#### Scrooge2

##### Senior member
Okay this problem has been driving me nuts for a while now:

The decay of a radioactive material may be modeled by assuming that the amount A(t) of material present (in grams) at time t(minutes) decays at a rate proportional to the amount present: that is

dA/dt =-kA

for some positive constant k.

1. Derive an equation for the amount A9t) pressent at time t in terms of the constant k and the amount a(0) present at time=0

2. if A(5)=1/3 A(3), find k

3 At what time t will the amount A(t) be 1/4 A(0)?

#### Viper GTS

##### Lifer
Yikes, differential equations. I dropped out of that class. Twice.

I won't be much help. Viper GTS

#### qIat

##### Senior member
Originally posted by: Scrooge2
Okay this problem has been driving me nuts for a while now:

The decay of a radioactive material may be modeled by assuming that the amount A(t) of material present (in grams) at time t(minutes) decays at a rate proportional to the amount present: that is

dA/dt =-kA

for some positive constant k.

1. Derive an equation for the amount A9t) pressent at time t in terms of the constant k and the amount a(0) present at time=0

2. if A(5)=1/3 A(3), find k

3 At what time t will the amount A(t) be 1/4 A(0)?
My physics professor would be ashamed of me. We did quite a lot of work with this in spring semester but I've forgotten it now #### rgwalt

##### Diamond Member
First, solve the ODE to obtain an expression for A(t).

It should be

A(t) = A(0)*exp(-kt)

Where A(0) is the starting amount and exp is equivalent to e, the base of the natural log function.

2. Get an expression for A(5) and A(3). Plug the expressions into the equation A(5)=1/3 A(3) and solve for k

3. Plug 1/4 A(0) in for A(t) in the equation, and solve for t. I assume you have to use the k found in number 2, unless there is a trick here that I am forgetting.

Ryan

EDIT: The ODE in this case is very simple and can be solved by multiplying both sides by dt and dividing both sides by A, then integrating. You have to play around with the constant from the integral to get it to work out, but remember, a constant is a constant and can be manipulated as such.

#### Scrooge2

##### Senior member
okay i got:

A=A(initial) e^(-kt)

as the equation. Now what were you saying about plugging in a5 and a3?

#### rgwalt

##### Diamond Member
You get A(5) = A(0)*exp(-5k) and A(3) = A(0)*exp(-3k). Plug those into your equation A(5) = 1/3 A(3) to get

A(0)*exp(-5k) = 1/3*A(0)*exp(-3k)

Cancel out A(0) and solve for k.

Ryan

Discussion Mobile Providers
Replies
13
Views
349
Replies
5
Views
987
Replies
1
Views
1K
Replies
21
Views
3K
Replies
25
Views
2K