The Nature of Exponential Growth Writing growth and decay problems with base e

Population Example At the beginning of the year 2000, the city of Oblique had a population that was approximately 600, and was growing at an annual rate of 15%. It was a very small place but it was growing very fast. The population, P, can be modelled by the equation P = 600(1.15) t, where t represents the time in years since When we use this equation, we assume that the growth rate is constant and that the growth continues to be exponential.

Graph of P=600(1.15) t What is the average growth rate per year? Let A be the point where t=1 and B be the point where t=2. What is the slope of the secant joining point A and point B? A B

Calculation of slope Point A is (1, 600(1.15)) which is A(1,690) Point B is (2, 600(1.15) 2 ) which is B(2, 793.5) The rise is – 690 = The run is 2 – 1 = 1 The slope is % of 690 is also The slope of the secant from one year to the next is the average growth rate from one year to the next. As a numerical value this will get higher each year. As a percentage it should always be the same – in this case 15%. Try it from t=2 to t=3.

Slope from t= 2 to t=3 The green graph is the secant from t= 2 to t=3. It has a slope of which is higher than the previous slope of As a percent of the population in 2002, the slope of the line is 15% of the population in 2002.

Conclusion – Average percentage increase As time passes the population grows faster. The average numerical growth rate will continue to increase over time. This is because the population is growing at the same average annual rate of 15%. We always add 15% of the previous years population. Since the population will grow by 15% each year, the population will increase and so will the value of the 15% of the previous year. We will always be taking 15% of a higher number as time increases. So the average numerical growth rate will always increase but the average percentage growth rate will be a constant. This average percentage growth rate (r) is visible in the formula of the form y=y 0 (1+r) x If r>0 there will be exponential growth If –1< r < 0 there will be exponential decay

Instantaneous Growth Rate It would be interesting to know what the instantaneous growth rate is at any time. To find this we will need to take the derivative of the exponential function. The easiest exponential function to differentiate is y=e x, so we will first change the base from 1.15 to base e.

Change to Base e Let 1.15 = e x Take the natural logarithm of both sides ln 1.15 = ln e x Think, e to what exponent is equal to e x ln 1.15 = x So, x= and 1.15 = e Therefore, the population equation P=600(1.15) t can be written as P = 600e t Notice that the coefficient of the exponent is just slightly less than the decimal part of the base in the previous equation. Now that we have the equation in base e it will be easy to find the derivative.

Find the derivative P=600e t P’=600e t x P’=83.86e t Notice that the instantaneous rate of population change is always times the population at that time. P’= P So the derivative of the population function is proportional to the function. In this case it is a vertical compression by a factor of about 0.14 The population growth rate is increasing but the instantaneous growth rate as a percent is always around 14% of the current population. Since the population is increasing, 14% of the population will also increase as time passes. P P’

Numerical instantaneous growth rate. The numerical instantaneous growth rate at t=1 would be P’=83.86e (1) = This is the slope of the tangent line to the population function at t=1. Note that the population at t=1 is P=600 e (1) and the slope is % of this. What would the numerical instantaneous growth rate at t=2 be? P’= 83.86e (2) = The instantaneous rate of change as a percent is %. (as a decimal ) P P’

Summary When a quantity grows exponentially, the numerical growth rate is proportional to the size of the quantity. For example, if we compare the growth rate when the population is 600 to when it is 1200, the numerical growth rate should double. Base e form: y=y o e kx k represents the instantaneous rate of change (a percent expressed in decimal form) For growth, k>0, for decay k<0 y’=k y o e kx Recall that in the basic form y=y 0 (1+r) x, r represents the average rate of change (a percent expressed in decimal form) The population graph (blue) has been compressed by the same factor at all values of t to produce the derivative graph (red).