Applications of Derivatives Chapter 6

Application of Derivatives

Rate of Change of Quantities

• The derivative of a function represents the rate of change of the function with respect to its variable.
• If y varies with x, then dy/dx represents the rate of change of y with respect to x.
• If two variables x and y are varying with respect to another variable t, then dy/dx can be calculated using the chain rule.

Examples of Rate of Change of Quantities

• The rate of change of the area of a circle with respect to its radius r is given by dA/dr = 2πr.
• The rate of change of the volume of a cube with respect to its edge length x is given by dV/dx = 3x^2.
• The rate of change of the surface area of a cube with respect to its edge length x is given by dS/dx = 6x.

Increasing and Decreasing Functions

• A function is said to be increasing on an interval if the function values increase as the input values increase.
• A function is said to be decreasing on an interval if the function values decrease as the input values increase.
• A function is said to be constant on an interval if the function values remain the same as the input values change.

First Derivative Test

• If f'(x) > 0 for all x in an interval, then f is increasing on that interval.
• If f'(x) < 0 for all x in an interval, then f is decreasing on that interval.
• If f'(x) = 0 for all x in an interval, then f is constant on that interval.

Examples of Increasing and Decreasing Functions

• The function f(x) = x^2 is increasing on the interval [0, ∞) and decreasing on the interval (-∞, 0].
• The function f(x) = 7x - 3 is increasing on the entire real line.
• The function f(x) = cos x is decreasing on the interval (0, π) and increasing on the interval (π, 2π).

Intervals of Increase and Decrease

• The function f(x) = x^2 - 4x + 6 is decreasing on the interval (-∞, 2) and increasing on the interval (2, ∞).
• The function f(x) = 4x^3 - 6x^2 - 72x + 30 is increasing on the intervals (-∞, -2) and (3, ∞), and decreasing on the interval (-2, 3).
• The function f(x) = sin 3x is increasing on the intervals (0, π/6) and (π/2, 5π/6), and decreasing on the intervals (π/6, π/2) and (5π/6, π).Here are the study notes for the text:

Increasing and Decreasing Functions

• A function f is increasing in an interval I if for every x1, x2 ∈ I, x1 < x2 implies f(x1) ≤ f(x2)
• A function f is decreasing in an interval I if for every x1, x2 ∈ I, x1 < x2 implies f(x1) ≥ f(x2)
• A function f is strictly increasing in an interval I if for every x1, x2 ∈ I, x1 < x2 implies f(x1) < f(x2)
• A function f is strictly decreasing in an interval I if for every x1, x2 ∈ I, x1 < x2 implies f(x1) > f(x2)

Maximum and Minimum Values

• A function f has a maximum value in an interval I if there exists a point c in I such that f(c) ≥ f(x) for all x ∈ I
• A function f has a minimum value in an interval I if there exists a point c in I such that f(c) ≤ f(x) for all x ∈ I
• An extreme value of a function f is either a maximum value or a minimum value of f

Local Maxima and Minima

• A point c is a point of local maxima of a function f if there is an h > 0 such that f(c) ≥ f(x) for all x in (c - h, c + h), x ≠ c
• A point c is a point of local minima of a function f if there is an h > 0 such that f(c) ≤ f(x) for all x in (c - h, c + h)
• The value f(c) is called the local maximum value or local minimum value of f

First Derivative Test

• If f'(x) changes sign from positive to negative as x increases through c, then c is a point of local maxima
• If f'(x) changes sign from negative to positive as x increases through c, then c is a point of local minima
• If f'(x) does not change sign as x increases through c, then c is neither a point of local maxima nor a point of local minima, and is called a point of inflection

Second Derivative Test

• If f'(c) = 0 and f"(c) < 0, then c is a point of local maxima
• If f'(c) = 0 and f"(c) > 0, then c is a point of local minima
• If f'(c) = 0 and f"(c) = 0, then the test fails, and we need to use the first derivative test

Examples

• The function f(x) = sin 3x has a local maximum value at x = π/6 and a local minimum value at x = π/2
• The function f(x) = x2 has a local minimum value at x = 0
• The function f(x) = 3 + |x| has a local minimum value at x = 0
• The function f(x) = 3x4 + 4x3 - 12x2 + 12 has a local maximum value and a local minimum value, which can be found using the first derivative test### Application of Derivatives

Example 21

• Find the local maxima and minima of the function f(x) = 2x³ - 6x² + 6x + 5
• f'(x) = 6x² - 12x + 6 = 6(x - 1)²
• f''(x) = 12(x - 1)
• f'(x) = 0 gives x = 1
• f''(1) = 0, so second derivative test fails
• Use first derivative test to find that x = 1 is a point of inflexion

Example 22

• Find two positive numbers whose sum is 15 and the sum of whose squares is minimum
• Let one of the numbers be x, then the other number is 15 - x
• S(x) = x² + (15 - x)² = 2x² - 30x + 225
• S'(x) = 4x - 30
• S'(x) = 0 gives x = 15/2
• S''(x) = 4 > 0, so x = 15/2 is a point of local minima
• The two numbers are 15/2 and 15/2

Example 23

• Find the shortest distance of the point (0, c) from the parabola y = x², where 1/2 ≤ c ≤ 5
• Let (h, k) be a point on the parabola, then D = (h - 0)² + (k - c)² = h² + (k - c)²
• D = k + (k - c)²
• D'(k) = 1 + 2(k - c)
• D'(k) = 0 gives k = (2c - 1)/2
• D'(k) is minimum at k = (2c - 1)/2
• The shortest distance is given by D = (2c - 1)/2 + ((2c - 1)/2 - c)²

Example 24

• Let AP and BQ be two vertical poles at points A and B, respectively
• RP² + RQ² is minimum when AP = 16, BQ = 22, and AB = 20
• Let AR = x, then RB = 20 - x
• RP² = AR² + AP² = x² + 16²
• RQ² = RB² + BQ² = (20 - x)² + 22²
• S(x) = RP² + RQ² = 2x² - 40x + 1140
• S'(x) = 4x - 40
• S'(x) = 0 gives x = 10
• S''(x) = 4 > 0, so x = 10 is a point of local minima
• The distance of R from A is AR = x = 10

Example 25

• Find the area of the trapezium when it is maximum
• The required trapezium has AP = x, QB = x, and DP = QC = √(100 - x²)
• A = (sum of parallel sides)(height)/2 = (2x + 10 + 10)(√(100 - x²))/2
• A'(x) = (x + 10)(√(100 - x²)) - (2x² + 10x - 100)/√(100 - x²)
• A'(x) = 0 gives x = 5
• A''(x) = (-2x² - 10x + 100)/√(100 - x²)
• A''(5) = -5 < 0, so x = 5 is a point of local maxima
• The area of the trapezium is maximum at x = 5