12 x1 t04 03 further growth & decay (2012)

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

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kt

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kt

kt

kteT 15501 ,10when

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31log

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kt

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kt

kt

46167.2t

kteT 15501 ,10when

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31log

31515

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kt

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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

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kt

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kt

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212After