A car depreciates at a rate of 30% per year. How much will a $20,000 car be worth after 5 years?

Printable View

- Apr 21st 2009, 08:47 PMmath123456exponential functions and depreciation
A car depreciates at a rate of 30% per year. How much will a $20,000 car be worth after 5 years?

- Apr 21st 2009, 08:57 PMChris L T521
In the problem you are given the rate of decrease.

Since the decay equation takes on the form $\displaystyle P=P_0e^{-kt}$, we can take note that $\displaystyle k=.3$, $\displaystyle P_0=20000$ and $\displaystyle t=5$. Plugging this into the equation, we have $\displaystyle P=20000e^{-.3(5)}\approx 4462.60$

So after 5 years, the car will be worth about $4,462.60

Does this make sense? - Apr 21st 2009, 09:42 PMpickslides
I would use the model

$\displaystyle V=A.P^t$

where V = current value, A = initial value, P = percentage change and t = time.

Therefore

$\displaystyle V=20000\times (1-.3)^t$

$\displaystyle V=20000\times .7^t$

then make t = 5

$\displaystyle V=20000\times .7^5= 3361.40$ - Apr 21st 2009, 10:15 PMearboth
I don't want to pick at you but in my opinion the equation you used is only applicable for rates of decrease smaller than 10%.

With the given data you know that the car's value after 1 year is $14.000. If you now calculate the constant k you'll get: $\displaystyle k\approx -0.356675$

Therefore your result will be too large. - Apr 21st 2009, 10:26 PMChris L T521
- Apr 22nd 2009, 03:53 AMpickslides
- Apr 22nd 2009, 04:27 AMearboth
If you use Chris L T 521's formula:

$\displaystyle A = A_0 \cdot e^{k \cdot t}$

then $\displaystyle k = \ln(1-p)$ where p is the percentage of decrease.

Now compare:

decrease by 2% ====> ln(0.98) = -0.020202...

decrease by 5% ====> ln(0.95) = -0.051293...

decrease by 10% ====> ln(0.9) = -0.10536...

...

decrease by 20% ====> ln(0.8) = -0.22314...

You see that the approximation of k = -p is only sufficient for p < 10%. - Apr 22nd 2009, 06:30 AMSoroban
Hello, math123456!

Quote:

A car depreciates at a rate of 30% per year.

How much will a $20,000 car be worth after 5 years?

Each year the car is worth only 70% of its value during the previous year.

. . Its value after $\displaystyle n$ years is: .$\displaystyle V \:=\:20,\!000(0.7)^n$

In 5 years, its value is: .$\displaystyle V \;=\;20,\!000(0.7)^5 \;=\;\$3,\!361.40 $

- Apr 22nd 2009, 08:25 AMSoroban
Seeing the way that Chris approached the problem,

I am obligated to ask: Is someone teaching that the $\displaystyle e$-formula

. . is theway to handle depreciation?*only*

If so, they are doing a great disservice.

While $\displaystyle P \:=\:P_oe^{-kt}$ is highly applicable to decay and population problems,

I feel that it requires more work when applied to depreciation problem,

. . especially when they*give us*the depreciation rate.

Consider the original problem . . .

. . Initial value: $20,000

. . Depreciation rate: 30% (annually)

We start with: .$\displaystyle P \:=\:P_oe^{-kt}$

When $\displaystyle t = 0,\:P = 20,\!000\!:\quad20,\!000 \:=\:P_oe^0 \quad\Rightarrow\quad P_o = 20,\!000$

. . Hence, the function (so far) is: .$\displaystyle P \;=\;20,\!000\,e^{-kt}$

When $\displaystyle t = 1,\;P = 14,\!000\!:\quad 14,\!000 \:=\:20,\!000e^{-k} \quad\Rightarrow\quad e^{-k} \:=\:0.7$

. . $\displaystyle -k \:=\:\ln(0.7) \quad\Rightarrow\quad k \:=\:-\ln(0.7) \:=\:-0.356674944$

Hence, the function is: .$\displaystyle \boxed{P \;=\;20,\!000\,e^{-0.356674944t}} $

This equation is correct . . . I have no complaint.

But simplify the equation: . $\displaystyle P \;=\;20,\!000\left(e^{-0.356674944}\right)^t $

Since $\displaystyle e^{-0.356674944} \:=\:0.7$, we have: .$\displaystyle \boxed{P \;=\;20,\!000\,(0.7)^t} $

We had this equation*way back then*, just "eyeballing" the problem.

- Apr 22nd 2009, 04:29 PMpickslides
I see the error is becomes quite significant, so why isn’t my solution (when given the depreciation level) always considered the better?

$\displaystyle V=20000\times (1-.3)^t$

$\displaystyle V=20000\times .7^t$

then make t = 5

$\displaystyle V=20000\times .7^5= 3361.40$ - Apr 23rd 2009, 01:39 AMearboth
The difficulty lies in the word "always" ...

With the given problem the variable t is a discret number, that means actually you are dealing with a geometric sequence: $\displaystyle a_n=20000 \cdot \left(\dfrac7{10} \right)^n$

If a value is increasing or decreasing**continuously**your method can't be applied, then you have to use the equation which was posted by ChrisLT521