key idea

Show it and say it

  • Every derivative, and the chain rule step.
  • Full accuracy, then 3 significant figures.
  • Units, and a sentence in context.
Key idea

Where the marks are

  • Achieved: both pieces, differentiated.
  • Merit: through to the rate, e.g. a sphere.
  • Excellence: build the link yourself.
Summary

What to take away

  • List the rates, link them in one variable, differentiate, then substitute.
Listgiven and wanted
Linka formula, one variable
Differentiatebefore substituting
Chainmultiply or divide
Signnegative is falling
Answerunits, in context
Your turn

Your turn: four questions

  • Pause here and answer each one.
  • Question 4: the cone has r = h/2.
  • Show every derivative, give units.
GivenFind
1circle, dr/dt = 2dA/dt at r = 5
2sphere, dV/dt = 36πdr/dt at r = 3
3cube, ds/dt = 0.5dV/dt at s = 4
4cone r = h/2, leak 2πdh/dt at h = 4
Show answer

1. dA/dt = 2πr × dr/dt = 20π ≈ 62.8 cm²/s. 2. dV/dr = 4πr² = 36π, so dr/dt = 36π ÷ 36π = 1 cm/s. 3. dV/ds = 3s² = 48, so dV/dt = 48 × 0.5 = 24 cm³/s. 4. V = πh³/12, dV/dh = πh²/4 = 4π, dV/dt = −2π, so dh/dt = −0.5 cm/s: the depth falls at 0.5 cm/s.

Worked example

The answers: questions 1 and 2

  1. 1. A = πr², dA/dr = 2πr = 10πr = 5
  2. dA/dt = dA/dr × dr/dtchain rule
  3. = 10π × 2 = 20π ≈ 62.8 cm²/s
  4. 2. V = (4/3)πr³, dV/dr = 4πr²
  5. r = 3: dV/dr = 36π
  6. dV/dt = dV/dr × dr/dtchain rule
  7. 36π = 36π × dr/dtsubstitute
  8. dr/dt = 1 cm/sdivide by 36π
Worked example

The answers: questions 3 and 4

  1. 3. V = s³, dV/ds = 3s² = 48s = 4
  2. dV/dt = dV/ds × ds/dtchain rule
  3. = 48 × 0.5 = 24 cm³/s
  4. 4. V = (1/3)π(h/2)²hsubstitute r = h/2
  5. = πh³/12(1/3)(1/4) = 1/12
  6. dV/dh = πh²/4 = 4πh = 4
  7. dV/dt = dV/dh × dh/dtchain rule
  8. −2π = 4π × dh/dtleaking, so negative
  9. dh/dt = −0.5 cm/sthe depth falls

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