Launching Calculus

A modern myth is that calculus began after an apple struck Sir Isaac Newton in the head. Whether this story is valid or not, one thing is certain—Newton’s curiosity into gravity and motion did in fact lead him to formulate what we now know as calculus.

Newton formulated three famous laws of motion, that applied to gravitational motion on earth. While exploring that area of study launches us into Newtonian mechanics, one thing we will mention is Newton’s insight into gravitational acceleration. At least on earth, and at least for sufficiently low heights, the gravitational acceleration $g \approx 9.81\ \text{m}/\text{s}$ is constant.

Let $a(t)$ denote the acceleration of an object in linear motion at time $t$ . Newton contemplated that the velocity $v(t)$ of the object can be obtained by summing up approximate accelerations in small units $\Delta t$ of time:

$\displaystyle v(t) \approx \sum a(t)\Delta t.$

Allowing the time intervals to approach $0$ , we obtain

$\displaystyle v(t) := v(0)+ \int_0^t a(u)\, \mathrm du.$

But velocity in turn can be conceptualised as small packets of changes in displacements (i.e. distances with direction), so that the displacement $s(t)$ can be defined by

$\displaystyle s(t) := s(0)+ \int_0^t v(u)\, \mathrm du.$

From these ideas we obtain the usual equivalent definitions of velocity and acceleration:

$\displaystyle \begin{aligned} v(t) &= s'(t) = \frac{\mathrm ds}{\mathrm dt}, \\ a(t) &= v'(t) = \frac{\mathrm d^2 s}{\mathrm dt^2}.\end{aligned}$

Theorem 1. Define $s,v$ as per the discussions above. Suppose $s(0) = s_0$ and $v(0) = v_0$ . Suppose $a(t) = a$ is constant. Then the following laws of kinematics hold:

$\begin{aligned}v(t) &= v_0 + at, \\ s(t) &= s_0 + v_0 t + \textstyle \frac 12 at^2 \\ v(t)^2 &= v_0^2 + 2a(s(t) - s_0). \end{aligned}$

Proof. By integration techniques,

$\begin{aligned} v(t) &= \int_0^t a(u)\, \mathrm du \\ &= \int_0^t a\, \mathrm du \\ &= v(0) + \left[au\right]_0^t \\ &= v_0 + at. \end{aligned}$

Integrating a second time,

$\begin{aligned} s(t) &= \int_0^t v(u)\, \mathrm du \\ &= \int_0^t (v_0 + au)\, \mathrm du \\ &= s(0) + \left[v_0 u + \textstyle \frac 12 au^2\right]_0^t \\ &= s_0 + v_0 t + \textstyle \frac 12 at^2. \end{aligned}$

Therefore,

$\begin{aligned}v(t)^2 &= (v_0 + at)^2 \\ &= v_0^2 + 2v_0 at + a^2 t^2 \\ &= \textstyle v_0^2 + 2a \left(v_0 t + \frac 12 a t^2 \right) \\ &= v_0^2 + 2a(s(t) - s_0).\end{aligned}$

It turns out that these laws of kinematics can be used to start wars, applied in the context of projectile motion. The question is simple: If we fired a cannonball with initial position $(0, y_0)$ , initial velocity $v_0$ , and initial angle $\theta$ to the horizontal ground, what will the cannon’s path look like?

Corollary 1. Assuming no air resistance, the horizontal displacement $x(t)$ and vertical displacement $y(t)$ are given by

$x(t) = v_0 \cos(\theta) t,\quad y(t) = y_0 + (v_0 \sin(\theta))t - \frac 12 gt^2,$

where $g \approx 9.81\ \text{m}/\text{s}^2$ denotes gravitational acceleration on earth. Furthermore, the path of the cannonball follows the shape of a parabola (i.e. the graph of a quadratic function).

Proof. The initial horizontal velocity is $v_0 \cos(\theta)$ and the horizontal acceleration is $0$ , so that by Theorem 1,

$x(t) = 0 + v_0 \cos(\theta) t + \frac 12 \cdot 0 \cdot t^2 = v_0 \cos(\theta) t.$

Similarly, the initial vertical velocity is $v_0 \sin(\theta)$ and the vertical acceleration is $-g$ , so that by Theorem 1,

$y(t) = y_0 + (v_0 \sin(\theta))t - \frac 12 gt^2.$

By algebraic manipulation,

$\displaystyle \begin{aligned} y &= y_0 + \frac{v_0 \sin(\theta)}{v_0 \cos(\theta)} x - \frac 12 g \left( \frac{x}{v_0 \cos(\theta)} \right)^2 \\ &= y_0 + \tan(\theta)x - \frac{g}{2 v_0^2 \cos^2(\theta)}x^2\end{aligned}$

is the graph of a quadratic equation.

Now, what about if air resistance is involved? Then we need to solve differential equations. Furthermore, since we need to account for both $x$ – and $y$ -directions, we’ll need a tinge of vector calculus to properly answer this question. However, for simplicity, let’s solve answer the question for vertical motion and assume $x(t) = 0$ .

Theorem 2. Let $v(t)$ denote the velocity of an object with mass $m$ dropped from rest after time $t$ (so that $v(0) = 0$ ). Let $k >0$ be a constant that quantifies the air resistance that opposes the object’s motion. Newton’s second law yields the equation