9-14 If "a fool and his money are soon parted," the rate at which it leaves is probably proportional to the amount remaining. If a certain fool starting with $20,000 starts gambling his money away and after 2 hours has lost $2000, how long will it take for him to loose 90% of the original amount?

Solution: The basic assumption in this problem is that the fool will loose in proportion to the amount he has at any time. Humans are a little harder to predict than bacteria, but this is a good assumption. Follow the procedural steps as written previously and be aware of the logic in the problem.

 Step1: The statement "the rate at which the fool looses money is proportional to the amount present" means that

Step 2: Rearrange, integrate, and evaluate the constant of integration with the initial data.

,

,

At t = 0, the fool has $20,000, so ln 20,000 = – k(0)and C = 20,000.

Now the equation reads

Step 3: Rearrange and switch to exponents.

,

and switching to exponents

or

.

Step 4: Use the given data to determine k.

At t = 2 hrs, A has declined to 18,000, so put these numbers into the amount statement and find k.

,

Switch to logarithms to solve this equation for k.

,

As you were following along this problem and "punching the numbers," so you would be very proficient at this logarithm and exponent calculating for the test on this topic, you may have noticed that your calculator displayed a negative number for ln(0.9). This is correct. In the original statement of the problem, dA/dt = – kAso that the calculation of k should produce a positive number. The reason for the ln of numbers less than 1 being negative has to do with one of those other definitions of the ln and will be taken up shortly.

Step 5: The specific equation for this situation is

The time for 10% remaining is the time for A to reach 2000. Substitute for A = 2000and solve for t.

Switching to logarithms,

or

.

Based on this model, it would take this particular fool 43 hours to loose 90% of an original amount of $20,000.