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Rate of change / Differentiation (3)

Learn about the rate of change and differentiation through finding gradients, equations of tangents, and equations of normals. Practice solving problems to improve your understanding.

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Rate of change / Differentiation (3)

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  1. Rate of change / Differentiation (3) Differentiating Using differentiating

  2. The Key Bit dy dx dy dx dy dx dy dx = nxn-1 dy dx E.g. if y = 5x4 = 5 x 4x3 = 20x3 E.g. if y = x2 = 2x E.g. if y = x3 = 3x2 The general rule (very important) is :- If y = xn

  3. Warm-up dy dx dy dx dy dx dy dx dy dx dy dx Find for these functions :- Gradient at x=-2 = 4x = 10x4 = 10x + 10 = 3x2 + 2x +1 = 8x3 - 8x = -8 = 10x(-2)4=160 = -20 + 10 = -10 = 3(-2)2+2(-2)+1 = 12 -4 +1 = 9 = 8(-2)3 -8(-2) = 8 x-8 – 8 x-2 = -64 +16 = -48 y= 2x2 y = 2x5 y = 5x2 + 10x + 5 y = x3 + x2 + x y = 2x4 - 4x2 + 7

  4. dy dx dy dx = 3ax2 + 8x -12 = 3a + 8 – 12 = 2 When x=1 A differentiating Problem The gradient of y = ax3 + 4x2 – 12x is 2 when x=1 What is a? 3a - 4 = 2 3a = 6 a = 2

  5. dy dx dy dx = 12x2 – 2ax + 10 = 12 -2a +10 = 6 When x=1 Try this The gradient of y = 4x3 - ax2 + 10x is 6 when x=1 What is a? 22 - 2a = 6 16 = 2a a = 8

  6. Rate of change / Differentiation (3 pt2) Equations of tangents Equations of normals

  7. y = x3 – 12x We have seen: if dy dx dy dx = 3x2 - 12 Function Notation Instead of ‘y’ we may use the function notation f(x) f(x)= x3 – 12x If then is replace by f’(x) f’(x)= 3x2 - 12 so f’(x) represent the differential/gradient function

  8. dy dx Increase in y Gradient = Increase in x Linear graphs y - intercept Gradient m and c will always be numbers in your examples y = 5x + 7 y = 2x - 1

  9. Definition Parallel lines are ones with the same slope/gradient. i.e.the number in front of the ‘x’ is the same y = 3x - 11.31 y = 3x + 2/3 y = 3x y = 3x + 8 y = 3x + 84 y = 3x - 3 y = 3x - 21 y = 3x + 1 y = 3x + 43 y = 3x - 1.5

  10. Equations of Tangents The tangent to the curve is the gradient at that point dy dx dy dx (3,9) When x=3; = 2x3 = 6 = 2x y=x2 y x What is the equation of the tangent? y=x2 y = mx + c Substitute gradient: 9 = 6x3 + c c = 9 - 18 = -9 y = 6x - 9

  11. Equations of Tangents - try me dy dx dy dx When x=1; = 6+4 = 10 = 6x + 4 y y=3x2 + 4x + 1 - differentiate - gradient at x =1 - y = mx + c - find c (1,8) x What is the equation of the tangent? y=3x2 + 4x + 1 y = mx + c Substitute gradient: 8 = 10x1 + c c = 8 - 10 = -2 y = 10x - 2

  12. Perpendicular Lines If two lines with gradients m1 and m2 are perpendicular, then:

  13. Equations of Normals The normal is always perpendicular to the tangent normal dy dx mT x mN = -1 When x=3; = 2x3 = 6 y=x2 y (3,9) x What is the equation of the normal? y = mx + c Substitute gradient: 9 = -1/6x 3 + c c = 9 - -3/6 =9 + 1/2 y = -1/6 x + 9 1/2 mT x mN = -1 6 x mN = -1 mN = -1/6

  14. Equations of Normals The normal is always perpendicular to the tangent normal dy dx mT x mN = -1 When x=3; = 2x3 = 6 y=x2 y (3,9) x What is the equation of the normal? y = mx + c Substitute gradient: 9 = -1/6x 3 + c c = 9 + 3/6 =9 + 1/2 y = -1/6 x + 9 1/2 mT x mN = -1 6 x mN = -1 mN = -1/6

  15. Equations of Normal - try me dy dx dy dx When x=1; = 6+4 = 10 = 6x + 4 y y=3x2 + 4x + 1 - differentiate - gradient at x =1 - mt x mn = -1 - y = mnx + c - find c (1,8) normal x What is the equation of the normal? y = mnx + c Substitute gradient: 8 = -1/10x 1 + c c = 8 + 1/10 =8 1/10 y = -1/10 x + 8 1/10 mT x mN = -1 10 x mN = -1 mN = -1/10

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