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Chapter 2 of 12
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Applications of Differential Calculus

Tamil Nadu Board · Class 12 · Mathematics

Flashcards for Applications of Differential Calculus — Tamil Nadu Board Class 12 Mathematics. Quick Q&A cards covering key concepts, definitions, and formulas.

45 questions25 flashcards5 concepts

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25 Flashcards
Card 1Derivatives as Rate of Change

Find the instantaneous rate of change of f(x) = 3x² + 2x at x = 1 using the derivative.

Answer

Step 1: Find the derivative: f'(x) = 6x + 2 Step 2: Evaluate at x = 1: f'(1) = 6(1) + 2 = 8 Step 3: The instantaneous rate of change at x = 1 is 8 units. Explanation: The derivative gives the slope of

Card 2Motion and Velocity

A particle's position is given by s(t) = t³ - 6t² + 9t + 1. Find when the particle is at rest.

Answer

Step 1: Velocity v(t) = ds/dt = 3t² - 12t + 9 Step 2: Set v(t) = 0: 3t² - 12t + 9 = 0 Step 3: Factor: 3(t² - 4t + 3) = 0 → 3(t - 1)(t - 3) = 0 Step 4: t = 1 or t = 3 Answer: The particle is at rest at

Card 3Related Rates

When do you use the related rates method? Provide a context example.

Answer

Use related rates when: - Two or more quantities are changing with respect to time - You know the rate of change of one quantity and need to find the rate of change of another - The quantities are rel

Card 4Equations of Tangent and Normal

Find the equation of the tangent line to y = x² - 3x + 2 at the point (1, 0).

Answer

Step 1: Find the derivative: dy/dx = 2x - 3 Step 2: Find the slope at x = 1: m = 2(1) - 3 = -1 Step 3: Use point-slope form: y - y₁ = m(x - x₁) Step 4: Substitute (1, 0) and m = -1: y - 0 = -1(x - 1)

Card 5Equations of Tangent and Normal

Find the equation of the normal line to the curve y = x³ - 2x at x = 1.

Answer

Step 1: Find dy/dx = 3x² - 2 Step 2: Find the slope of tangent at x = 1: m_tangent = 3(1)² - 2 = 1 Step 3: Slope of normal is perpendicular: m_normal = -1/m_tangent = -1/1 = -1 Step 4: Find y-coordina

Card 6Angle Between Two Curves

Find the angle between the curves y = x² and y = (x - 3)² at their point of intersection.

Answer

Step 1: Find intersection points: x² = (x - 3)² → x² = x² - 6x + 9 → x = 3/2 Step 2: At x = 3/2: y = (3/2)² = 9/4, so point is (3/2, 9/4) Step 3: Find slope of first curve: dy/dx = 2x → m₁ = 2(3/2) =

Card 7Rolle's Theorem

Use Rolle's Theorem to verify that f(x) = x² - 4x has a critical point in (0, 4).

Answer

Rolle's Theorem conditions: Step 1: Check if f is continuous on [0, 4] → Yes, polynomial Step 2: Check if f is differentiable on (0, 4) → Yes, polynomial Step 3: Check if f(0) = f(4): - f(0) = 0² - 4(

Card 8Lagrange's Mean Value Theorem

Use the Mean Value Theorem to find c for f(x) = x² on [1, 3].

Answer

Mean Value Theorem: ∃c ∈ (a, b): f'(c) = [f(b) - f(a)]/(b - a) Step 1: Calculate f(3) - f(1) = 9 - 1 = 8 Step 2: Calculate (b - a) = 3 - 1 = 2 Step 3: Average rate of change = 8/2 = 4 Step 4: Find f'(

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Frequently Asked Questions

What are the important topics in Applications of Differential Calculus for Tamil Nadu Board Class 12 Mathematics?
Key topics in Applications of Differential Calculus include Applications of Differential Calculus - Concept Overview, Shows the relationship between secant and tangent slopes, leading to the derivative concept, Problem-solving strategy for related rates problems. These are the concepts Tamil Nadu Board Class 12 examiners draw on most — study them first, then practise related questions.
How to score full marks in Applications of Differential Calculus — Tamil Nadu Board Class 12 Mathematics?
Understand the core concepts first, then work through the 45 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 Differential Calculus?
There are 25 flashcards for Applications of Differential Calculus covering key definitions, formulas, and concepts. Use them daily for 10–15 minutes for best results.

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