Examples from Section 6.5
Exponential Growth and Decay

p. 488

Growth of Bacteria

8.

A colony of bacteria is grwon under ideal conditions in a laboratory so that the population increases exponentially with time. At the end of 3 h therre are 10,000 bacteria. At the end of 5 h there are 40,000. How many bacteria were present initially.

Solution. The number of bacteria is given by y  =  (Subscript[y, 0]) (e^(k t))where (Subscript[y, 0])is the initial amount and  k is some constant and t is in hours. To determine k we use the two given points of data:
10000  =  (Subscript[y, 0]) (e^(k 3))
40000  =  (Subscript[y, 0]) (e^(k 5))
Divide the second equation by the first to get
4  =  (e^(k (5  -  3))) =  (e^(2 k)).
Taking logarithms gives
ln (4)  =  2 k or k  =  (1/2) ln (4)  =  ln (2)
Therefore,
y  =  (Subscript[y, 0]) (e^(ln (2) t)) = (Subscript[y, 0]) (2^t).
We can now determine (Subscript[y, 0]) by plugging in one of the original data points:
10000  =  (Subscript[y, 0]) (2^3)
which gives (Subscript[y, 0]) =  10000/8  =  12500.

Solve[{10000  ==  (Subscript[y, 0]) (E^(k 3)), 40000  ==  (Subscript[y, 0]) (E^(k 5))}, 
       {k,  (Subscript[y, 0])}]

Radioactive Decay

17.

The decay equation for radon-222 gas is known to be y (t)  =  (Subscript[y, 0]) (e^(-0.18 t)) with t in days. About how long will it take the radon in a sealed sample of air to fall to 90% of its original value?

Solution. We must solve for (Subscript[t, 1]) in the equation:
0.90 (Subscript[y, 0])  =   y ((Subscript[t, 1])) = (Subscript[y, 0]) (e^(-0.18 (Subscript[t, 1]))).
The (Subscript[y, 0])'s cancel and we take logarithms to get
ln (0.90)  =  -0.18 t or t  =  -ln (0.90)/0.18  =  0.585336198099034898`.

Solve[0.90  == (E^(-0.18 t))]

Newton's Law of Cooling

23.

A pan of warm water (46°C) was put in a refrigerator. Ten minutes later the water's temperature was 39°C; 10 minutes after that, it was 33°C. Use Newton's Law of Cooling to estimate how cold the refrigerator was.

Solution. The equation for the temperature T (t) at time t (in minutes) is given on p.487:
T (t)  -  (Subscript[T, s])  =  ((Subscript[T, 0]) - (Subscript[T, s])) (e^(-k t)),
where (Subscript[T, s]) is the temperature of the refrigerator, (Subscript[T, 0]) is the initial temperature of the water (46°C), and k is some constant. Plugging in the given data T (10)  =  39 and T (20)  =  33 gives two equations:
39  -  (Subscript[T, s]) =  (46  -  (Subscript[T, s])) (e^(-k 10)),
33  -  (Subscript[T, s]) =  (46  -  (Subscript[T, s])) (e^(-k 20)).
Solving for k in both equations gives:
k  =  
        -(1/10) ln (((39 - (Subscript[T, s]))/(46 - (Subscript[T, s]))))  =  
          -(1/20) ln (((33 - (Subscript[T, s]))/(46 - (Subscript[T, s]))))
The last equation can be exponentiated to yield
(((((39 - (Subscript[T, s]))/(46 - (Subscript[T, s])))))^(1/10)) =  (((((33 - (Subscript[T, s]))/(46 - (Subscript[T, s])))))^(1/20))
(note that we could have arrived at this point by isolating (e^(-k))in the original equations) or
(((((39 - (Subscript[T, s]))/(46 - (Subscript[T, s])))))^2) =  (((33 - (Subscript[T, s]))/(46 - (Subscript[T, s]))))

Solve[
      (((((39 - (Subscript[T, s]))/(46 - (Subscript[T, s])))))^2) ==  
        (((33 - (Subscript[T, s]))/(46 - (Subscript[T, s]))))]

So the temperature of the refrigerator was -3. The constant k is
k  =  
        -(1/10) ln 
            (((39 - (-3))/(46 - (-3))))  =  
          -ln ((6/7)) = 0.0154151

The tempature equation now reads
T (t)  -  (-3)  =  (46 - (-3)) (e^(t  (1/10) ln ((6/7))))  =  
        49 ((((6/7)))^(t/10))
T (t)  =  -3  +  49 ((((6/7)))^(t/10)).