# Interpretation of Differentials


We frequently solve geometrical and physical problems by obtaining an approximate expression for differential $$\da$$ in terms of differential $$\db$$ and then integrating $$\da$$ to obtain $$\qa$$. We assume that the expression for $$\qa$$ is exact even though we used an approximate formula for $$\da$$. This is justified by saying that the differentials are infinitely small quantities. For example, when we derive an expression for the area of a circular disc (see example 1) we set $$dA = 2 \pi r dr$$ which is an approximate expression when the diffentials are interpreted as real numbers. In this article we try to define a method for computing $$\qa$$ so that we don't need approximate expressions in the derivation.

Theorem 1

Let $$a,b \in \realnumbers$$ and $$a < b$$. Suppose that $$\Delta f = g(x) \Delta x + h(x, \Delta x)$$ for all $$x \in \realnumbers$$ and $$\Delta x \in \positiverealnumbers$$ for which $$x, x + \Delta x \in [a, b]$$. Suppose also that $$g$$ is Riemann integrable and for all $$\varepsilon \in \positiverealnumbers$$ there exists $$R \in \positiverealnumbers$$ so that $$\left\vert \frac{h(x, \Delta x)}{\Delta x} \right\vert < \varepsilon \tag{1}$$ for all $$x, x + \Delta x \in [a, b]$$, $$0 < \vert \Delta x \vert < R$$. Let $$t \in [a, b]$$. Let $$n \in \positiveintegers$$, and define $$\Delta' x := \frac{t - a}{n}$$ and $$x_i := a + \frac{i}{n} \Delta' x$$ $$i \in \naturalnumbers$$, $$i \leq n$$. Define also $$\Delta f_i := f(x_{i+1}) - f(x_i)$$ where $$i \in \naturalnumbers$$, $$i < n$$. Define $$f(t) := \lim_{n \to \infty} \sum_{i=0}^{n-1} \Delta f_i$$ Now $$f(t) = \int_a^t g(x) dx$$ and $$df = g(x) dx$$.

Proof

Let $$f(t) := \lim_{n \to \infty} \sum_{i=0}^{n-1} \Delta f_i = \lim_{n \to \infty} \sum_{i=0}^{n-1} g(x_i) \Delta'x + \lim_{n \to \infty} \sum_{i=0}^{n-1} h(x_i, \Delta'x).$$ Now $$\lim_{n \to \infty} \sum_{i=0}^{n-1} g(x_i) \Delta'x = \int_a^t g(x) dx$$ by the definition of the Riemann integral.

Let $$\varepsilon \in \positiverealnumbers$$. Choose $$R_1 \in \positiverealnumbers$$ so that $$\left\vert \frac{h(x, \Delta x)}{\Delta x} \right\vert < \frac{\varepsilon}{t-a}$$ for all $$x, x + \Delta x \in [a, b]$$, $$0 < \vert \Delta x \vert < R_1$$. Let $$n \in \positiveintegers$$ so that $$n > \frac{t-a}{R_1} .$$ Now \begin{align*} \left\vert \sum_{i=0}^{n-1} h(x_i, \Delta'x) \right\vert & < \sum_{i=0}^{n-1} \frac{\varepsilon}{t-a} \vert \Delta'x \vert = n \frac{\varepsilon}{t-a} \frac{t-a}{n} = \varepsilon . \end{align*} Hence $$\lim_{n \to \infty} \sum_{i=0}^{n-1} h(x_i, \Delta'x) = 0 .$$

$$\tag*{\blacksquare}$$

Theorem 2

A sufficient condition for inequality (1) is that there exist $$S, C \in \positiverealnumbers$$ so that $$\vert h(x, \Delta x) \vert < C \vert \Delta x \vert^2$$ for all $$x, x + \Delta x \in [a, b]$$ and $$0 < \vert \Delta x \vert < S$$.

Proof

Set $$R := \min \{ S, \varepsilon / C \}$$. Now $$\left\vert \frac{h(x, \Delta x)}{\Delta x} \right\vert < C \vert \Delta x \vert < \varepsilon .$$

$$\tag*{\blacksquare}$$

Note that we can't prove Theorem 1 directly by the Fundamental Theorem of Calculus because we would need to \textit{assume} that there exists a function $$f : [a,b] \to \realnumbers$$ for which $$f(x + \Delta x) - f(x) = \Delta f(x, \Delta x)$$ in order to use it.

Example 1

Derive an expression for the area of a disc whose inner radius is $$r_a$$ and outer radius $$r_b$$.

The area to be computed in example 1

Solution

Define $$\Delta A$$ to be the area of a disc with inner radius $$r$$ and width $$\Delta r$$. We have $$2 \pi r \Delta r \leq \Delta A \leq 2 \pi (r + \Delta r) \Delta r$$ By setting $$g(r) := 2 \pi r$$ and $$h(r, \Delta r) := 2 \pi (\Delta r)^2$$ we get $$A = \pi r_b^2 - \pi r_a^2$$ by Theorems 1 and 2.

Example 2

Suppose that a particle is moving under influence of a constant force $$F = ma$$ for time $$T$$ and the particle is initially at rest. Derive an expression for the kinetic energy of the particle. Assume that the work done by a constant force $$F$$ is $$W = F s$$ where $$s$$ is the distance that the particle moves in the direction of the force. Assume also that the kinetic energy of a particle at rest is $$0$$.

Solution

We define $$\Delta s$$ to be the distance that the particle moves in the time interval $$[t, t + \Delta t]$$. We have $$v = at$$, $$a t \Delta t \leq \Delta s \leq a (t + \Delta t) \Delta t ,$$ and $$a (t + \Delta t) \Delta t = a t \Delta t + a (\Delta t)^2 .$$ Set $$g(t) := a t$$ and $$h(t, \Delta t) := a (\Delta t)^2$$ and it follows from Theorems 1 and 2 that the distance that the particle moves in time $$T$$ is $$s = \int_0^T a t dt = \frac{1}{2} a T^2$$ By setting $$v_f = a T$$ we obtain $$E_k = W = \frac{1}{2} F a T^2 = \frac{1}{2} m a^2 T^2 = \frac{1}{2} m v_f^2 .$$

Do you find this formalism useful?

Tommi Höynälänmaa

• I'm curious about the \newcommands you have in the beginning of your post. Do you really find it more convenient to type \qa than to type P? And is \Bbb R_+ (or even \mathbb R_+) actually simplified by \positiverealnumbers? Feb 16 '19 at 9:20
• I view $d$ as an operator, almost the same as $\frac{d}{dx}$. Both operations act on functions, and the resulting $df$ and $\frac{d}{dx}f$ are also functions. $df := \frac{d}{dx}f \Delta x$. So the definition of $df$ requires you to know what $\frac{d}{dx}f$ means. The two are essentially the same operation. There is one function where the differential of it, is just an increment. consider $g: x \mapsto x$, then $dg = 1 \Delta x$. When you see $\Delta x$, it's often written as $dx$ . Or independent variable $\Delta r$ often as $dr$. When you turn the increment of an independent variable Feb 16 '19 at 19:42
• into a differential like $dx$ or $dr$, you are viewing the independent variable as an identity function (you can only apply $d$ to functions). $\int$ means anti-derivative, it can also mean anti-differential. That's how I view differentials. I don't add anything new to it besides what a derivative is (though it still is slightly different: tangent height as opposed to tangent slope) Feb 16 '19 at 19:46

For the application in example 2, What are taking the formal definition of work? that your assuming monotonicity in the very first inequality appearing in the solution of example 2 ? If you are taking somewhat an axiomatic definition of work as given in Apostol Calculus volume 1 chapter on Application of Integral calculus then the question comes why to remember such set of axioms rather than taking the definition of work as the integral $$\large\int_a^{a'} \vec F.d\vec s$$ in general, when they are equivalent in the special case of real line?