A Real Life Problem Exponential Growth Swine Flu Epidemic 2009 A Real Life Problem Exponential Growth

Part 1: Days 0 and 1 of the Swine Flu Epidemic in the U.S. On April 23, 2009, the CDC (Center for Disease Control) confirmed 6 cases of swine flu in the U.S, the first reported in the U.S. On April 24, the CDC confirmed 7 cases.

Date (d) Number of cases (c) Ordered pair 4/23 6 (0, 6) 4/24 7 (1, 7) Plot the data shown in the table on graph paper for day 0 and day 1. Date (d) Number of cases (c) Ordered pair 4/23 6 (0, 6) 4/24 7 (1, 7) 4/25 ????? (2, ?)

Answer the following questions based on the data in your graph. Is c = d + 6 a reasonable model for this data?____________. Justify your answer. Using this model, what would be the number of confirmed cases on April 25? _______ If you graphed the number of predicted cases on 4/23 as the ordered pair (1,6) and the data for 4/24 as (2, 7), write the new equation:__________________________ What kind of model is d = c + 6? Circle one. Linear Quadratic Cubic Exponential Click to see answers. YES 8 d = c + 5

Part 2: Days 2 and 3 of the Swine Flu Epidemic in the U.S. Add the data points to your graph for April 25 and April 26 from the table below. Date (d) Number of cases (c) Ordered pair 4/23 6 (0, 6) 4/24 7 (1, 7) 4/25 11 (2, 11) 4/26 20 (3, 20)

Answer the following questions based on the new data in your graph. Did the equation y = x + 6 correctly predict the number of cases for day 2 and day 3 of the epidemic? Explain your reasoning. Draw a smooth curve through your data points and extend the curve to April 27. Based on your graph, predict the number of cases on April 27:____________ NO Answers vary between 30 and 50 Click to see answers.

Predict the number of cases for April 27 for each model below: Model #1: # of cases: Model #2: # of cases: Model #3: # of cases: The number of cases on April 27 was 40. Which model from #3 best predicted the number of cases on 4/27? Explain your answer: 38 43 36 Quadratic Predicted value closest to 40 Click to see answers.

5. Match the following equations with the name of the parent function: B C A

Part 3: Days 4, 5, and 6 of the Swine Flu epidemic in the U.S. Add the data points to your graph for April 27, April 28, and April 29 from the table below.

1. Predict the number of cases for April 29 for each model below: Model #1: # of cases:_____ Model #2: # of cases:_____ Model #3: # of cases:_____ 2. Which model best predicted the actual number of cases for April 29? 90 100 96 Quadratic Click to see answers.

If the number of predicted cases for April 30 is 157, which equation from #1 is the best model? If the number of predicted cases for May 1 is 257, which equation from #1 is the best model? Exponential Exponential Click to see answers.

Part 4: How many cases of swine flu in the U.S. on May 22, 2009? Use the exponential model: and predict the number of infected cases of swine flu in the U.S. after 30 days from the first confirmed cases: d = 30, c = 13,661,913 or approx. 13,700,000 Click to see answers.

Part 5: What factors could change the number of predicted cases of swine flu based on this math model? Measures to prevent the spread of the disease Medical treatment List other factors

Exponential Growth The following equation can be used to model exponential growth. Exponential growth equations often model real life situations like population growth and the spread of contagious diseases like the swine flu. y = a(b)x a is the initial value (beginning population or beginning number of cases of a contagious disease like swine flu). b is the growth factor that is equal to 1 + growth rate (rate of increase). y is the new value (number in population or number of infected cases) after x intervals (usually periods of time) x is the power of the base, b.

Day (d) Cases (c) Number of added cases each day 50% percent of previous cases added each day (growth rate = 0.50) ------------- ----------- 1 2 3 4 6 9 3 6 + .50 (6) 13.5 9 + .50 (9) 4.5 6.75 20.25 13.5 + .50 (13.5) 20.25 + .50 (20.25) 30.375 10.125 Click to see answers.

c = 6(1.5)x What does 6 represent in this situation? Example 1: The following equation predicts the number of infected cases, c, for the number of days, d, from the beginning of a contagious disease. c = 6(1.5)x What does 6 represent in this situation? What does 1.5 (1 + ??) represent in this Initial number of infected cases Exponential growth with a growth rate of 50% or 0.50 Click to see answers.

c = 10(1.75)d Example 2: Write the exponential growth equation for a contagious disease that starts with 10 reported cases and increases at a rate of 75% each day. c = 10(1.75)d Click to see answer.

A rancher had 400 square-miles to use as a black bear habitat. REMEMBER the bear problem in SAS 6: Rates of Change in Exponential Models? A rancher had 400 square-miles to use as a black bear habitat. He brought in 10 young bears to grow the population. The annual growth rate is about 0.8. The population density is no more than 1.5 black bears per square mile. How many bears could the the land support? 400 * 1.5 = 600 bears LandSize Population Density

This type of problem represents classic LOGISTIC GROWTH.

Logistic Growth The following equation can be used to model logistic growth. Logistic growth equations often model real life situations like population growth and the spread of contagious diseases like the swine flu. Note: Population density is the number a certain area can hold. Carrying capacity is the MAXIMUM number that area can hold.

Bear problem: (600) (growth rate) 400 square miles P0 = 10 k = 0.8 population density = 1.5/sq mile carrying capacity = 400*1.5 = 600 bears

The S-shaped graph of a Logistic Model is a sigmoidal curve.

Sigmoidal Graph

So…..what if you don’t know the growth rate, k?

Logistic Growth Model Practice: Ten grizzly bears were introduced to a national park 10 years ago. There are 23 bears in the park at the present time. The park can support a maximum of 100 bears. Assuming a logistic growth model, when will the bear population reach 50? 75? 100?

Ten grizzly bears were introduced to a national park 10 years ago Ten grizzly bears were introduced to a national park 10 years ago. There are 23 bears in the park at the present time. The park can support a maximum of 100 bears. Assuming a logistic growth model, when will the bear population reach 50? 75? 100?

Note: The value of A can also be found algebraically by substituting P(0)=10.

Ten grizzly bears were introduced to a national park 10 years ago Ten grizzly bears were introduced to a national park 10 years ago. There are 23 bears in the park at the present time. The park can support a maximum of 100 bears. Assuming a logistic growth model, when will the bear population reach 50? 75? 100?

p We can graph this equation and use “trace” to find the solutions. Years Bears We can graph this equation and use “trace” to find the solutions. y=50 at 22 years We can also solve algebraically. y=75 at 33 years y=100 at 75 years p