Growth & Decay Problems



~Most basic growth or decay problems use the basic equation P=Poekt, where P gives the amount of the quantity at any time t, Po is the amount present initially (t=0), and k is the growth or decay constant

~We usually start a problem with this equation or by the differential equation which gives it (will show this derivation in this article)

~Po is usually known, however, there are certain types where it is not required for a solution. However, the constant k must be found in order to solve the problem. Information given enables us to do this

~Either the equation form is given or you may have to supply it. In either case, for basic growth/decay, the one used is the one mentioned above

~There are problem types for which this basic equation form does not approximate the growth/decay acturately. For these, the wording yields a differential equation which gives a different equation form. A good example of this is Newton's Law of Cooling. I will mention why in this article

~Let's focus on the basic types first & how this equation form is derived

~For these types (most problems covered in Calculus), we need to translate the wording of the problem into a simple differential equation. The equation form will come from the solution

~The wording goes like this: "The rate of change of a quantity at any given time t is proportional to the amount present at that time"

~If I let P represent the quantity, this translates to: dP/dt proportional to P

~This gives the following differential equation: dP/dt = kP, where k is the proportionality constant (also called the growth or decay constant)

~Separating the variables we get dP/P=kdt. Integrating both sides gives ln(P)=kt+C

~Assuming P=Po at t=0, we get ln(Po)=C. Replace C by this value gives ln(P)=kt+ln(Po)

~Getting the ln's on the left side & using a property of logs, we get ln(P/Po)=kt, which gives the basic growth/decay form P=Poekt

~To find the value of k in a given problem, more information is needed

~Usually, this comes form a measurement of P at some other time

~An example of this is information on how long it takes for P to double or triple. In that case, when substituting in 2Po or 3Po into the basic equation form, the Po's divide out & it is not needed for the calculation of the constant k

~For example, if the information is that P triples in 2 hours, substituting into our basic equation form will give 3Po=Poe3k. The Po will divide out of our problem leaving a free path to finding k

~In Newton's Law of Cooling, the rate of change of Temperature, T is not proportional the the temperature at any given time t, instead, it will be proportional to the difference in the temperature & the surroundings (ambient temp) at any given time t

~This yields a different differential equation that gives a different growth/decay model

~For example, if a hot cup a coffee is placed on the kitchen table and the temperature of the room is 72 degrees F, then the coffee will cool proportional to the difference in the temperature of the coffee & 72

~In radio-carbon dating (which is a decay), the half-life of the substance is use to find the constant, but the equation form is the basic one. Half-life gives the time value for which 1/2 of the substance decays. For many radioactive materials, this time can be in the thousands of years.

~See my article on radio-carbon dating for a complete example

~Also, study solutions of assigned problems & the ones given in the class presentations