Functional method and several variables
Consider now the functional method for several proportionalities and select as an example the determination of the pressure of a fluid on the bottom of a reservoir. This pressure depends on the height h of the water level, the size g of the compressed area as well as the kind of fluid - its specific weight s. It is a function of h, g and s.
The dependence on several independent variables is never examined simultaneously, but one after another. You let only one of the variables vary and keep the others constant. In each of these cases, you try to find out the characteristic property of the functional dependence with respect to that one variable. Therefore we will start with water, leave the basic area constant and only vary h. The pressure at the bottom in the presence of twice or half the height of the fluid surface is twice or half as large, respectively, etc.; in other words, the pressure at the bottom is proportional to the height h. Now, still considering water and keeping the height h constant, we will vary the bottom area g. Here again we find proportionality between the pressure b at the bottom and the area of the bottom. If you keep g and h constant, the bottom pressure is also proportional to the specific weight s of the fluid. You learn about these functional properties of fluids in physics; our task is now to derive the function, when you are given these characteristic functional properties.
If the bottom pressure y is directly proportional to the height of the fluid surface, we must have y = bx or, in the notation just introduced and writing c instead of m, as it is often done, b = c·h. What is the meaning of c in this equation? As long as all other variables are constant, c is constant. However, if b varies because of another independent variable, then the equation b = c·h, since h is not affected by the new variable, can only remain valid, if also c is a function of this new variable. If for twice the bottom area also b becomes twice as large, then in b = c·h also c must become twice as large, etc. Hence c must be directly proportional to g, that is c = c1·g, whence the equation becomes b=c·h=c1·gh. You can continue in this manner. The direct proportionality which exists between b and s is transferred to c1 and s, whence c1 = c2·s, that is b = c1·g·h = c2·s·g·h. The possible functional relations between the given variables are thus exhausted, and since there are no new variables, c2 is constant. A system of measured values allows to compute c2. For s = 1 (water), g = 1 cm² and h = 1 cm, the bottom pressure is 1g. Inserting these values into b = c2·s·g·h, you find 1 = c2·1·1·1 or c2 = 1, so that the required formula for the bottom pressure is b = ghs.
As has been pointed out several times before, a formula only contains numbers and, if you use it, you must confirm the units in which the numbers are measured. If the units change, the constant may no longer be 1.
If s = 1, g = 1 m² and h = 4 dm, you must write s = 1, g = 10000 and h = 40, in order to find b in grams. Hence b = 1·10000·40m whence the bottom pressure is 400000 g or 400 kg.
Task 1: Use the functional method to derive the formula for capital interest.
The interest of capital is directly proportional to 1) the amount of capital, 2) the magnitude of the interest rate and 3) the duration of the investment, whence z=ck, and, since z is proportional to the interest rate, so is c, whence c = c1p and, finally, c1 = c2n, that is z = ck = c1p·k = c2·npk. The constant c2 is determined by means of a given pair of values. At 1%, 100 Baht yields 1 Baht interest in 1 year, whence z = c2·npk yields 1 = c2·1·1·100, that is c2 = 1/100. Writing the factors in the established sequence, the formula for interest becomes z = kpn/100. Note that you must take into account the units in which the variables are measured.
Task 2: Use the functional method to derive the formula for the area of a rectangle.
At first, let b be constant. If you place such a rectangle with base a next to one with the same base, the area is doubled, that is,the doubled base yields twice the area. If the base is n·a, the area is n times as large as that with base a. The area of a rectangle with base a/2 is half that of the rectangle with base a, etc. The area of a rectangle is proportional to its base, that is, A = ca. If you now keep a constant and vary b, for the same reasons, A and hence also c is directly proportional to b, whence c = c1b or A = c1·a·b. The constant c1 is determined by a given set of values. If a = 1cm and b = 1cm, the area is 1 cm², whence 1 = c1·1·1 or c1= 1 and A = ab. The formula for the volume of a cube is obtained in the same manner.
Exercise: Find by means of the above reasoning the formula for the volume of a box with sides a, b, c.