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Calculus

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Calculus 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.

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

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y = 2x y = 4 x = 2

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differentiation

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**Y = x2 dy/dx =2x Video The normal is at right angles to the tangent**

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**The derivative of ex is ex (itself)**

dy/dx = ex

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**The derivative of ex is ex (itself)**

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**The table of derivatives**

y = f(x) dy/dx f′(x) k, any constant x x x x x 2 xn, any constant n nxn−1

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**The table of derivatives**

y = f(x) dy/dx f′(x) ex ex ekx kekx ln x = logex 1/x

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**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 sec2 x tan kx k sec2 kx Watch the minus sign

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differentiation y = axn dy/dx = axnxn-1

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differentiation y = 4 x3 + 2x2 +3 dy/dx = 12x2 +4x

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differentiation y = (x3 + x) / x2 x3/x2 + x/x2 = x + 1/x = x + x-1 dy/dx = 1 - x-2 = 1 - 1/x2

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differentiation y = √x + 1/√x = x1/2 + x-1/2 dy/dx = 1/2 x-1/2 - 1/2x-3/2) =1/(2√x) - 1/(2√x3)

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**Finding a tangent the gradient of the curve at the points quoted**

Find the tangent of y = 3x2 – 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)

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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

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Finding a tangent Find the tangent of y = 3x2 – 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)

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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

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Chain rule Differentiate y=sin(x2 +3) let: u = x 2 +3 So du/dx = 2x let y = sin u dy/du =cos u dy/dx = 2xcos(x2 +3)

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**Differentiate with respect to x:**

Chain rule Differentiate with respect to x: y= 3√e1-x = e(1-x)/3 = e1/3-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

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**Derivatives of sine cosine and tangent**

Example y = sin 3x (dy)/(dx) = 3 cos 3x Example y = 5 sin 3x . ` (dy)/(dx)` = 15 cos 3x

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**Derivatives of sine cosine and tangent**

Example y =cos3x. dy/dx = - 3sin 3x Example y= tan 3x dy/dx = 3sec23x

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**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(x2 +3)

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**The product rule: if y = uv then dy/dx = u.dv/dx + v.du/dx**

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The product rule y = x3sin2x let u = x3; and v = sin2x dy/dx = v.du/dx + u.dv/x du/dx = 3x2 and dv/dx = 2Cos2x Therefore dy/dx = sin2x.3x2 + x3 2cos2x

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**The quotient rule: if y = u/v then dy/dx = (v.du/dx - u.dv/dx)/v2**

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**dy/dx = (v.du/dx - u.dv/x) / v2**

The quotient rule Minus when divided y = (e4x)/(x2 +2) let u = e4x; and v = x2 + 2 dy/dx = (v.du/dx - u.dv/x) / v2 du/dx = 4e4x and dv/dx = 2x Therefore dy/dx = [(x2+2).4e4x] - [e4x.2x]/(x2 +2)2 = e4x (2x2 +8 – 2x) /(x2 +2)2 Don’t forget

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**Velocity and acceleration**

An object travels a distance s = 2t3 - 2t2 +8t + 6 metres in t seconds. Find both velocity and acceleration of the body at time 5 seconds. velocity = ds/dt = 6t2 – 4t +8 When t =5 velocity = m/s continue

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**Velocity and acceleration**

Acceleration = dv/dt = 12t – 4 From v = 6t2 -4t +8 (last slide) Acceleration at 5t sec = 60-4 = 56m/s2

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Turning points The equation of a curve is y = x3 – 4.5x2 - 12x +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 = 3x2 - 9x – 12 = 0 = x2 - 3x – 4 (dividing by common factor 3) = 0 Factorising x2 - 3x - 4; then (x - 4)(x+1) = 0. Turning points are at x = + 4 and x = -1 continue

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Turning points Check for max and min. d2(3x2-9x-12)y/dx2 = 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

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Turning points Therefore ymin = x3 – 4.5x2 - 12x +15 = = - 41 (as x = +4) Therefore ymax = x3 – 4.5x2 - 12x +15 = -1 – = (x = -1)

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**Integration (opposite of differentiation)**

∫ xn dx = [(xn+1)/n+1] + c ∫ x2 dx = [(x2+1)/2+1] + c = x3/3 +c

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**Integration (opposite of differentiation)**

∫1/x dx = (ln x) + c ∫ 1/ax + b dx = 1/a ln |ax + b| + c ∫ ex dx = ex + c

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**Integration (opposite of differentiation)**

∫ emx dx =1/memx + c ∫cos x dx = sin x + c ∫ cos nx dx =1/n(sin nx) + c

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**Integration (opposite of differentiation)**

∫ sin x dx = −cos nx + c ∫ sin nx dx = −1/n cos nx + c

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**Integration (opposite of differentiation)**

∫ sec2 x dx = tan x + c ∫ sec2 nx dx =1/n tan nx + c

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**IntEgraTIon BETWEEN LIMITS**

Example Evaluate ∫ ( 6x2 - 2x + 2) dx, between the limits +4 and +1 ∫ ( 6x2 - 2x + 2) dx = 6x3/3 – 2x2/2 +2x +c = 2x3 – x2 +2x +c continue

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**IntEgraTING BETWEEN LIMITS**

When x = +4 2x3 – x2 +2x +c = 128 – c = 120 +c (equation 1) When x =+1 2x3 – x2 +2x +c = 2 – c =3 + c (equation 2) continue

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**IntEgraTIon BETWEEN LIMITS**

To find ∫ ( 6x2 - 2x + 2) dx, between the limits +4 and +1 subtract equation 2 from equation c c =117

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Core 3 Differentiation Learning Objectives:

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