Lifetime of oil with exponential consumption rate

[doombanana]

Homework Statement

The total amount of oil in a well is 24000 barrels. The present rate of consumption is 100 barrels per year. How long will the gas supply last if the present yearly rate of consumption increases by 1% per year?

Homework Equations

We can use the approximation (1+x)^i = 1+xi (1)
\sum i = \frac{n(n+1)}{2} (2)
and the quadratic formula

The Attempt at a Solution

Using (1), the amount of oil left at any given year is y=24000-100(1+.01t).

The sum of (2) should equal the total number of barrels (24000).

let n = 100+t where t is the final year

substituting n into (2) gives
\frac{(100+t)^2 + (100+t)}{2} = 24000

which gives
0= t^2-201t-37900

this gives a value of t= 119 years, but when I plug in my equation for y into Excel I get 142 years so I know I'm doing something wrong. Thank you for your help.

 

[SammyS]

Homework Statement

The total amount of oil in a well is 24000 barrels. The present rate of consumption is 100 barrels per year. How long will the gas supply last if the present yearly rate of consumption increases by 1% per year?

Homework Equations

We can use the approximation (1+x)^i = 1+xi (1)

This is not a good approximation for an exponential function. After approximately 70 years, rate of consumption will have doubled to 200 barrels per year.

\sum i = \frac{n(n+1)}{2} (2)
and the quadratic formula

The Attempt at a Solution

Using (1), the amount of oil left at any given year is y=24000-100(1+.01t).

The sum of (2) should equal the total number of barrels (24000).

let n = 100+t where t is the final year

substituting n into (2) gives
\frac{(100+t)^2 + (100+t)}{2} = 24000

which gives
0= t^2-201t-37900

this gives a value of t= 119 years, but when I plug in my equation for y into Excel I get 142 years so I know I'm doing something wrong. Thank you for your help.

Try modeling this as a geometric progression.

 

[doombanana]

This is not a good approximation for an exponential function. After approximately 70 years, rate of consumption will have doubled to 200 barrels per year.

We were told that the approximation does overestimate the lifetime by about 20 years. The equations I listed in the relevant equations section were given as hints as to how to solve the problem. By doing this he kind of explicitly walked us through the solution of the problem, but for some reason I'm not getting it.

Try modeling this as a geometric progression.

I've done this for both the approximation (24000 - 100(1+.01i) gives 142 years) and the exact equation (24000 - 100(1+.01)^i gives 122 years). This was a test question, though, so I should be able to calculate this by hand and get a lifetime of 142 years.

I assume I'm doing the sum wrong, as I can't see any other place that I could have made an error.

 

[Ray Vickson]

We were told that the approximation does overestimate the lifetime by about 20 years. The equations I listed in the relevant equations section were given as hints as to how to solve the problem. By doing this he kind of explicitly walked us through the solution of the problem, but for some reason I'm not getting it.



I've done this for both the approximation (24000 - 100(1+.01i) gives 142 years) and the exact equation (24000 - 100(1+.01)^i gives 122 years). This was a test question, though, so I should be able to calculate this by hand and get a lifetime of 142 years.

I assume I'm doing the sum wrong, as I can't see any other place that I could have made an error.

The total amount used by time t is
S = 100 \sum_{i=0}^{t -1}(1.01)^i.
Note: this formula reckons the timing of consumption as follows. It assumes that consumption in year 1 (= interval from times t = 0 to t = 1) is 100, that consumption in year 2 (= interval from times t = 1 to t = 2) is 101, etc. If you use the formula for a finite geometric sum and solve the equation S = 24000, you will find out how many years the oil will last.

 

[SammyS]

Using the approximation you've been instructed to use, along with what Ray V gave you leaves you with

\displaystyle S = 100 \sum_{i=0}^{t -1}(1.01)^i.

\displaystyle \quad\ = 100 \sum_{i=0}^{t -1}\left(1+(0.01)i\right)\ .​