The derivative considered as the ratio of two rates

Let

$$y = f(x)$$

be the equation of a curve generated by a moving point P. Its coordinates $x$ and $y$ may then be considered as functions of the time, as explained in §6.13. Differentiating with respect to $t$, by the chain rule (Formula XXV in §5.1), we have

$$\frac{dy}{dt} = f'(x) \frac{dx}{dt}.$$ (10.1)

At any instant the time rate of change of $y$ (or the function) equals its derivative multiplied by the time rate change of the independent variable.

Or, write (10.1) in the form

$$\frac{\frac{dy}{dt}}{\frac{dx}{dt}} = f'(x) = \frac{dy}{dx}.$$

The derivative measures the ratio of the time rate of change of $y$ to that of $x$.

Figure 10.1: Geometric visualization of the derivative the arc length.

$\frac{ds}{dt}$ being the time rate of change of length of arc, we have from (6.26),

$$\frac{ds}{dt} = \sqrt{\left ( \frac{dx}{dt} \right )^2 + \left ( \frac{dt}{dt} \right )^2}.$$ (10.2)

which is the relation indicated by Figure 10.1.

As a guide in solving rate problems use the following rule:

• FIRST STEP. Draw a figure illustrating the problem. Denote by $x$, $y$, $z$, etc., the quantities which vary with the time.

• SECOND STEP. Obtain a relation between the variables involved which will hold true at any instant.

• THIRD STEP. Differentiate with respect to the time.

• FOURTH STEP. Make a list of the given and required quantities.

• FIFTH STEP. Substitute the known quantities in the result found by differentiating (third step), and solve for the unknown.