Exponential Growth and Decay (1 Viewer)

[nick1048]

This is pathetic, but I don't get these concepts, if anyone gets what to do can u please answer these questions for me? Thanks in advance, it's really irritating me atm:

1) A heated body is cooling and the excess of its temperature, @, above that of its surroundings is @ = A . e^-kt where t is time in minutes.
a) If the temperature of the surroundings is 20 degrees Celcius and the body cools to 70 degrees celcius in 10 minutes, find:
i) its temperature after 20 minutes,
ii) The time taken to reach 60 degrees celcius.

2) The rate of decay of a radioactive element is proportional to the amount of the element present at any time. If one half of a given quantity of the element disintegrates in 1600 years, what percentage will disintegrate in 100 years?

3) The number of bacteria (N) in a colony after t minutes is given by
N = 10000 e ^0.05t
Find the rate at which the colony increases when N = 20000

For some reason these things seem ridiculously easy and I'm just missing a crucial component. If there are any hints or tricks to doing these things effectively I would be more than greatful to know them. Thanks for your effort.

nick1048

EDIT: Ok judging from the response maybe these arent so so easy... *sigh*

 

[SaHbEeWaH]

haven't done this topic for a long time ...

2) The rate of decay of a radioactive element is proportional to the amount of the element present at any time. If one half of a given quantity of the element disintegrates in 1600 years, what percentage will disintegrate in 100 years?

N = Ae^kt
the initial quantity is a whole unit
after one year, 1/2 disintegrates, so, N = 0.5, A = 1

0.5 = 1e^1600k
k = ln0.5/1600
N = e^{100(ln0.5/1600)} = 0.95760328
therefore 95.76% remains after 100 years, 4.24% disintegrates in 100 years

 

[SaHbEeWaH]

3) The number of bacteria (N) in a colony after t minutes is given by
N = 10000 e ^0.05t
Find the rate at which the colony increases when N = 20000

find the time when N = 20000
20000 = 10000e^0.05t
2 = e^0.05t
t = ln2/0.05 = 13.86

N = 10000e^0.05t
dN/dt = 500e^0.05t

when t = 13.86,
dN/dt = 500e^0.693 = 999.85

therefore when N = 20000, the colony is increasing at 1000 bacteria/min

 

[Slidey]

1) A heated body is cooling and the excess of its temperature, @, above that of its surroundings is @ = A . e^-kt where t is time in minutes.
a) If the temperature of the surroundings is 20 degrees Celcius and the body cools to 70 degrees celcius in 10 minutes, find:
i) its temperature after 20 minutes,
ii) The time taken to reach 60 degrees celcius.

Sure you're not missing something? Like the initial temperature?

 

[monique66]

i suppose you could assume it was at 100 when t=0

 

[Slidey]

You could try that to get at least some marks, however I highly doubt that'd get you full marks.

 

[monique66]

possibly, but how else do you do it? You have no intial tempterature to approximate A with, without that what are you mean to do? Unless there is something wrong with the question, that or me...

 

[Slidey]

I believe there's something wrong with the question.

 

[Jago]

I really enjoyed exponential growth and decay last year, it just made sense to me

 

[Slidey]

Ah. Got a worked solution?

 

[nick1048]

sure slide:

b i)
@ = 70 - 20 (We want the excess temperature minus the surrounding temperature)
= 50

.'. 50 = 80 e^10k
5/8 = e^10k
(ln 5/8) / 10 = k
.'. k = -0.047...... (store all the decimals)

Now, when t = 20

@ = 80e^20k
= 31.25 (Now add back the surrounding temperature of 20 degrees.)
= 31.25 + 20
= 51.25 degrees celcius.

ii) @ = 60 - 20 (following that same principle of subtracting the surrounding temperature)
= 40

40 = 80e^kt
(ln0.5) / k = t
.'. t = 14.8 minutes

Thanks everyone for your help with this section, I understand the key to these questions now

 

[acmilan]

sure slide:

b i)
@ = 70 - 20 (We want the excess temperature minus the surrounding temperature)
= 50

.'. 50 = 80 e^10k
5/8 = e^10k
(ln 5/8) / 10 = k
.'. k = -0.047...... (store all the decimals)

Now, when t = 20

@ = 80e^20k
= 31.25 (Now add back the surrounding temperature of 20 degrees.)
= 31.25 + 20
= 51.25 degrees celcius.

ii) @ = 60 - 20 (following that same principle of subtracting the surrounding temperature)
= 40

40 = 80e^kt
(ln0.5) / k = t
.'. t = 14.8 minutes

Thanks everyone for your help with this section, I understand the key to these questions now

Where did you get 80 from (im assuming thats the initial temp?). It wasnt in your initial question?

 

[nick1048]

=\ fuck yeah 80 was the initial temp from like the a section or something... sorry for wasting everyones time, I get it now lol thanks for all your help