# Section 8A

### exercises

...

#### Linear or Exponential Exercise 10

The population of MeadowView is increasing at a rate of 3% per year. If the population is 100,000 today, what will it be in three years?

This is exponential growth: the population increases by a certain percentage every year.

One year later the population becomes $$100000 \times (1+0.03) = 100000 \times 1.03$$.

Two years later it becomes $$100000 \times 1.03 \times (1+0.03) = 100000 \times 1.03^2$$

Three years later it becomes $$100000 \times 1.03^2 \times (1+0.03) = 100000 \times 1.03^3 = 109273$$

#### Linear or Exponential Exercise 10 ... continued

The pattern $$n$$ years later the population is $$100000 \times 1.03^n$$

is pretty clear.

For example, 30 years later, the population becomes

$$100000 \times 1.03^{30} = 242726$$

#### Linear or Exponential Exercise 12

The price of a gallon of gasoline is increasing by 4 cents per week. If the price is $$\ 3.10$$ per gallon today, what will it be in ten weeks?

This is linear growth: the price increases by a certain amount (4 cents) every week independently on what it was.

During ten weeks the price will increase by
$$4 \times 10=40$$,
and will become $$3.10 + 40 = \ 3.50$$

Here the pattern is: $n$ weeks later the price is
$$310 + 4 \times n$$ (cents).

#### Linear or Exponential Exercise 14

The value of your car is decreasing by 10% per year. If the car is worth $$\ 12,000$$ today, what will it worth in two years?

This is exponential decay: the value decreases by a certain percentage (10%) every year.

One year from now the car is worth
$$12000 - 12000 \times 0.1 = 12000 \times (1-0.1) = 12000 \times 0.9 = \ 10800$$

Two years from now the car is worth
$$10800 - 10800 \times 0.1 = 10800-1080 = \9720$$

Linear or Exponential Exercise 14 ... variation ... $$n$$ years

One year from now the car is worth
$$12000 - 12000 \times 0.1 = 12000 \times (1-0.1) = 12000 \times 0.9 = \ 10800$$

Two years from now the car is worth
$$(12000 \times (1-0.1))\times(1-0.1) = 12000 \times (1-0.1)^2$$

Three years from now the car is worth
$$(12000 \times (1-0.1)^2)\times(1-0.1) = 12000 \times (1-0.1)^3$$

Now the pattern becomes clear: $n$ years from now the car is worth
$$12000 \times (1-0.1)^n = 12000 \times 0.9^n$$

For example, $20$ years from now, the car is worth
$$12000 \times 0.9^{20} \approx \ 1,459$$

##### Chessboard parable Exercise 18

How many grains of wheat should be placed on square 32 of the chessboard?

Solution. Recall that there is 1 grain on the first square.

there are $$2=2^1$$ on the second square, $$4=2^2$$ on the third square, $$8=2^3$$ on the fourth, and so on...

The pattern is clear: square number $$n$$ has $$2^{n-1}$$ grains.

In particular, square 32 has
$$2^{31} =2147483648\approx 2.1 \times 10^9$$ grains.

##### Chessboard parable Exercise 18 ... continued

Find the total number of grains and their weight at this point assuming that a grain of wheat weights 1/7000 pound.

The pattern for the total number of grains on board is only a bit more complicated, and is given on p475 of the textbook.

After square $$n$$ is full, there are $$2^n-1$$ grains on the board.

For $$n=32$$, this is $$2^{32}-1 = 4,294,967,295 \approx 4.2 \times 10^9$$

The weight in question is
$$4294967295 \times 1/7000 \approx 613,567$$ pounds.

##### Magic Penny Parable Exercise 24

Suppose that you could keep making a single stack of the pennies. After how many days would the stack be long enough to reach the nearest star (beyond the Sun), which is $$4 \times 10^{13}$$ km away?

There are $$2^t$$ pennies after $$t$$ days.

We need the thickness of a penny

We thus want that
$$2^t \approx \frac{4 \times 10^{16}}{1.52 \times 10^{-3}} \approx 2.6 \times 10^{19}$$

After some trial and error we find that
$$2^{65} \approx 3.7 \times 10^{19}$$ so 65 days should suffice.

##### Bacteria in a Bottle Parable Exercise 26

How many bacteria are there in the bottle at 11:15?

Solution. Recall that $$t$$ minutes past 11:00, there are $$2^t$$ bacteria in the bottle.

Thus there are $$2^{15}$$ of them at 11:15.

##### Exercise 26 ... continued

What fraction of the bottle is full at that time?

The bottle is full at noon with $$2^{60}$$ bacteria.

At 11:15, $$2^{15}$$ bacteria constitute
$$\frac{2^{15}}{2^{60}} = \frac{1}{2^{45}} = 2^{-45} \approx 2.8 \times 10^{-14}$$