Very confusing problem!! Please help me!!!?

Question

Consider the initial value problem
dA/dt = kA, A(0) = A_0 (where '_' means subscript)
as the model for the decay of a radioactive substance.

(i) Show that, in general, the half-life T of the substance is
T = - ln 2 / k.
(ii) Show that the solution of the initial value problem in part (1) can be written as
A(t) = A_(0) * 2^(e*(-t / T)).
(iii) If a radioactive substance has the half-life T given in part (i), how long will it take an initial amount A_0 of the substance to decay to 1/8 A_0?

I am at a loss as to what to do for all parts of this question. If someone who knew how to do this, would maybe express their ideas (even solve the problem!!) - well that would be wonderful!!

Thank-you!

Answer 1

let us determine the general solution to dA/dt = kA.

dA/ A = k dt

ln A = kt + C

A = e^(kt+C) otherwise since e^C is still a constant...

A(t) = B e^(kt) .... now since were given that A(0) = A0

A(t) = A0 e^(kt) .... i'll not write the underline here....



i) To show the half life, you need the value of t such that A(t) is 1/2A0.

Then 1/2A0 = A0 e^(kt) ... 1/2 = e^(kt) ... kt = ln(1/2)=-ln2.

Thus T = -ln2/k. (the variable t was simply made capital to show that it is a specific value already.)



ii) just return k to the equation...



iii) replace A(t) by 1/8A0, then solve for such t. Follow the solution steps in (i) because there it was simply replaced by 1/2A0.



Actually if you just replace 2 by 8 it is clear that you will arrive at -ln8/k = -3ln2/k.



d: (if you have any questions still, dont hesitate to ask me.)

Answer 2

Nicely answered! well done! although I must point out a tiny mistake in the second part; it is actually correct!

you don't really need to take LN of the whole thing!

simply when you replace K by -Ln2/T you can just consider it like A=A0*(e^ln2)*(e^-t/T)

e^ln2 = 2

therefore:

A=2*A0*e^-t/T

Answer 3

dA/dt = kA

dA/A = kdt

Integrating from A0 to A and 0 to t,

ln(A) - ln(A0) = kt

A = A0e^kt

(i)

A/2 = A0e^kT

1/2 = (e^kT) / (e^0)

2 = e^-kT

- kT = ln(2)

T = - (ln(2))/k

(ii)

A(t) = A_(0) * 2^(e*(-t / T)) is incorrect:

k = - (ln(2))/T

A = A0e^kt

A = A0e^(- (ln(2))/T)t

ln(A) = ln(A0) - (ln(2))/T)t

ln(A) = ln(A0) - (t/T)ln(2)

ln(A) = ln(A0) + (-t/T)ln(2)

ln(A) = ln(A0) + ln(2^(-t/T))

ln(A) = ln(A0*2^(-t/T))

A = A0*2^(-t/T)

(iii)

if A/A0 = 1/8 = 2^(-t/T)

2^(t/T) = 8 = 2^3

t/T = 3

t = 3T