Question

Derivative question, picture posted with question. Please help figure out how to do this. Grade 12 calc.

Answer 1

Part B)ii) is the part I can't solve

Answer 2

I got a(1/2)^5 as the amount for the substance to reach 1/32

Answer 3

after 0 days 1/1th
after 20 days 1/2
after 40 days 1/4
etc

Answer 4

So that is 20*(1/2)^n

To get the rate of decay you have to derive the function and then use the answer of question B)i) and enter it in the derivative function

Answer 5

I got 5a/16

Answer 6

\[-5\log(2) \times 2^{2-n}\]

Answer 7

which rule did you apply there?

Answer 8

hey jdoe

Answer 9

Really struggling with this question :/

Answer 10

for b.i. my answer is 100 years

Answer 11

but i fugred that out because (1/2)^x=(1/32)

Answer 12

and x=5 multipled by half life of 20

Answer 13

hey agent smith

Answer 14

Did you get the equation for the rate of decay first? You'll prob need it for part b ii

Answer 15

no i didnt

Answer 16

i figured it would be a(1/2)^n/20

Answer 17

but im kinda stumped now i have been looking at this question for over an hour

Answer 18

\[\Large A = A_o e^{-kx}\] You should be using an equation something like this. k is your rate of decay.

After 20 days (ie x=20), A is equal to 0.5Ao, so plug those in and find k.

Answer 19

In our book it says A=A(initial amount)^(t/(half life rate))

Answer 20

Oh i think i was thinking of exponential decay. \[\Large A = A_o ^{t/20}\] find the derivative of this function then

Answer 21

actually it should probably be\[\Large A = A_o ^{t/r}\] where r is the half life rate, which you find with the 20 days.

Answer 22

so i get a(1/2)^5

Answer 23

which becomes 5a/16

Answer 24

is this correct?

Answer 25

Are you sure this is the correct equation?
\[\huge A = A_o ^{\frac{ t }{ r }} \]

Answer 26

A(1/2)^100/20

Answer 27

And then i took a derivative of that

Answer 28

Is r actually the half-life of 20 days, or is it a half life rate?

Answer 29

half life rate is 20 days

Answer 30

i need the rate of decay at 100 days

Answer 31

Then you can't use half A., only 1/32 remains.

A=A(initial amount)^(t/(half life rate)) and this formula seems to not be the one you're using... when you post things like A(1/2)^100/20, that is not the same as (0.5A)^100/20. A(1/2)^100/20 means A*(1/2)^100/20, ie A doesn't have the exponent.

Answer 32

ah thats starting to make sense

Answer 33

(a/2)^5 would make more sense

Answer 34

but wait i still get the same aswer

Answer 35

That still doesn't look right...

Answer 36

or no i shouldnt actually

Answer 37

Man why is this so difficult lol

Answer 38

suri, any ideas?

Answer 39

\[\huge A= A_o \left( \frac{ 1 }{2 } \right) ^ {t/r}\] is this the equation in the book?

Answer 40

B.ii) is the question

Answer 41

yes
identical

Answer 42

OH wait so its 20/100???

Answer 43

Well that's nothing like what you said earlier...A=A(initial amount)^(t/(half life rate)) is nothing like that.

Answer 44

no no its not 100/20 is correct

Answer 45

it is tho

Answer 46

half life rate=r=20

Answer 47

t=100

Answer 48

A=intitial amount

Answer 49

A(1/2) to the exponent of 100 over 20

Answer 50

Those are two completely diff equations...
the first is A=A(initial amount)^(t/(half life rate)) or \[\huge A = (A_o)^{t/r}\]
this is the one i just posted\[\huge A= A_o \left( \frac{ 1 }{2 } \right) ^ {t/r} \]

Answer 51

Oh sorry that was my mistake.

Answer 52

The second one is the one I'm talking about

Answer 53

\[A(1/2)^5 \]

Answer 54

I need the derivative of that

Answer 55

This is why you gotta try learning to use the equation editor... hand written equations don't read well.

\[\huge A= A_o \left( \frac{ 1 }{2 } \right) ^ {t/r} \] looks like you'll have to differentiate this then. Post it as a new question, this is way too long now. And running too slow.

Answer 56

k thx

Answer 57

and medals help too :P @unprovoked

Answer 58

i have tried twice every time sight crashes.

Answer 59

\[assume that x _{0}be the initial mass and x be the mass after time t.\]

Answer 60

Then rate of decay of substance
\[\frac{ dx }{dt }\alpha x or \frac{ dx }{dt }=kx\]

Answer 61

separating the variables and integrating
\[\int\limits \frac{ dx }{x }=\int\limits k dt or \ln x=kt+c\]

Answer 62

when t=0,
\[x=x _{0}\]
\[\ln x _{0}=0+c\]