In connection with the differential quotient we had introduced the notion of Differential Equations and shortly described the particularly simple and important differential equations for trigonometric and exponential functions.

In this chapter we want to deal extensively with this kings discipline of infinitesimal calculus, which provides the key to a deeper understanding of physical relationships.

Which concrete meaning can be associated in ones imagination with the first and second differential quotients ${y}^{\prime}$ and ${y}^{\u2033}$ of a function $y$?(Higher differential quotients barely play a role.)

We consider a graphical presentation of the function $y=f\left(x\right)$ in a plane coordinate system.

The first derivative ${y}^{\prime}\left(x\right)=\frac{dy}{dx}\left(x\right)$ then is the slope or steepness of the curve describing the function at position $x$. It indicates how strongly $y$ changes for a given $x$ as a function of $x$. Positive values signify an increase, negative values a decrease.

The second derivative ${y}^{\u2033}\left(x\right)=\frac{{d}^{2}y}{d{x}^{2}})\left(x\right)=\frac{d{y}^{\prime}}{dx}\left(x\right)$ describes the change of slope and thus the local curvature. Positive values mean an increase of the slope and thus concave curvature, while negative values signify a decrease of the slope and thus convex curvature.

We now want to interpret especially the variable x as time t; we thus consider changes of the quantity $y$ as function of time. An example would be a driving car, for which $y$ is the distance traveled during the time interval $t$: $y=f\left(t\right)$. Thus at time $t=0$ the position of the car is $y\left(0\right)=f\left(0\right)$

The first derivative ${y}^{\prime}\left(t\right)=\frac{dy}{dt}$ then is the change of the distance traveled per infinitesimal time interval, measured at a certain time point $t$, and thus has the meaning of the instantaneous velocity $v$ of the car.

The second derivative ${y}^{\u2033}\left(t\right)=\frac{{d}^{2}y}{d{t}^{2}}\left(t\right)=\frac{d{y}^{\prime}}{dt}$ describes the change of velocity and thus the instantaneous acceleration $a$ of the car. Positive acceleration means an increase of the velocity, negative acceleration means a decrease of the velocity, i.e. deceleration.

Thus, for the illustrative description of differential quotients (derivatives) the designations slope or steepness are equivalent as well as the designations curvature and acceleration.

How strong the predictive power of an extremely simple differential equation is, we demonstrate here and now for the example of a driving car. In school we learn with great effort the formula for the time dependence of the traveled distance $s$ with ${v}_{0}$ as initial velocity and ${v}_{0}$ as initial value of $s$:

$$s\left(t\right)=\frac{a}{2}{t}^{2}+{v}_{0}t+{s}_{0}$$

We also learn the restrictive condition, that the acceleration $a$ must be constant for the equation to be correct at all (every child knows however, that this does not happen in reality). The simple differential equation

$${s}^{\u2033}=a$$

does not only the same situation, but is also valid, if the acceleration is not constant, but is an arbitrary function of time $a\left(t\right)$. To distinguish between all individual events, that satisfy the differential equation, it is sufficient to know the respective initial values ${v}_{0}$ and ${s}_{0}$.

To calculate the time dependence from the differential equation is a routine job that is identical for all differential equations, for which one can use the analytical tools of integral calculus or which is just left to a numerical computational code.

In physics we often do not want calculate the values that result in a special case, but to primarily understand which causal relationships are behind a certain phenomenon. The example of distance traveled simply provides the answer: the acceleration is important. Newton

This statement in its formal simplicity is also valid if we examine a “curved” three-dimensional orbit in space under the influence of different forces! For the force vector $\mathbf{F}$ that acts on the object with mass $m$ we have

$$\mathbf{F}=m\mathbf{a}\phantom{\rule{0em}{0ex}},$$

where $\mathbf{a}$ is the acceleration vector. That was the great insight of Isaac Newton.