The rate at which an item depreciates is proportional to its value at that instant. If an item is presently values at $70,000 and 10 months later it is valued at $22,222, when in years will it be valued at $19,999?
I'm not sure what formula to use...
Ah hah, a real calculus problem :)
dV/dt = - k V
dV/V = -k dt
ln V = -k t + C
e^ln V = V = e^(-kt+C) = c e^-kt
so here at t = 0
e^0 = 1
so V = c = 70,000
and
V = 70,000 e^-kt
now when t = 10/12 years, V = 22,222
22,222 = 70,000 e^-k(.83333)
ln (22,222/70,000) = - .83333 k
-1.1474= - .83333 k
k = 1.377
so
V = 70,000 e^-1.377 t
19,999 = 70,000 e^-1.377 t
ln(19,999/70,000) = -1.377 t
t = .9098 years
By the way, are you sure it was 10 months and not ten yars?
thank you and yes it was months
To solve this problem, we can use the concept of exponential decay, where the rate of depreciation is proportional to the current value of the item.
The formula for exponential decay is:
V(t) = V0 * e^(-kt)
Where:
V(t) = the value of the item at time t
V0 = the initial value of the item
k = the decay constant
t = time (in years)
In this case, we are given that the item is presently valued at $70,000 (V0) and 10 months later it is valued at $22,222 (V(t)). We need to find the time (t) when the item will be valued at $19,999.
To find the value of the decay constant (k), we can rearrange the formula to solve for k:
k = -ln(V(t) / V0) / t
Using the given values, we have:
k = -ln(22,222 / 70,000) / (10/12)
Now, we substitute the given values and calculate the decay constant (k).