4.1 Exponential Growth and Decay

The formula for an exponential function is: P(t) = (Subscript[P, 0]) (a^t).  Here (Subscript[P, 0]) is the initial population, a>0 is the base, and t is the real independent variable.

P[t_]:=(Subscript[P, 0])*(a^t);

P[2]

P[1]

P[2]/P[1]

P[4]/P[3]

TableForm[Table[{t,1.0/(3^t),(3.0^t),(0.9^t),(1.1^t)},{t,0,15}],TableHeadings->{None,{"t","1/(3^t)","(3^t)","(0.9^t)","(1.1^t)"}}]

Plot[{1/(3^t),(3^t),(0.9^t), (1.1^t)},{t,-5,5}, PlotLegends->{"y=1/(3^t)","y=(3^t)","y=(0.9^t)","y=(1.1^t)"}]

Example 4.1

Radioactive decay -- G grams of tritium decay after 1 year to 0.945G grams of tritium.  Let’s start with 50 grams of tritium and answer some questions!

Part 1

We know P(t) = (Subscript[P, 0])(a^t), where (Subscript[P, 0])= 50 grams.  We do not know a, but we are given information:

Clear[a]

G[a_,t_]:=50*(a^t)

NSolve[G[a,1]==0.945*50,a]

So now we know that P(t) = 50 * (0.945^t).  The initial amount is P(0) = 50 grams, the exponential base (decay factor) is a = 0.945.

G[0.945,0]

Part 2

There are 37.68 grams of tritium left after 5 years.

G[0.945,5]

Part 3

Since the base a = 0.945< 1, we know that the amount present decays.  Each successive year, we know that only 94.5% of the previous years amount remains, so we would expect a faster decrease in total amount early on, with a slower decrease in the reduction in the amount of tritium as time progresses.  This can be seen from the graph as it is decreasing and concave up!

Plot[G[0.945,t],{t,0,30},AxesLabel->{"t years", "grams of tritium"}, AxesOrigin->{0,0}]

Part 4

This is known as the half-life, and we need to solve P(t) = (1/2)(Subscript[P, 0])  for t.  Using the NSolve command below, we see that the half-life is approximately 12.25 years.  We can graph P(t) and 25 on the same graph, and note that it does indeed occur near t=12.

NSolve[G[0.945,t] == G[0.945,0]*0.5,t, Reals]

Plot[{G[0.945,t],25},{t,0,30},AxesLabel->{"t years", "grams of tritium"}, AxesOrigin->{0,0}, PlotLegends->{"grams of tritium", "half initial amount"}]

G[0.945,12.25*2]

Example 4.2

The worlds population in 1900 was approximately 1.6 billion, and on average, to find next year’s population, we multiply the current year’s population 1.014

1900*1.014

%*1.014

1900*(1.014^2)

Part 1

This is exponential growth as each successive year’s population is proportional to the previous years by a fixed multiplicative factor.

Part 2

P[t_]:=1.6*(1.014^t)

Part 3

Plot[P[t],{t,0,120}, AxesOrigin->{0,0}, AxesLabel->{"years after 1900","population billions"}]

Part 4

Here we simply solve P(t) = 5.0 for t.  This is done below, and it appears to be approximately 82 years after 1900, or in 1982.

NSolve[P[t]==5.0,t, Reals]

Example 4.3

We are given that the half-life of C-14 is 5770 years, and it decays exponentially.

Part 1

This can be done without any real calculations.  After 5770 years, 1/2 the original amount will be left.  After another 5700 years, 1/2 of the 1/2 will be left, which is 1/4 of the original amount.  So 5770*2 = 11,540 years is the correct answer here.

Part 2

The formula is still given by P(t) = (Subscript[P, 0])(a^t).  We can solve for a if we set P(t) = (1/2)(Subscript[P, 0]) for any (Subscript[P, 0]) since we know the half-life.  We can set the initial amount to 1 gram and use 1/2 gram for half the amount.  We see two solutions below, clearly we want the positive solution of a = 0.99988.

Clear[a]

NSolve[0.5 == 1.0*(a^5770.0),a, Reals]

Part 3

We are given that (Subscript[P, 0]) = 25 grams, and we know a = 0.99988 from Part 3 so P(t) = 25*(0.99988^t)

P[t_]:=25.0*(0.99988^t)

Plot[{P[t],12.5},{t,0,10000}, AxesOrigin->{0,0}]

Exercise 5

A person weighs 104 pounds after an illness and they hope to multiply their weight by 1.03 each week

Part a.

Here (Subscript[P, 0]) = 104 lbs, and the growth factor is a=1.03.  Time is measured in weeks here...

w[t_]:=104*(1.03^t)

Part b.

It appears that after 9 weeks they will reach 135 lbs.

NSolve[w[t]==135,t, Reals]

Exercise 7

We are given the function P = 300*(0.86^m), where m is measured in months.

Part a.

If we wish to convert the variable m to weeks w instead, assuming there are 4 weeks per month, then we need to find a new base b for this variable w.  In particular, we know that (0.86^1) = (b^4), which is arrived at by letting m=1 and w=4 and setting (a^1) = (b^4) where we already know a = 0.86.  We see that the base for the weekly variable w is b = 0.963.  So P(w) = 300*(0.963^w).

NSolve[0.86==(b^4),b,Reals]

300*0.86^2

300*0.962996^8

300*0.86^0.25

300*0.963

Part b.

In a similar fashion, we want to solve for a new base b in the exponential function for the exponent variable y measured in years.  In this case, we want (0.86^12) = (b^1) since there are 12 months in a year.  Thus P(y) = 300*(0.164^y).

NSolve[(0.86^12)==b,b,Reals]

(0.86^12)