12x1 t04 03 further growth & decay (2011)
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Further Growth & Decay
Further Growth & DecayIn order to take into account other conditions (e.g. the temperature of an object will never be lower than the temperature of its surroundings), we need to change the equation.
Further Growth & Decay
NPkdtdP
In order to take into account other conditions (e.g. the temperature of an object will never be lower than the temperature of its surroundings), we need to change the equation.
Further Growth & Decay
NPkdtdP
In order to take into account other conditions (e.g. the temperature of an object will never be lower than the temperature of its surroundings), we need to change the equation.
NP
the solution to the equation is;
(trivial case)
Further Growth & Decay
NPkdtdP
In order to take into account other conditions (e.g. the temperature of an object will never be lower than the temperature of its surroundings), we need to change the equation.
NP
P = population at time t
N + A = initial population
k = growth(or decay) constant
t = time
the solution to the equation is;
(trivial case)OR
ktAeNP
e.g.(1993) Let T be the temperature inside a room at time t and let A be the constant outside air temperature.Newton’s law of cooling states that the rate of change of the temperature is proportional to (T – A).
e.g.(1993) Let T be the temperature inside a room at time t and let A be the constant outside air temperature.Newton’s law of cooling states that the rate of change of the temperature is proportional to (T – A).
cooling. of law sNewton' satisfies constants) are and (where that Show a) kCCeAT kt
e.g.(1993) Let T be the temperature inside a room at time t and let A be the constant outside air temperature.Newton’s law of cooling states that the rate of change of the temperature is proportional to (T – A).
cooling. of law sNewton' satisfies constants) are and (where that Show a) kCCeAT kt
ktCeAT
e.g.(1993) Let T be the temperature inside a room at time t and let A be the constant outside air temperature.Newton’s law of cooling states that the rate of change of the temperature is proportional to (T – A).
cooling. of law sNewton' satisfies constants) are and (where that Show a) kCCeAT kt
ktCeAT
ktkCedtdT
e.g.(1993) Let T be the temperature inside a room at time t and let A be the constant outside air temperature.Newton’s law of cooling states that the rate of change of the temperature is proportional to (T – A).
cooling. of law sNewton' satisfies constants) are and (where that Show a) kCCeAT kt
ktCeAT
ktkCedtdT
but ktT A Ce
e.g.(1993) Let T be the temperature inside a room at time t and let A be the constant outside air temperature.Newton’s law of cooling states that the rate of change of the temperature is proportional to (T – A).
cooling. of law sNewton' satisfies constants) are and (where that Show a) kCCeAT kt
ktCeAT
ktkCedtdT
dT k T Adt
but ktT A Ce
e.g.(1993) Let T be the temperature inside a room at time t and let A be the constant outside air temperature.Newton’s law of cooling states that the rate of change of the temperature is proportional to (T – A).
cooling. of law sNewton' satisfies constants) are and (where that Show a) kCCeAT kt
ktCeAT
ktkCedtdT
ATkdtdTCeAT kt satisfies
dT k T Adt
but ktT A Ce
hour.an half in 17 to20 from drop tore temperaturoom inside thecauses
breakdown system heating a and 5 is atureair temper outside The b)CC
C
?10 toequal re temperaturoom inside theis hoursmany howAfter C
ktCeT 5
hour.an half in 17 to20 from drop tore temperaturoom inside thecauses
breakdown system heating a and 5 is atureair temper outside The b)CC
C
?10 toequal re temperaturoom inside theis hoursmany howAfter C
ktCeT 5
kteTC
Tt
15515
20,0when
hour.an half in 17 to20 from drop tore temperaturoom inside thecauses
breakdown system heating a and 5 is atureair temper outside The b)CC
C
?10 toequal re temperaturoom inside theis hoursmany howAfter C
ktCeT 5
kteTC
Tt
15515
20,0when
ke
Tt5.015517i.e.17,5.0when
hour.an half in 17 to20 from drop tore temperaturoom inside thecauses
breakdown system heating a and 5 is atureair temper outside The b)CC
C
?10 toequal re temperaturoom inside theis hoursmany howAfter C
ktCeT 5
kteTC
Tt
15515
20,0when
ke
Tt5.015517i.e.17,5.0when
1215 5.0 ke
hour.an half in 17 to20 from drop tore temperaturoom inside thecauses
breakdown system heating a and 5 is atureair temper outside The b)CC
C
?10 toequal re temperaturoom inside theis hoursmany howAfter C
ktCeT 5
kteTC
Tt
15515
20,0when
ke
Tt5.015517i.e.17,5.0when
1215 5.0 ke
1512log2
1512log5.0
15125.0
k
k
e k
hour.an half in 17 to20 from drop tore temperaturoom inside thecauses
breakdown system heating a and 5 is atureair temper outside The b)CC
C
?10 toequal re temperaturoom inside theis hoursmany howAfter C
kteT 15501 ,10when
kteT 15501 ,10when
31log1
31log
31515
kt
kt
e
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kt
kt
kteT 15501 ,10when
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31log
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46167.2t
kteT 15501 ,10when
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kt
46167.2tC10 todropped has re temperatu thehours
212After
Exercise 7H; 2, 4, 5, 8, 9, 10
kteT 15501 ,10when
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31515
kt
kt
e
e
kt
kt
46167.2tC10 todropped has re temperatu thehours
212After