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Suppose someone has a lake with an area of $A = 1240~m^2$ in this example). It is covered by an area of $c = 10~m^2$) of algae at the beginning ($t = 0$), which doubles each day. This gives us the exponential formula

$A(t) = c \cdot a^t$

$A(t) = 1240 \cdot 2^t$

Now, suppose the owner of the lake removes a constant area of algae at the end of each day ($b$, in this case $= 8 m^2$). The tasks are the following:

  1. find a formula to describe this removal.
  2. calculate after how many days (real number, not integer) the lake will be fully covered.
  3. calculate how much area the owner has to remove so that it takes exactly a) 14 days and b) one year for the lake to be fully covered.

(Note that this is a 12th-grade task we were supposed to solve during the lesson. The teacher showed me the solution but couldn't explain it.)

My initial approach to this was to use a recursive formula

$A(0) = c - b$

$A(t) = A(t - 1) \cdot a - b$

which expands to

$A(0) = 10 - 8 = 2$

$A(t) = A(t - 1) \cdot 2 - 8$

However, this couldn't be used to find the solution for task 2 and 3b.

The apparent formula my teacher gave me was the following:

$A(t) = c \cdot a^t - b\cdot\frac{a^t - 1}{a - 1}$

This expands to

$A(t) = 10 \cdot 2^t - b\cdot\frac{2^t - 1}{2 - 1}$

$A(t) = 10 \cdot 2^t - b\cdot(2^t - 1)$

Can anyone explain this solution?

Winther
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Clashsoft
  • 117

4 Answers4

2

Hint:

Day 1: $A(1)=2c-b$

Day 2: $A(2)=A(1)\cdot2-b=2(2c-b)-b=4c-3b$

Day 3: $A(3)=A(2)\cdot2-b=2(4c-3b)-b=8c-7b$

and so on. Now, I observe that I may write $A(1), A(2)$ and $A(3)$ as follows:

\begin{align}A(1)&=2^1c-(2^1-1)b\\A(2)&=2^2c-(2^2-1)b\\A(3)&=2^3c-(2^3-1)b\end{align}

So, I make the guess (which I need to prove, possibly with induction) that

Day t: $A(t)=2^tc-(2^t-1)b$

Jimmy R.
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Another way to look at the problem is to say that the area $b$ does not take part in the doubling (as if we pick some area $b$ it first gets doubled and we way assume that the $b$ removed is the area tha tgot added by this. So we might just as well look at a lake of area $A-b$ where algae double each day, starting at $c-b$. Thus after $t$ days, we have $A(t)-b = 2^t(c-b)$, so $A(t)=2^t(c-b)+b$.

Hagen von Eitzen
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Suppose $A(t)=aA(t-1)$ and $B(t)=aB(t-1)+b$. Define $C=A+B$. Then $C(t)=aC(t-1)+b$. On the other hand, suppose $D(t)=aD(t-1)+b$ and $E(t)=aE(t-1)+b$. Define $F=D-E$. Then $F(t)=aF(t-1)$. This shows a general phenomenon in linear recurrence relations: the general solution to an inhomogeneous linear recurrence relation consists of one particular solution, plus the general solution to the homogeneous equation.

You know the solution to your homogeneous equation, it is $A(t)=c a^t$ where $c$ is an arbitrary constant. We can guess a particular solution to the inhomogeneous equation. It turns out that there is a constant solution in this case, i.e. a solution with $A(t)=A(t-1)$. In this case they are both equal to some $x$ and you have $x=ax+b$. Thus $x=\frac{b}{1-a}$ provided $a \neq 1$ (which holds in your case).

Overall then, if $a \neq 1$, the general solution to $A(t)=aA(t-1)+b$ is

$$\frac{b}{1-a} + c a^t$$

for any real number $c$. You can then identify the right $c$ for your initial condition.

I apologize that this may be a bit more general than intended for this assignment.

Ian
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You start with the behavior after first second third... day and then generalize

  • First day it doubles $2c$ and is reduced by $b$, $2c-b$
  • Second day it doubles $2(2c-b)$ and is reduced by $b$, $2(2c-b)-b$
  • Third day it is $2(2(2c-b)-b)$ and then it is reduced by $b$, $2(2(2c-b)-b)-b$

You can write now for $4^{th}$ day without calculating anything $2(2(2(2c-b)-b)-b)-b$

You have $c\cdot2^4$ and then how many times you reduce by b: $-2\cdot 2\cdot 2\cdot b-2\cdot 2\cdot b-2\cdot b-b$

It is clear that in general case after n days you are going to have

$c\cdot 2^n-2^{n-1} b-2^{n-2}b-...-2b-b = c\cdot 2^n-(2^{n-1}+2^{n-2}+...+2+1)b$

This part $2^{n-1}+2^{n-2}+...+2+1$ is a sum of geometric series, simply multiply by $1=(2-1)$ and telescope to find $2^{n-1}+2^{n-2}+...+2+1=2^n-1$

The solution is then $c\cdot2^n-b\cdot (2^n-1)$

As you can see the solution is the same as if we have $(c-b)$ area from the start plus one additional $b$, $(c-b)\cdot2^n+b$