Question

The half-life of a radioactive substance is 3200 years. Find the quantity Q(t) of the substance left at time t>0 if Q(0)=20 g.

Answer 1

@SithsAndGiggles

Answer 2

How do I separate this?

Answer 3

So I got Q=kt^2/2+C, Q(0)=20, C=20, Q=kt^2/2+20.

Answer 4

Oh hold on, I'd made a mistake with the ODE, it's actually
\[\frac{dQ}{dt}=kQ\]
and so separating the variables gives
\[\frac{dQ}{Q}=k\,dt\]
Sorry about that

Answer 5

Note this equation is also linear and can be solved using the other methods I mentioned.

Answer 6

So I got ln abs(Q)=k^2/2+C, Q=Ce^(k^2/2) and C=20, Q=20e^(k^2/2).

Answer 7

Check the RHS again. You're integrating with respect to \(t\), not \(k\).

Answer 8

So Q=20e^(kt)?

Answer 9

Yes!

Answer 10

But the answer is Q=20e^(-(t*ln2)/3200) g.

Answer 11

Right, you still have to solve for \(k\). You're implicitly given another "initial" condition, that the half-life is 3200 years. What this means is that given a starting amount of the radioactive material, it will decay to half this amount in 3200 years. If the starting amount is 20 grams, then after 3200 years you have 10 grams remaining:
\[\large10=20e^{3200k}\]
Solve for \(k\).

Answer 12

Thanks!