Exponential Models
Example 1 - Assume that in the 1940's a cup of coffee sold for about $0.10 and that in the year 2000 that same cup of coffee sold for $1.25. Further assume that the growth of the price of cup of coffee can be modeled as an exponential process. We want to determine the exponential relationship that gives the price of coffee as a function of time. P(t) = a bt. In addition we will predict the price of a cup of coffee in 2010 using our exponential function.
This model involves a growth process.
- When t = 0 the price is $0.10 so P(0) = $0.10 = a b0 = a (1) = a. So a is $0.10
- When t = 60 the price is $1.25 so P(60) = 1.25 = 0.10 (b)60 or b60 = 1.25/0.10 = 12.5
- solving for b we have b = (12.5)1/60 = 1.0430
- the exponential equation can be written as P(t) = 0.10 (1.0430)t
- To predict the price in 2010 (t=70) we use P(70) = 0.10 (1.0430)70=1.91 or $1.91 (rounded)
Example 2 - Assume that a new radioactive element has been discovered that has a half-life of 30 seconds and we want to model the decay process of this element. We will assume that radioactive decay can be modeled as a continuous process with an equation of the form Q(t) = Qoe-kt. Once we have the equation we will use it to predict when only 10% of the amount we have at t=0 is left.
This model involves a decay process.
- When t = 30 seconds we have half the material we started with left so 0.5 Qo = Qoe-k 30
- Since we have Qo on both sides of the equation we can cancel it out leaving 0.5 = e-k 30
- Taking the natural logarithm (ln) of both sides of the equation gives us: ln(0.5) = -30k ln(e) [we moved the power in front of ln(e)]
- Finally k = ln(0.5)/-30 = 0.0231 which is the constant that we wanted.
- The equation we were seeking can be written as: Q(t) = Qo e-0.0231(t)t
- Now let's address the question of when we will have 10% (or 0.10 left)
- 0.10 Qo = Qo e-0.0231(t)t (What ever we started with we only have 10% now)
- The Qo's cancel out leaving us with: 0.10 = e-0.0231(t)t
- Taking the natural logarithm of both sides of the equation yields: ln(0.10) = ln(e-0.0231(t)t)
- Pulling the exponent in front of ln e and recognizing that ln(e) = 1 we have: ln(0.10) = -0.0231t
- Solving for t gives: t = ln(0.10)/-0.0231 = 99.7 seconds (rounded).
Logarithmic Models
Example 1 - In modeling the shape of the land that we live on some research has indicated a relationship between the size of the landform and its duration. In one case the reported equation was log L = 0.6 log T - 2.8 where L is the size of the landform (in km) and T is the time span of its existence in years. The landforms discussed range from raindrop impact craters to mountains and continential shields. To illustrate the use of this formula let's consider two cases:
- Consider a landform that has a lifespan of about 1 minute years (10-6 years) (a raindrop crater).
- To estimate its size we would use log L = 0.6 log 10-6 - 2.8
- log L = 0.6(-6)log 10 -2.8
- log L = -3.6 (1) -2.8 = -6.4
- L = 10-6.4 = 0.000000398 km or 0.000398 m or 0.398 mm
- This gives us a rough estimate of the size of a raindrop crater.
Example 1 - Now lets turn the question around and specify the size of the landform and determine how long we would expect it to exist (recognize that this equation gives a rough estimate)
- Suppose we have a longform that has a characteristic length of 1 km (a valley or an individual mountain)
- The equation would be log (1) = 0.6 log T - 2.8
- 0 = 0.6 log T - 2.8
- 2.8 = 0.6 log T
- log T = 2.8/0.6 = 4.6667
- T = 104.6667 = 46,420 years (rounded)
- This says that we would expect a valley or a mountain to have a typical life span of about 46,420 years.