SaitechAI Worksheet — Target Practice (TP)

Differentiation: Numerical word problems on rate of change (single variable applications)

Keys are hidden.
Print note: Keys are intentionally hidden in print view.
Instructions: Answer each problem in the space provided. Use correct units. Use π = 3.1416 where needed.
Q1

The radius of a sphere is increasing at 0.5 cm/s. Find the rate of increase of its volume when the radius is 10 cm. (Use V = (4/3)πr³)

Key: dV/dt = 4πr²(dr/dt) = 4π(10²)(0.5) = 200π ≈ 628.32 cm³/s.
Q2

The volume of a cube is increasing at 24 cm³/s. How fast is the edge length increasing when the edge is 3 cm? (Use V = a³)

Key: dV/dt = 3a²(da/dt) ⇒ da/dt = 24 / (3·3²) = 24/27 = 0.8889 cm/s.
Q3

The side of a square is decreasing at 0.2 m/s. Find the rate of change of its area when the side is 5 m. (Use A = s²)

Key: dA/dt = 2s(ds/dt) = 2·5·(-0.2) = -2 m²/s.
Q4

The radius of a circular field is expanding at 0.1 m/s. Find how fast its area is increasing when the radius is 20 m. (Use A = πr²)

Key: dA/dt = 2πr(dr/dt) = 2π·20·0.1 = 4π ≈ 12.5664 m²/s.
Q5

Water is poured into a cylindrical tank of fixed radius 2 m so that the water level rises at 0.05 m/s. Find the rate at which volume is increasing. (Use V = πr²h)

Key: r constant ⇒ dV/dt = πr²(dh/dt) = π·(2²)·0.05 = 0.2π ≈ 0.62832 m³/s.
Q6

A cube-shaped ice block has edge length 12 cm and is melting so that its volume decreases at 36 cm³/min. Find the rate at which the edge length is decreasing at that instant. (Use V = a³)

Key: dV/dt = 3a²(da/dt) ⇒ da/dt = (-36)/(3·12²) = -36/432 = -0.08333 cm/min.
Q7

The radius of a sphere is 8 cm and is shrinking at 0.03 cm/s. Find the rate at which its surface area is changing at that moment. (Use S = 4πr²)

Key: dS/dt = 8πr(dr/dt) = 8π·8·(-0.03) = -1.92π ≈ -6.0319 cm²/s.
Q8

The side length of a square is increasing at 0.04 m/s. How fast is the diagonal increasing when the side is 2 m? (Diagonal d = s√2)

Key: d = √2·s ⇒ dd/dt = √2·ds/dt = √2·0.04 ≈ 0.05657 m/s.
Q9

A circular plate has area increasing at 3 cm²/s. Find the rate of increase of its radius when the radius is 5 cm. (Use A = πr²)

Key: dA/dt = 2πr(dr/dt) ⇒ dr/dt = 3/(2π·5) = 3/(10π) ≈ 0.09549 cm/s.
Q10

A right circular cone has fixed height 12 cm and its radius is increasing at 0.2 cm/s. Find the rate of increase of its volume when the radius is 6 cm. (Use V = (1/3)πr²h)

Key: h constant ⇒ dV/dt = (1/3)π·(2r·dr/dt)·h = (2/3)πrh(dr/dt) = (2/3)π·6·12·0.2 = 9.6π ≈ 30.1594 cm³/s.
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