Modeling Depreciation: Understanding the Relationship Between Time and Value

[marmot]

so my bro ahd an algebra 2 midterm and a question was this:

) A printer costs $35,000 (how old are these questions lol) but it depreciates 5% a year. What is the value by the 4th year?

ok so being an overtly complex person i tried to model a differential equation of this just for the kicks.

At first I thought

dP/dt=-0.05P

where P is price

and the solution is Po*exp(-0.5t) where Po=35000

however this is wrong. So I assumed I did not know r from dP/dt=-rP and worked the problem by finding the initial values.

so the solution gives me P=exp(-.051293t)Po which is correct.

I don't grasp intuitively the answer. why is dP/dt=.051293P when the problem says it goes down 0.5 each year so I assume P changes over time by -0.05P per year?

 

[csprof2000]

The correct equation is, of course.

P = P_0 * (0.95)^t.

Another way of writing this is:

P = P_0 * exp(ln(0.95)*t).

As you probably have already guessed, ln(0.95) is the number you need.

ln(0.95) = -.051293, approximately.

 

[marmot]

yeah, that's true. i already knew that but, why isn't dP/dt=-0.5P! it does not make sense in my mind. it seems that the changing rate of P is -0.5 P per year

 

[csprof2000]

Almost certainly because 0.5 isn't the instantaneous rate, but the "average" yearly rate.

What you should say is that the integral from 0 to 1 of e^r equals -0.05...

then e^r = 0.95 and r = ln(0.95).

In fact, that's exactly it. You can't substitute the average (or cumulative, to put it in better terms) rate for the instantaneous rate.

 

[marmot]

that makes a lot of sense! because i can still integrate from 0 to .1 and that wouldn't obviously be right.