01 · Explore
Rate of Change of Quantities
The derivative represents how one quantity changes in relation to another. If a variable y depends on x according to a rule y = f(x), then the derivative dy/dx is the rate of change of y with respect to x.
When we find dy/dx at a specific point x = x₀, we are calculating the instantaneous rate of change at that exact moment. This is a fundamental concept in physics, where the derivative of distance with respect to time (ds/dt) is defined as velocity.
If two variables x and y both depend on a third variable t (often time), we use the Chain Rule to relate their rates of change. Specifically, dy/dx = (dy/dt) / (dx/dt), provided that dx/dt is not zero. This allows us to find how y changes with respect to x even if they are only connected through their mutual dependence on t.
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Question
If the volume of a cube increases at 9 cm³/s, find the rate of change of its surface area S when the edge length x = 10 cm.
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02 · Explore
Increasing and Decreasing Functions
Calculus provides a precise way to determine if a function's value is rising or falling over an interval by examining the sign of its first derivative.
A function f is called increasing on an interval I if for any two points x₁ and x₂ in I, x₁ < x₂ implies f(x₁) ≤ f(x₂). If f(x₁) < f(x₂), it is strictly increasing. Conversely, it is decreasing if x₁ < x₂ implies f(x₁) ≥ f(x₂).
The First Derivative Test for monotonicity states that if f is continuous on [a, b] and differentiable on (a, b), then: (i) f is increasing if f'(x) ≥ 0 for all x in (a, b); (ii) f is decreasing if f'(x) ≤ 0 for all x in (a, b); and (iii) f is constant if f'(x) = 0.
| Condition | Sign of f'(x) | Nature of Function |
|---|---|---|
| f'(x) ≥ 0 | Non-negative | Increasing |
| f'(x) ≤ 0 | Non-positive | Decreasing |
| f'(x) = 0 | Zero | Constant |
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Question
Determine the intervals where f(x) = x² - 4x + 6 is increasing or decreasing.
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03 · Explore
Concepts of Maxima and Minima
Points where a function reaches its highest or lowest values locally are known as local maxima and local minima. These occur at 'turning points' where the function changes its direction.
A point c is a local maximum if f(c) ≥ f(x) for all x in a small neighborhood around c. Similarly, c is a local minimum if f(c) ≤ f(x) in that neighborhood. These points, along with points where the derivative does not exist, are called critical points.
At a local maximum or minimum of a differentiable function, the derivative f'(c) must be zero. Geometrically, this means the tangent to the curve at these points is horizontal. However, f'(c) = 0 is a necessary but not sufficient condition; for example, f(x) = x³ has f'(0) = 0 but is neither a maximum nor a minimum at x = 0.
Identifying Local Extrema
- 1
Find f'(x)
Differentiate the function to find the rate of change.
- 2
Solve f'(x) = 0
Identify critical points where the slope is zero.
- 3
Test Intervals
Check the sign of f'(x) to the left and right of each critical point.
- 4
Conclude Extrema
Determine if the point is a local maximum, minimum, or point of inflection.
The sequential process of locating and classifying local maximum and minimum values using the first derivative.
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What is a point of inflection?
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04 · Explore
The First Derivative Test
The First Derivative Test classifies critical points by observing how the sign of the derivative changes as we move through the point.
If f'(x) changes from positive to negative at c, then f has a local maximum at c. This indicates the function was climbing and then started falling.
If f'(x) changes from negative to positive at c, then f has a local minimum at c. This indicates the function was falling and then started climbing. If the sign does not change, the point is a point of inflection.
Finding Extrema for f(x) = x³ - 3x + 3
f'(x) = 3(x - 1)(x + 1)
Critical points are x = 1 and x = -1. At x = -1, f'(x) changes from positive to negative, so it is a local maximum with value f(-1) = 5. At x = 1, f'(x) changes from negative to positive, so it is a local minimum with value f(1) = 1.
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Question
If f'(x) > 0 for x < c and f'(x) < 0 for x > c, what is the nature of point c?
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05 · Explore
The Second Derivative Test
The Second Derivative Test is often a quicker way to classify critical points by checking the concavity of the function at those points.
Let f be twice differentiable at a critical point c where f'(c) = 0. If f''(c) < 0, the function is concave down, making c a local maximum. If f''(c) > 0, the function is concave up, making c a local minimum.
If f''(c) = 0, the test is inconclusive. In such cases, you must revert to the First Derivative Test to determine the nature of the critical point.
Applying Second Derivative Test
f(x) = 3x⁴ + 4x³ - 12x² + 12
f'(x) = 12x(x-1)(x+2). Critical points are 0, 1, -2. f''(x) = 36x² + 24x - 24. f''(0) = -24 (Max, value 12). f''(1) = 36 (Min, value 7). f''(-2) = 72 (Min, value -20).
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When does the Second Derivative Test fail?
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06 · Explore
Absolute Maxima and Minima
On a closed interval [a, b], a continuous function is guaranteed to reach an absolute maximum and an absolute minimum value.
Absolute extrema can occur either at the critical points within the interval or at the endpoints of the interval. This is different from local extrema, which only consider the immediate neighborhood of a point.
To find absolute extrema: 1. Find all critical points in (a, b). 2. Calculate the function values at these critical points and at the endpoints a and b. 3. The largest value is the absolute maximum, and the smallest is the absolute minimum.
- 1
Identify the function f(x) and the closed interval [a, b].
- 2
Find f'(x) and solve f'(x) = 0 to find critical points within (a, b).
- 3
List the endpoints a and b along with the critical points.
- 4
Evaluate f(x) for every point in the list.
- 5
Compare the results: the highest is the absolute maximum; the lowest is the absolute minimum.
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Question
Find absolute extrema of f(x) = x³ on [-2, 2].
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07 · Explore
Applied Optimization Problems
Derivatives are used to solve real-world problems where we need to maximize profit, minimize cost, or find the most efficient dimensions for a shape.
In optimization, we translate a word problem into a mathematical function. We often use a secondary equation (a constraint) to express the function in terms of a single variable before differentiating.
Common applications include finding the maximum area of a rectangle with a fixed perimeter, the minimum surface area of a can with a fixed volume, or the shortest distance between a point and a curve.
Maximizing Area of a Trapezium
A(x) = (x + 10)√(100 - x²)
For a trapezium with three sides of 10 cm, let the base extension be x. Area A = (x+10)√(100-x²). Setting A'(x) = 0 gives x = 5. The maximum area is A(5) = 15√75 = 75√3 cm².
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Question
Two positive numbers have a sum of 15. Find them if the sum of their squares is minimum.
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08 · Explore
Miscellaneous Applications
Derivatives also apply to geometry and motion, such as calculating the rate at which a water level rises or the speed of a shadow.
In problems involving shadows, similar triangles are used to relate the man's distance from a lamp post to the length of his shadow. Differentiating this relationship with respect to time allows us to find the rate of shadow growth.
For filling tanks (like inverted cones), the volume is related to the height. Since the radius and height maintain a constant ratio (based on the semi-vertical angle), we can express volume solely in terms of height to find the rate at which the water level rises.
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Question
A car's distance is x = t²(2 - t/3). When does it stop?
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