Question

every month, john buys $10.00 of paper per month for his printer. Originally the price was $0.005 per sheet, but the price per sheet increases at a constant rate of $0.0001 per year. How long will it take John to buy his first 500,000 sheets of paper? Set up the differential equation for above problem

Answer 1

\[P(t) = .005+\frac{.0001}{12}t\]
where t is num of months, p is price of sheet
\[N(t) = \frac{10}{P(t)}\]
N is num of sheets purchased with $10 at month t
\[500,000 = \sum_{t=0}^{T} N(t)\]
find T
where does the diff equ come in?

Answer 2

@dumbcow i agree with you. i was thinking of the same.

Answer 3

because i am taking differential equation class, and this is one of the first order differential equation modeling problem

Answer 4

I guess that we consider the $10.00 per month is the initial condition

Answer 5

is he buying only once a month or do you assume hes buying continuously
usually a diif equ model deals with continuous change

Answer 6

continuous as \[t \rightarrow \]

Answer 7

\[t \rightarrow \infty \]

Answer 8

differential equation can be used to understand that there is a linear increase of the cost of the sheets per year.

Answer 9

make sense

Answer 10

sorry?

Answer 11

i mean if we can write the DE as function of time from the amount of sheets of paper with given IVP (initial value problem) to find C constant.

Answer 12

then with the condition of 500,000 sheets of paper, we can use that equation to find out how long it take for john to buy it.

Answer 13

look, cost of the sheet.=0.0001+[y/12], where y is the no. of months.
[.] denotes the integral part.

Answer 14

http://www.wolframalpha.com/input/?i=+sum+10%2F%28.005+%2B+%28.0001%2F12%29t%29+from+t%3D0+to+309

looks like 25 yrs 9 months

Answer 15

\[500,000=10.12(\frac{ 1 }{ 0.005 }+\frac{ 1 }{ 0.0051 }+..)\]
now, if you can find the sum of the harmonic progession, then you can find the time taken.

Answer 16

awesome, thank you guys

Answer 17

you're welcome! :)