exponential models

# exponential models - MAC1105: Exponential Models 1 x We are...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MAC1105: Exponential Models 1 x We are going to develop formulas of the form y = C(a) , where a = ∆x x 0 5 6 20 y C 16 20 ? x 0 5 7 20 y 2 y 2 ∆x = , and fill-in the blanks. y1 y1 y C 16 20 ? ∆x = 1 ( 6 − 5) ∆x = 2 (7 − 5) 20 = 1.25 16 y = C(1.25) x 16 = C(1.25) 5 16 C= ≈ 5.24 (1.25)5 y = 5.24(1.25) x y = 5.24(1.25) 20 ≈ 454.50 20 ≈ 1.12 16 y = C(1.12) x 16 = C(1.12) 5 16 C= ≈ 9.08 (1.12)5 y = 9.08(1.12) x y = 9.08(1.12) 20 ≈ 87.59 a= a= x 0 5 8 20 y C 20 16 ? ∆x = 3 ( 8 − 5) 16 ≈ 0.93 20 y = C(0.93) x 20 = C(0.93)5 20 C= ≈ 28.75 (0.93)5 y = 28.75(0.93) x y = 28.75(0.93) 20 ≈ 6.73 a= 3 Example: In the year 2000 an exponentially growing population equaled 50,000 people, and by 2005 the population had grown to 60,000 people. Describe this population with a formula of the form P = P0 (a)t , where t measures years since 2000. Use your formula to compute the population in 2010, and predict the population in 2013. Solution: 60000 = 1.2 , which is a five year growth factor between years 2000 and 2005. Therefore the population 50000 in 2010 (another five year period) can easily be computed by multiplying the population in 2005 by 1.2: 60,000(1.2) = 72,000 people. To compute the population in 2013, let’s compute a one year growth factor. a = 5 1.2 ≈ 1.037 . So the formula becomes P = 50000(1.037)t , and if we let t = 13 we get P = 50000(1.037)13 ≈ 80,185 ≈ 80, 000 people ...
View Full Document

## This document was uploaded on 11/16/2011.

Ask a homework question - tutors are online