Depreciating Printer Value: Algebra 2 Midterm Question Answered

[marmot]

o 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?

 

[John Creighto]

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?

P does change by 0.5 per year. Look up continuously compounded interest vs annually compounded interest.

http://en.wikipedia.org/wiki/Compound_interest

 

[csprof2000]

That's weird. Where did my posts go? I swear I answered this question already...