Typesetting Exercises – Physicspupil

September 5, 2024September 5, 2024

202409051606 Pastime Exercise 012

About the figure below, tell some stories as probable as probable can be.

Roughwork.

Ignoring gravitational effect, denote the initial and the final velocity before and after impact by $\mathbf{u}=(u_x,u_y)$ and $\mathbf{v}=(v_x,v_y)$ . Taking the positive an up and a right, then $u_x,u_y,v_x<0$ and $v_y>0$ . Under conservation of momentum,

$\begin{aligned} m|\mathbf{u}|+M|\mathbf{U}| & =m|\mathbf{v}|+M|\mathbf{V}| \\ m|(u_x\,\hat{\mathbf{i}}+u_y\,\hat{\mathbf{j}})| + M|(\mathbf{0})| & = m|(v_x\,\hat{\mathbf{i}}+v_y\,\hat{\mathbf{j}})| + M|(\mathbf{0})| \\ |u_x| & = |v_x| \\ |u_y| & = |v_y| \\ \textrm{( } |u_i| & = |v_i| \textrm{ )}\\ \end{aligned}$

and conservation of energy

$\begin{aligned} \frac{1}{2}mu^2 + \frac{1}{2}MU^2 & = \frac{1}{2}mv^2 + \frac{1}{2}MV^2 \\ m(u_x^2+u_y^2) + M(0)^2 & = m(v_x^2+v_y^2) + M(0)^2 \\ u_x^2+u_y^2 & = v_x^2 + v_y^2 \\ \sum u_i^2 & = \sum v_i^2 \\ \end{aligned}$

Therefore

$\begin{aligned} \frac{|u_x|}{|u_y|} & = \frac{|v_x|}{|v_y|} \\ \tan\theta_i & = \tan\theta_r \\ \theta_i & = \theta_r \\ \end{aligned}$

(to be continued)

August 19, 2024August 20, 2024

202408191743 Geometry 0

$\forall$ : for all/any/each/every
$\exists$ : there exist(s)/is/are

If, in any event $(x,y,z,t)$ :

i. no two points $a$ and $b$ occupy the same position. I.e.,

$\forall\, t\, \nexists\, a(\neq b)\enspace\textrm{s.t. }(x_a,y_a,z_a)=(x_b,y_b,z_b)$

ii. none any point $a$ occupies more than one position. I.e.,

$\forall\,a,t\,\exists\, !\mathbf{r}\enspace\textrm{s.t. }a\stackrel{\mathbf{r}(t)}{\longmapsto}(x_a,y_a,z_a)$

such points are said to be mass particles; or else massless particles.

(to be continued)

May 9, 2024May 10, 2024

202405091824 Pastime Exercise 011

About the graph below, tell some stories as probable as probable can be.

Roughwork.

Assuming linear (/rectilinear) motion in a single direction.

Assuming uniform acceleration, there are two cases: i. zero acceleration and ii. non-zero (constant) acceleration; in the former velocity being (a) constant and the latter (b) non-constant.

Assuming a flat spacetime metric of which the spatial part does not expands with the temporal part.

Imagine a man walking along a straight line from point $(x_1,y_1)$ to point $(x_2,y_2)$ . From

$\displaystyle{\textrm{Speed }(v)=\frac{\textrm{Distance }(s)}{\textrm{Time }(t)}}$

as the path is fixed, i.e., $\Delta s=\textrm{Const.}$ , we see speed $v$ and time $t$

$\begin{aligned} v\uparrow\enspace \Leftrightarrow \enspace& t\downarrow \\ v\downarrow\enspace\Leftrightarrow\enspace& t\uparrow \\ \end{aligned}$

in an inverse relationship. As the man keeps his own fair pace and makes himself a good timekeeper, we can treat speed $v$ as an independent variable, and time $t$ a dependent variable.

$\textrm{\scriptsize{CASE} \textbf{\texttt{(a)}}}$ : velocity being constant

Parameterize the Cartesian equation $y=mx+c$ by the parameter $t'$ (*as distinguished from the natural/unit-speed/arc-length parameter time $t$ ) so as to write a set of parametric equations:

$\begin{aligned} & \begin{cases} s_x(t')=t' \\ s_y(t')=mt'+c \\ \end{cases} \\ & \\ & \begin{cases} v_x(t') =s_x'(t') = 1 \\ v_y(t') = s_y'(t') = m \\ \end{cases} \\ \end{aligned}$

$\begin{aligned} v^2(t') & = v_x^2(t')+v_y^2(t') \\ v(t') & = \displaystyle{\sqrt{\bigg(\frac{\mathrm{d}s_x(t')}{\mathrm{d}t'}\bigg)^2 +\bigg(\frac{\mathrm{d}s_y(t')}{\mathrm{d}t'}\bigg)^2}} \\ |\mathbf{v}(t')| & = \sqrt{1+m^2} \\ t & = \int_{0}^{t'}|\mathbf{v}(t)|\,\mathrm{d}t \\ & = (\sqrt{1+m^2})t'\\ \end{aligned}$

$\textrm{\scriptsize{CASE} \textbf{\texttt{(b)}}}$ : velocity being non-constant

The man begins with initial speed $v_1$ at start point $(x_1,y_1)$ and ends with final speed $v_2$ at finish point $(x_2,y_2)$ . We have

$\displaystyle{\textrm{Acceleration }(a)=\frac{\textrm{Change in speed }(v-u)}{\textrm{Time }(t)}}$

Write by SUVAT equations of motion:

$\begin{aligned} \mathbf{u} & = (u_x,u_y) \\ \mathbf{v} & = (v_x,v_y) \\ \mathbf{a} & = (a_x,a_y) \\ & = \bigg( \frac{v_x-u_x}{t}, \frac{v_y-u_y}{t}\bigg) \\ \mathbf{s} & = (s_x,s_y) \\ & = \bigg( u_xt+\frac{1}{2}a_xt^2, u_yt+\frac{1}{2}a_yt^2\bigg) \\ & = \bigg(\frac{1}{2}(u_x+v_x)t, \frac{1}{2}(u_y+v_y)t\bigg) \\ & = (x_2-x_1,y_2-y_1) \\ \end{aligned}$

Is this time $t$ also a natural parameter?

For a given parametric curve, the natural parametrization is unique up to a shift of parameter.

_{Wikipedia on Differentiable curve}

(to be continued)

May 8, 2024May 8, 2024

202405081402 Pastime Exercise 010

About the graph below, tell some stories as probable as probable can be.

Roughwork.

We naturally assume there are no forms of negative energy. That is, kinetic energy $\textrm{KE: } K\geqslant 0$ , potential energy $\textrm{PE: }U\geqslant 0$ , and (total) mechanical energy $K+U=E=\textrm{Const.}\geqslant 0$ .

The graph is divided into Zone ①, Zone ②, and Zone ③.

$\begin{aligned} V(x) & = \begin{cases} +\infty & \textrm{for }x\leqslant 1 \\ 2 & \textrm{for }1<x\leqslant 2 \\ -2|x-3|+4& \textrm{for }2\leqslant x\leqslant 5 \\ \frac{1}{2}(x^2-10x+25) & \textrm{for }5\leqslant x\\ \end{cases} \\ \end{aligned}$

For any conservative system, total energy $E$ takes the form of a horizontal line $y=\textrm{Const.}$ as in a graph. Take $E=2$ as an example:

$\begin{aligned} & \qquad E: y=2 \\ & \Longrightarrow \begin{cases} (E-U=)K> 0\Rightarrow x\in (4,7) \\ (E-U=)K = 0\Rightarrow x \in (1,2]\cup\{ 4\}\cup\{ 7\} \\ (E-U=)K< 0\Rightarrow x\in (-\infty ,1]\cup (2,4)\cup (7,\infty) \\ \end{cases} \\ \end{aligned}$

$\begin{aligned} & \qquad K=\frac{1}{2}mv^2 = \frac{p^2}{2m} \\ & \Longrightarrow \begin{cases} K(x)>0\Rightarrow \textrm{a particle moves at }x \\ K(x)=0\Rightarrow\textrm{a particle stays at }x \\ K(x)<0\Rightarrow \textrm{no particle exists at }x \\ \end{cases} \\ \end{aligned}$

(to be continued)

March 1, 2024March 1, 2024

202403011713 Revision Paper IV Q8

Ball $A$ (mass $3\,\mathrm{kg}$ ) is travelling due east in a straight line with speed $5\,\mathrm{m\,s^{-1}}$ when it collides with ball $B$ (mass $4\,\mathrm{kg}$ ) which was at rest. After the collision both balls move east with relative speed $2.5\,\mathrm{m\,s^{-1}}$ . Find their separate speeds.

_{Extracted from A. Godman & J. F. Talbert. (1973). Additional Mathematics Pure and Applied in SI Units.}

Roughwork.

Take eastward positive.

$\begin{aligned} \textrm{CoM:}\qquad p_{\textrm{initial}} & = p_{\textrm{final}} \\ \sum_{i=A}^{B}m_iu_i & = \sum_{i=A}^{B}m_iv_i \\ m_Au_A+m_Bu_B & = m_Av_A+m_Bv_B \\ (3)(5)+(4)(0) & = (3)(v_A)+(4)(v_B) \\ \textrm{Eq. (1):}\qquad\quad 15 & = 3v_A+4v_B \\ \textrm{Eq. (1)':}\qquad\quad v_A & = 5-\frac{4}{3}v_B \\ \end{aligned}$

$\begin{aligned} \textrm{CoE:}\qquad\textrm{KE}_{\textrm{initial}} & = \textrm{KE}_{\textrm{final}} \\ \sum_{i=A}^{B}\frac{1}{2}m_iu_i^2 & = \sum_{i=A}^{B}\frac{1}{2}m_iv_i^2 \\ m_Au_A^2+m_Bu_B^2 & = m_Av_A^2+m_Bv_B^2 \\ (3)(5)^2 + (4)(0)^2 & = (3)(v_A)^2 + (4)(v_B)^2 \\ \textrm{Eq. (2):}\qquad\qquad 75 & = 3v_A^2 + 4v_B^2 \\ \end{aligned}$

$\begin{aligned} 75 & = 3\bigg( 5-\frac{4}{3}v_B\bigg)^2 + 4v_B^2 \\ 25 & = \bigg( 25 - \frac{40}{3}v_B + \frac{16}{9}v_B^2\bigg) + \frac{4}{3}v_B^2 \\ 0 & = v_B\bigg( -\frac{40}{3} + \frac{28}{9}v_B \bigg) \\ v_B & = 0\textrm{ \scriptsize{OR} }\frac{30}{7} \\ v_A & = \bigg[ 5-\frac{4}{3}(0)\bigg] \textrm{ \scriptsize{OR} }\bigg[ 5-\frac{4}{3}\bigg(\frac{30}{7}\bigg)\bigg] \\ & = 5\textrm{ \scriptsize{OR} }-\frac{5}{7} \\ \end{aligned}$

$(v_A,v_B) = \bigg( \displaystyle{-\frac{5}{7} , \frac{30}{7}}\bigg) \qquad \text{}^*\textrm{(5,0) is rejected.}$

This contradicts with $v_B-2.5=v_A>0$ . As the law of conservation of momentum is always observed, herein has energy $\textrm{\scriptsize{NOT}}$ been conserved.

This problem is not to be attempted.

February 15, 2024February 15, 2024

202402151329 Solution to 1985-CE-AMATH-II-5

If the equation

$x^2+y^2+kx-(2+k)y=0$

represents a circle with radius $\sqrt{5}$ ,

(a) find the value(s) of $k$ ;
(b) find the equation(s) of the circle(s).

As if sitting an exam in additional mathematics, it certainly not being showing off, we should perhaps involve ourselves with some sort of calculus.

Roughwork.

As always, have in mind some pictures. Hence, we draw:

and also

after the stage got set, we write

$\begin{aligned} 0 & = x^2+y^2+kx-(2+k)y \\ 0 & = 2x\,\mathrm{d}x+2y\,\mathrm{d}y + k\,\mathrm{d}x - (2+k)\,\mathrm{d}y \\ y' & = \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x+k}{k-2y+2} \\ \end{aligned}$

and furthermore,

$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x} & = 0 \\ & \Rightarrow 2x+k=0 \\ & \Rightarrow x = -\frac{k}{2} \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \infty \\ & \Rightarrow k-2y+2 = 0 \\ & \Rightarrow y = \frac{k}{2}+1 \\ \end{aligned}$

In the case of $x=-\frac{k}{2}$ :

$\begin{aligned} 0 & = \bigg( -\frac{k}{2}\bigg)^2+y^2+k\bigg( -\frac{k}{2}\bigg) -(2+k)y \\ 0 & = y^2 - (2+k)y - \frac{k^2}{4} \\ \Delta & = \big( -(2+k)\big)^2 - 4(1)\bigg( -\frac{k^2}{4}\bigg) \\ & = 2(k^2+2k+2)\qquad (> 0) \\ y_1,y_3 & = \frac{-\big( -(2+k)\big) \pm \sqrt{\Delta}}{2(1)} \\ 2\sqrt{5} & = y_3-y_1 \\ \sqrt{\Delta} & = 2\sqrt{5} \\ \Delta & = 20 \\ 20 & = 2(k^2+2k+2) \\ 0 & = k^2+2k-8 \\ 0 & = (k-2)(k+4) \\ k & = 2,-4 \\ \end{aligned}$

and the case of $y=\frac{k}{2}+1$ :

$\begin{aligned} 0 & = x^2 + \bigg( \frac{k}{2}+1\bigg)^2 + kx - (2+k)\bigg( \frac{k}{2}+1\bigg) \\ 0 & = x^2+kx - \bigg(\frac{k}{2}+1\bigg)^2\\ \Delta & = (k)^2 - 4(1)\bigg( -\bigg(\frac{k}{2}+1\bigg)^2\bigg) \\ & = 2(k^2+2k+2) \qquad (> 0) \\ x_1,x_3 & = \frac{-(k)\pm\sqrt{\Delta}}{2(1)} \\ 2\sqrt{5} & = x_3 - x_1 \\ \sqrt{\Delta} & = 2\sqrt{5} \\ \Delta & = 20 \\ 20 & = 2(k^2+2k+2) \\ 0 & = k^2+2k-8 \\ 0 & = (k-2)(k+4) \\ k & = 2,-4 \\ \end{aligned}$

Targeting the centre $C(a,b)$ :

$\begin{aligned} a_{i=1,2} & = \frac{x_3 - x_1}{2} = -\frac{k}{2}\bigg|_{k=2,-4} = -1,2 \\ b_{i=1,2} & = \frac{y_3-y_1}{2} = \frac{-\big( -(2+k)\big)}{2}\bigg|_{k=2,-4} = 2,-1 \\ (a,b) & = (-1,2)\cup (2,-1) \\ \end{aligned}$

Of the equation of circle, the standard form is

$\left\{ \begin{aligned} (x+1)^2 + (y-2)^2 & = (\sqrt{5})^2 \\ (x-2)^2 + (y+1)^2 & = (\sqrt{5})^2 \\ \end{aligned}\right\}$

and the general form

$\left\{ \begin{aligned} x^2+y^2+2x-4y & = 0 \\ x^2+y^2-4x+2y & = 0 \\ \end{aligned}\right\}$

This problem is not to be attempted.

February 5, 2024February 5, 2024

202402051319 Pastime Exercise 009

The blogger claims no originality of his problem below.

On a rollover road,

a vehicle performs circular motion:

Discuss how the driver could prevent a traffic accident.

Roughwork.

Take positive both, the angle $\theta$ in an anti-clockwise direction, and the displacement $\mathbf{s}$ in a rightward and an upward direction.

$\begin{aligned} \mathbf{W} & = m\mathbf{g} = -mg\,\hat{\mathbf{j}} \\ \mathbf{N}(\theta ) & = -N(\theta )\,\hat{\mathbf{r}} \\ \mathbf{N}(\theta ) & = N_x(\theta )\,\hat{\mathbf{i}} + N_y(\theta )\,\hat{\mathbf{j}} \\ N^2(\theta ) & = N_x^2(\theta )+N_y^2(\theta ) \\ N_x(\theta ) & = N\sin\theta \\ N_y(\theta ) & = N\cos\theta \\ \mathbf{f}(\theta ) & = -\mu N(\theta )\,\hat{\boldsymbol{\theta}} \\ \mathbf{f}(\theta ) & \perp \mathbf{N}(\theta ) \\ \end{aligned}$

In order for the vehicle not to leave the track, the radius of curvature $r$ $\textrm{\scriptsize{MUST}}$ be kept constant, i.e., $r=R(=\textrm{Const.})$ .

Recall, in uniform circular motion there isn’t any (angular) acceleration in angular velocity, i.e.,

$\begin{aligned} \alpha (t) & = 0 \\ \frac{\mathrm{d}}{\mathrm{d}t}\big(\omega (t)\big) & = 0 \\ \frac{\mathrm{d}}{\mathrm{d}t}\bigg( \frac{\mathrm{d}}{\mathrm{d}t}\big(\theta (t)\big)\bigg) & = 0 \\ \frac{\mathrm{d}}{\mathrm{d}t}\big(\theta (t)\big) & = a \\ \theta (t) & = at+b\qquad (\exists\, a,b\in\mathbb{R}) \\ \end{aligned}$

If uniform circular motion is $\textrm{\scriptsize{NOT}}$ assumed,

i.e., $\alpha \neq 0\textrm{ \scriptsize{OR} }r\neq\textrm{Const.}$ ,

then angular displacement $\theta (t)$ wrt time $t$ will be a polynomial in an indeterminate $t$ of degree $2$ or higher,

i.e., $O(t)=O(t^c)\qquad\exists\, c\, (\in\mathbb{Z}^+)\geqslant 2$ .

Angular acceleration $\boldsymbol{\alpha}$ is less often mentioned than is its parallel tangential acceleration $\mathbf{a}_{\parallel}\, (=r\boldsymbol{\alpha})$ as the pair to centripetal (/centrifugal) acceleration $\mathbf{a}_{\perp}$ .

By Newton 2^nd Law, write

$\begin{aligned} \mathbf{F}_{\textrm{net}} & = m\mathbf{a} \\ \mathbf{W}+\mathbf{N}+\mathbf{f} & = m(\mathbf{a}_{\parallel}+\mathbf{a}_{\perp}) \\ \end{aligned}$

The presence of a centripetal acceleration, i.e., $\exists\,\mathbf{a}_{\perp}\neq\mathbf{0}$ , is necessary for any circular motion, be it uniform or not; but not sufficient for the absence of a tangential acceleration $\forall\,\mathbf{a}_{\parallel}= 0$ (why?). One knows instinctively, that the vehicle should possess a minimum angular speed $\min (\omega )$ to do this dangerous stunt. And so much the better if beyond this bound $\min (\omega )\leqslant \omega \leqslant \max (\omega ) \ll c$ one has all degrees of freedom.

By definition,

$\begin{aligned} \mathbf{a}_{\parallel} & = R\dot{\omega}\,\hat{\boldsymbol{\theta}} = R\ddot{\theta}\,\hat{\boldsymbol{\theta}} \\ \mathbf{a}_{\perp} & = -R\omega^2\,\hat{\mathbf{r}} = -R\dot{\theta}^2\,\hat{\mathbf{r}} \\ \end{aligned}$

Stepping on and off the accelerator (/gas pedal), however delicately, is $\textrm{\scriptsize{NOT}}$ designed to keep a constant acceleration for the gear.

This problem is not to be attempted.

January 31, 2024January 31, 2024

202401311548 Pastime Exercise 008

Let there be a cone $R\,\mathrm{(unit)}$ in radius of its base and $h\,\mathrm{(unit)}$ in height.

Find the volume $V\,\mathrm{(cubic\, unit)}$ of the cone, using what are so-called i. the disc (/washer) method and ii. the (cylindrical) shell method, if the names my memory serves me right.

Answer. $\displaystyle{V=\frac{1}{3}\pi R^2h}$ .

Roughwork.

Visualize the cone.

i. By disc method,

$\begin{aligned} \frac{R}{h} & = \frac{r}{h-z} = \tan\theta \\ r & = R\bigg( 1-\frac{z}{h}\bigg) \\ V & = \pi R^2\int_{0}^{h}\bigg( 1-\frac{z}{h}\bigg)^2\,\mathrm{d}z \\ \cdots & \cdots\cdots\cdots\cdots\cdots\cdots\cdots \\ \textrm{let }& z'=1-\frac{z}{h} \\ \frac{\mathrm{d}z'}{\mathrm{d}z} & = -\frac{1}{h} \\ \mathrm{d}z & = -h\,\mathrm{d}z' \\ z=0 & \Leftrightarrow z'=1 \\ z=h & \Leftrightarrow z'= 0 \\ \cdots & \cdots\cdots\cdots\cdots\cdots\cdots\cdots \\ V& = \pi R^2\int_{1}^{0} z'^2(-h\,\mathrm{d}z') \\ & = -\pi R^2h \bigg[ \frac{z'^3}{3}\bigg]\bigg|_{1}^{0} \\ & = -\pi R^2h \bigg(\frac{(0)^3}{3} - \frac{(1)^3}{3}\bigg) \\ & = \frac{\pi R^2h}{3}\\ \end{aligned}$

ii. By shell method,

we are obtaining the volume of a solid of revolution about the $z$ -axis by integrating the slices (/cross sections) ranging between $\theta\in [0,2\pi ]$ .

It is left the reader as an exercise.

This problem is not to be attempted.

January 3, 2024January 3, 2024

202401031807 Pastime Exercise 007

The blogger claims no originality of his problem below.

A ball of mass $m$ is being passed through a smooth extensible light string, the two ends of which being attached to walls a distance $d$ apart.

When time $t=t_0$ :

when $t=t_1$ :

when $t=t_2$ :

when $t=t_3$ :

when $t=t_4$ :

when $t=t_5$ :

when $t=t_6$ :

such that stroboscopically for $t\in [t_0,t_6]$ , we see the ball’s trajectory fits into a parabola as can be described by some quadratic equation:

The tension $T(t)$ of the string and the speed $v(t)$ of the ball are time-varying variables dependent on mass $m$ and distance $d$ as well. By considering the free-body diagrams of the ball and of the string in discrete time frames $t_i$ ‘s (where $i\in\{ 0,1,2,3,4,5,6\}$ ), find the equation of motion at continuous time intervals of $\Delta t=t_6-t_0$ .

This problem is not to be attempted.

November 8, 2023December 10, 2023

202311081445 Pastime Exercise 006

The figure below is a representation of the $\mathbf{E}$ -field by electric field lines in a family $R$ of infinitely many quadratic functions

$y_t(x_{t'})=a_tx_{t'}^2+b_tx_{t'}+c_t$

Roughwork.

A tangent to any quadratic curve at some point in the locus gives the $\mathrm{\pm ve}$ direction of the electric force experienced by a (positive) test charge placed there. I.e.,

$\displaystyle{T(x_{t'},y_{t})=\frac{\mathrm{d}}{\mathrm{d}x}\big( y_t(x_{t'})\big) = 2a_tx_{t'}+b_{t}}$ ,

the slope of tangent.

WLOG we work with the first quadrant. First, begin with the repulsive force $\mathrm{}_Q\mathbf{F}_{q}$ acting on the test charge $+q$ due to point charge $+Q$ .

$\begin{aligned} \mathrm{}_QF_{q,x}^2+\mathrm{}_QF_{q,y}^2 & = \mathrm{}_QF_{q}^2 \\ \mathrm{}_QF_{q,y} & = \bigg(\frac{y_t}{x_{t'}+d/2}\bigg)\mathrm{}_QF_{q,x} \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_QF_{q,x} & = \frac{\mathrm{}_QF_{q}}{\sqrt{1+\Big(\frac{y_t}{x_{t'}+d/2}\Big)^2}} \\ \mathrm{}_QF_{q,y} & = \bigg(\frac{y_t}{x_{t'}+d/2}\bigg) \Bigg(\frac{\mathrm{}_QF_{q}}{\sqrt{1+\Big(\frac{y_t}{x_{t'}+d/2}\Big)^2}}\Bigg) \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_Q\mathbf{F}_{q} & = \mathrm{}_QF_{q,x}\,\hat{\mathbf{i}}+ \mathrm{}_QF_{q,y}\,\hat{\mathbf{j}} \\ \end{aligned}$

Next, continue with the attractive force $\mathrm{}_{-Q}\mathbf{F}_{q}$ acting on the test charge $+q$ due to point charge $-Q$ .

$\begin{aligned} \mathrm{}_{-Q}F_{q,x}^2+\mathrm{}_{-Q}F_{q,y}^2 & = \mathrm{}_{-Q}F_{q}^2 \\ \mathrm{}_{-Q}F_{q,y} & = \bigg(\frac{y_t}{d/2-x_{t'}}\bigg)\mathrm{}_{-Q}F_{q,x} \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_{-Q}F_{q,x} & = \frac{\mathrm{}_{-Q}F_{q}}{\sqrt{1+\Big(\frac{y_t}{d/2-x_{t'}}\Big)^2}} \\ \mathrm{}_{-Q}F_{q,y} & = \bigg(\frac{y_t}{d/2-x_{t'}}\bigg) \Bigg(\frac{\mathrm{}_{-Q}F_{q}}{\sqrt{1+\Big(\frac{y_t}{d/2-x_{t'}}\Big)^2}}\Bigg) \\ \cdots\cdots & \cdots\cdots\\ \mathrm{}_{-Q}\mathbf{F}_{q} & = \mathrm{}_{-Q}F_{q,x}\,\hat{\mathbf{i}}+ \mathrm{}_{-Q}F_{q,y}\,\hat{\mathbf{j}} \\ \end{aligned}$

Adding $\mathrm{}_{Q}\mathbf{F}_{q}$ and $\mathrm{}_{-Q}\mathbf{F}_{q}$ will give the resultant electric force $\mathbf{F}_{E}(x_{t'},y_t)$ .

Note that

$\begin{aligned} \mathrm{}_QF_q & = k\frac{Q}{\mathrm{}_Qr_{q}^2} \\ \mathrm{}_{-Q}F_q & = k\frac{-Q}{\mathrm{}_{-Q}r_{q}^2} \\ \end{aligned}$

where

$\begin{aligned} \mathrm{}_Qr_q & = \sqrt{(d/2+x_{t'})^2+y_t^2} \\ \mathrm{}_{-Q}r_q & = \sqrt{(d/2-x_{t'})^2+y_t^2} \\ & \\ & \\ \end{aligned}$

are the distances of a test charge at $q(x_{t'},y_t)$ from point charges $+Q$ at $A(-d/2,0)$ and $-Q$ at $B(d/2,0)$ .

The smallest angle between $\mathbf{F}_{E}(x_{t'},y_{t})$ and the level should be equal to the slope of tangent $2a_tx_{t'}+b_{t}$ at point $q(x_{t'},y_t)$ .