Half life with calculus

1. Aug 17, 2011

ttttrigg3r

1. The problem statement, all variables and given/known data
This is the equation
y = y0e-kt

2. Relevant equations

b) Show using your expression for λ that if at time t1 the amount is y1, then at time t1 + λ it will be y1/2, no matter what t1 is.

3. The attempt at a solution

y1=y0ekt1

that part I got. Then the answer key goes into the next step saying:λ = (−ln2/k)y0ek(t1+λ)

That is where I got lost. What is the step to go from the y1 equation into the lambda equation? I know that for half life, y1=y0/2 so that makes t=ln2/k . The time it takes for an element to decompose to half its mass is t=ln2/k . How do I make the connection to the very last lambda equation?

2. Aug 17, 2011

ehild

The statement "at time t1 the amount is t1" is the same as y1=y0e-kt1. At any time, the amount is y=y0e-kt.
At time t=t1+λ, the amount is y=y0e-k(t1+λ)=y0e-kt1-kλ. You know the identity ax+y=ax ay, don't you? And λ is the half-life, so y0e-kλ=y0/2.
Can you proceed from here?

ehild

3. Aug 17, 2011

ttttrigg3r

using your method and identity, I got: y=y0e-kt1*e-k$\lambda$
so
y=y1*e-k$\lambda$

and then I am stuck again. lambda = half life, but what does that mean for me to get to y0e-kλ=y0/2.

4. Aug 17, 2011

ehild

y0e-kλ=y0/2, that is e-kλ=1/2.

ehild

5. Aug 17, 2011

ttttrigg3r

I still do not see it. is lambda in this equation being treated as a variable or constant? is lambda the same as time?

6. Aug 17, 2011

SammyS

Staff Emeritus
λ is the time it takes for a sample of an element to decompose so that the mass of the original element in the resulting sample is half of the original mass. Thus, λ, must be the half-life.

7. Aug 18, 2011

ttttrigg3r

Ok I think I got it. This is my understanding, tell me if I am correct.

e-k$\lambda$ is the multiplier it takes to make an element exactly 1/2 of itself therefore we can set e-k$\lambda$=1/2
So when I come to y=y0e-kt1*e-kλ That is the same thing as y=y0e-kt1*(1/2) and knowing that y0e-kt1=y1 we can say y=y1*(1/2)

am I right?

8. Aug 18, 2011

ehild

Yes, You are right, well done!

ehild

9. Aug 18, 2011

ttttrigg3r

THank you all those who helped.