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Applications of Derivatives - I

ICSE · Class 12 · Mathematics

Flashcards for Applications of Derivatives - I — ICSE Class 12 Mathematics. Quick Q&A cards covering key concepts, definitions, and formulas.

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A graph of a function y=f(x) with a tangent line at a point (x, y), illustrating dy/dx as the slope of the tangent.
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37 Flashcards
Card 1Rate of change

Solve: A side of a square sheet is increasing at 4 cm/min. Find the rate at which the area increases when the side is 5 cm.

Answer

Step 1: Let side = x cm and area = A cm². Step 2: A = x². Step 3: Differentiate w.r.t. time t: dA/dt = 2x · dx/dt. Step 4: Substitute x = 5 and dx/dt = 4. Step 5: dA/dt = 2 × 5 × 4 = 40 cm²/min. Answe…

Card 2Rate of change

Solve: A side of a square is increasing at 0.5 cm/s. Find the rate of increase of its perimeter.

Answer

Step 1: Let side = x cm and perimeter = P cm. Step 2: P = 4x. Step 3: Differentiate w.r.t. time t: dP/dt = 4 · dx/dt. Step 4: Substitute dx/dt = 0.5. Step 5: dP/dt = 4 × 0.5 = 2 cm/s. Answer: 2 cm/s.

Card 3Rate of change

Solve: If x and y are the sides of two squares and y = x - x², find dA₂/dA₁, where A₁ = x² and A₂ = y².

Answer

Step 1: A₁ = x², so dA₁/dx = 2x. Step 2: A₂ = y² = (x - x²)². Step 3: dA₂/dx = 2(x - x²)(1 - 2x). Step 4: dA₂/dA₁ = (dA₂/dx)/(dA₁/dx). Step 5: dA₂/dA₁ = [2(x - x²)(1 - 2x)]/(2x). Step 6: Simplify: (1 …

Card 4Rate of change

Solve: The length of a rectangle decreases at 5 cm/min and width increases at 4 cm/min. Find the rate of change of perimeter when x = 8 cm and y = 6 cm.

Answer

Step 1: Let length = x, width = y, perimeter = P. Step 2: P = 2(x + y). Step 3: Differentiate: dP/dt = 2(dx/dt + dy/dt). Step 4: Substitute dx/dt = -5 cm/min and dy/dt = 4 cm/min. Step 5: dP/dt = 2(-5…

Card 5Rate of change

Solve: The same rectangle has length decreasing at 5 cm/min and width increasing at 4 cm/min. Find the rate of change of area when x = 8 cm and y = 6 cm.

Answer

Step 1: Let area = A = xy. Step 2: Differentiate: dA/dt = x(dy/dt) + y(dx/dt). Step 3: Substitute x = 8, y = 6, dy/dt = 4, dx/dt = -5. Step 4: dA/dt = 8(4) + 6(-5) = 32 - 30 = 2 cm²/min. Answer: 2 cm²…

Card 6Rate of change

Solve: A stone is dropped into a quiet lake and waves move in a circle at 5 cm/s. Find the rate at which the enclosed area changes when radius is 8 cm.

Answer

Step 1: Let radius = R and area = A. Step 2: A = πR². Step 3: Differentiate: dA/dt = 2πR · dR/dt. Step 4: Substitute R = 8 cm and dR/dt = 5 cm/s. Step 5: dA/dt = 2π × 8 × 5 = 80π cm²/s. Answer: 80π cm…

Card 7Rate of change

Solve: Find dA/dr for a circle when r = 5 cm, where A = πr².

Answer

Step 1: A = πr². Step 2: Differentiate w.r.t. r: dA/dr = 2πr. Step 3: Put r = 5. Step 4: dA/dr = 2π(5) = 10π cm. Answer: 10π cm.

Card 8Rate of change

Solve: Find dA/dC for a circular disc when radius is 8 cm, where A = πR² and C = 2πR.

Answer

Step 1: A = πR² and C = 2πR. Step 2: Write A in terms of C: R = C/(2π), so A = C²/(4π). Step 3: Differentiate: dA/dC = 2C/(4π) = C/(2π). Step 4: Since C = 2πR, dA/dC = R. Step 5: For R = 8, dA/dC = 8 …

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Frequently Asked Questions

What are the important topics in Applications of Derivatives - I for ICSE Class 12 Mathematics?
Key topics in Applications of Derivatives - I include First Derivative Test for Monotonicity, Important Geometric Problems. These are the concepts ICSE Class 12 examiners draw on most — study them first, then practise related questions.
How to score full marks in Applications of Derivatives - I — ICSE Class 12 Mathematics?
Understand the core concepts first, then work through the 55 practice questions available for this chapter. Revise formulas and definitions regularly, and use flashcards for quick recall before the exam.
How many flashcards are available for Applications of Derivatives - I?
There are 37 flashcards for Applications of Derivatives - I covering key definitions, formulas, and concepts. Use them daily for 10–15 minutes for best results.

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