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CALCULUS. Calculus (differentiation and integration) was developed to improve this understanding. Differentiation and integration can help us solve many.

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Presentation on theme: "CALCULUS. Calculus (differentiation and integration) was developed to improve this understanding. Differentiation and integration can help us solve many."— Presentation transcript:

1 CALCULUS

2 Calculus (differentiation and integration) was developed to improve this understanding. Differentiation and integration can help us solve many types of real-world problems. We use the derivative to determine the maximum and minimum values of particular functions (e.g. cost, strength, amount of material used in a building, profit, loss, etc.). Derivatives are met in many engineering and science problems, especially when modelling the behaviour of moving objects.

3 DIFFERENTIATION dy/dx = 1 y = 3 x = 3

4 Y = 2X y = 4 x = 2

5 DIFFERENTIATION

6 Y = X 2 y = 2 x =1 tangent normal The normal is at right angles to the tangent dy/dx =2x Video dy/dx =-1/2x

7 The derivative of e x is e x (itself) dy/dx = e x

8 The derivative of e x is e x (itself)

9 THE TABLE OF DERIVATIVES y = f(x) dy/dx f′(x) k, any constant 0 x 1 x 2 2x x 3 3x 2 x n, any constant n nx n−1

10 THE TABLE OF DERIVATIVES y = f(x) dy/dx f′(x) e x e x e kx ke kx l n x = log e x 1/x

11 THE TABLE OF DERIVATIVES y = f(x) dy/dx f′(x) sin x cos x sin kx k cos kx cos x −sin x cos kx −k sin kx tan x sec 2 x tan kx k sec 2 kx Watch the minus sign

12 DIFFERENTIATION y = ax n dy/dx = a x nx n-1

13 DIFFERENTIATION y = 4 x 3 + 2x 2 +3 dy/dx = 12x 2 +4x

14 DIFFERENTIATION y = (x 3 + x) / x 2 x 3 /x 2 + x/x 2 = x + 1/x = x + x -1 dy/dx = 1 - x -2 = 1 - 1/x 2

15 DIFFERENTIATION y = √x + 1/√x = x 1/2 + x -1/2 dy/dx = 1/2 x -1/2 - 1/2x -3/2) =1/(2√x) - 1/(2√x 3 )

16 FINDING A TANGENT THE GRADIENT OF THE CURVE AT THE POINTS QUOTED Find the tangent of y = 3x 2 – 2x + 4 when x = 0 and 3 1) When x = 0 y = +4 (substituting in the y equation) Gradient dy/dx = 6x -2 = -2 (when x = 0) Also (y – 4)/(x – 0) = m (gradient)

17 FINDING A TANGENT So (y – 4)/x = -2 (y – 4) = - 2x y = -2x +4 The equation for the gradient at x = 0, y=4 is y = -2x + 4

18 FINDING A TANGENT Find the tangent of y = 3x 2 – 2x + 4 when x = 0 and 3 When x = 3 y = = 25 (substituting in the y equation) Gradient dy/dx = 6x -2 = 16 (when x = 3) Also (y – 25)/(x – 3) = m (gradient)

19 FINDING A TANGENT So (y – 25)/(x -3) =16 (y – 25) = 16(x – 3) = 16x – 48 y = 16x – The equation for the gradient at x = 0, y=4 y = 16x - 23

20 CHAIN RULE Differentiate y=sin(x 2 +3) let: u = x 2 +3 So du/dx = 2x let y = sin u dy/du =cos u dy/dx = 2xcos(x 2 +3)

21 CHAIN RULE Differentiate with respect to x: y= 3 √e 1-x = e (1-x)/3 = e 1/3-x/3 let: u = 1/3-x/3 du/dx -1/3 dy/du = e (1-x)/3 dy/dx = -1/3 e (1-x)/3 Chain rule

22 DERIVATIVES OF SINE COSINE AND TANGENT Example 1 y = sin 3x (dy)/(dx) = 3 cos 3x Example 2 y = 5 sin 3x. ` (dy)/(dx)` = 15 cos 3x

23 DERIVATIVES OF SINE COSINE AND TANGENT Example 3 y =cos3x. dy/dx = - 3sin 3x Example 4 y= tan 3x dy/dx = 3sec 2 3x

24 DERIVATIVES OF SINE COSINE AND TANGENT Example: Differentiate y=sin(x 2 +3) let: u = x 2 +3 So du/dx = 2x let y = sin u dy/du =cos u dy/dx = 2xcos(x 2 +3)

25 THE PRODUCT RULE The product rule: if y = uv then dy/dx = u.dv/dx + v.du/dx

26 THE PRODUCT RULE y = x 3 sin2x let u = x 3 ; and v = sin2x dy/dx = v.du/dx + u.dv/x du/dx = 3x 2 and dv/dx = 2Cos2x Therefore dy/dx = sin2x.3x 2 + x 3 2cos2x

27 THE QUOTIENT RULE The quotient rule: if y = u/v then dy/dx = (v.du/dx - u.dv/dx)/v 2

28 THE QUOTIENT RULE y = (e 4x )/(x 2 +2) let u = e 4x ; and v = x dy/dx = (v.du/dx - u.dv/x) / v 2 du/dx = 4e 4x and dv/dx = 2x Therefore dy/dx = [(x 2 +2).4e 4x ] - [e 4x.2x]/(x 2 +2) 2 = e 4x (2x 2 +8 – 2x) /(x 2 +2) 2 Minus when divided Don’t forget

29 VELOCITY AND ACCELERATION An object travels a distance s = 2t 3 - 2t 2 +8t + 6 metres in t seconds. Find both velocity and acceleration of the body at time 5 seconds. velocity = ds/dt = 6t 2 – 4t +8 When t =5 velocity = m/s continue

30 VELOCITY AND ACCELERATION Acceleration = dv/dt = 12t – 4 From v = 6t 2 -4t +8 (last slide) Acceleration at 5t sec = 60-4 = 56m/s 2

31 TURNING POINTS The equation of a curve is y = x 3 – 4.5x x +15. Find by calculation the value of the maximum and minimum turning points on the curve. Turning points occur at dy/dx = 0 Therefore:- dy/dx = 3x 2 - 9x – 12 = 0 = x 2 - 3x – 4 (dividing by common factor 3) = 0 Factorising x 2 - 3x - 4; then (x - 4)(x+1) = 0. Turning points are at x = + 4 and x = -1 continue

32 TURNING POINTS Check for max and min. d 2 (3x 2 -9x-12)y/dx 2 = 6x - 9 Substituting for x = +4: 6x – 9 = = 15. Since this is positive it must be a minimum y point Substituting for x = -1: 6x + (-9) = (-9) = -15. Since this is negative it must be a maximum y point continue

33 TURNING POINTS Therefore y min = x 3 – 4.5x x +15 = = - 41 (as x = +4) Therefore y max = x 3 – 4.5x x +15 = -1 – = (x = -1)

34 INTEGRATION (OPPOSITE OF DIFFERENTIATION) ∫ x n dx = [(x n+1 )/n+1] + c ∫ x 2 dx = [(x 2+1 )/2+1] + c = x 3 /3 +c

35 INTEGRATION (OPPOSITE OF DIFFERENTIATION) ∫ 1 / x dx = (l n x) + c ∫ 1 / ax + b dx = 1 / a ln |ax + b| + c ∫ e x dx = e x + c

36 INTEGRATION (OPPOSITE OF DIFFERENTIATION) ∫ e mx dx = 1 / m e mx + c ∫ cos x dx = sin x + c ∫ cos nx dx = 1 / n (sin nx) + c

37 INTEGRATION (OPPOSITE OF DIFFERENTIATION) ∫ sin x dx = −cos nx + c ∫ sin nx dx = − 1 / n cos nx + c

38 INTEGRATION (OPPOSITE OF DIFFERENTIATION) ∫ sec 2 x dx = tan x + c ∫ sec 2 nx dx = 1 / n tan nx + c

39 INTEGRATION BETWEEN LIMITS Example Evaluate ∫ ( 6x 2 - 2x + 2) dx, between the limits +4 and +1 ∫ ( 6x 2 - 2x + 2) dx = 6x 3 /3 – 2x 2 /2 +2x +c = 2x 3 – x 2 +2x +c continue

40 INTEGRATING BETWEEN LIMITS When x = +4 2x 3 – x 2 +2x +c = 128 – c = 120 +c (equation 1) When x =+1 2x 3 – x 2 +2x +c = 2 – c =3 + c (equation 2) continue

41 INTEGRATION BETWEEN LIMITS To find ∫ ( 6x 2 - 2x + 2) dx, between the limits +4 and +1 subtract equation 2 from equation c c =117


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