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Applications of Calculus - Contents
1. Rates 0f Change
2.Exponential Growth &
Decay
3.Motion of a particle
4.Motion & Differentiation
5.Motion & Integration
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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 Change
constant.
Rate of Change
varies.
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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
i.e dV = 4 + 3t2
0=0+0+C C=0
dt
3
@ t = 12
V
=
4t
+
t
2
 V = (4 + 3t ).dt
3
=
4
x
12
+
12
= 4t + t3 + C
= 1776 units3
∫
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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
i.e dV = 4 + 3t2
0=0+0+C C=0
dt
3
@ t = 12
V
=
4t
+
t
2
 V = (4 + 3t ).dt
3
=
4
x
12
+
12
= 4t + t3 + C
= 1776 units3
∫
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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)
B = 2t4 - t2 + 2000
= 2(0)4 – (0)2 + 2000
= 2000 bacteria
c) Rate of growth after 5
hours. (t =5)
dB = 8t3 -2t
dt
= 8(5)3 -2(5)
b) Bacteria after 5
= 990 bacteria/hr
hours. (t =5)
= 2(5)4 – (5)2 + 2000
= 3225 bacteria
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Exponential Growth and Decay
A Special Rate of Change.
Eg Bacteria, Radiation, etc
It can be written as dQ = kQ
dt
Initial Quantity
dQ = kQ can be solved as Q = Aekt
dt
Growth
Time
Growth Constant k
(k +ve = growth, k -ve = decay)
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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
9000 = 6000 e8k
e8k = 9000
6000
e8k = 1.5
1.5 =e8k
loge1.5 = loge e8k
= 8k loge e
= 8k
k = loge1.5
8
k ≈ 0.0507
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Growth and Decay – Example
Number of Bacteria given by N = Aekt
A = 6000, t = 48 hours, k ≈ 0.0507
b) 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 = 68 344
c) Rate bacteria increasing after 2 days
dN = kN = 0.0507 N
dt
= 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 ≈ 0.0507
d) When will the bacteria reach 1 000 000.
N = A ekt
1 000 000 = 6000 e0.0507t
e0.0507t =
1 000 000
6 000
e0.0507t = 166.7
logee0.0507t = loge166.7
0.0507t logee = loge166.7
0.0507t = loge166.7
t = loge166.7
0.0507
≈ 100.9 hours
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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.
dN = kN
dt
k is the growth constant
k = 0.0507 x 100%
= 5.07%
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Motion of a particle 1
Displacement (x)
Measures the distance from a point.
To the left is negative - To the right is positive
The Origin implies x = 0
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Motion of a particle 2
Velocity (v)
Measures the rate of change of displacement.
dx
v =
dt
To the left is negative - To the right is positive
Being Stationary implies v = 0
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Motion of a particle 3
Acceleration (a)
Measures the rate of change of velocity.
2x
dv
d
a =
= 2
dt dt
To the left is negative - To the right is positive
+v –a
-v +a
}
Slowing
Down
+v +a
-v -a
}
Speeding
Up
Having Constant Velocity implies a = 0
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Motion of a particle - Example
x
When is the particle at rest?
t 2 & t5
t1
t2
t3
t4
t5
t6
t7
t
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Motion of a particle - Example
x
When is the particle at the origin?
t 3 & t6
t1
t2
t3
t4
t5
t6
t7
t
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Motion of a particle - Example
x
Is the particle faster at t1 or t7?
Why?
t1
t2
t3
t4
t5
t6
t7
t
t1
Gradient
Steeper
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Motion of a particle - Example
x
Is the particle faster at t1 or t7?
Why?
t1
t2
t3
t4
t5
t6
t7
t
t1
Gradient
Steeper
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Motion and Differentiation
Displacement
Velocity
Acceleration
x
dx
x 
dt
2
dv d x

x 
dt dt 2
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Motion and Differentiation - Example
Displacement x = -t2 + t +2 in cm.
Find initial velocity (in cm/s).
x  t  t  2
dx
v  x 
dt
 2t  1
2
Initially t = 0
v  2  0  1
 1cm s
1
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Motion and Differentiation - Example
Displacement x = -t2 + t + 2
Show acceleration is constant.
v  2t  1
dv
a  x 
dt
 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.
x  t  t  2
Origin @ x = 0
2
0  t  t  2
2
0  (t  t  2)
0  (t  2)(t  1)
t  2 &t  1
2
@ Origin when
t = 2 sec
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Motion and Differentiation - Example
Displacement x = -t2 + t +2
Find the maximum displacement from origin.
Maximum Displacement
dx
when v =
0
dt
dx
 2t  1  0
dt
 2t  1
x  t 2  t  2
x  (0.5)2  0.5  2
x  2.25cm
t  0.5
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Motion and Differentiation - Example
Displacement x = -t2 + t +2
Sketch the particles motion
x
Maximum
Displacement
2
x=2.25
Initial
Displacement @ t=0.5
x=2 @ t=0
1
1
Return to
2
Origin ‘0’
x=0 @ t=2
t
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Motion and Integration
Acceleration
a
Velocity
v   a .dt
Displacement
x  v .dt
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Motion and Integration - Example
Velocity v= 3t2 + 2t + 1, xo=-2cm
Find displacement after 5 secs.
3
2

x

t

t
t  2
x  v .dt
When
t=5
2
  (3t  2t  1).dt
3
2

5

5
5 2
3
2
x  t t t  C
 125  25  5  2
When t=0, x=-2
 153
3
2
2  0  0  0 C
Displacement is 153cm
C  2
to right of origin.
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Motion and Integration - Example
2
Acceleration a= 6 - (t + 1)2, vo=0, xo=+1m
Find displacement after 9 secs.
v   a .dt
2 

  6 
.dt
2 
t  1 

2
  6  2t  1 .dt
1
t  1
 6t  2
C
1
2
 6t 
C
(t  1)
When t=0, v=0
2
0  60  
C
(0  1)
0  2  C  C  2
2
v  6t 
2
(t  1)
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Motion and Integration - Example
2
Acceleration a= 6 - (t + 1)2, vo=0, xo=+1m
Find displacement after 9 secs.
2
v  6t 
2
(t  1)
x  v .dt
2


   6t 
 2 .dt
(t  1)


 3t 2  2loge (t  1)  2t  c
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Motion and Integration - Example
2
Acceleration a= 6 - (t + 1)2, vo=0, xo=+1m
Find displacement after 9 secs.
When t=0, x=1
x  3t 2  2loge (t  1)  2t  c
1  3(0)2  2loge (0  1)  2(0)  c  c  1
x  3t 2  2loge (t  1)  2t  1 When t=9
x  3(9)2  2loge (9  1)  2(9)  1
 226  2loge (10)
 2113  loge (10) Displacement
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