APPLICATIONS OF CALCULUS
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Subject: Additional Mathematics
Class: SHS 2
Term: 2nd Term
Week: 17
Grade code: 2.3.2.LI.2
Strand code: 3
Sub-strand code: 2
Content standard code: 2.3.2.CS.1
Indicator code: 2.3.2.LI.2
Theme: CALCULUS
Subtheme: APPLICATIONS OF CALCULUS
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This lesson explores one of the most powerful applications of differential calculus: optimisation and curve sketching. We will learn how to determine the "turning points" of a function—the peaks and valleys of its graph. This is not just an abstract mathematical exercise. In Ghana, a small business owner might want to know the production level that gives the maximum profit. An engineer might need to find the design that provides maximum strength with minimum material. A farmer wants to know the optimal amount of fertilizer to use for maximum crop yield. Calculus gives us the tools to solve these real-world problems.
A. Recap: What are Stationary Points?
A stationary point (or turning point) on a curve is a point where the gradient is zero. The tangent to the curve at this point is a horizontal line. To find the x-coordinate of a stationary point for a function `y = f(x)`, we solve the equation: `dy/dx = f'(x) = 0`
There are three main types of stationary points: Local Maximum: The peak of a "hill" on the curve. The function's value is higher at this point than at all nearby points. Local Minimum: The bottom of a "valley" on the curve. The function's value is lower at this point than at all nearby points. Point of Inflection (or Saddle Point): A point where the curve flattens out before continuing in the same general direction. The curve changes its concavity here. B. The Idea of Concavity and the Second Derivative
The first derivative, `dy/dx`, tells us the gradient of the curve. The second derivative, `d²y/dx²`, tells us the rate of change of the gradient. This helps us understand the *shape* or *concavity* of the curve. Concave Up (Like a cup: ∪): If `d²y/dx² > 0`, the gradient is increasing. The curve is bending upwards. Any stationary point in a concave-up region must be a local minimum. Concave Down (Like a cap: ∩): If `d²y/dx² 0` (positive), the point `(c, f(c))` is a local minimum. If `f''(c) 0` (positive), the point at `x = 4` is a local minimum. Find the corresponding y-coordinates. When `x = 2`, `y = (2)³ - 9(2)² + 24(2) = 8 - 9(4) + 48 = 8 - 36 + 48 = 20`. So, the local maximum is at the point (2, 20). When `x = 4`, `y = (4)³ - 9(4)² + 24(4) = 64 - 9(16) + 96 = 64 - 144 + 96 = 16`. So, the local minimum is at the point (4, 16). Sketch the curve. Find the y-intercept (where the curve cuts the y-axis). This happens when `x = 0`. `y = (0)³ - 9(0)² + 24(0) = 0`. The curve passes through the origin (0, 0). Plot the key points: (0, 0), the maximum (2, 20), and the minimum (4, 16). Draw a smooth curve connecting these points, ensuring it has a "cap" shape (∩) at the maximum and a "cup" shape (∪) at the minimum.