Pure Mathematics — Second: Differential & Integral Calculus · Mathematics. These are 10 questions from the Magmo3 bank on this lesson, from easy to hard. Solve each one, then open "Show answer and explanation". Magmo3 has 66 questions on this lesson, each with an explanation.
The side of a square increases at a rate of 3 cm/s. Find the rate of increase of its area when the side length is 10 cm.
Correct answer: (A) 60 cm²/s
Explanation: Step 1: (relation) Area A = x², where x is the side length. Step 2: Differentiate w.r.t. t: dA/dt = 2x · dx/dt. Step 3: Substitute x = 10, dx/dt = 3: dA/dt = 2 · 10 · 3 = 60. ∴ dA/dt = 60 cm²/s.
A spherical balloon is inflated so that its radius increases at 1 cm/s. Find the rate of increase of its volume when the radius is 5 cm.
Correct answer: (D) 100π cm³/s
Explanation: Step 1: (relation) Volume V = (4/3)π r³. Step 2: Differentiate w.r.t. t: dV/dt = 4π r² · dr/dt. Step 3: Substitute r = 5, dr/dt = 1: dV/dt = 4π · 25 · 1 = 100π. ∴ dV/dt = 100π cm³/s.
The radius of a circle increases at a rate of 3 cm/s. Find the rate of increase of its circumference.
Correct answer: (B) 6π cm/s
Explanation: Step 1: (relation) Circumference C = 2π r. Step 2: Differentiate w.r.t. t: dC/dt = 2π · dr/dt. Step 3: Substitute dr/dt = 3: dC/dt = 2π · 3 = 6π (independent of r). ∴ dC/dt = 6π cm/s.
Water is poured into a cylindrical tank of base radius 4 cm at a rate of 32π cm³/s. Find the rate at which the water level rises.
Correct answer: (D) 2 cm/s
Explanation: Step 1: (relation) Volume V = π r² h = π · 16 · h (radius is fixed). Step 2: Differentiate w.r.t. t: dV/dt = 16π · dh/dt. Step 3: Substitute dV/dt = 32π: 32π = 16π · dh/dt, so dh/dt = 2. ∴ dh/dt = 2 cm/s.
A man 2 m tall walks away from the base of a lamppost 6 m high at 1.5 m/s. Find the speed of the tip of his shadow along the ground.
Correct answer: (C) 2.25 m/s
Explanation: Step 1: (relation) Tip distance from post = x + s, and from similar triangles s = x/2, so tip = x + x/2 = (3/2)x. Step 2: Differentiate w.r.t. t: d(tip)/dt = (3/2) · dx/dt. Step 3: Substitute dx/dt = 1.5: d(tip)/dt = 1.5 · 1.5 = 2.25. ∴ the tip of the shadow moves at 2.25 m/s.
A stone dropped into a still pond makes a circular ripple whose radius increases at 4 cm/s. Find the rate at which the enclosed area increases when the radius is 10 cm.
Correct answer: (D) 80π cm²/s
Explanation: Step 1: (relation) Area A = π r². Step 2: Differentiate w.r.t. t: dA/dt = 2π r · dr/dt. Step 3: Substitute r = 10, dr/dt = 4: dA/dt = 2π · 10 · 4 = 80π. ∴ dA/dt = 80π cm²/s.
A ladder 10 m long leans against a vertical wall. Its base slides away from the wall at 2 m/s. Find the rate at which the top slides down the wall when the base is 6 m from the wall.
Correct answer: (D) 1.5 m/s
Explanation: Step 1: (relation) x² + y² = 10² = 100; at x = 6, y = √(100 − 36) = 8. Step 2: Differentiate w.r.t. t: 2x · dx/dt + 2y · dy/dt = 0, i.e. x · dx/dt + y · dy/dt = 0. Step 3: Substitute x = 6, y = 8, dx/dt = 2: 6 · 2 + 8 · dy/dt = 0, so dy/dt = −12/8 = −1.5. ∴ the top slides down at 1.5 m/s.
A rectangle has length 10 cm and width 6 cm. At the same moment its length starts to increase at 2 cm/s and its width starts to decrease at 0.5 cm/s. After how many seconds does its area stop increasing?
Correct answer: (C) 3.5 s
Explanation: Step 1: After t seconds: length x = 10 + 2t and width y = 6 − 0.5t. Step 2: Area A = xy, so dA/dt = y dx/dt + x dy/dt = 2(6 − 0.5t) − 0.5(10 + 2t). Step 3: dA/dt = 12 − t − 5 − t = 7 − 2t. Step 4: dA/dt is positive while t is less than 3.5 and becomes 0 at t = 3.5 (width then 4.25 cm, still positive). Step 5: The area stops increasing after 3.5 s.
A kite flies at a constant height of 80 m and moves horizontally away from the person holding the string at 5 m/s. Find the rate at which the string is let out when its length is 100 m.
Correct answer: (A) 3 m/s
Explanation: Step 1: (relation) L² = x² + 80² (x = horizontal distance); at L = 100, x = √(100² − 80²) = 60. Step 2: Differentiate w.r.t. t: 2L · dL/dt = 2x · dx/dt, i.e. L · dL/dt = x · dx/dt. Step 3: Substitute L = 100, x = 60, dx/dt = 5: 100 · dL/dt = 60 · 5 = 300, so dL/dt = 3. ∴ dL/dt = 3 m/s.
A spherical ice ball melts so that its volume decreases at 36π cm³/s. Find the rate at which its surface area decreases when the radius is 3 cm.
Correct answer: (C) 24π cm²/s
Explanation: Step 1: (relation) V = (4/3)π r³, so dV/dt = 4π r² · dr/dt; at r = 3: −36π = 4π · 9 · dr/dt = 36π · dr/dt, giving dr/dt = −1. Step 2: (relation) S = 4π r², so differentiate w.r.t. t: dS/dt = 8π r · dr/dt. Step 3: Substitute r = 3, dr/dt = −1: dS/dt = 8π · 3 · (−1) = −24π. ∴ the surface area decreases at 24π cm²/s.
The first 10 questions in every subject are free, each with an explanation.