The numbers of polio cases in the world are shown in the table for various years.

Question

Year | Number of Polio cases (thousands)

1988- 350

1992-138

1996- 36

2000- 5

2005- 3.2

2007- 1.3

1.Find an equation of f.

f

2.Predict the number of polio cases in 2016.

3.Predict in which year there will be 1 case of polio.

4.Find the approximate half-life of the number of polio cases.

Answer 1

Fitting a function data works better when the years are shifted to smaller numbers. So let t be the number of years since 1988 and y be the number of polio cases. Then our data points are

(0,350), (4,138), (8,36), (12,5), (17,3.2), and (19,1.3).

Question 1

Question 4, about half-life, suggests fitting an exponential curve y=a·b^t to the data. Take the logarithm of both sides and use the properties of logarithms to rearrange the function to a linear equation:

y = a·b^t

ln(y) = ln(a·b^t)

ln(y) = ln(a) + t·ln(b)

If we let Y=ln(y), A=ln(a), and B=ln(b) then we have the linear equation Y=A+t·B.

Now we can do linear regression and fit the line of least squares to the data. You can use a calculator (using STAT>CALC>LinReg(a+bx) on a TI-84), Excel (using =LINEST()), or similar software, or do the calculations manually by using the formulas

slope: B=(n∑t_iy_i – ∑t_i∑y_i)/(n∑t_i² – (∑t_i)²)

Y-intercept: A = (∑y_i – b∑t_i)/n

where n is the number of data points (so n=6 in this question).

We find B=-0.2978 and A=5.8783. So b=e^B=e^-0.2978≈0.74 and a=e^A=e^5.8783≈357.20. Therefore the exponential curve y=357.2(0.74)^t is a reasonable fit for the data.

Question 2

The year 2016 corresponds to t=28. There will be y=357.2(0.74)²⁸ ≈ 0.08 cases in 2016. (Or we could reason that this implies there will be zero cases.)

Question 3

y=1 when 357.2(0.74)^t=1. To solve for t, use a logarithm:

0.74^t = 1/357.2

ln(0.74^t) = ln(1/357.2)

t·ln(0.74) = ln(1/357.2)

t = ln(1/357.2)/ln(0.74) ≈ 19.5

There will be 1 case of polio about 19.5 years after 1988, so sometime in 2007.

Question 4

When t=0, y=357.2. The half-life is the time it takes for the number of polio cases to be cut in half. We want to solve for t when y=357.2/2.

357.2(0.74)^t = 357.2/2

0.74^t = 1/2

ln(0.74^t) = ln(1/2)

t·ln(0.74) = ln(1/2)

t = ln(1/2)/ln(0.74) ≈ 2.3

The half-life is 2.3 years.