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Differentiation

2.6 Related Rates

Focus on changing area, linked distances, changing volume and surface area.

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Ideas in this lesson

01Changing areaWorked steps · five questions02Linked distancesWorked steps · five questions03Changing volume and surface areaWorked steps · five questions
Idea 1 of 3

Changing area

Write the geometric relation before differentiating with respect to time. Substitute current values afterward.

Worked example 1

A circle has radius 22 growing at 11 units/s. Find the rate of change of its area.
  1. A=πr2A = \pi r^{2}; C=2πrC = 2\pi r
  2. dAdt=2πr(drdt)\dfrac{dA}{dt} = 2\pi r(\dfrac{dr}{dt})
  3. dAdt=2π(2)(1)=4π≈12.57\dfrac{dA}{dt} = 2\pi (2)(1) = 4\pi \approx 12.57

Answer: 4π≈12.574\pi \approx 12.57

Worked example 2

A circle has radius 33 growing at 22 units/s. Find the rate of change of its circumference.
  1. A=πr2A = \pi r^{2}; C=2πrC = 2\pi r
  2. dCdt=2π(drdt)\dfrac{dC}{dt} = 2\pi (\dfrac{dr}{dt})
  3. dCdt=2π(2)=4π≈12.57\dfrac{dC}{dt} = 2\pi (2) = 4\pi \approx 12.57

Answer: 4π≈12.574\pi \approx 12.57

Idea 2 of 3

Linked distances

A fixed-length constraint links horizontal and vertical motion. Signs distinguish motion toward and away from a reference.

Worked example 1

A5−A 5- unit ladder has its foot 33 units from the wall. The foot moves away at 11 units/s. Find the signed speed of the top along the wall.
  1. x2+y2=25  ⟹  y=4x^{2} + y^{2} = 25 \implies y = 4
  2. 2x(dxdt)+2y(dydt)=02x(\dfrac{dx}{dt}) + 2y(\dfrac{dy}{dt}) = 0
  3. dydt=−3⋅14=−34≈−0.75\dfrac{dy}{dt} = -\dfrac{3 \cdot 1}{4} = \dfrac{-3}{4} \approx -0.75

Answer: −34≈−0.75\dfrac{-3}{4} \approx -0.75

Worked example 2

A5−A 5- unit ladder has its foot 44 units from the wall. The foot moves away at 22 units/s. Find the signed speed of the top along the wall.
  1. x2+y2=25  ⟹  y=3x^{2} + y^{2} = 25 \implies y = 3
  2. 2x(dxdt)+2y(dydt)=02x(\dfrac{dx}{dt}) + 2y(\dfrac{dy}{dt}) = 0
  3. dydt=−4⋅23=−83≈−2.67\dfrac{dy}{dt} = -\dfrac{4 \cdot 2}{3} = \dfrac{-8}{3} \approx -2.67

Answer: −83≈−2.67\dfrac{-8}{3} \approx -2.67

Idea 3 of 3

Changing volume and surface area

Differentiate the geometric formula and multiply by the rate of the changing dimension. Include rate units.

Worked example 1

The radius of a sphere is 22 and drdt=1\dfrac{dr}{dt} = 1. Find dVdt\dfrac{dV}{dt}.
  1. V=43πr3V = \dfrac{4}{3}\pi r^{3}; S=4πr2S = 4\pi r^{2}
  2. dVdt=4πr2(drdt)\dfrac{dV}{dt} = 4\pi r^{2}(\dfrac{dr}{dt}); dSdt=8πr(drdt)\dfrac{dS}{dt} = 8\pi r(\dfrac{dr}{dt})
  3. dVdt=4π(2)2(1)=16π≈50.27\dfrac{dV}{dt} = 4\pi (2)^{2}(1) = 16\pi \approx 50.27

Answer: 16π≈50.2716\pi \approx 50.27

Worked example 2

The radius of a sphere is 33 and drdt=2\dfrac{dr}{dt} = 2. Find dVdtπ\dfrac{\dfrac{dV}{dt}}{\pi }.
  1. V=43πr3V = \dfrac{4}{3}\pi r^{3}; S=4πr2S = 4\pi r^{2}
  2. dVdt=4πr2(drdt)\dfrac{dV}{dt} = 4\pi r^{2}(\dfrac{dr}{dt}); dSdt=8πr(drdt)\dfrac{dS}{dt} = 8\pi r(\dfrac{dr}{dt})
  3. dVdtπ=4(3)2(2)=72\dfrac{\dfrac{dV}{dt}}{\pi } = 4(3)^{2}(2) = 72

Answer: 7272

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §2.6,p2.6, p. 148148. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.