June 2023

Problem:

Shooting an object into the air at an angle $\theta$

where

$\begin{cases} x = v_0\cos(\theta)t \quad\quad\quad\quad\quad\quad(1-1)\\ y = v_0\sin(\theta)t - \frac{1}{2}gt^2\quad\quad\quad(1-2)\end{cases}$

With $v_0$ fixed, find the angle $\theta \;\; (0 < \theta <\frac{\pi}{2})$ so that the object lands the maximum distance on the ground.

Solution:

When the object lands at time $t, y = 0.$ By (1-2)

$0 = v_0\sin(\theta) t - \frac{1}{2}gt^2.\quad\quad\quad(1-3)$

Solving (1-1) for $t$ yields

$t = \frac{x}{v_0 cos(\theta)}.$

Substituting it into (1-3) gives $0 = v_0\sin(\theta)\frac{x}{v_0\cos(\theta)}-\frac{1}{2}g\left(\frac{x}{v_0\cos(\theta)}\right)^2.$ i.e.,

$0 = v_0^2\sin(\theta)\cos(\theta)x-\frac{1}{2}gx^2.$

$x = \frac{2v_0^2\sin(\theta)\cos(\theta)}{g} = \frac{v_0^2\sin(2\theta)}{g}.$

Since $v_0, g$ are positive quantities, $x>0$ requires $\sin(2\theta) > 0.$ That is,

$\sin(2\theta) = |\sin(2\theta)| \le 1.\quad\quad\quad(1-4)$

Hence,

$x = \frac{v_0^2\sin(2\theta)}{g} \overset{(1-4)}{\le}\frac{v_0^2}{g} \implies x_{max}=\frac{v_0^2}{g}$

and it occurs when $\sin(2\theta) = 1$ or $2\theta = \frac{\pi}{2} \implies \theta = \frac{\pi}{4}.$

Consider any repeatable experiment whose outcome is non-deterministic.

Suppose with $n$ trials, the occurrence of outcome $A$ is $n_A$ . The ratio $\frac{n_A}{n}$ lies between $0$ and $1$ inclusive. i.e.,

$0 \le \frac{n_A}{n} \le 1.\quad\quad\quad(1)$

If we define the probability of outcome $A$ as

$P(A) = \lim\limits_{n \to \infty}\frac{n_A}{n},\quad\quad\quad(2)$

we must have

$0 \le P(A) \le 1\quad\quad\quad(*)$

as a result.

To show that (*) is true, we first prove

$\left(a_n \le b_n, \lim\limits_{n\to \infty}a_n = a, \lim\limits_{n \to \infty} b_n =b\right) \implies a\le b\quad\quad\quad(3)$

by reductio ad abusrdum:

If $a > b$ then by definition,

$\exists n_1 \ni n>n_1 \implies |a_n-a| <\frac{a-b}{2};$

$\exists n_2 \ni n>n_2 \implies |b_n-b| <\frac{a-b}{2}.$

It means for $n > n_*= max(n_1, n_2)$ ,

$|a_n-a| < \frac{a-b}{2} \implies \underline{-\frac{a-b}{2} < a_n-a }< \frac{a-b}{2}\implies a_n > \frac{a+b}{2};$

$|b_n-b| < \frac{a-b}{2} \implies -\frac{a-b}{2} < \underline{b_n-b < \frac{a-b}{2}} \implies b_n < \frac{a+b}{2}.$

And so $b_n < a_n$ which contradicts $a_n \le b_n.$

Express $a_n, b_n$ in (3) as $c_n, a_n$ respecitvely, we have

$\left(c_n \le a_n, \lim\limits_{n\to \infty}c_n = c, \lim\limits_{n \to \infty} a_n =a\right) \implies c\le a$

Hence,

$\left(c_n \le a_n \le b_n, \lim\limits_{n\to \infty}c_n =c, \lim\limits_{n\to \infty}a_n=a, \lim\limits_{n\to \infty}b_n=b\right)\implies c\le a\le b.\quad(4)$

Now, let $c_n = 0, a_n=\frac{n_A}{n}, b_n=1.$

We have

$c = \lim\limits_{n\to \infty} c_n =\lim\limits_{n\to \infty} 0 = 0,$

$a = \lim\limits_{n \to \infty}a_n = \lim\limits_{n \to \infty} \frac{n_A}{n} = P(A),$

$b = \lim\limits_{n\to \infty} b_n =\lim\limits_{n\to \infty} 1 = 1.$

It follows that by $(1)$ and $(4)$ ,

$0 \le P(A) \le 1.$

Month: June 2023

Solving Minimization and Maximization Problems without Calculus [2]

Why probability lies between 0 and 1 inclusive