• Aug 23rd 2010, 03:20 PM
bookworm247
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
• Aug 23rd 2010, 03:30 PM
pickslides
Your working looks good, I get.

$\displaystyle \displaystyle \frac{dA}{dt}= -kA$

First you separate the variables

$\displaystyle \displaystyle \frac{dA}{A}= -k ~dt$

now integrate

$\displaystyle \displaystyle \int \frac{dA}{A}= \int-k ~dt$

$\displaystyle \displaystyle \ln A= -kt +C$

$\displaystyle \displaystyle A= e^{-kt +C}$

$\displaystyle \displaystyle A= e^{-kt} e^C$

$\displaystyle \displaystyle e^C$ is just a constant so $\displaystyle \displaystyle e^C = A_0$

$\displaystyle \displaystyle A= A_0e^{-kt}$

For $\displaystyle \displaystyle A(5)= \frac{1}{4}$

$\displaystyle \displaystyle A= A_0e^{-kt}\implies \frac{1}{4}= A_0e^{-5k}$

now solve for $\displaystyle k$
• Aug 24th 2010, 01:46 PM
bookworm247
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 :)
• Aug 24th 2010, 01:51 PM
pickslides
Quote:

Originally Posted by pickslides

$\displaystyle \displaystyle A= A_0e^{-kt}$

For $\displaystyle \displaystyle A(5)= \frac{1}{4}$

$\displaystyle \displaystyle A= A_0e^{-kt}\implies \frac{1}{4}= A_0e^{-5k}$

now solve for $\displaystyle k$

Here's k

$\displaystyle \displaystyle \frac{1}{4}= A_0e^{-5k}$

$\displaystyle \displaystyle \frac{1}{4A_0}= e^{-5k}$

$\displaystyle \displaystyle \ln \frac{1}{4A_0}= -5k$

$\displaystyle \displaystyle k= \frac{-1}{5}\ln \frac{1}{4A_0}$
• Aug 24th 2010, 02:09 PM
bookworm247
So this is the answer to part b and c?

Thank you so much for the help :)
• Aug 25th 2010, 02:06 PM
bookworm247
I get that for part c.) of my question the answer is that when t is 5 it will be 1/4 (that is correct right?)

But for part b.) is the answer k=-1/5 ln 1/4Ao (the o is below the A) Is this correct?

Thanks
• Aug 25th 2010, 03:16 PM
pickslides
The $\displaystyle A_0$ is all one number. It means the initial amount of $\displaystyle A$

Your understanding in part c) is correct.