Unit III: Application of Derivatives
Application of Derivatives as a Rate Measurer
Derivatives represent the instantaneous rate of change of one quantity with respect to another. If a variable y depends on time t according to the function y = f(t), then the rate of change of y with respect to t is given by dy / dt.
Definition: Let y = f(x) be a function. The rate of change of y with respect to x at a given point is defined as dy / dx = f'(x).
Key Concepts and Steps
- Identify the given variables and what needs to be found with respect to time t or another variable.
- Formulate an equation connecting the variables based on the geometric or physical problem.
- Differentiate both sides of the equation implicitly with respect to time t.
- Substitute the given numerical values at the specific instant to find the required rate.
Practical Example: The radius of a circle is increasing at the rate of 3 cm/sec. Find the rate at which its area is increasing when the radius is 10 cm.
Step-by-Step Explanation: Let the radius of the circle be r and the area be A. We are given dr / dt = 3 cm/sec. The formula for the area of a circle is A = πr2. Differentiating both sides with respect to time t: dA / dt = 2πr × (dr / dt). Substituting r = 10 and dr / dt = 3: dA / dt = 2 × π × 10 × 3 = 60π cm2/sec.
Common Mistakes: Forgetting to apply the chain rule when differentiating functions with respect to time t instead of the variable itself.
Simple Problems on Area, Perimeter, and Volume
Derivatives are widely used to solve optimization and rate problems involving geometric shapes such as rectangles, circles, spheres, cylinders, and cones. Changes in dimensions directly affect perimeter, surface area, and volume.
Important Formulas
- Rectangle: Area A = l × w, Perimeter P = 2(l + w)
- Circle: Area A = πr2, Circumference C = 2πr
- Sphere: Volume V = (4 / 3)πr3, Surface Area S = 4πr2
- Cylinder: Volume V = πr2h, Curved Surface Area S = 2πrh
Exam-Oriented Note: Always include proper units in your final answer (e.g., cm/sec for linear rates, cm2/sec for area rates, and cm3/sec for volume rates).
Maximum and Minimum of a Function (Using 2nd Derivative Only)
A function can reach peak values (peaks and valleys) within its domain. The second derivative test helps determine whether a critical point corresponds to a local maximum or a local minimum.
Definition: Let f(x) be a differentiable function defined on an interval. A point x = c is called a critical point if f'(c) = 0 or if f'(c) does not exist.
Step-by-Step Procedure for Second Derivative Test
- Find the first derivative of the function, f'(x).
- Set f'(x) = 0 and solve for x to find the critical points (let's call them x = a).
- Find the second derivative of the function, f''(x).
- Evaluate the second derivative at each critical point, f''(a).
- Apply the decision rules based on the sign of f''(a).
| Condition at x = a | Conclusion |
|---|---|
| f'(a) = 0 and f''(a) < 0 | x = a is a point of Local Maximum. The maximum value is f(a). |
| f'(a) = 0 and f''(a) > 0 | x = a is a point of Local Minimum. The minimum value is f(a). |
| f'(a) = 0 and f''(a) = 0 | The test fails; further investigation (higher-order derivatives) is required. |
Practical Example: Find the local maximum and minimum values of the function f(x) = 2x3 - 3x2 - 12x + 5 using the second derivative test.
Step-by-Step Solution:
1. Find f'(x): f'(x) = 6x2 - 6x - 12.
2. Set f'(x) = 0: 6(x2 - x - 2) = 0 ⇒ 6(x - 2)(x + 1) = 0. Critical points are x = 2 and x = -1.
3. Find f''(x): f''(x) = 12x - 6.
4. Test at x = 2: f''(2) = 12(2) - 6 = 18 > 0. Since f''(2) > 0, f(x) has a local minimum at x = 2. Minimum value = 2(2)3 - 3(2)2 - 12(2) + 5 = -15.
5. Test at x = -1: f''(-1) = 12(-1) - 6 = -18 < 0. Since f''(-1) < 0, f(x) has a local maximum at x = -1. Maximum value = 2(-1)3 - 3(-1)2 - 12(-1) + 5 = 12.
Monotone Functions and Intervals of Monotonicity
Understanding the direction in which a function increases or decreases helps analyze its overall behavior across a given domain.
Definition of Monotone Function: A function f(x) is called a monotone function on an interval if it is either entirely increasing or entirely decreasing throughout that interval.
Types of Monotonicity
- Strictly Increasing Function: A function f(x) is strictly increasing on an interval (a, b) if for any x1 < x2 in the interval, we have f(x1) < f(x2). Mathematically, f'(x) > 0.
- Strictly Decreasing Function: A function f(x) is strictly decreasing on an interval (a, b) if for any x1 < x2 in the interval, we have f(x1) > f(x2). Mathematically, f'(x) < 0.
- Constant Function: If f'(x) = 0 for all x in an interval, the function is constant.
Steps to Find the Intervals of Monotonicity
- Find the given function f(x).
- Compute the first derivative f'(x).
- Set f'(x) = 0 to find the critical points. These points divide the domain into various disjoint intervals.
- Select a test point from each interval and substitute it into f'(x).
- Determine the sign of f'(x) in each interval to conclude whether the function is increasing (positive derivative) or decreasing (negative derivative).
Important Observation: If f'(x) ≥ 0 for all x in an interval, the function is increasing. If f'(x) ≤ 0, it is decreasing.