Mastering Differential Calculus: Optimizing Rate of Change and Minima

1. Rate of change / Differentiation Gradient of curves Differentiating Maxima and Minima Optimisation

2. Topic 7 : Introduction to Differential Calculus

3. Lesson 1 Do Now #1 #2 #3 #4 Topic 7 : Introduction to Differential Calculus

4. Answer #1 Answer #2 Answer #3 Answer #4 Topic 7 : Introduction to Differential Calculus

5. dy dx dy dx Recall: A bit of symbology y x = “difference in y” “difference in x” = gradient of line PRONOUNCED “dee-why by dee-ex”

6. s s s t t t ds dt ds dt ds dt Graphs of gradient vs time “y=ax3+bx2+cx+d” “y=ax2+bx+c” “y=mx+c” t t t “y=const.” “y=ax2+bx+c” “y=mx+c”

7. Introduction to Applications Topic 7 : Introduction to Differential Calculus

8. Maxima, minima and stationary points Minimum dy dx dy dx Maximum Stationary points =0 B Terminology y B is a local maximum D is a local minimum Points where dy/dx=0 are call stationary points B & D are also known as turning points (where they gradient changes from +ve to -ve) D x x

9. At a minimum At a maximum dy dx dy dx dy dx dy dx + - - + =0 =0 > 0 < 0 Finding Stationary Points

10. -ve +ve +ve -ve Minimum Maximum Example f(x)= x3 - 12x + 1 “Find stationary points and determine their types” f’(x)= 3x2 - 12 When x = +2 When x = -2 Stationary points occur when gradient [derivative] is 0 Pick point to left (x = -2.1) f`(x) = 3x(-2.1)2 -12 = 1.23 [+ve] Pick point to left (x = 1.9) f`(x) = 3x(1.9)2 -12 = -1.17 [-ve] 3x2 - 12 = 0 3x2 = 12 x2 = 4 x = 2 or x = -2 Pick point to right (x = 2.1) f`(x) = 3x(2.1)2 -12 = 1.23 [+ve] Pick point to right (x = -1.9) f`(x) = 3x(-1.9)2 -12 = -1.17 [-ve]

11. -ve +ve +ve -ve Minimum Maximum Example (continued) f’(x)= 3x2 - 12 f(x)= x3 - 12x + 1 “… and sketch the graph” Stationary points When x=0 [y-axis] y = 03 -12x0 +1 = 1 (0,1) When x = +2 When x = -2 y = (2)3 -12(2) +1 = 8 - 24 + 1 = -15 Minimum at (2,-15) y = (-2)3 -12(-2) +1 = -8 + 24 + 1 = 17 Maximum at (-2,17)

12. Crosses y-axis at (0,1) Cubic function 20 Maximum at (-2,17) 2 -2 Minimum at (2,-15) -20 Example (continued) f(x)= x3 - 12x + 1 Y X Try Page 125 Ex A Q1

13. First derivative f’(x) = 3x2 -12 f’(x) [=0 at maximum and minimum] x 2 -2 Second derivative f’’(x) = 6x f’’(x) Gives the “rate of change of gradient” x f(x) The second derivative f(x)= x3 - 12x + 1 x -ve at maximum +ve at minimum

14. The second derivative The second derivative is denoted by .. for a function p(x) it is denoted by p’’(x) d2y d2y d2y dx2 dx2 dx2 If is negative then the stationary point is a maximum If is positive then the stationary point is a minimum KEY POINTS It is used at a simple way of telling if a stationary point is a maximum or a minimum

15. The second derivative - example “Find stationary points and determine their nature” f(x) = 1/3x3 - 2x2 + 3x +1 f’(x) = x2 - 4x + 3 Stationary points when f’(x)=0 x2 - 4x + 3 = 0 (x - 3)(x - 1) = 0 x = 3 and x = 1 are maxima or minima Look at the second derivative to determine their nature f’’(x) = 2x - 4 When x=1 f’’(x) = 2x1 - 4 = -2 which is negative … hence a maximum When x=3 f’’(x) = 2x3 - 4 = 2 which is positive … hence a minimum

16. The second derivative - Exam Q “Find stationary points and determine their nature” f(x) = 4x3 - 9x2 - 30x +1 f’(x) = 12x2 - 18x - 30 12x2 - 18x - 30 = 0 2x2 - 3x - 5 = 0 (2x - 5)(x + 1) = 0 x = 2.5 and x = -1 Stationary points when f’(x)=0 are maxima or minima Look at the second derivative to determine their nature f’’(x) = 24x - 18 When x=-1 f’’(x) = 24x-1 - 18 = -42 which is negative … hence a maximum When x=2.5 f’’(x) = 24x2.5 - 18 = 42 which is positive … hence a minimum