Application of differentiation

A certain radioactive element decays with time t (in days). The amount (in grams) of the element present A(t) can be modelled by
A(t) = A0 * e^(-kt),
where A0 is the amount present at time t = 0 and k is a constant.
Suppose the element will decay to one quarter of its original amount in 8 days.

(a) Find the value of k.
When t = 8, A(t) = A0 1/8
A0 1/8 = A0 * e^(-8k)
-ln8 = -8k
k = (ln8)/8
(b) If the initial amount of the element present is 80g, find the rate of decay of the element after 16 days.
dA(t)/dt = -kA0*e^(-kt)
dA(t)/dt = -(ln8)/8 (80) * e^( (ln8)/8 * 16)
= -10ln8*e^2ln8
= -10ln8*8^2
= -640ln8
= -1920ln2 g/day


2011-03-21 20:32:35 補充：
Correction: (Yes you're right, a quarter = 1/4)
Since i always remind myself on the time Orz~_~ Sorry"
When t = 8, A(t) = A0 1/4
A0 1/4 = A0 * e^(-8k)
-ln4 = -8k
k = (ln4)/8

2011-03-21 20:33:52 補充：
(b.)dA(t)/dt = -kA0*e^(-kt)
dA(t)/dt = -(ln4)/8 (80) * e^( (ln4)/8 * 16)
= -10ln4*e^2ln4
= -10ln4*4^2
= -160ln4
= -320ln2 g/day

2011-03-21 21:00:39 補充：
(b.)dA(t)/dt = -kA0*e^(-kt)
dA(t)/dt = -(ln4)/8 (80) * e^( -(ln4)/8 * 16)
= -10ln4*e^-2ln4
= -10ln4*4^-2
= -5ln4 /8
= (5/8)(ln1/4) g/day