Download - Applications of Calculus - Contents
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Applications of Calculus - Contents
1.Rates 0f Change2.Exponential Growth &
Decay3.Motion of a particle4.Motion & Differentiatio
n5.Motion & Integration
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Rates of ChangeThe 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 + 3t2The 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 + 3t2The 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 DecayA 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 = loge166.70.0507t 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 - Examplex
tt1 t2 t4 t5 t6 t7t3
When is the particle at the origin?
t3 & t6
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Motion of a particle - Examplex
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 - Examplex
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
Velocity
Accelerationdtdvx =&& 2
2
dtxd=
x
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Motion and Differentiation - ExampleDisplacement x = -t2
+ t +2 in cm.Find initial velocity (in cm/s).
Initially t = 0
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Motion and Differentiation - ExampleDisplacement x = -t2
+ t + 2Show acceleration is constant.
Acceleration-2 units/s2
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Motion and Differentiation - ExampleDisplacement x = -t2
+ t +2Find when the particle is at the origin.
Origin @ x = 0
&
@ Origin when
t = 2 sec
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Motion and Differentiation - Example
0=dtdx
Displacement x = -t2 + t +2
Find the maximum displacement from origin.Maximum
Displacementwhen v = 012 =+−= t
dtdx
12 −=− t
5.0=t
22 ++−= ttx
25.0)5.0( 2 ++−=x
cmx 25.2=
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Motion and Differentiation - ExampleDisplacement x = -t2
+ t +2Sketch the particles motion
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1
2Initial
Displacementx=2 @ t=0
MaximumDisplacement
x=2.25@ t=0.5
Return toOrigin ‘0’x=0 @ t=2
x
t
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Motion and Integration
Acceleration
Velocity ∫= dtav .
Displacement
a
∫= dtvx .
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Motion and Integration - ExampleVelocity v= 3t2
+ 2t + 1, xo=-2cmFind displacement after 5 secs.
∫= dtvx .
∫ ++= dttt ).123( 2
Ctttx +++= 23
When t=0, x=-2C+++=− 0002 23
2−=C
223 −++= tttxWhen t=5
2555 23 −++=2525125 −++=
153=Displacement is 153cm
to right of origin.
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Motion and Integration - Example
Find displacement after 9 secs.Acceleration a= 6 - , vo=0, xo=+1m2
(t + 1)2
∫= dtav .
( )∫ ⎟⎠⎞⎜
⎝⎛
+−= dtt
.126 2
( )( )∫ −+−= dtt .126 2
( ) Ctt +−+−=
−
1126
1
When t=0, v=0
Ct
t ++
+=)1(
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( ) C+++= )10(2060
C+=20 2−=⇒ C
2)1(26 −+
+=t
tv
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Motion and Integration - Example
Find displacement after 9 secs.Acceleration a= 6 - , vo=0, xo=+1m2
(t + 1)2
2)1(26 −+
+=t
tv
∫= dtvx .
dtt
t .2)1(26∫ ⎟
⎠⎞⎜
⎝⎛ −
++=
cttt e +−++= 2)1(log23 2
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Motion and Integration - Example
Find displacement after 9 secs.Acceleration a= 6 - , vo=0, xo=+1m2
(t + 1)2
ctttx e +−++= 2)1(log23 2
When t=0, x=1
ce +−++= )0(2)10(log2)0(31 2 1=⇒ c
12)1(log23 2 +−++= tttx e When t=91)9(2)19(log2)9(3 2 +−++= ex
)10(log2226 e−=( ))10(log1132 e−= Displacement