Here is a seedling. It has two leaves which collect
sunlight for photosynthesis. Some of the
sugar created is used for the plant’s metabolism and some is used to create
DNA, protein, cellulose, et cetera to make new cells and hence new leaves.
The plant at time ‘zero’.
The plant at week 1.
The plant at week 2.
The plant at week 3.
The plant at week 4.
Time (weeks)
|
Leaf area (arbitrary units)
|
0
|
2
|
1
|
3
|
2
|
4
|
3
|
5
|
4
|
6
|
The plant is adding one new leaf every week. As it originally had two leaves, its growth
rate can be considered to be 50% per week.
To create a formula we will
·
Use the letter P to represent the original number of
leaves. Therefore P = 2.
·
Use the letter R to represent the growth rate. Therefore R = 50% = 0.5
·
Use the letter T to represent the number of weeks. Therefore
0 <= T <= 4.
·
Use the letter I to represent the number of new leaves.
·
Use the letter A to represent the total number of
leaves.
Our
formula for plant growth is I = PRT. Consequently, A = P + I.
T
|
I = PRT
|
A = P + I
|
0
|
I = 2 × 0.5 × 0 = 0
|
A = 2 + 0 = 2
|
1
|
I = 2 × 0.5 × 1 = 1
|
A = 2 + 1 = 3
|
2
|
I = 2 × 0.5 × 2 = 2
|
A = 2 + 2 = 4
|
3
|
I = 2 × 0.5 × 3 = 3
|
A = 2 + 3 = 5
|
4
|
I = 2 × 0.5 × 2 = 2
|
A = 2 + 2 = 4
|
n
|
I = 2 × 0.5 × n = n
|
A = 2 + n
|
As you
can see, the formula matches perfectly with the observed growth. Some of you may be saying those formulas look
suspiciously like the simple interest formulas for the growth of money in a
bank account. Well, yes. Actually, these are valuable plants which
produce an anti-malarial drug.
Consequently, each leaf can be sold for exactly $1. 00.
End of
story? Not really. A biologist examines the data. He notices
that in week 0 the plant produces 200 units of sugar, i.e. 100 units per
leaf. From the 200 units, the plant used 150 units for metabolism and 50
units to grow a new leaf. In succeeding
weeks the plant produced much more sugar. However every week the plant only
allocated the same 50 units to growth.
T
|
Sugar produced
|
Sugar used for metabolism
|
Sugar used for growth of a new leaf
|
Percentage of sugar used for growth
|
0
|
200
|
150
|
50
|
25%
|
1
|
300
|
250
|
50
|
17%
|
2
|
400
|
350
|
50
|
13%
|
3
|
500
|
450
|
50
|
10%
|
4
|
600
|
550
|
50
|
8%
|
n
|
200 + 100 n
|
150 + 100n
|
50
|
|
The biologist notes growth is tending towards 0% as n
becomes larger. Does this not contradict
the previous statement that the growth rate R remains at a constant 50% every
week? No, because we are ‘comparing apples and oranges.’ The 50% growth rate is
in comparison to the original size of the plant. The declining growth rate
noted by the biologist is recalculated on a week-to-week basis.
The owner of the plant asks the biologist to use DNA
technology to increase the plant’s growth.
After months, or possibly years, of failed experiments, the genetically modified
plant grows as follows.
Time (weeks)
|
Leaf area
|
Growth on a week by week basis
|
0
|
2
|
n/a
|
1
|
3
|
(3-2)/2
× 100 = 50%
|
2
|
4.5
|
(4.5-3)/3
× 100 = 50%
|
3
|
6.75
|
(6.75-4.5)/4.5
× 100 = 50%
|
4
|
10.125
|
(10.125
- 6.75)/6.75 × 100 = 50%
|
Some readers
may object to the leaf area being given as 4.5 in week 2. Does this imply ½ a leaf? A moment’s
consideration will show there is no cause for concern. The original plant’s
leaves were all 10 cm2 in size. In the genetically modified plant, most
of the new leaves are smaller than the original two leaves, but each leaf still
contributes some growth. So in some sense, “Yes, ½ a leaf.”
Financially
sophisticated readers will see that the biologist has improved the plant’s
yield by causing the plant to switch from a linear growth model to an
exponential growth model.
The
sequence 2, 4, 6, 8, 10, … exhibits
linear growth. The equation of a line is y = mx + c. This equation is of exactly the same form as A = PI + P. We
match y with A, m with P, x with I and c also with P.
The
sequence 2, 4, 8, 16, 32, … exhibits
exponential growth. The equation of an exponential curve is y = k×bx.
This equation is of exactly the same
form as A = P(1 + r)n. We match y with A, k with P, (1 + r) with
b and x with n. In conclusion, not only is it the case that “time is money” but
plant growth is also money!
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