Question

initial population is 500 increases by 15% in ten years

Answer 1

just like in 162 your job is to find \(A(t)=A_0e^{kt}\)
\(A_0\) you are told, only need to find \(k\)

Answer 2

\[A_0=500\] the initial amount

Answer 3

ok

Answer 4

to find \(k\) note that in 10 years you increase by 15% so have \(1.15\times 500=575\) then to find \(k\) set \[500e^{10k}=575\] and solve for \(k\)

Answer 5

note that the 500 didn't really matter because first step is to divide by 500 and get \[e^{10k}=1.15\]

Answer 6

so \[10k=\ln(1.15\] making \[k=\frac{\ln(1.15)}{10}\]

Answer 7

not that we did not mention differential equations at all

Answer 8

oh ok

Answer 9

so anytime there is a percent then it should be like 1.25 if 25%?

Answer 10

right

Answer 11

does this ring a bell?

Answer 12

YES

Answer 13

now here is the diffeq part at the beginning
saying the population grows at a rate proportional to the present amount is the same as saying \[\frac{dA}{dt}=kA\]

Answer 14

separating we get \[\frac{dA}{A}=kdt\] integrating we get \[\int \frac{dA}{A}=\int kt\]\[\ln(A)=kt+c\]
so \[A(t)=e^{kt+c}=Ce^{kt}\]

Answer 15

we couldn't say that in 162, just took a different approach

Answer 16

Right so its just like a separable equation?

Answer 17

exactly

Answer 18

we still didn't answer the question but now that we have the mathematical model we can do it

Answer 19

\[A(t)=500e^{.01376t}\] making \[A(30)=500e^{.01376\times 30}\]

Answer 20

and then we solve right?

Answer 21

solve aka use a calculator

Answer 22

Right right

Answer 23

btw if the grows rate was 8% in 12 years then \[k=\frac{\ln(1.08)}{12}\] and so on

Answer 24

ohhh that pattern is easy

Answer 25

next

Answer 26

What about decay? is it just a minus sign?

Answer 27

hell no

Answer 28

oop.

Answer 29

if it decreases by 12% every 5 years then \[k=\frac{\ln(88)}{5}\]

Answer 30

actually \[\frac{\ln(.88)}{5}\]

Answer 31

because 100% - 12% = 88% = 0.88

Answer 32

Right right. Next

Answer 33

5?

Answer 34

Sure

Answer 35

half life is 3.3 hours start with one gram
model is \[P=1\times e^{kt}\] we need k

Answer 36

in 3.3 hours you have one half of what you started with
set \[e^{3.3k}=.5\] solve for \(k\)

Answer 37

so \[k=\frac{\ln(.5)}{3.3}\]

Answer 38

if the half life what 28 minutes then it would be \[k=\frac{\ln(.5)}{28}\]

Answer 39

half life means your left with half, 50% decay

Answer 40

notice again no diffeq, but the beginning part is identical to the beginning part of question 3, i see no real reason to repeat it

Answer 41

we still did not answer the question though
it it clear how to proceed?

Answer 42

So its always going to be .5?

Answer 43

for half life yes because... \(.5=\frac{1}{2}\)
but only for half life

Answer 44

lol thanks

Answer 45

do you know how to finish the problem?

Answer 46

Plug and chug?

Answer 47

how long will it take for 90% of it to decay?

Answer 48

i am going to spend 60 seconds to get a snack
when i come back you will have the answer

Answer 49

back

Answer 50

Ce^((ln(.1))*t)

Answer 51

hell no

Answer 52

You didn't give me parameters

Answer 53

model is \[P(t)=e^{kt}\] we found k already , what is it?

Answer 54

scroll up

Answer 55

k=ln(.5)/28

Answer 56

?

Answer 57

keep scrolling up

Answer 58

k=ln(.5)/3.3

Answer 59

right
what is that as a decimal?

Answer 60

-.21

Answer 61

ok so model is \[P(t)=e^{-.21t}\] when 90%is gone what percent is left?

Answer 62

.1

Answer 63

right solve \[e^{-.21t}=.1\] and you are done

Answer 64

Wait so this is half life and decay?

Answer 65

same thing half life is just a specific kind of decay , where you are told how long something takes to decay by one half

Answer 66

oh ok

Answer 67

you could decay by 10% every 8 year but half life means you decay by 50% in say 8 years

Answer 68

oh ok

Answer 69

you still didn't answer the question

Answer 70

oh!

Answer 71

\[e^{-.21t}=.1\] solve for \(t\)

Answer 72

t = (ln(.1))/-.21

Answer 73

ok next

Answer 74

One more

Answer 75

and then i'll annoy you tomorrow

Answer 76

which one?

Answer 77

You pick

Answer 78

9 takes two seconds
it is going to be \ where \[e^{3k}=.25\] solve for k lets try another

Answer 79

k = ln(.25)/3

Answer 80

next

Answer 81

11 do yourself

Answer 82

look it up real easy

Answer 83

13 is newton ,but it it identical to decay with one slight difference, the quantity that decays is the difference in the temperatures, not the actual temperature

Answer 84

it is 70 degrees put in a 10 degree environment, then 1/2 a minute later it is 50 degrees

translate as:
the initial difference was 60 1/2 a minute later the difference is 40

Answer 85

how is it 40?

Answer 86

Did you solve for it?

Answer 87

you are confused let me make another thread

Answer 88

ok