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Applications of Calculus - Contents
1. Rates 0f Change2. Exponential Growth & Dec
ay3. Motion of a particle4. Motion & Differentiation
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Rates of Change
The gradient of a line is a measure of the Rates 0f Change of y in relation to x.
Rate of Changeconstant.
Rate of Changevaries.
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Rates of Change – Example 1/2
R = 4 + 3t2
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 + 3t2
i.e dVdt
= 4 + 3t2
V = (4 + 3t2).dt∫= 4t + t3 + C
When t = 0, V = 0
0 = 0 + 0 + C C = 0
V = 4t + t3 @ t = 12
= 4 x 12 + 123
= 1776 units3
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Rates of Change – Example 1/2
R = 4 + 3t2
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 + 3t2
i.e dVdt
= 4 + 3t2
V = (4 + 3t2).dt∫= 4t + t3 + C
When t = 0, V = 0
0 = 0 + 0 + C C = 0
V = 4t + t3 @ t = 12
= 4 x 12 + 123
= 1776 units3
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Rates of Change – Example 2/2
B = 2t4 - t2 + 2000 a) The initial number
of bacteria. (t =0)
The number of bacteria is given by
B = 2t4 - t2 + 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)
dBdt
= 8t3 -2t
= 8(5)3 -2(5)
= 990 bacteria/hr
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Exponential Growth and Decay
A Special Rate of Change.Eg Bacteria, Radiation, etc
It can be written as dQdt
= kQ
dQdt
= kQcan be solved as Q = Aekt
Growth
Initial Quantity
Growth Constant k(k +ve = growth, k -ve = decay)
Time
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Growth and Decay – Example
Number of Bacteria given by N = Aekt
N = 9000, A = 6000 and t = 8 hoursa) Find k (3 significant figures)
N = A ekt
9000 = 6000 e8k
e8k =90006000
e8k = 1.5
loge1.5 = loge e8k
1.5 =e8k
= 8k loge e= 8k
k =loge1.58
k ≈ 0.0507
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Growth and Decay – Example
Number of Bacteria given by N = Aekt
A = 6000, t = 48 hours, k ≈ 0.0507b) Number of bacteria after 2 days
N = A ekt
= 6000 e0.0507x48
= 68 344 bacteria
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Growth and Decay – Example
Number of Bacteria given by N = Aekt
k ≈ 0.0507, t = 48 hours, N = 68 344c) Rate bacteria increasing after 2 days
dNdt
= kN = 0.0507 N
= 0.0507 x 68 3444
= 3464 bacteria/hr
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Growth and Decay – Example
Number of Bacteria given by N = Aekt
A = 6000, k ≈ 0.0507d) When will the bacteria reach 1 000 000.
N = A ekt
1 000 000 = 6000 e0.0507t
e0.0507t = 1 000 0006 000
e0.0507t = 166.7
logee0.0507t = loge166.7
0.0507t logee = loge166.7
0.0507t = loge166.7
t =loge166.70.0507
≈ 100.9 hours
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Growth and Decay – Example
Number of Bacteria given by N = Aekt
A = 6000, k ≈ 0.0507e) The growth rate per hour as a percentage.
dNdt
= kN k is the growth constant
k = 0.0507x 100%
= 5.07%
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Motion of a particle 1
Displacement (x)
Measures the distance from a point.
To the right is positiveTo the left is negative -
The Origin implies x = 0
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Motion of a particle 2
Velocity (v)
Measures the rate of change of displacement.
To the right is positiveTo the left is negative -
Being Stationary implies v = 0
v =dxdt
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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 = =dvdt
d2xdt2
To the right is positiveTo the left is negative -
SlowingDown
+v + a-v - a}Speeding
Up
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Motion of a particle - Example
When is the particle at rest?
t2 & t5
x
tt1 t2 t4 t5 t6 t7t3
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Motion of a particle - Example
x
tt1 t2 t4 t5 t6 t7t3
When is the particle at the origin?
t3 & t6
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Motion of a particle - Example
x
tt1 t2 t4 t5 t6 t7t3
Is the particle faster at t1 or t7?Why?
t1
GradientSteeper
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Motion of a particle - Example
x
tt1 t2 t4 t5 t6 t7t3
Is the particle faster at t1 or t7?Why?
t1
GradientSteeper
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Motion and Differentiation
Displacement
Velocitydtdx
x
Accelerationdtdv
x 2
2
dt
xd
x
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Motion and Differentiation - Example
Displacement x = -t2 + t +2
Find initial velocity.
22 ttx
dt
dxv
12 t
Initially t = 0
102 v11 sunit
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Motion and Differentiation - Example
Displacement x = -t2 + t + 2
Show acceleration is constant.
12 tv
dt
dva
2
Acceleration
-2 units/s2
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Motion and Differentiation - Example
Displacement x = -t2 + t +2
Find when the particle is at the origin.
22 ttxOrigin @ x = 0
20 2 tt
)1)(2(0 tt1t & 2t
@ Origin when
t = 2 sec)2(0 2 tt
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Motion and Differentiation - Example
0dtdx
Displacement x = -t2 + t +2
Find the maximum displacement from origin.
Maximum Displacement
when v = 012 tdtdx
12 t
5.0t
22 ttx
25.0)5.0( 2 x
25.2x
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Motion and Differentiation - Example
Displacement x = -t2 + t +2
Sketch the particles motion
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1
2Initial
Displacement‘2’
MaximumDisplacement
x=2.25@ t=0.5
Return toOrigin ‘0’
@ t=2
x
t
