Your working looks good, I get.
First you separate the variables
now integrate
is just a constant so
For
now solve for
The problem I am having with this is that I learned how to solve this type of problem a long time ago and when I was taught it they used another type of formula with different letters so I am kind of lost: (Can you tell me what I got write and what I did wrong if the answer is incorrect-Thanks)
Here is the question:
I am given the equation dA/dt=-kA
A(t) = the amount of material present(in grams)
t = time (minutes)
k= some positive constant
a.) Derive an equation for the amount A(t) present at time t in terms of the constant k and the amount A(0) present at time t=0
This is what I did: (However I used the letters I learned and so would this be correct)
(- dN/N)= λ * dt
So,
dN(t)/dt =-λ N(t)
dN(t)/N(t)= -λ dt
In N(t)=-λ dt
In N(t)= -λ t+C
N(t)=e^C e^-λt= Noe^-λ t (the o is between the N and e but on the bottom)
(did I do the problem correctly- if not what steps am I missing - Thanks)
b.) If A(5)= 1/3 A(3), find k
(What would I do to find the solution to this problem- could you tell me the steps/formula to take- Thanks)
c.) At what time t will the amount A(t) be 1/4 A(0)
(Could you help me with this problem as well I don't now where ton start- could you show me the steps to take- Thanks)
Thank you for all the help I really appreciate it
I get that for a.) the equation will be A= Aoe^-kt (The o is meant to be between the A and e)
However, I am still confused about part b.) If A(5)= 1/3 A(3), find k
and
c.)At what time t will the amount A(t) be 1/4 A(0)
(What part was the A(5) = 1/4 for and the A= Aoe^-kt --> 1/4 = Aoe^-5k for?)
Are you saying that for question c.) the answer is that when t is 5 it will be 1/4
How do you solve for k and what part is this question to b.) or c.)
Thanks for the help