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Calculus

Calculus. Calculus. Calculus (differentiation and integration) was developed to improve this understanding. Differentiation and integration can help us solve many types of real-world problems .

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Calculus

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  1. Calculus

  2. 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.

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

  4. y = 2x y = 4 x = 2

  5. differentiation

  6. Y = x2 The normal is at right angles to the tangent tangent normal dy/dx =-1/2x dy/dx =2x y = 4 x =1

  7. The derivative of ex is ex (itself) dy/dx = ex

  8. The derivative of ex is ex (itself)

  9. The table of derivatives y = f(x) dy/dx f′(x) k, any constant 0 x 1 x2 2x x3 3x 2 xn, any constant n nxn−1

  10. The table of derivatives • y = f(x) dy/dx f′(x) • ex ex • ekxkekx • lnx = logex 1/x

  11. The table of derivatives • y = f(x) dy/dx f′(x) • sin x cos x • sin kxk coskx • cos x −sin x • coskx −k sin kx • tan x sec2x • tan kxk sec2kx Watch the minus sign

  12. differentiation • y = axn • dy/dx = axnxn-1

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

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

  15. 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)

  16. 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) continue

  17. Finding a tangent the gradient of the curve at the points quoted • 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 continue

  18. Finding a tangent • Find the tangent of y = 3x2 – 2x + 4 when x = 0 and 3 • When x = 3 y = 27 - 6 + 4 = 25 (substituting in the y equation) • Gradient dy/dx = 6x -2 = 16(when x = 3) • Also (y – 25)/(x – 3) = m (gradient) continue

  19. Finding a tangent the gradient of the curve at the points quoted • So (y – 25)/(x -3) =16 • (y – 25) = 16(x – 3) = 16x – 48 y = 16x – 48 + 25 • The equation for the gradient at x = 0, y=4 • y = 16x - 23

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

  21. 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

  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 4y= tan 3x • dy/dx = 3sec23x

  24. Derivatives of sine cosine and tangent • Example: Differentiate y=sin(x 2 +3) • let: u = x2 +3  • So du/dx = 2x • let y = sinu • dy/du =cos u • dy/dx = 2xcos(x2 +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 = x3sin2x • let u = x3; and v = sin2x • dy/dx = v.du/dx + u.dv/dx • du/dx = 3x2 and dv/dx = 2Cos2x • Therefore dy/dx = sin2x.3x2 + x3 2cos2x

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

  28. The quotient rule Minus when divided • y = (e4x)/(x2 +2) • let u = e4x; and v = x2 + 2 • dy/dx = (v.du/dx -u.dv/dx) / 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

  29. 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 = 150 -20 + 8 • 138 m/s continue

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

  31. 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

  32. Turning points • Check for max and min. d2(3x2-9x-12)y/dx2= 6x - 9 • Substituting for x = +4: 6x – 9 = 24 - 9 = 15. • Since this is positive it must be a minimum y point • Substituting for x = -1: 6x + (-9) = - 6 + (-9) = -15. • Since this is negative it must be a maximum y point continue

  33. Turning points • Therefore ymin= x3– 4.5x2 - 12x +15 • = 64 - 72 - 48 +15 = - 41 (as x = +4) • Therefore ymax = x3– 4.5x2 -12x +15 • = -1 – 4.5 + 12 +15 =+ 21.5 (x = -1)

  34. Integration (opposite of differentiation) ∫ xn dx = [(xn+1)/n+1] + c ∫ x2dx = [(x2+1)/2+1] + c = x3/3 +c

  35. Integration (opposite of differentiation) • ∫1/xdx= (ln x) + c • ∫1/ax+ b dx = 1/aln|ax + b| + c • ∫ex dx = ex + c

  36. Integration (opposite of differentiation) • ∫emxdx =1/memx+ c • ∫cosx dx = sin x + c • ∫ cosnx dx =1/n(sin nx) + c

  37. Integration (opposite of differentiation) • ∫sin x dx = −cosnx + c • ∫sin nx dx = −1/ncosnx + c

  38. Integration (opposite of differentiation) • ∫sec2 x dx = tan x + c • ∫sec2nx dx =1/n tan nx + c

  39. 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

  40. IntEgraTING BETWEEN LIMITS • When x = +4 • 2x3 – x2 +2x +c = 128 – 16 +8 +c • = 120 +c (equation 1) • When x =+1 • 2x3 – x2 +2x +c = 2 – 1 +2 +c • =3 + c (equation 2) continue

  41. IntEgraTIonBETWEEN LIMITS • To find ∫ ( 6x2 - 2x + 2) dx, between the limits +4 and +1 • subtract equation 2 from equation 1 • 120 +c - 3 + c • =117

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