N(t)=14.54+slope*Y
Y = year-2009
slope= (15.37-14.54)/5
b. N(t)=14.54 e^(kY)
y= year-2009
k= 1/5 * ln(15.37/14.54)
There were 15.37 million licensed drivers in Texas in 2009 and 14.54 million in 2004. Find a formula for the number, N, of licensed drivers in the US as a function of t, the number of years since 2004, assuming growth is
(a) Linear
N(t)= million drivers
(b) Exponential
N(t)= million drivers
2 answers
If we consider 2004 to be represented by 0,
then 2009 would correspond with 5
and we have two ordered pairs (0,14.54) and (5,15.37)
slope = (15.37-14.54)/(5-0) = .83
N = .83t + c
at (0,14.54)
14.54 = 0 + c , c = 14.54
N = .83t + 14.54 , where t is in years since 2004 and N is in millions of drivers
You don't specify what type of exponential, (what base), so I will use base e
N = a e^kt, were a is the initial number and t is the time since 2004
N = 14.54 e^kt
when t = 5
15.37=14.54 e^(5k)
1.057... = e^5k
using ln
5k = ln 1.057..
k = .0111028.. or appr .0111
N(t) = 14.54 e^(.0111 t)
then 2009 would correspond with 5
and we have two ordered pairs (0,14.54) and (5,15.37)
slope = (15.37-14.54)/(5-0) = .83
N = .83t + c
at (0,14.54)
14.54 = 0 + c , c = 14.54
N = .83t + 14.54 , where t is in years since 2004 and N is in millions of drivers
You don't specify what type of exponential, (what base), so I will use base e
N = a e^kt, were a is the initial number and t is the time since 2004
N = 14.54 e^kt
when t = 5
15.37=14.54 e^(5k)
1.057... = e^5k
using ln
5k = ln 1.057..
k = .0111028.. or appr .0111
N(t) = 14.54 e^(.0111 t)
2 answers