1. Jan 21, 2013

### x0hkatielee

I've just been really thrown off by what this problem is asking me.

Given:
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. Every subpart of this question refers to exactly the same radioactive material.

Question:
a. derive an equation for the amount A(t) present at time t in terms of the constant k and the amount A(o) present at time t=0

b. if A(5) = 1/3A(3), find K

c. at what time t will the amount A(t) be 1/4A(0)

My attempt (?):
for part a I wasnt sure if it was referring to just giving the equation A(t)=A(0)e^-kt
and I have no idea how to go about b or c. please help me if possible ! I think i'm over thinking the problem. I'm just really thrown off by the lack of numbers. Thank you!

2. Jan 21, 2013

### rock.freak667

For part a, you were given dA/dt = -kA, you need to solve this first order DE to derive the equation you want. Are you able to do this?

For part b), once you have the formula from part a, you just need to substitute the numbers.

For example A(1) means the value of A when t=1.

3. Jan 21, 2013

### Staff: Mentor

This is fine, except you need some parentheses for the exponent.
A(t)=A(0)e^(-kt)

What is A(5)? What is A(3)?
Substitute these expressions into the the equation A(5) = (1/3)A(3) and solve for k.

For the last one, you should have gotten the value of k in part b, so you should be able to do it by then.

4. Jan 21, 2013

### x0hkatielee

Thanks for the reply! I'm a little rusty at calc because I haven't taken it in a while. I'm not sure exactly how to get the formula from the equation given.

5. Jan 21, 2013

### x0hkatielee

Thanks for the reply! What is throwing me off is not having numbers or any values for A(3) or A(5).

6. Jan 21, 2013

### Staff: Mentor

For part a, it appears that you need to derive the equation you showed, not just merely write it down. This entails solving the diff. eqn. dA/dt = -kA.

This equation is fairly simple to solve, using separation of variables, one of the first techniques that you learn in diff. equations.

dA/dt = -kA => dA/A = -k dt

Part b assumes that you have a formula (function) for A(t). Use the formula to find A(5) and A(3), and substitute these values into the equation A(5) = (1/3) A(3).

7. Jan 21, 2013

### x0hkatielee

I got the equation the way you said to, but the problem doesn't give me any values for A(0), A(3), or A(5). And this is really confusing me. This problem has been stumping me for a long time because it doesn't give me much to work with. Sorry if i'm bugging you!

8. Jan 21, 2013

### Staff: Mentor

It doesn't need to. Show us what you did.
It gives you all the information you need.

9. Jan 21, 2013

### x0hkatielee

I had dA/dt=-kA and rearranged it and integrated both sides to get to the equation A(t)=A(0)e^(-kt). now I know I'm supposed to plug the values for A(5) and A(3) into part b to solve for k, but how do I get those values if I don't know A(0) or k.

10. Jan 21, 2013

### Staff: Mentor

What is A(5)?
What is A(3)?

11. Jan 21, 2013

### x0hkatielee

the only thing that they give me towards figuring that out is that A(5)=(1/3)A(3)... ? Not sure if that's what you meant. I wouldn't be able to solve for them using the equation for part a without knowing K or A(0).

12. Jan 21, 2013

### Staff: Mentor

That's not what I asked, and it's not what I meant.

You have a formula for A(t). Evaluate your formula at t = 5 and t = 3. What do you get?
Both expressions (A(5) and A(3)) will have k and A(0) in them. You don't need to know these values to write A(5) and A(3), though.

13. Jan 21, 2013

### x0hkatielee

Okay so I think I understand ! If I set A(5)=(1/3)A(3) in the equations then the A(0) values cancel and I can solve for k=(ln 3)/2.. hope that's right!

Last edited: Jan 21, 2013
14. Jan 21, 2013

### LCKurtz

Of course it will help. Once you have A(5) too, maybe you can use the given fact that A(5) = (1/3)A(3) to solve for k. Did you try that?

15. Jan 21, 2013

### x0hkatielee

I think I finally got it right. so like A(0)e^-(5k)=(1/3)A(0)e^-(3k)... and the A(0)'s cancel out. So I get e^-(5k)=(1/3)e^-(3k) . and then once I simplify more I can find k

16. Jan 21, 2013

### x0hkatielee

okay so it looks like what I'm getting for part b is k= (ln 3)/-2 ....? Is this right or am I completely off.

17. Jan 21, 2013

### Staff: Mentor

You have a mistake, but you're on the right track.

18. Jan 21, 2013

### x0hkatielee

it's the negative sign, right? I noticed that.

19. Jan 21, 2013

### Staff: Mentor

Show how you got k...