Class 12 Mathematics Notes Chapter 6 (Application of derivatives) – Mathematics Part-I Book

Alright students, let's get straight into Chapter 6: Application of Derivatives from your NCERT Class 12 Mathematics Part-I book. This chapter is crucial not just for your board exams but also forms the basis for many questions in government entrance examinations. It shows us the practical power of calculus.

We'll break down the key concepts you need to master.

Chapter 6: Application of Derivatives - Detailed Notes for Exam Preparation

Core Idea: The derivative, dy/dx or f'(x), represents the instantaneous rate of change of y (or f(x)) with respect to x. It also represents the slope of the tangent to the curve y = f(x) at the point (x, y). This chapter explores various applications stemming from these fundamental ideas.

1. Rate of Change of Quantities:

• Concept: If a quantity 'y' depends on another quantity 'x' such that y = f(x), then dy/dx represents the rate of change of 'y' with respect to 'x'.
• Related Rates: Often, quantities change with respect to time ('t'). If y depends on x, and both x and y depend on t, we use the Chain Rule:
  dy/dt = (dy/dx) * (dx/dt)
• Common Examples:
  • Finding the rate of change of the area of a circle when its radius is changing (dA/dt given dr/dt).
  • Finding the rate of change of the volume of a cube/sphere/cone when its side/radius/height is changing.
  • Problems involving moving objects (related rates of change of distances).
• Key Steps:
  1. Identify the quantities involved and what rates are given/required.
  2. Establish a relationship (formula) between the quantities (e.g., Area A = πr², Volume V = x³).
  3. Differentiate the relationship with respect to time 't' (using Chain Rule).
  4. Substitute the known values to find the required rate.
• Units: Pay close attention to units (e.g., cm²/s, m³/min).

2. Increasing and Decreasing Functions:

• Concept: A function f(x) is:
  • Strictly Increasing on an interval (a, b) if f'(x) > 0 for all x in (a, b).
  • Increasing on an interval [a, b] if f'(x) ≥ 0 for all x in (a, b).
  • Strictly Decreasing on an interval (a, b) if f'(x) < 0 for all x in (a, b).
  • Decreasing on an interval [a, b] if f'(x) ≤ 0 for all x in (a, b).
  • Constant on an interval (a, b) if f'(x) = 0 for all x in (a, b).
• Critical Points: Points where f'(x) = 0 or f'(x) is undefined. These points partition the domain into intervals.
• Finding Intervals:
  1. Find the derivative f'(x).
  2. Find the critical points by setting f'(x) = 0 and identifying points where f'(x) is undefined.
  3. Plot these critical points on a number line.
  4. Test the sign of f'(x) in each interval created by the critical points.
  5. Conclude whether the function is strictly increasing or strictly decreasing in each interval based on the sign of f'(x).

3. Tangents and Normals:

• Slope of Tangent: The slope 'm' of the tangent to the curve y = f(x) at a point (x₀, y₀) is given by m = f'(x₀) = [dy/dx] at (x₀, y₀).
• Equation of Tangent: Using the point-slope form, the equation of the tangent at (x₀, y₀) is:
  y - y₀ = m (x - x₀) => y - y₀ = f'(x₀) (x - x₀)
• Slope of Normal: The normal line is perpendicular to the tangent line at the point of tangency. Its slope 'm_normal' is:
  m_normal = -1 / m = -1 / f'(x₀) (provided f'(x₀) ≠ 0)
• Equation of Normal: The equation of the normal at (x₀, y₀) is:
  y - y₀ = m_normal (x - x₀) => y - y₀ = [-1 / f'(x₀)] (x - x₀)
• Special Cases:
  • If the tangent is parallel to the x-axis, then m = f'(x₀) = 0.
  • If the tangent is parallel to the y-axis (vertical tangent), then f'(x₀) is undefined (denominator of f'(x) is zero), and the tangent equation is x = x₀. The normal is parallel to the x-axis (y = y₀).

4. Approximations:

• Concept: We can use derivatives to approximate the value of a function near a known point.
• Let y = f(x). Let Δx be a small change in x, and Δy be the corresponding change in y.
  Δy = f(x + Δx) - f(x)
• We know dy ≈ Δy for small Δx.
• Also, dy = f'(x) dx. Taking dx = Δx, we get dy = f'(x) Δx.
• Therefore, Δy ≈ f'(x) Δx
• Substituting Δy: f(x + Δx) - f(x) ≈ f'(x) Δx
• Approximation Formula: f(x + Δx) ≈ f(x) + f'(x) Δx
• Application: Used to approximate values like √25.3, (3.98)³, log(10.1), etc.
  • Example: To find √25.3: Let f(x) = √x. Choose x = 25 (known square root) and Δx = 0.3.
    f'(x) = 1/(2√x). f'(25) = 1/10 = 0.1.
    √25.3 = f(25 + 0.3) ≈ f(25) + f'(25) * (0.3) = √25 + (0.1)(0.3) = 5 + 0.03 = 5.03.
• Error Calculation: dy ≈ Δy can also be interpreted as the approximate error in y (dy) resulting from a small error in x (Δx).

5. Maxima and Minima:

• Concept: Finding the maximum or minimum values of a function, either locally or over its entire domain (or a specific interval).
• Critical Points: Points where f'(x) = 0 or f'(x) is undefined. These are potential locations for local maxima or minima.
• Local Maxima/Minima:
  • First Derivative Test:
    1. Find critical points c (where f'(c) = 0 or undefined).
    2. Examine the sign of f'(x) just to the left and just to the right of c.
    3. If f'(x) changes sign from +ve to -ve as x increases through c, then f(c) is a local maximum value.
    4. If f'(x) changes sign from -ve to +ve as x increases through c, then f(c) is a local minimum value.
    5. If f'(x) does not change sign, then c is a point of inflection (neither max nor min).
  • Second Derivative Test: (Often easier if f''(x) is easy to compute)
    1. Find critical points c by solving f'(c) = 0.
    2. Calculate the second derivative f''(x).
    3. Evaluate f''(c):
       • If f''(c) < 0, then f(c) is a local maximum value.
       • If f''(c) > 0, then f(c) is a local minimum value.
       • If f''(c) = 0, the test fails. Use the First Derivative Test.
• Absolute Maxima/Minima in a Closed Interval [a, b]:
  1. Find all critical points of f(x) within the open interval (a, b).
  2. Evaluate the function f(x) at these critical points.
  3. Evaluate the function f(x) at the endpoints of the interval, i.e., find f(a) and f(b).
  4. The largest value among those calculated in steps 2 and 3 is the absolute maximum value.
  5. The smallest value among those calculated in steps 2 and 3 is the absolute minimum value.
• Word Problems (Optimization): Many practical problems involve maximizing or minimizing a quantity (like area, volume, profit, cost) subject to certain constraints.
  1. Read the problem carefully, identify the quantity to be optimized and the constraints.
  2. Express the quantity to be optimized as a function of a single variable using the constraints.
  3. Determine the domain (interval) for this variable based on the problem's context.
  4. Use the methods for finding local or absolute extrema (derivative tests) to find the required maximum or minimum value.

Exam Focus:

• Rate of change problems are common.
• Finding intervals of increasing/decreasing functions.
• Finding equations of tangents and normals.
• Maxima/Minima problems, especially word problems involving optimization and finding absolute extrema in a closed interval.

Multiple Choice Questions (MCQs)

Here are 10 MCQs based on the concepts discussed:

1. The radius of a circle is increasing uniformly at the rate of 3 cm/s. The rate at which the area of the circle is increasing when the radius is 10 cm is:
   (a) 30π cm²/s
   (b) 60π cm²/s
   (c) 90π cm²/s
   (d) 10π cm²/s

2. The interval in which the function f(x) = 2x³ - 3x² - 36x + 7 is strictly decreasing is:
   (a) (-∞, -2)
   (b) (3, ∞)
   (c) (-2, 3)
   (d) (-∞, -2) U (3, ∞)

3. The slope of the tangent to the curve y = x³ - x + 1 at the point whose x-coordinate is 2 is:
   (a) 6
   (b) 7
   (c) 11
   (d) 12

4. The equation of the normal to the curve y = x² at the point (1, 1) is:
   (a) x + 2y = 3
   (b) x - 2y = -1
   (c) 2x + y = 3
   (d) 2x - y = 1

5. Using differentials, the approximate value of √36.6 is:
   (a) 6.00
   (b) 6.05
   (c) 6.08
   (d) 6.10

6. The local maximum value of the function f(x) = x³ - 3x + 3 is:
   (a) 1
   (b) 3
   (c) 5
   (d) 0

7. For the function f(x) = x + 1/x, x > 0, the local minimum value occurs at x = ?
   (a) -1
   (b) 1
   (c) 0
   (d) 2

8. The absolute maximum value of the function f(x) = x³ in the interval [-2, 2] is:
   (a) -8
   (b) 0
   (c) 8
   (d) 16

9. If the tangent to the curve y = f(x) at (x₀, y₀) is parallel to the y-axis, then:
   (a) f'(x₀) = 0
   (b) f'(x₀) = 1
   (c) f'(x₀) is undefined
   (d) f'(x₀) = -1

10. The side of a cube is increasing at the rate of 0.5 cm/s. The rate of increase of the volume of the cube when the side length is 4 cm is:
    (a) 8 cm³/s
    (b) 12 cm³/s
    (c) 16 cm³/s
    (d) 24 cm³/s

Answer Key for MCQs:

1. (b) [A = πr², dA/dt = 2πr (dr/dt) = 2π(10)(3) = 60π]
2. (c) [f'(x) = 6x² - 6x - 36 = 6(x² - x - 6) = 6(x-3)(x+2). f'(x) < 0 for x in (-2, 3)]
3. (c) [dy/dx = 3x² - 1. At x=2, dy/dx = 3(2)² - 1 = 12 - 1 = 11]
4. (a) [dy/dx = 2x. At (1,1), slope of tangent m = 2(1) = 2. Slope of normal = -1/2. Eq: y - 1 = (-1/2)(x - 1) => 2y - 2 = -x + 1 => x + 2y = 3]
5. (b) [f(x) = √x, x=36, Δx=0.6. f'(x)=1/(2√x). f(36+0.6) ≈ f(36) + f'(36)Δx = √36 + (1/(2√36))(0.6) = 6 + (1/12)(0.6) = 6 + 0.05 = 6.05]
6. (c) [f'(x) = 3x² - 3 = 3(x-1)(x+1). Critical points x=1, x=-1. f''(x) = 6x. f''(-1) = -6 < 0 => Local Max at x=-1. f(-1) = (-1)³ - 3(-1) + 3 = -1 + 3 + 3 = 5]
7. (b) [f'(x) = 1 - 1/x². f'(x)=0 => x²=1 => x=1 (since x>0). f''(x) = 2/x³. f''(1) = 2 > 0 => Local Min at x=1]
8. (c) [f'(x) = 3x². Critical point x=0. Evaluate f(-2)=(-2)³=-8, f(0)=0³=0, f(2)=2³=8. Absolute Max is 8]
9. (c) [Tangent parallel to y-axis means it's a vertical line, slope is undefined. f'(x₀) represents the slope.]
10. (d) [V = x³, dV/dt = 3x² (dx/dt) = 3(4)²(0.5) = 3 * 16 * 0.5 = 48 * 0.5 = 24]

Study these concepts thoroughly, practice problems from your NCERT book and previous year question papers. Good luck!