Precalculus question 5

Question

A vehicle purchased for $22,400 depreciates at a constant rate of 8%. Determine the approximate value of the vehicle 13 years after purchase. Round to the nearest whole dollar.

Answer 1

A=Pe^rt, where r= annual rate of change, t= years,

P= initial amount, A=future Amount, e=an irrational number

A = 22400e^-.08(13)

= 22,400e^-1.4

= 22,400/e^1.4

= about $5,524 in 13 years

for continuous depreciation

in one year it depreciates to

22400(.92) = 20608

in two years

22400(.92^2) = 18959

in 13 years

22400(.92^13) = $7,577

A=P(1+r)^t

=22400(1-.08)^13

=22400(.92^13) = $7,577

with annual compounding depreciation

realistically, it's not clear why it would

depreciate just once a year and

not continually

but odds are you want the latter value

although a precalculus course usually does get into logs, natural logs and exponential functions

although maybe annual compounding might make sense in the auto industry, if there is

a sudden drop in value as the model year gets one year older? Have to talk to auto dealers

to know, but it's possible

general formula for continuously compounded growth (or decay) is A=Pe^rt

if r <0 it's decay or decpreciation. if r>0 it's growth or appreciation

if r=0, there's no change

general formula for compounding for finite periods is

A=P(1+r/n)^nt, n= the number of compounding periods per year

r=rate of change, either positive or negative, t=years when r is the annual rate of change

Answer 2

Use the exponential equation:

The value (V) as a function of the number of years (y):

V(y) = V₀(1 - r)^y

where V₀ is the initial amount (value at year zero) and "r" is the depreciation rate in decimal form (8% = 0.08):

V(y) = 22400(1 - 0.08)^y

V(y) = 22400(0.92)^y

V(13) = 22400(0.92)¹³

Plug this in your calculator to get the final answer