Lesson Notes By Weeks and Term v5 - Grade 12

Lesson Video

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Performance objectives

• Apply various differentiation rules to find derivatives.
• Determine the equation of a tangent and a normal to a curve.
• Solve optimization problems to find maximum or minimum values.
• Use derivatives to analyze the shape of a graph.
• Apply calculus to solve rate of change problems.

Lesson summary

This week's focus is on consolidating our understanding of Calculus, specifically Differentiation and Applications of Differentiation. Calculus is a cornerstone of further mathematical studies and has direct applications in various fields, including engineering, economics, and computer science. Understanding rates of change and optimization is crucial for solving real-world problems, from designing efficient infrastructure to predicting market trends. In South Africa, understanding these concepts can contribute to addressing challenges related to resource management, urban planning, and economic development.

Lesson notes

2.1 Differentiation Rules: A Refresher Differentiation is the process of finding the derivative of a function. The derivative represents the instantaneous rate of change of the function. We need to be fluent in the following rules: Power Rule: If `f(x) = x^n`, then `f'(x) = nx^(n-1)`.

Constant Rule: If `f(x) = c` (where c is a constant), then `f'(x) = 0`.

Constant Multiple Rule: If `f(x) = cg(x)`, then `f'(x) = c*g'(x)`.

Sum/Difference Rule: If `f(x) = u(x) ± v(x)`, then `f'(x) = u'(x) ± v'(x)`.

Product Rule: If `f(x) = u(x) v(x)`, then `f'(x) = u'(x)v(x) + u(x)v'(x)`.

Quotient Rule: If `f(x) = u(x) / v(x)`, then `f'(x) = [u'(x)v(x) - u(x)v'(x)] / [v(x)]^2`.

Chain Rule: If `f(x) = g(h(x))`, then `f'(x) = g'(h(x)) h'(x)`. Think of it as "derivative of the outside, evaluated at the inside, times derivative of the inside." Example 1 (Power Rule & Constant Multiple): Find the derivative of `f(x) = 3x^4 - 2x^2 + 5x - 7`. `f'(x) = 3 4x^(4-1) - 2 2x^(2-1) + 5 1x^(1-1) - 0` `f'(x) = 12x^3 - 4x + 5` Example 2 (Product Rule): Find the derivative of `f(x) = (x^2 + 1)(2x - 3)`. Let `u(x) = x^2 + 1` and `v(x) = 2x - 3`. Then `u'(x) = 2x` and `v'(x) = 2`.

Using the product rule: `f'(x) = (2x)(2x - 3) + (x^2 + 1)(2)` `f'(x) = 4x^2 - 6x + 2x^2 + 2` `f'(x) = 6x^2 - 6x + 2` Example 3 (Quotient Rule): Find the derivative of `f(x) = (x + 1) / (x - 1)`. Let `u(x) = x + 1` and `v(x) = x - 1`. Then `u'(x) = 1` and `v'(x) = 1`.

Using the quotient rule: `f'(x) = [(1)(x - 1) - (x + 1)(1)] / (x - 1)^2` `f'(x) = (x - 1 - x - 1) / (x - 1)^2` `f'(x) = -2 / (x - 1)^2` Example 4 (Chain Rule): Find the derivative of `f(x) = (2x + 1)^3`. Let `g(u) = u^3` and `h(x) = 2x + 1`. Then `g'(u) = 3u^2` and `h'(x) = 2`.

Using the chain rule: `f'(x) = 3(2x + 1)^2 2` `f'(x) = 6(2x + 1)^2` 2.2 Equations of Tangents and Normals The derivative `f'(x)` at a point `x = a` gives the gradient of the tangent line to the curve `f(x)` at that point. The tangent is a straight line that touches the curve at a single point (locally). The normal is a line perpendicular to the tangent at that same point. Gradient of tangent at x = a: `m_tangent = f'(a)` Gradient of normal at x = a: `m_normal = -1 / f'(a)` (since the product of the gradients of perpendicular lines is -1) The equation of a straight line is given by `y = mx + c`, or alternatively, `y - y1 = m(x - x1)`.

Example 5: Find the equation of the tangent to the curve `f(x) = x^2 - 3x + 2` at the point where `x = 1`. Find the y-coordinate when x = 1: `f(1) = (1)^2 - 3(1) + 2 = 0`. So, the point is (1, 0).

Find the derivative: `f'(x) = 2x - 3`. Find the gradient of the tangent at x = 1: `f'(1) = 2(1) - 3 = -1`.

Use the point-slope form of a line: `y - 0 = -1(x - 1)` Simplify: `y = -x + 1` The equation of the tangent is `y = -x + 1`. 2.3 Optimization Problems Optimization involves finding the maximum or minimum value of a function, often subject to certain constraints. The key is to use the derivative to find stationary points (where `f'(x) = 0`) and then determine whether these points are maxima, minima, or points of inflection. You might need to use the second derivative test (if `f''(x) > 0` at a stationary point, it's a minimum; if `f''(x) `l + w = 200` => `l = 200 - w` Area: `A = l w = (200 - w) w = 200w - w^2` Find the derivative of A with respect to w: `dA/dw = 200 - 2w` Set the derivative equal to zero and solve for w: `200 - 2w = 0` => `w = 100` Find the length: `l = 200 - w = 200 - 100 = 100` Check that this is a maximum by finding the second derivative: `d^2A/dw^2 = -2` which is negative, so it's a maximum. The dimensions that maximize the area are 100m x 100m (a square). 2.4 Analyzing the Shape of a Graph Increasing/Decreasing Intervals: If `f'(x) > 0`, the function is increasing. If `f'(x) 0`, the function is concave up (shaped like a cup). If `f''(x) `x^2 = 1` => `x = ±1` Find the second derivative: `f''(x) = 6x` At x = 1, `f''(1) = 6 > 0`, so it's a local minimum. `f(1) = 1 - 3 = -2`. Minimum at (1, -2). At x = -1, `f''(-1) = -6 `x = 0`. `f(0) = 0`. Point of inflection at (0, 0). For x 0, increasing. For -1 1, f'(x) > 0, increasing. For x 0, f''(x) > 0, concave up. 2.5 Rate of Change Problems Calculus is used to calculate related rates. These problems typically involve finding the rate at which one quantity is changing in terms of the rate at which another quantity is changing.

Example 8: Water is poured into a conical container at a rate of 10 cm³/s. The cone is 30 cm high and the radius of the top is 10 cm. How fast is the water level rising when the water is 8 cm deep? (Volume of a cone: V = (1/3)πr²h) We are given `dV/dt = 10 cm³/s` and we want to find `dh/dt` when `h = 8 cm`. We have `V = (1/3)πr²h`. We need to relate r and h. Using similar triangles, `r/h = 10/30 = 1/3` => `r = h/3`.