Posted by **Jen** on Friday, October 20, 2006 at 9:51pm.

The amount A(gms) of radioactive plutonium remaining in a 20gm sample after t days is given by: A = 20 * (1/2)^t/140

At what rate is the plutonium decaying when t = 2 days.

I think you need to find dA/dt and evaluate it at t=2

This should be a fairly straight forward differentiate and evaluate problem, which you've already done.

But how do I evaluate 20 * (1/2)^t/140

(It is 20 multiplied by (1/2)over(t/140)

Is it like we do for a^x?

So d/dx(20 * (1/2)^t/140 ) is 20 * (1/2)^(t/140)* ln(1/2)?

At t=2, 20 * (1/2)^(2/140)* ln(1/2)

I get a negative number for this.

I am supposed to get .098gms/day.

:(

I'm not sure your derivative is complete. The function is

f(t) = 20 * (1/2)

^{(t/140)}
then

f'(t)=20 * (1/2)

^{(t/140)} * ln(1/2) * 1/140

Yes, it's like d/dx a

^{x}, but you might want to think of it as

d/dx a

^{u} where u is a function of x, so

d/dx a

^{u} = a

^{u} * ln(a) * du/dx

Yes, the derivative is negative because this is a decreasing funtion on it's entire domain. In this case the function is 'losing' radioactivity per day. If you were to look at f(2) you'd see it's less than f(1).

If we look at this as a chemistry problem, and evaluate A for 2 days, A = 19.803 g remaining. How much decayed in the two days?

20.000 g = 19.803 = 0.197 g.

How much is that per day? 0.197/2 = 0.985 g/day. This won't give the exact amount (I don't think) that you will get mathematically because this procedure averages the amount lost on day one with that lost on day two. But since the half life of Pu-239 is so LARGE, it will be close.

where is the question

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