A function which models exponential growth or decay can be ..., Lecture notes of Physics

Download A function which models exponential growth or decay can be ... and more Lecture notes Physics in PDF only on Docsity! A function which models exponential growth or decay can be written in either the form P (t) = P0b t or P (t) = P0e kt. In either form, P0 represents the initial amount. The form P (t) = P0e kt is sometimes called the continuous exponential model. The constant k is called the continuous growth (or decay) rate. In the form P (t) = P0b t, the growth rate is r = b− 1. The constant b is sometimes called the growth factor. The growth rate and growth factor are not the same. It is a simple matter to change from one model to the other. If we are given P (t) = P0e kt, and want to write it in the form P (t) = P0b t, all that is needed is to note that P (t) = P0e kt = P0 ( ek )t, so if we let b = ek, we have the desired form. If we want to switch from P (t) = P0b t to P (t) = P0e kt, it again is just a matter of noting that b = ek, and solving for k in this case. That is, k = ln b. Template for solving problems: Given a rate of growth or decay r, • If r is given as a constant rate of change (some fixed quantity per unit), then the equation is linear (i.e. P = P0 + rt) • If r is given as a percentage increase or decrease, then the equation is exponential – if the rate of growth is r, then use the model P (t) = P0b t, where b = 1 + r – if the rate of decay is r, then use the model P (t) = P0b t, where b = 1− r – if the continuous rate of growth is k, then use the model P (t) = Pekt, with k positive – if the continuous rate of decay is k, then use the model P (t) = Pekt, with k negative Thus, the form P (t) = P0e kt should be used if the rate of growth or decay is stated as a continuous rate. Example: A trash dumpster starts with 5 pounds of garbage. Write a function which represents the amount of garbage in the dumpster after t days given the following rates a) The amount of garbage increases by 3 lbs per day ∗ Since the rate is stated as a constant 3 pounds per day, the equation is linear. So, the model is Q(t) = 5 + 3t, where Q represents the amount of trash, and t is measured in days. b) The amount of garbage increases by 3% per day ∗ Since the rate is now given as percentage increase, we need to use the exponential model P (t) = P0b t. Since the growth rate is r = .03, the base of our model should be b = 1 + .03. So, if we again let Q represent the amount of trash, we have Q(t) = 5(1.03)t. · Note that the growth rate is .03, and the growth factor is 1.03 c) The amount of garbage increases continuously by 3% per day ∗ Since the rate is given as a continuous percentage increase, we need to use the exponential model P (t) = P0e kt. We have k = .03, so our model is Q(t) = 5e.03t Example: Suppose the population of ants in a colony grows by 4.2% per month. a) Determine a model which represents the population of the colony after t months. What is the growth rate? What is the continuous growth rate? ∗ The model is simply P (t) = P0(1.042)t. The growth rate is r = .042 (or 4.2%). In order to find the continuous growth rate, we need to convert the model to the form P (t) = P0e kt. So, we need to solve for k in 1.042 = ek. Taking the natural log of both sides, we get k = ln(1.042) ≈ .04114. Thus the continuous growth rate is approximately .04114 (or about 4.114%).