## Exponential Growth: Doubling Time: Exponential Decay

exponential function: a function of the form y = Abx

doubling time: the time it takes for a population to double

step function: a function whose graph increases in discrete amounts and is constant between increments

exponential growth: growth that can be modelled by an exponential function

• N(t) is the amount of substance after time, t
• No represent the original amount of substance
• t = elapsed time
• d = doubling period(time) or generation time

What to Learn

Help to Remember What to Learn

The exponential function N(t) = No(r)t/d, No > 0, r > 1, d > 0, where No, r, t and d are real numbers.

 Exponential Growth: r >1 the function is increasing. As the value of r increases, the slope of the function gets steeper. When r = 1 linear, N(t) = r. (Not shown) Remember for exponential function r1 horizontal asymptote No = 0 (x-axis) vertical asymptote none x-intercept none y-intercept No domain t > 0, t R range N(t) > 0, N(t) R

Interactive Activity

• Click on the applet to activate
• Move the a, d and r sliders. Observe the change(s) have on the graph of the exponential function.
• Move the t slider and observe the N(t) value of the data point on the exponential curve.

Solving Exponential Growth Equations

Note: When finding r, you will get and undefined display for r when t = 0. Remember that r0 = 1, regardless of r as long a r0. When t = 0, we cannot determine the value r with N(t) = No(r)t/d.

Note: When finding d, you will get and undefined display for d when t = 0. Remember at t = 0, N(t) = No. When N(t) and No are equal, (log N(t) - log No) = 0. Dividing by 0 is undefined. When t = 0, t/d = #doubling periods = 0 regardless of the value of d. When t = 0, we cannot determine the value d with N(t) = No(r)t/d.

Note: When finding t or d, you may get a roundoff error due to the rounding of N(t) in the first calculation.

Note: Some resources use P = Po(r)n. If you use a formula in this form, be remember when you solve for n that you are calculating the number of population periods.

If you are asked for the time(t) it takes the population to reach a certain level:

t = (# population periods)(population period length)

Example

A population doubles every 10 years. If the population is currently 5000, how long will it take to grow to 30 000.

Solution

P = Po(r)n, P = 30 000, Po = 5000, r = 2, population period = 10

30 000 = 5000(2)n

6 = 2n

log 6 = log (2n)

log 6 = nlog 2

n = log 6
log 2

n2.585, but remember it takes 10 years for population to double

time(t) = (# population periods)(population period length)

2.585(10)

25.9

It will take the population 25.9 years to reach 30 000.

 Example 1: Doubling Time The number of bacteria, N(t), in a a culture is given by the formula , where t is the elapsed time (in minutes) and d is the doubling time. How many bacteria were in the initial account. Solution Method 1 The formula for doubling time is , No is the initial account. According to the equation given, No = 300 Method 2 300 bacteria were in the initial count. Example 2: Doubling Time Initially a bacteria culture contains 1250 bacteria. The count increased to approximately 80 000 bacteria 1 1/2 hours later. What is the doubling period for this bacterium? Solution Let N(t) represent the number of bacteria after t hours; let N0 represent the initial bacteria count; and let d represent the doubling time. The doubing period for this bacterium is 0.25 h or 15 min. Example 2: Doubling Time A single bacterium has a mass of 1.0 × 10-15 kg. If its doubling time is 30 min and if there are sufficient nutrients and the wastes they produced do not inhibit growth, what would the total mass of the bacteria be after 3 days? How would that mass compare to the mass of Earth, which is approximately 6.0 × 1024 kg? Solution Let y be the number of bacteria at any time, t (in hours). Since the initial population is 1 bacterium and the number is doubling every 30 min (or 0.5 h), the function that models the population after t hours is Convert days into hours. 3d × 24 h/d = 72 h Now, find the number of bacteria after 3 days. Multiply the number of bacteria by the mass of 1 bacterium. The total mass of the bacteria would be about 2.23 × 1028 kg. Compare the mass of the bacteria to the mass of the Earth. The mass of the bacteria would be approximately 3700 time the mass of the Earth. In the preceding example you investigated exponential growth in terms of its doubling time. For bacteria, the time required for the population to double is also called the generation time. Example 4: Doubling Time The initial bacteria culture doubles in 3 hours. Write an equation that represents the time elapsed for the culture? Solution

## Exponential Growth     Exponential Decay: Half Life:

half-life: the time it takes for a radioactive substance to decay to one-half its original amount
• A(t) is the amount of substance remaining after time, t
• Ao represent the original amount of substance
• t = elapsed time
• h = half-life period

What to Learn

Help to Remember What to Learn

The exponential function N(t) = No(r)t/h, No > 0, 0 < r < 1, h > 0, where No, r, t and h are real numbers.

 Exponential Decay: 0 < r < 1 the function is decreasing. When r = 0 linear: N(t) = 0 (Not shown) Remember for exponential function r > 0. When r < 0 alternating geometric sequence. Example: N(t) = -(-2)t/1, t N {2, -4, 8, -16, ...} Remember for exponential function r > 0. horizontal asymptote No = 0 (x-axis) vertical asymptote none x-intercept none y-intercept No domain t > 0, t R range N(t) > 0, N(t) R

Interactive Activity

• Click on the applet to activate
• Move the a, d and r sliders. Observe the change(s) have on the graph of the exponential function.
• Move the t slider and observe the N(t) value of the data point on the exponential curve.

Solving Exponential Decay Equations

 Example 1 The half-life of strontium-90 is 25 years. A 100 g sample of strontium-90 is purchased. How much strontium-90 will remain in 60 years. In 60 years there will be 19.95 g remaining in the strontium-90 sample. Example 2 The half-life of carbon-14 is 5730 years. This is the time it takes for 50% of the original carbon-14 atoms to decay to nitrogen. The percentage, P(t), of the original carbon-14 that remains after t years can be modelled by the following exponential function: If the remaining percentage is 25%, then the age, t (in years), of the artifact can be found by solving the following exponential equation: When there is 25% of the carbon-14 remaining, the artifact will be 11 460 years old. Example 3 The half-life of of iodine-131 is 8.1 days. How long will it take for 100 g of iodine-131 to decay to 77.4 g. Solution The general equation is , where A(t) represents the amount present after t days, A0 is the original amount, and H is the half-life of the substance. A(t) = 77.4 g, A0 = 100 g, and H = 8.1 days. It will take approximately 3.0 days for 100 g of iodine-131 to decay to 77.4 g. Example 4 The half-life of strontium-90 is 25 years. How much time has elapsed if 1/4 remains in a sample. Solution Let A(t) represent the amount of strontium-90 after t years, and let A0 represent the original amount. Since the substance is decreasing by one-half every 25 years, the equation is If ¼ remains after t years, then Therefore, 50 years have elapsed. Example 5 What is the half life of plutonium-243 if 1/64 of the original amount remains after 30 h? Solution Let A(t) represent the amount of plutonium-243 after t hours; let A0 represent the original amount; and let h be the half-life. Since the substance is decreasing by one-half every h hours, the equation is If 1/64 of the original amount remains after 30 h, then Therefore, the half-life of plutonium-243 is 5 h. Example 6 The half life of cobalt-60 is 5.3 years. After 10.6 years what fraction of the cobalt-60 will remain? How long will it take until 12.5% of the original cobalt-60 remains. Solution After 10.6 years, t = 10.6. After 10.6 years, only ¼ of the original amount remains. After 15.9 years, only 12.5% of the original amount remains.

Parts of this work has been adapted from a Math 30 Pure learning resource originally produced and owned by Alberta Education

Comments to:  Jim Reed - Homepage