2. 1 Applications of Calculus - Contents 1.Rates 0f ChangeRates 0f Change 2.Exponential Growth & DecayExponential Growth & Decay 3.Motion of a particleMotion of a particle 4.Motion & DifferentiationMotion & Differentiation

3. 2 Rates of Change The gradient of a line is a measure of the Rates 0f Change of y in relation to x. Rate of Change constant. Rate of Change varies.

4. 3 Rates of Change – Example 1/2 R = 4 + 3t 2 The rate of flow of water is given by When t =0 then the volume is zero. Find the volume of water after 12 hours. R = 4 + 3t 2 i.e dV dt = 4 + 3t 2  V = (4 + 3t 2 ).dt ∫ = 4t + t 3 + C When t = 0, V = 0  0 = 0 + 0 + C  C = 0 V = 4t + t 3 @ t = 12 = 4 x 12 + 12 3 = 1776 units 3

5. 4 Rates of Change – Example 1/2 R = 4 + 3t 2 The rate of flow of water is given by When t =0 then the volume is zero. Find the volume of water after 12 hours. R = 4 + 3t 2 i.e dV dt = 4 + 3t 2  V = (4 + 3t 2 ).dt ∫ = 4t + t 3 + C When t = 0, V = 0  0 = 0 + 0 + C  C = 0 V = 4t + t 3 @ t = 12 = 4 x 12 + 12 3 = 1776 units 3

6. 5 Rates of Change – Example 2/2 B = 2t 4 - t 2 + 2000 a) The initial number of bacteria. (t =0) The number of bacteria is given by B = 2t 4 - t 2 + 2000 = 2(0) 4 – (0) 2 + 2000 = 2000 bacteria b) Bacteria after 5 hours. (t =5) = 2(5) 4 – (5) 2 + 2000 = 3225 bacteria c) Rate of growth after 5 hours. (t =5) dB dt = 8t 3 -2t = 8(5) 3 -2(5) = 990 bacteria/hr

7. 6 Exponential Growth and Decay A Special Rate of Change. Eg Bacteria, Radiation, etc It can be written as dQ dt = kQ dQ dt = kQ can be solved as Q = Ae kt Growth Initial Quantity Growth Constant k (k +ve = growth, k -ve = decay) Time

8. 7 Growth and Decay – Example Number of Bacteria given by N = Ae kt N = 9000, A = 6000 and t = 8 hours a) Find k (3 significant figures) N = A e kt 9000 = 6000 e 8k e 8k = 9000 6000 e 8k = 1.5 log e 1.5 = log e e 8k 1.5 =e 8k = 8k log e e = 8k k = log e 1.5 8 k ≈ 0.0507

9. 8 Growth and Decay – Example Number of Bacteria given by N = Ae kt A = 6000, t = 48 hours, k ≈ 0.0507 b) Number of bacteria after 2 days N = A e kt = 6000 e 0.0507x48 = 68 344 bacteria

10. 9 Growth and Decay – Example Number of Bacteria given by N = Ae kt k ≈ 0.0507, t = 48 hours, N = 68 344 c) Rate bacteria increasing after 2 days dN dt = kN= 0.0507 N = 0.0507 x 68 3444 = 3464 bacteria/hr

11. 10 Growth and Decay – Example Number of Bacteria given by N = Ae kt A = 6000, k ≈ 0.0507 d) When will the bacteria reach 1 000 000. N = A e kt 1 000 000 = 6000 e 0.0507t e 0.0507t = 1 000 000 6 000 e 0.0507t = 166.7 log e e 0.0507t = log e 166.7 0.0507t log e e = log e 166.7 0.0507t = log e 166.7 t =t = log e 166.7 0.0507 ≈ 100.9 hours

12. 11 Growth and Decay – Example Number of Bacteria given by N = Ae kt A = 6000, k ≈ 0.0507 e) The growth rate per hour as a percentage. dN dt =kN k is the growth constant k = 0.0507x 100% = 5.07%

13. 12 Motion of a particle 1 Displacement (x) Measures the distance from a point. To the right is positive To the left is negative - The Origin implies x = 0

14. 13 Motion of a particle 2 Velocity (v) Measures the rate of change of displacement. To the right is positive To the left is negative - Being Stationary implies v = 0 v = dx dt

15. 14 Motion of a particle 3 Acceleration (a) Measures the rate of change of velocity. +v – a -v + a } Having Constant Velocity implies a = 0 a = = dv dt d 2 x dt 2 To the right is positive To the left is negative - Slowing Down +v + a -v - a } Speeding Up

16. 15 Motion of a particle - Example When is the particle at rest? t2 & t5t2 & t5 x t t1t1 t2t2 t4t4 t5t5 t6t6 t7t7 t3t3

17. 16 Motion of a particle - Example x t t1t1 t2t2 t4t4 t5t5 t6t6 t7t7 t3t3 When is the particle at the origin? t3 & t6t3 & t6

18. 17 Motion of a particle - Example x t t1t1 t2t2 t4t4 t5t5 t6t6 t7t7 t3t3 Is the particle faster at t 1 or t 7 ? Why? t1t1 Gradient Steeper

19. 18 Motion of a particle - Example x t t1t1 t2t2 t4t4 t5t5 t6t6 t7t7 t3t3 Is the particle faster at t 1 or t 7 ? Why? t1t1 Gradient Steeper

20. 19 Motion and Differentiation Displacement Velocity Acceleration

21. 20 Motion and Differentiation - Example Displacement x = -t 2 + t +2 Find initial velocity. Initially t = 0

22. 21 Motion and Differentiation - Example Displacement x = -t 2 + t + 2 Show acceleration is constant. Acceleration -2 units/s 2

23. 22 Motion and Differentiation - Example Displacement x = -t 2 + t +2 Find when the particle is at the origin. Origin @ x = 0 & @ Origin when t = 2 sec

24. 23 Motion and Differentiation - Example Displacement x = -t 2 + t +2 Find the maximum displacement from origin. Maximum Displacement when v =

25. 24 Motion and Differentiation - Example Displacement x = -t 2 + t +2 Sketch the particles motion 21 1 2 Initial Displacement ‘2’ Maximum Displacement x=2.25 @ t=0.5 Return to Origin ‘0’ @ t=2 x t