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 5.Motion & IntegrationMotion & Integration

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.

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 = C  C = 0 V = 4t + t t = 12 = 4 x = 1776 units 3

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 = C  C = 0 V = 4t + t t = 12 = 4 x = 1776 units 3

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

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

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 = e 8k = 1.5 log e 1.5 = log e e 8k 1.5 =e 8k = 8k log e e = 8k k = log e k ≈

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

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

10 Growth and Decay – Example Number of Bacteria given by N = Ae kt A = 6000, k ≈ d) When will the bacteria reach N = A e kt = 6000 e t e t = e t = log e e t = log e t log e e = log e t = log e t =t = log e ≈ hours

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

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

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

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

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

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

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

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

19 Motion and Differentiation Displacement Velocity Acceleration

20 Motion and Differentiation - Example Displacement x = -t 2 + t +2 in cm. Find initial velocity (in cm/s). Initially t = 0

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

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

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

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

25 Motion and Integration Acceleration Velocity Displacement

26 Motion and Integration - Example Velocity v= 3t 2 + 2t + 1, x o =-2cm Find displacement after 5 secs. When t=0, x=-2 When t=5 Displacement is 153cm to right of origin.

27 Motion and Integration - Example Find displacement after 9 secs. Acceleration a= 6 -, v o =0, x o =+1m 2 (t + 1) 2 When t=0, v=0

28 Motion and Integration - Example Find displacement after 9 secs. Acceleration a= 6 -, v o =0, x o =+1m 2 (t + 1) 2

29 Motion and Integration - Example Find displacement after 9 secs. Acceleration a= 6 -, v o =0, x o =+1m 2 (t + 1) 2 When t=0, x=1 When t=9 Displacement