specialist maths differential equations week 3. problem solving with separable differential...
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SPECIALIST MATHSDifferential Equations
Week 3
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Problem Solving with Separable Differential Equations
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Example 1 (Ex 8E1)• Fluid is flowing through thick walled tube.
• It maintains the temperature on the inner wall of the pipe at 600oC.
• Heat is lost through the tube per unit length according to the formula:
• Calculate the external temperature of the tube.
2.0 and 680 where2 kqdr
dTkrq
0.02m0.04m
r
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Solution to Example 1
dr
dTkrq 2
kr
q
dr
dT
2
rk
q
dr
dT 1
2
rdr
dT 1
2.02
680
rdr
dT 11700
drr
T11700
crT ln1700
2.0 ,680 kq
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Solution to Example 1 continued
crT ln1700
02.0 when 600 rCT o
c 02.0ln1700
600
02.0ln1700
600
c
02.0ln1700
600ln1700
rT
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Solution to Example 1 continued again
02.0ln1700
600ln1700
rT
04.0 when ? rT
02.0ln1700
60004.0ln1700
T
CT o9.224
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Example 2 (Ex 8E2)• The rate of decay of a radioisotope is
proportional to the number of particles present.
• If the number of particles decreases by 10% after 50 years, find the percentage left after 200 years.
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Solution to example 2present particles ofnumber toalproportion isdecay of Rate
kNdt
dN therefore
kdt
dN
N
1
kdtdtdt
dN
N
1
kdtdN N
1
cktN lnckteN
ckteeN ktceeN
c
kt
eA
AeN
where
,
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Solution to example 2 continued0 be particles ofnumber original let the N
0 ,0 when ie NNt ktAeN
00 AeN
AN 0
kteNN 0 therefore
09.0 ie 10%,by
decreases ,50when
NN
Nt
50000.9 therefore keNN
ke500.9
500.9 ke
501
(0.9)ke
50)(0.9 ke
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Solution to example 2 continued again
500 )9.0( therefore
tNN
kteNN 0
200when t
40 9.0NN
656.00NN
0656.0 NN
remaining 65.6% ie
501
(0.9)ke
50200
0 )9.0(NN
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Example 3 (Ex 8E2)• A man falls out of a plane and accelerates
toward the earth with an acceleration given by
• Terminal velocity is a constant velocity he will eventually reach.
• Calculate his terminal velocity.
1-ms 5
8.9v
a
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Solution to Example 3
5
8.9v
dt
dv
495 vdt
dv
149
5
dt
dv
v
149
5
dtdt
dt
dv
v
149
5
dtdv
v
ct49ln1
51
v
1 ct49ln5 v
c5
49ln t
v
49 5c
t
ev
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Solution to Example 3 continued
49 5c
t
ev
v49 5
t
Ae
49 5
tceev
49 5
t
Aev
0 when 0 tv
490 0Ae
490 A
49A
4949 5
t
ev
0 , tas
occurs velocity terminal
5
t
e
049 v
49v
-1ms 49 velocity terminal
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Example 4 (Ex 8E2)• The rate at which temperature changes is
proportional to the differences between its temperatures.
• A hot metal bar is 700oC at 12:00pm when taken out of the furnace. Its temperature falls to 200oC at 2:00pm.
• If the air temperature is 20oC, calculate its temperature at 3:00pm.
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Solution to Example 4
rTTdt
dT
rTTdt
dTk
k1
dt
dT
TT r
k1
dt
dT
TT r
kdt1
dtdt
dT
TT r
kdt1
dTTT r
cktTT r ln
cktr eTT
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Solution to Example 4ckt
r eeTT ktc
r eeTT kt
r AeTT
700 ,0when
and 20
Tt
Tr
020700 Ae
680A
kteT 68020
200 ,2when Tt
268020200 ke
ke2680180
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Solution to Example 4 continued
ke2680180
kteT 68020
ke2
680
180
234
9 ke
21
34
9
ke
tkeT
68020 2
34
968020
t
T
? ,3when Tt
23
34
968020
T
CT 061.112
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Growth and Decay Problemsky
dx
dy form theof equations aldifferenti are These
kdx
dy
y
1
kdxdxdx
dy
y
1
kdxdyy
1
ckxy ln
ckxey
ckxeey
kxAey
problem.decay 0
problem,growth 0,For
k
k
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Growth & Decay GraphskxAey kxAey
growth. 0,For k decay. 0,For ky
x
y
x0y
0y
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Logistic Differential Equations• The function starts off growing exponentially,
then flattens out approaching a limiting value “A”.
• This happens with populations as the recourses available to support it are limited.
A
0P
P
t
)(P t
P)(' P kt
A
P1)(' P kt
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Logistic Differential Equations• Logistic differential equations are of the form:
• When P is small, an exponential growth.
• When a horizontal line.
A
PkP
dt
dP1
kPdt
dP
A
P ,0
0 ,0P
A-1 ,
dt
dPAP
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A Handy Algebraic Manipulation
PAP
11 Proof
PAP
PPA
PAP
A
PAP
11
PAP
A
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Example 5 (Ex 8F)
The relative growth rate of snakes on Groute Island is given by:
(a) What is the environments carrying capacity?
(b) Solve the differential equation if there are 2280 snakes originally.
(c) When would you expect the population to reach 4000 snakes.
51601
100
1 PP
dt
dP
A
PkP
dt
dP1
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Solution to Example 5
. is population limiting 1For AA
PkP
dt
dP
snakes 5160 is population limiting the
51601
100
1For (a)
PP
dt
dP
51601
100
1 (b)
PP
dt
dP
5160
5160
100
1 PP
dt
dP
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5160
5160
100
1 PP
dt
dP
5160
5160
100
11 P
dt
dP
P
Pdt
dP
P 5160
100
15160
100
1
5160
5160
dt
dP
PP
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100
1
5160
5160
dt
dP
PP
PP
5160
11
PP
PP
5160
5160
PP
5160
5160
100
1
5160
11 therefore
dt
dP
PP
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dtdtdt
dP
PP 100
1
5160
11
dtdPPP 100
1
5160
11
ctP
P
100
1
1
5160lnln
ctPP 100
15160lnln
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ctP
P
100
1
5160ln
cte
P
P
100
1
5160
ctee
P
P 100
1
5160
ctee
P
P
100
1
15160
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ctee
P
P
100
1
15160
ctee
P
P
P
100
15160
tbe
P100
1
15160
tbe
P100
1
15160
tbe
P100
1
15160
tbe
P
100
1
1
1
5160
tbe
P100
1
1
5160
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tbe
P100
1
1
5160
2280 ,0when Pt
01
51602280
be
b
1
51602280
516012280 b
516012280 b
516022802280 b
228051602280 b
28802280 b
2280
2880b
263.1b
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tbe
P100
1
1
5160
te
P100
1
263.11
5160
? ,4000 when (c) PP
te 100
1
263.11
51604000
te 100
1
263.11
51604000
5160263.114000 100
1
te
4000
5160263.11 100
1
t
e
29.1263.11 100
1
t
e
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29.0263.1 100
1
t
e
263.1
29.0100
1
t
e
23.0100
1
t
e
23.0ln100
1 t
23.0ln100
1 t
23.0ln100t
years 147t
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Example 6 (Ex 8F)
Returning from a visit to a remote planet, Captain Boss brings with him a rare disease that infects the crew of the ship, spreading according to a logistic equation.
(1) If there are 126 crew members write a differential equation governing the spread.
(2) After 3 days 15 members have the disease, write a formula for the spread in terms of t.
(3) When is the rate of infection greatest.
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(1)
1261
NkN
dt
dN
cktNN 126lnln
126
126 NkN
dt
dN
kdt
dN
NN
126
126
kdt
dN
NN
126
11
kdtdtdt
dN
NN 126
11
kdtdNNN 126
11
cktN
N
126ln
ckteN
N 126
ckteN
N 126
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ckteN
N
1126
ckteN
N
N
126
ckteeN
1126
ktbeN
1126
ktbe
N
1
1
126
ktbeN
1
126
(2) When t = 0, N = 1
01
1261
be
1261 b125b
kteN
1251
126
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When t = 3, N = 15
ke 31251
12615
126125115 3 ke
4.81251 3 ke
4.7125 3 ke
0529.03 ke
0529.03 ke
0529.0ln3 k
0529.0ln3
1k
942.0k
teN
942.01251
126
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(3) When is the rate of infection greatest.
126
2kNkN
dt
dN
dt
dNkN
dt
dNk
dt
Nd
126
22
2
dt
dNkNk
dt
Nd
632
2
1261
NkN
dt
dN
126
2kNkN
dt
dN
0 when occurs this2
2
dt
Nd
dt
dNkNk
630
630
kNk
kkN
63
k
kN
63
63N
63hen greatest w is rateinfection N
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te 942.01251
12663
126125163 942.0 te
21251 942.0 te
1125 942.0 te
008.0942.0 te
008.0ln942.0 t
008.0ln942.0 t
008.0ln942.0
1t
days 1.5t
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This Week
• Text book pages 294 – 307.
• Exercise 8E2 Q1 – 12
• Exercise 8F Q1 - 4.
• Review Sets 8A -8D