Calculus question? Half-lives? Please help!!

Question

A building is found to contain a radioactive gas. Thirty hours later, 60% of the initial amount of gas is still present. (Note: Use the concepts of half-life or doubling time.)

 

(a) Write a model for the percentage of the initial amount of the radioactive gas present after t hours. (Round all numerical values to three decimal places.)

 

(b) Calculate the half-life of the radioactive gas. (Round your answer to three decimal places.)

Answer 1

The basic formula is A=Pe^rt   This works for compounding continuously as well as all growth and decay problems.

 

The trick with half life ( or any percentage) is that no matter how much you start with, after a certain time you'll have that percentage left.

 

So if you started with 100 lbs of gas and it dissipated to 60 pounds after 30 hours, you can use those numbers in the formula, even though you made them up!!!

 

60 = 100 *e^r*30

.6 = e^30r

ln.6 = 30r

ln .6 / 30 = r which is approximately r=-.017 note that it's negative which means it's a decaying situation.

 

So your general formula for this problem is A = P e^{-.017 t}.

 

b) So you can use this formula to see when your 100 lbs becomes 50 pounds....again, the made up number don't make a difference since when you divide them, you'll get .5

 

50=100e^{-.017 t}

.5 = e^{-.017 t}

ln .5 = -.017 t

ln .5 / -.017 = t   note that the number should be larger than 30 since in 30 hours, it decayed to 60 % and now you are looking for a 50% decay.

 

So your answer is approximately t = 40.773 hours.

 

Hope this helped.

Answer 2

Let A(t) = amount remaining after t hours

 

       A₀ = initial amount

 

A(t) = A₀e^kt

 

A(30) = 0.60A₀ = A₀e^30k

 

               0.60 = e^30k

 

               30k = ln(0.60) = -0.5108

 

                  k = -0/017

 

A(t) = A₀e^-0.017t

 

To find the half-life, Set A(t) = 0.50A₀ and solve for t.

 

0.50A₀ = A₀e^-0.017t

 

  0.50 = e^-0.017t

 

  -0.017t = ln(0.50)

 

          t ≈ 40.733 hours