Tom’s old car has a major oil leak, losing 25% of the oil in the engine every week. Tom adds a quart of oil weekly. The capacity of the engine is 6 quarts of oil. In the long run, what will the oil level (in quarts) be at the end of every week before each quart of oil is added?
Difference Equations
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0I don'$t$ think the oil is added all at once--it is added once a week at the end of the week. – 2012-12-07
2 Answers
Let's write out a few test cases and see if we can find a pattern:
At the end of week: Capacity (quarts): After filling (quarts): 1 4.5 5.5 2 4.125 5.125 3 3.84375 4.84375
Well, that wasn't too useful for pattern finding, but we can at least use it to check our work...
Let $a_i$ represent the amount of oil in the tank at then end of a given week $i$ (before filling the tank):
$a_n = (a_{n-1}+1)\cdot\frac{3}{4}$ for $n >= 0$ where $a_0 = 5$
Expanding: $a_{n+2} = (((a_{n}+1)\cdot\frac{3}{4})+1)\cdot\frac{3}{4}$ $a_{n+2} = (((a_{n}+1)\cdot\frac{3^2}{4^2})+\frac{3}{4})$ $a_{n+2} = (((\frac{3^2}{4^2}\cdot a_{n}+\frac{3^2}{4^2}))+\frac{3}{4})$ I will generalize this to: $a_{n} = \left(\frac{3}{4}\right)^n\cdot a_{0}+\sum_{k=1}^{n}\frac{3^k}{4^k}$ By geometric series: $a_{n} = \left(\frac{3}{4}\right)^n\cdot a_{0}+\left(\frac{3}{4}\cdot\frac{1-\left(\frac{3}{4}\right)^n}{1-\left(\frac{3}{4}\right)}\right)$
$a_{n} = \left(\frac{3}{4}\right)^n\cdot a_{0}+\left(3\frac{2^{2n}-3^n}{2^{2n}}\right)$
Computing $a_3$ yields $3.84375$, thus I assume the formula is correct.
EDIT: I now see you want what happens in the "long term." Taking the limit of the series as $n\to\infty$ evaluates to 3 quarts.
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0Just checked 1st ten cases--the explicit formula is correct... – 2012-12-07
Let's first find the answer, on the assumption there is an answer. Let $a_n$ be the amount at the end of the $n$-th week, before adding the quart.
Then, as at the beginning of anorton's answer, we have $a_{n+1}=\dfrac{3}{4}\left(1+a_n\right)\tag{$1$}$
Assume that the limit of $a_n$ as $n\to\infty$ exists. Let that limit be $a$. Then $a=\frac{3}{4}(1+a).$ Solve for $a$: we get $a=3$.
We still owe a debt, to show existence. Since we "know" that the answer is $3$, it is natural to let $a_n=b_n+3$. Substituting in $(1)$, we obtain $b_{n+1}=\frac{3}{4}b_n.$ This finishes things, the "error" $b_n$ gets multiplied by $\dfrac{3}{4}$ each time, so $b_n$ has limit $0$. It follows that $a_n$ has limit $3$.
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0wow... I would have never thought of that approach. Much better than mine (with regards to time usage). – 2012-12-07