Applications of Calculus

Applications of Calculus - Contents • Rates 0f Change • Exponential Growth & Decay • Motion of a particle • Motion & Differentiation

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

Rates of Change – Example 1/2 The rate of flow of water is given by R = 4 + 3t2 When t =0 then the volume is zero. Find the volume of water after 12 hours. R = 4 + 3t2 When t = 0, V = 0 dV dt i.e = 4 + 3t2  C = 0  0 = 0 + 0 + C @ t = 12 V = 4t + t3 ∫  V = (4 + 3t2).dt = 4 x 12 + 123 = 4t + t3 + C = 1776 units3

Rates of Change – Example 2/2 The number of bacteria is given by B = 2t4 - t2 + 2000 a) The initial number of bacteria.(t =0) c) Rate of growth after 5 hours.(t =5) B = 2t4 - t2 + 2000 = 2(0)4 – (0)2 + 2000 dB dt = 8t3 -2t = 2000 bacteria = 8(5)3 -2(5) b) Bacteria after 5 hours.(t =5) = 990 bacteria/hr = 2(5)4 – (5)2 + 2000 = 3225 bacteria

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

Growth and Decay – Example Number of Bacteria given by N = Aekt N = 9000, A = 6000 and t = 8 hours a) Find k (3 significant figures) N = A ekt loge1.5 = loge e8k = 8k loge e 9000 = 6000 e8k = 8k 9000 6000 e8k = loge1.5 8 k = e8k = 1.5 k ≈ 0.0507 1.5 =e8k

Growth and Decay – Example Number of Bacteria given by N = Aekt A = 6000, t = 48 hours, k ≈ 0.0507 b) Number of bacteria after 2 days N = A ekt = 6000 e0.0507x48 = 68 344 bacteria

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

Growth and Decay – Example Number of Bacteria given by N = Aekt A = 6000, k ≈ 0.0507 d) When will the bacteria reach 1 000 000. logee0.0507t= loge166.7 N = A ekt 1 000 000 = 6000 e0.0507t 0.0507t logee= loge166.7 1 000 000 6 000 0.0507t= loge166.7 e0.0507t = loge166.7 0.0507 t= e0.0507t= 166.7 hours ≈ 100.9

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

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

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

Motion of a particle 3 Acceleration (a) Measures the rate of change of velocity. dv dt d2x dt2 a = = To the left is negative - To the right is positive } } +v –a -v +a Slowing Down +v +a -v -a Speeding Up HavingConstant Velocityimpliesa = 0

x t t1 t2 t3 t4 t5 t6 t7 Motion of a particle - Example When is the particle at rest? t2&t5

x t t1 t2 t3 t4 t5 t6 t7 Motion of a particle - Example When is the particle at the origin? t3&t6

x t t1 t2 t3 t4 t5 t6 t7 Motion of a particle - Example Is the particle faster at t1or t7? Why? t1 Gradient Steeper

Motion and Differentiation Displacement Velocity Acceleration

Motion and Differentiation - Example Displacement x=-t2+t+2 Find initial velocity. Initially t = 0

Motion and Differentiation - Example Displacement x=-t2+t+2 Show acceleration is constant. Acceleration -2 units/s2

Motion and Differentiation - Example Displacement x=-t2+t+2 Find when the particle is at the origin. Origin @ x = 0 @ Origin when t = 2 sec &

Motion and Differentiation - Example Displacement x=-t2+t+2 Find the maximum displacement from origin. Maximum Displacement when v =

2 1 1 2 Motion and Differentiation - Example Displacement x=-t2+t+2 Sketch the particles motion x Maximum Displacement x=2.25 @ t=0.5 Initial Displacement ‘2’ Return to Origin ‘0’ @ t=2 t

Applications of Calculus - Contents