Write an exponential function modeling the situation and state if it’s decay or growth. Evaluate f(5) and f(12)

Question

Population of species at 1800 cut by a third each year

amount of 50g of an isotope remaining after m minutes if it’s half life is 8 min

show each step

Answer 1

Hi Alyssa N.,

For exponential growth you use A(t) = A_oe^kt, and for exponential decay you use A(t) = A_oe^-kt. The difference between the two is, growth is positive k, and decay is negative -k. To set up these functions, you are usually given an initial amount of something (A_o), and a final amount (A) at some time (t), and then you find (k or -k). Once you have found (k), you can set up the function for any point in time, f(t). You must decide whether you are dealing with growth or decay to decide which function to use to set up.

For the species portion of the question, 1800 is cut by a third each year, so this is decay. The 50g isotope has a half life so this is also decay. So both will use the function A(t) = A_oe^-kt.

For the species you are given an initial amount of species of 1800, (A_o = 1800). You are also given the species is cut to 1/3 after 1 year, (A = 1/3 * 1800 = 600, and t = 1, with t in years).

We can now solve for (-k):

A = A_oe^-kt,

600 = 1800e^-k(1),

1/3 = e^-k,

ln(1/3) = -k,

-1.099 = -k.

We can now write a function for any time (t), A = 1800e^-1.099t, or we can write it as A = 1800e^t*ln(1/3), either way A is now a function of (t), A = f(t). I'll let you plug in f(5) and f(12), but for f(5) I get 7.4 [f(5) = 7.4], but since we can't have .4 of a species I get f(5) = 7, and for f(12) = 0.

Can you try and set up the isotope? (Hint: A_o = 50g, and A = 25g at t = 8).

Let me know if you need more help, Joe.