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 2000.

• 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?

AB

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 793.5 – 690 = 103.5

• The run is 2 – 1 = 1

• The slope is 103.5

• 15% of 690 is also 103.5

• 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

119.025 which is higher than the previous slope of 103.5

• 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=y0(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=ex , so we will first change the base from 1.15 to base e.

Change to Base e• Let 1.15 = ex

• Take the natural logarithm of both sides• ln 1.15 = ln ex

• Think, e to what exponent is equal to ex

• ln 1.15 = x• So, x=0.13976 and 1.15 = e0.13976

• Therefore, the population equation P=600(1.15)t can be written as P = 600e0.13976t

• 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=600e0.13976t

• P’=600e0.13976tx0.13976• P’=83.86e0.13976t

• Notice that the instantaneous rate of population change is always 0.13976 times the population at that time. P’=0.13976P

• 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’=83.86e0.13976(1) = 96.4388

•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 e0.13976(1) and the slope is 13.976% of this.

•What would the numerical instantaneous growth rate at t=2 be?

•P’= 83.86e0.13976(2) = 110.9044

•The instantaneous rate of change as a percent is 13.976 %. (as a decimal 0.13976)

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=yoekx

• k represents the instantaneous rate of change (a percent expressed in decimal form)

• For growth, k>0, for decay k<0• y’=k yoekx

• Recall that in the basic form y=y0(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).