Derivatives measure the rate of change. This chapter applies the power of differentiation to practical problems: finding tangents and normals, determining where functions increase or decrease, locating maximum and minimum values, and approximating values.
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Key Concepts
Rate of Change: If y = f(x), then dy/dx represents the rate of change of y with respect to x. When x represents time t, dy/dt is the rate of change with time.
- Tangents and Normals:
- Slope of tangent at (x0, y0): m = f'(x0).
- Equation of tangent: y - y0 = m(x - x0).
- Slope of normal: -1/m (negative reciprocal of tangent slope).
- Equation of normal: y - y0 = (-1/m)(x - x0).
- Increasing and Decreasing Functions:
- f is increasing on an interval if f'(x) > 0 on that interval.
- f is decreasing on an interval if f'(x) < 0 on that interval.
- At a point where f'(x) = 0, f is called stationary.
- Maxima and Minima:
- First Derivative Test: At a critical point x = c where f'(c) = 0:
- If f' changes from + to -, then f has a local maximum at c.
- If f' changes from - to +, then f has a local minimum at c.
- If f' does not change sign, then c is a point of inflection.
- Second Derivative Test: At x = c where f'(c) = 0:
- If f''(c) < 0, then f has a local maximum at c.
- If f''(c) > 0, then f has a local minimum at c.
- If f''(c) = 0, the test is inconclusive.
Absolute (Global) Maxima/Minima on [a, b]: Evaluate f at all critical points in (a, b) and at endpoints a, b. The largest is the absolute maximum and the smallest is the absolute minimum.
Approximations using Derivatives:
If deltax is a small change in x, then:
deltay approx dy/dx · deltax (approximate change in y).
Increasing Function on a Closed Interval: Check f'(x) ≥ 0 for all x in [a, b].
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Worked Examples
A ladder 5 m long leans against a wall. If its foot slides away at 2 m/s, how fast is the top sliding down when the foot is 4 m from the wall?
Let x = distance of foot, y = height. x2 + y2 = 25. When x = 4, y = 3.
2x(dx/dt) + 2y(dy/dt) = 0. 2(4)(2) + 2(3)(dy/dt) = 0. dy/dt = -8/3 m/s (downward).
Find the equation of the tangent to y = x3 - 2x + 1 at x = 1.
y(1) = 1 - 2 + 1 = 0. y' = 3x2 - 2; y'(1) = 1.
Tangent: y - 0 = 1(x - 1), i.e., y = x - 1.
Find intervals where f(x) = 2x3 - 9x2 + 12x - 5 is increasing.
f'(x) = 6x2 - 18x + 12 = 6(x2 - 3x + 2) = 6(x - 1)(x - 2).
f'(x) > 0 when x < 1 or x > 2. So f is increasing on (-inf, 1) and (2, inf).
Find local maxima and minima of f(x) = x3 - 6x2 + 9x + 15.
f'(x) = 3x2 - 12x + 9 = 3(x - 1)(x - 3) = 0. Critical points: x = 1, x = 3.
f''(x) = 6x - 12. f''(1) = -6 < 0: local max. f''(3) = 6 > 0: local min.
Local max value = f(1) = 1 - 6 + 9 + 15 = 19. Local min value = f(3) = 27 - 54 + 27 + 15 = 15.
Find the absolute maximum of f(x) = 2x3 - 15x2 + 36x on [1, 5].
f'(x) = 6x2 - 30x + 36 = 6(x - 2)(x - 3). Critical points in (1,5): x = 2, x = 3.
f(1) = 23, f(2) = 28, f(3) = 27, f(5) = 55. Absolute max = 55 at x = 5.
Using approximation, find the approximate value of √(25.2).
Let f(x) = √(x), a = 25. f(a) = 5. f'(x) = 1/(2 · √(x)), f'(25) = 1/10.
deltax = 0.2. deltay approx (1/10) · 0.2 = 0.02. √(25.2) approx 5.02.
A farmer has 200 m of fencing. She wants to enclose a rectangular plot. What dimensions give maximum area?
Let length = x, width = y. Perimeter: 2x + 2y = 200, so y = 100 - x.
Area A = x(100 - x) = 100x - x2. dA/dx = 100 - 2x = 0 => x = 50.
d2A/dx2 = -2 < 0: maximum. So 50 x 50 m square gives max area = 2500 sq m.
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Common mistakes
- Finding local maxima/minima but not checking endpoints for absolute extrema on a closed interval.
- Forgetting to check the sign change of f' (not just setting f'(c) = 0) in the first derivative test.
- Confusing the tangent slope with the normal slope; the normal has slope -1/m.
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Summary
Derivatives are applied to find rates of change, slopes of curves, and optimal values. The first and second derivative tests identify maxima/minima. Optimisation problems reduce to finding the absolute extremum on a feasible domain.