# Mathematical Models and Idea of Direct and Inverse Variation

In the later years of his life, the Italian scientist Galileo Galilei (1564 – 1642) wrote about his experiments with motion in a treatise called Dialogues Concerning Two New Sciences. Here he described his wonderful discovery that distances covered in consecutive equal time intervals by balls rolling down inclined planes are proportional to the successive odd positive integers. Thus, if $$d$$ denotes the distance covered during the $$nth$$ time interval, then $$d = k(2n – 1)$$.

Here, is $$k$$ a suitable constant? Galileo determined that the constant $$k$$ depends only on the incline and not on the mass of the ball or the material which it is composed of. He reasoned that the same result should hold for freely falling bodies if air resistance is neglected.

Galileo’s discovery of a mathematical model for uniformly accelerated motion is considered to be the beginning of the science of dynamics. A **mathematical model** is an equation or a set of equations in which letters represent real world quantities, for instance, distance and time intervals in the equation $$d = k(2n – 1)$$. A mathematical model is often an idealization of the real world situation it supposedly describes. For instance, Galileo’s mathematical model assumes a perfectly smooth inclined plane, no friction, and no air resistance, and it neglects the rotational movement of the inertia of the balls as they roll down the inclined plane.

We study mathematical models related to the concept of **variation**.

Thus, the letters $$y$$ and $$x$$ in the following definition could represent variable quantities in a mathematical model.

__Direct Variation__

If there is a constant $$k$$ such that

\[y = kx\]

holds for all values of $$x$$, we say that $$y$$ **is directly proportional to **$$x$$ or that $$y$$ **varies directly as **$$x$$ (or with $$x$$). The constant $$k$$ is called the **constant of proportionality **or the **constant of variation**.

For instance, if an automobile is moving at a constant speed of $$55$$ miles per hour, then the distance $$d$$ it travels in $$t$$ hours is given by the formula

\[d = 55t\]

Here $$d$$ and $$t$$ are variables in the mathematical model $$d = 55t$$ describing the motion of the automobile; the distance $$d$$ is directly proportional to the time $$t$$, and the constant of proportionality is $$55$$ miles per hour.

Suppose that $$x$$ and $$y$$ are variables and that

\[y = kx\]

so that $$y$$ is directly proportional to $$x$$ with $$k$$ as the constant of proportionality.

Note that $$y = 0$$ when $$x = 0$$. Also, if $$x \ne 0$$, we have

\[ \frac{y}{x} = k\]

So the fraction, or ratio, $$y/x$$ maintains the constant value $$k$$ as $$x$$ varies through nonzero values. Thus, if a nonzero value of $$x$$ and the corresponding value of $$y$$ are known, the value of $$k$$ can be determined.

__Example__:

Suppose that the rate $$r$$ at which impulses are transmitted along a nerve fiber is directly proportional to the diameter $$d$$ of the fiber. Given that $$r = 20$$ meters per second when $$d = 6$$ micrometers:

**(a)** Write a formula that express $$r$$ in terms of $$d$$.

**(b)** Find $$r$$ if $$d = 4$$ micrometers.

__Solution__:

Since $$r$$ is directly proportional to $$d$$, we have

\[r = kd\]

where $$k$$ is the constant of proportionality.

**(a) **For $$d \ne 0$$, we have

\[k = \frac{r}{d}\]

Substituting $$r = 20$$ meters per second and $$d = 6$$ micrometers into this equation, we find that $$k = 20/6 = 10/3$$. Therefore, for all values of $$d$$, we have $$r = \frac{{10}}{3}d$$.

**(b) **When $$d = 4$$ micrometers, we have $$r = \frac{{10}}{3}(4) = \frac{{40}}{3}$$ meters per second.

If we say that $$y$$ **varies directly as the square of **$$x$$**, **we mean that $$x$$ and $$y$$ are related by an equation of the form

\[y = k{x^2}\]

for some constant of variation $$k$$. For instance, the formula $$A = \pi {r^2}$$ for the area of a circle states that $$A$$ is directly proportional to the square of the radius $$r$$, and that $$\pi $$ is the constant of variation. More generally:

__Variation with the nth power__

If there are constants $$n$$ and $$k$$ such that

\[y = k{x^n}\]

This holds for all values of $$x$$, so we say that $$y$$** is directly proportional to the nth power of **$$x$$ or that $$y$$ **varies directly as **(or with) ** the nth power of **$$x$$.

__Example__:

The formula $$V = \frac{4}{3}\pi {r^3}$$ for the volume of a sphere states that $$V$$ is directly proportional to the cube of the radius $$r$$ and the constant of variation is $$\frac{4}{3}\pi $$. Find the volume of a sphere of radius $$2.71$$ centimeters and round off your answer appropriately.

__Solution__:

$$V = \frac{4}{3}\pi {x^3} = \frac{4}{3}\pi {(2.71)^3} \approx 83.4$$ cubic centimeters.

Assuming that the radius $$r = 2.71$$ centimeters is correct to three significant digits, we have rounded off our answer to three significant digits.

The situation in which $$y$$ is proportional to the reciprocal of $$x$$ arises so frequently that it is given a special name: **inverse variation**.

__Inverse Variation__

If there is a constant $$k$$ such that

\[y = \frac{k}{x}\]

For all nonzero values of $$x$$, we say that $$y$$ **is inversely proportional to **$$x$$ or that $$y$$ **varies inversely as **$$x$$ (or with $$x$$).

Naturally, if $$y = k/{x^n}$$ for some constant $$k$$ and some positive rational number $$n$$, we say that $$y$$ **is inversely proportional to the nth power of **$$x$$**.**

__Example__:

A company is informed by a shopkeeper that the number $$S$$ of units of a certain item sold per month seems to be inversely proportional to the quantity $$p + 20$$, where $$p$$ is the selling price per unit in dollars. Suppose that $$800$$ units per month are sold when the price is $$10$$ per unit. How many units per month would be sold if its price were dropped to $$5$$ per unit?

__Solution__:

Since $$S$$ is inversely proportional to $$p + 20$$, there is a constant $$k$$ such that

\[S = \frac{k}{{p + 20}}\]

We know that $$S = 800$$ when $$p = 10$$, so we have

\[800 = \frac{k}{{10 + 20}} = \frac{k}{{30}}\]

or

\[k = 30(800) = 24,000\]

Therefore, the formula

\[S = \frac{{24000}}{{p + 20}}\]

gives $$S$$ in terms of $$p$$. For $$p = 5$$ dollars, we have

\[S = \frac{{24000}}{{5 + 20}} = \frac{{24000}}{{25}} = 960\]

So $$960$$ units per month would be sold at a price of $$\$$ 5$$ per unit.