Trigonometry & Calculus

Introduction to Calculus and Vectors

Grade 12

✓By the end of this lesson students will be able to understand the concept of a limit and evaluate limits algebraically.
✓By the end of this lesson students will be able to define the derivative as the instantaneous rate of change and the slope of the tangent.
✓By the end of this lesson students will be able to apply differentiation rules to find derivatives of polynomial functions.
✓By the end of this lesson students will be able to perform operations with vectors, including addition, subtraction, and scalar multiplication.
✓By the end of this lesson students will be able to solve problems involving optimization and kinematics using calculus and vector concepts.

Key concepts

Limits

A limit describes the value that a function 'approaches' as the input (x) approaches some value. It's about the behaviour of the function near a point, not necessarily at the point itself.

lim (x->a) f(x) = L

The Derivative

The derivative of a function represents its instantaneous rate of change at any given point. Geometrically, it is the slope of the tangent line to the function's graph at that point. It is formally defined using the limit of the difference quotient.

f'(x) = lim (h->0) [f(x+h) - f(x)] / h

Differentiation Rules (Power Rule)

Differentiation rules provide shortcuts for finding derivatives without using the limit definition. The Power Rule is fundamental for polynomial functions.

If f(x) = x^n, then f'(x) = n*x^(n-1). For a constant c, d/dx(c) = 0. For a constant k, d/dx(k*f(x)) = k*f'(x).

Vectors

A vector is a quantity that has both magnitude (size) and direction. This contrasts with a scalar, which only has magnitude. Vectors can be represented geometrically as directed line segments or algebraically using components (e.g., (x, y) in 2D or (x, y, z) in 3D).

For a vector v = (vx, vy), its magnitude is |v| = sqrt(vx^2 + vy^2).

Vector Operations

Vectors can be added, subtracted, and multiplied by scalars. Geometrically, vector addition follows the triangle or parallelogram rule. Algebraically, operations are performed component-wise.

If u = (ux, uy) and v = (vx, vy):\nVector Addition: u + v = (ux+vx, uy+vy)\nVector Subtraction: u - v = (ux-vx, uy-vy)\nScalar Multiplication: k*v = (k*vx, k*vy)

Applications of Derivatives (Kinematics)

Derivatives are crucial in physics and engineering. For motion along a line, if s(t) is the position function of an object at time t, then its velocity v(t) is the first derivative of position, and its acceleration a(t) is the second derivative of position (or the first derivative of velocity).

Velocity: v(t) = s'(t)\nAcceleration: a(t) = v'(t) = s''(t)

Applications of Derivatives (Optimization)

Optimization involves finding the maximum or minimum value of a function. This is achieved by finding the critical points where the derivative is zero or undefined, and then testing these points along with endpoints of the domain.

To find local extrema, set f'(x) = 0 and solve for x.

Key facts to remember

1A limit describes the value a function approaches, not necessarily its value at that point.
2The derivative f'(x) represents the instantaneous rate of change of f(x) and the slope of the tangent line to f(x) at x.
3The Power Rule for differentiation states that d/dx(x^n) = n*x^(n-1).
4A vector has both magnitude and direction, while a scalar has only magnitude.
5Vector operations (addition, subtraction, scalar multiplication) are performed component-wise.
6The magnitude of a 2D vector v = (vx, vy) is |v| = sqrt(vx^2 + vy^2).
7In kinematics, velocity is the derivative of position (v(t) = s'(t)), and acceleration is the derivative of velocity (a(t) = v'(t) = s''(t)).
8Optimization problems often involve finding critical points where the derivative of a function is zero or undefined.

Worked examples

Example 1

Find the derivative of the function f(x) = 4x^3 - 2x^2 + 5x - 1 and determine the slope of the tangent line to the curve at x = 1.

IApply the Power Rule and Sum/Difference Rule to differentiate f(x):

IIf'(x) = d/dx(4x^3) - d/dx(2x^2) + d/dx(5x) - d/dx(1)

IIIf'(x) = 4 * (3x^(3-1)) - 2 * (2x^(2-1)) + 5 * (1x^(1-1)) - 0

IVf'(x) = 12x^2 - 4x + 5

VTo find the slope of the tangent at x = 1, substitute x = 1 into f'(x):

VIf'(1) = 12(1)^2 - 4(1) + 5

VIIf'(1) = 12 - 4 + 5

VIIIf'(1) = 13

Answer

The derivative is f'(x) = 12x^2 - 4x + 5. The slope of the tangent line at x = 1 is 13.

Remember that the derivative gives the slope of the tangent line at any point x.

Example 2

Given vectors u = (5, -3) and v = (-1, 4), find the resultant vector w = 3u - 2v and its magnitude.

IFirst, perform scalar multiplication for 3u:

II3u = 3 * (5, -3) = (3*5, 3*(-3)) = (15, -9)

IIINext, perform scalar multiplication for 2v:

IV2v = 2 * (-1, 4) = (2*(-1), 2*4) = (-2, 8)

VNow, perform vector subtraction to find w = 3u - 2v:

VIw = (15, -9) - (-2, 8)

VIIw = (15 - (-2), -9 - 8)

VIIIw = (15 + 2, -9 - 8)

9w = (17, -17)

10Finally, calculate the magnitude of w:

11|w| = sqrt(wx^2 + wy^2)

12|w| = sqrt((17)^2 + (-17)^2)

13|w| = sqrt(289 + 289)

14|w| = sqrt(578)

15|w| = sqrt(289 * 2)

16|w| = 17 * sqrt(2)

Answer

The resultant vector is w = (17, -17) and its magnitude is |w| = 17*sqrt(2).

Ensure to perform operations component-wise for vector addition and subtraction.

Example 3

The position of a particle moving along a straight line is given by the function s(t) = t^3 - 9t^2 + 15t, where s is in metres and t is in seconds. Find the velocity and acceleration of the particle at t = 4 seconds.

ITo find the velocity function, differentiate the position function s(t) with respect to t:

IIv(t) = s'(t) = d/dt(t^3 - 9t^2 + 15t)

IIIv(t) = 3t^2 - 18t + 15

IVTo find the acceleration function, differentiate the velocity function v(t) with respect to t:

Va(t) = v'(t) = d/dt(3t^2 - 18t + 15)

VIa(t) = 6t - 18

VIINow, substitute t = 4 into the velocity function to find the velocity at t = 4s:

VIIIv(4) = 3(4)^2 - 18(4) + 15

9v(4) = 3(16) - 72 + 15

10v(4) = 48 - 72 + 15

11v(4) = -9 m/s

12Substitute t = 4 into the acceleration function to find the acceleration at t = 4s:

13a(4) = 6(4) - 18

14a(4) = 24 - 18

15a(4) = 6 m/s^2

Answer

At t = 4 seconds, the velocity of the particle is -9 m/s and its acceleration is 6 m/s^2.

Negative velocity indicates movement in the opposite direction (e.g., left or down).

Common mistakes

✗Confusing the value of a function at a point with its limit as x approaches that point.
✗Incorrectly applying differentiation rules, especially forgetting to multiply by the original exponent or subtracting 1 from the exponent.
✗Forgetting to include units in application problems (e.g., m/s for velocity, m/s^2 for acceleration).
✗Mixing up scalar and vector quantities or operations (e.g., trying to add a scalar to a vector).
✗Errors in calculating vector magnitudes, particularly with negative components (e.g., (-3)^2 = -9 instead of 9).

Exam tips

★Always show your steps clearly when evaluating limits or finding derivatives, especially when using the limit definition.
★For vector problems, drawing a diagram can help visualize the vectors and their resultant, especially for geometric addition/subtraction.
★Pay close attention to units in application problems; they are crucial for interpreting your answers correctly.
★Practice a variety of problems, including those that combine calculus and vector concepts, to build confidence and understanding.

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