Applied Mathematics – Hong Kong Advanced Level (HKAL)

December 7, 2022December 7, 2022

202212071606 Solution to 1976-AL-AMATH-I-8

During an epidemic in a country, the rate of spread of the disease is proportional to the number of people who are healthy; the rate of cure is proportional to the number currently sick. Let $P$ be the total population:

(a) Show that

$\displaystyle{\frac{\mathrm{d}W}{\mathrm{d}t}=kP-(k+k_1)W}$

where $W$ is the number of sick people; $k$ and $k_1$ are constants.
(b) Let the total population be growing at a steady rate, so that (independent of the epidemic)

$P=a+bt$

where $a$ and $b$ are positive. Show that the number of sick people is given by

$W=\alpha +\beta t+ce^{\mu t}$ .

Calculate $\alpha$ , $\beta$ , and $\mu$ in terms of the given constants $a$ , $b$ , $k$ , and $k_1$ .
What further information would you require to determine $c$ ?
(c) After a very long time, what is the proportion of people who will be in a state of sickness?

Roughwork.

$\begin{aligned} \frac{\mathrm{d}W}{\mathrm{d}t}+(k+k_1)W & = k(a+bt) \\ \dots\,\textrm{ by integrating factor } I=e^{\int (k+k_1)\,\mathrm{d}t} & =e^{(k+k_1)t}\,\dots \\ e^{(k+k_1)t}\bigg(\frac{\mathrm{d}W}{\mathrm{d}t}\bigg) + \Big( e^{(k+k_1)t} (k+k_1)\Big) W & = e^{(k+k_1)t} k(a+bt) \\ e^{(k+k_1)t}W & = \int e^{(k+k_1)t}k(a+bt)\,\mathrm{d}t \\ \end{aligned}$

$\begin{aligned} \textrm{RHS} & = \int e^{(k+k_1)t}k(a+bt)\,\mathrm{d}t \\ & = ka\int e^{(k+k_1)t}\,\mathrm{d}t + kb\int te^{(k+k_1)t}\,\mathrm{d}t \\ & = \frac{ka}{k+k_1}\Big( e^{(k+k_1)t}\Big) + \frac{kb}{(k+k_1)^2}\int t'e^{t'}\,\mathrm{d}t' \\ & = \frac{ka}{k+k_1}\Big( e^{(k+k_1)t}\Big) + \frac{kb}{(k+k_1)^2}\Big( e^{(k+k_1)t}((k+k_1)t-1)+C\Big) \\ W &= \textrm{RHS}\Big/ e^{(k+k_1)t} \\ & = \bigg(\frac{ka}{k+k_1}-\frac{kb}{(k+k_1)^2}\bigg) + \bigg(\frac{kb}{k+k_1}\bigg) t + \bigg(\frac{kbC}{(k+k_1)^2}\bigg) e^{(-(k+k_1))t} \\ & = \alpha +\beta t+ce^{\mu t} \\ \end{aligned}$

$\begin{aligned} \frac{W}{P} & = \frac{\alpha +\beta t+ce^{\mu t}}{a+bt} \\ \lim_{t\to \infty}\frac{W}{P} & = \lim_{t\to\infty}\frac{\beta t}{a+bt} \\ & = \frac{\beta}{b} \\ & = \frac{k}{k+k_1} \\ \end{aligned}$

This problem is not to be attempted.

December 5, 2022December 5, 2022

202212051136 Solution to 1978-AL-AMATH-II-3

A smooth homogeneous hollow right circular cylinder with open, flat ends stands freely on smooth horizontal ground so that its axis is vertical. Two spheres $A$ and $B$ of radii $a$ and $b$ ( $a<b$ ) and weights $W_a$ and $W_b$ respectively rest in equilibrium inside the cylinder as shown in the diagram. Suppose that the internal and external radii of the cylinder are $c$ and $d$ respectively, where $c<(a+b)$ .

i. Show that the vertical forces acting on the spheres reduce to a couple. Determine the moment of the couple.
ii. Determine the minimum weight of the cylinder such that it will not overturn.
iii. A third sphere $C$ , identical to $A$ , is then placed on top of $B$ in contact with the cylinder. Determine the minimum weight of the cylinder so that it will not overturn for the two possible equilibrium positions of $C$ .

You may assume that the cylinder is tall enough to hold all the spheres.

Roughwork.

WLOG reduce the problem from three-dimensional to two. Begin with three free-body diagrams as follow:

So many unknowns I don’t know how to get set.

(to be continued)

November 25, 2022November 26, 2022

202211251510 Solution to 1982-AL-AMATH-I-3

A rocket fully loaded with fuel has total mass $M$ including mass $F$ of fuel. The rocket is fired vertically upwards at $t=0$ . During its journey, the fuel is allowed to burn at a constant rate $b$ such that the relative backward velocity of the exhaust gases is $u$ .

(a) If $m(t)$ is the mass of the rocket plus fuel and $v(t)$ its velocity at time $t$ , show that, if air resistance is neglected, the equation of motion is

$\displaystyle{-mg=m\frac{\mathrm{d}v}{\mathrm{d}t}+u\frac{\mathrm{d}m}{\mathrm{d}t}}$ .

(b) Show that, the speed of the rocket at time $\displaystyle{t\leqslant \frac{F}{b}}$ is given by

$\displaystyle{-gt-u\ln \bigg( 1-\frac{b}{M}t\bigg)}$ .

(c) The height $H$ which is reached by the rocket at the instant when the fuel is all burnt depends on the rate of burning $b$ . Determine the rate of burning $b_0$ such that the height reached will be maximal. (Hint: For the stationary value to be maximal, first show that

$\displaystyle{\frac{\mathrm{d}^2H}{\mathrm{d}b^2}=-\frac{gF^2}{b_0^4}}$

at $b=b_0$ )

Roughwork.

(a) Jot $m(t)\big|_{t=0}=M$ . Take $\big\uparrow \textrm{(+ve)}$ . By Newton’s 2^nd law,

$\begin{aligned} \textrm{Net }F & =\frac{\mathrm{d}p}{\mathrm{d}t} \\ -mg & = \frac{\mathrm{d}}{\mathrm{d}t}\big( mv\big) \\ -mg & = m\frac{\mathrm{d}v}{\mathrm{d}t} + u\frac{\mathrm{d}m}{\mathrm{d}t} \\ \end{aligned}$

(b)

$\begin{aligned} \int (-g)\,\mathrm{d}t & = \int\bigg(\frac{\mathrm{d}v}{\mathrm{d}t}\bigg)\,\mathrm{d}t+\int\bigg(\frac{u}{m}\frac{\mathrm{d}m}{\mathrm{d}t}\bigg)\,\mathrm{d}t \\ -gt & = \int\mathrm{d}v + u\int\frac{\mathrm{d}m}{m}\\ \end{aligned}$

and the result follows.

(c) From

$\begin{aligned} v(t) & = -gt-u\ln \bigg( 1-\frac{b}{M}t\bigg) \\ H(t) & =\int_0^t v(t')\,\mathrm{d}t' \\ & = \int_0^t \bigg[ -gt'-u\ln \bigg( 1-\frac{b}{M}t'\bigg) \bigg]\,\mathrm{d}t' \\ & = -\frac{g}{2}t'^2\bigg|_0^t +\frac{uM}{b}\int_{t'=0}^{t'=t} \ln T\,\mathrm{d}T \\ \dots\enspace & \textrm{as }\int \ln x\,\mathrm{d}x=x\ln x-x+C\enspace \dots \\ \end{aligned}$

This problem is not to be attempted.

April 8, 2021October 31, 2022

202104080551 Solution to 1973-AL-AMATH-I-5

A particle of mass $m$ moving in a straight line in a medium with a velocity $v$ is subjected to a force $mk(v^3+a^2v)$ in the direction opposite to the velocity, where $k$ and $a$ are constants. Show that, at any subsequent time $t$ , the velocity $v$ is related to the initial velocity $v_0$ by

$\displaystyle{\frac{v}{(v^2+a^2)^{1/2}}=\frac{v_0}{(v_0^2+a^2)^{1/2}}e^{-a^2kt}}$ .

Show that for any value of the initial velocity, no matter how large, the particle will never travel a distance greater than $\pi /(2ka)$ .

Background.

Newton's second law:

$\begin{aligned} F = -mk(v^3+a^2v) & = ma \\ -mk(\dot{x}^3+a^2\dot{x}) & = m\ddot{x} \\ \frac{\mathrm{d}\dot{x}}{\mathrm{d}t} & = -k(\dot{x}^3+a^2\dot{x}) \\ \frac{\mathrm{d}\dot{x}}{\dot{x}^3+a^2\dot{x}} & = -k\,\mathrm{d}t \\ \dots \textrm{integrating from }& (v=v_0,\, t=0)\textrm{ to }(v=v,\, t=t) \\ \int_{v_0}^{v} \frac{1}{\dot{x}(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x} & = \int_{0}^{t}-k\,\mathrm{d}t \end{aligned}$

(to be continued)

Roughwork.

Let . Then,

$\begin{aligned} 1 & \equiv A(u^2+a^2)+Bu^2+Cu \\ 1 & \equiv (A+B)u^2+Cu+Aa^2 \\ A+B & = 0 \\ C & = 0 \\ A & = \frac{1}{a^2} \\ \Rightarrow B & = -\frac{1}{a^2} \end{aligned}$
Thus,
$\displaystyle{\int \frac{1}{u(u^2+a^2)}\,\mathrm{d}u = \int \bigg(\frac{1}{a^2u} - \frac{u}{a^2(u^2+a^2)}\bigg) \,\mathrm{d}u}$

(continue)

$\begin{aligned} & \quad \int_{v_0}^{v} \frac{1}{\dot{x}(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x} \\ & = \int_{v_0}^{v} \bigg(\frac{1}{a^2\dot{x}} - \frac{\dot{x}}{a^2(\dot{x}^2+a^2)}\bigg) \,\mathrm{d}\dot{x} \\ & = \int_{v_0}^v\frac{1}{a^2\dot{x}}\,\mathrm{d}\dot{x} - \int_{v_0}^v \frac{\dot{x}}{a^2(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x} \\ \end{aligned}$

The first term is

$\begin{aligned} \int_{v_0}^v\frac{1}{a^2\dot{x}}\,\mathrm{d}\dot{x} & = \bigg[ \frac{\ln \dot{x}}{a^2} \bigg]\bigg|_{v_0}^{v} \\ & = \bigg( \frac{\ln v}{a^2} \bigg) - \bigg( \frac{\ln v_0}{a^2} \bigg) \\ & = \frac{1}{a^2}\ln \bigg( \frac{v}{v_0} \bigg) \end{aligned}$

and the second term is

$\begin{aligned} &\quad -\int_{v_0}^v \frac{\dot{x}}{a^2(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x}\\ \dots & \textrm{ let } \dot{x}=a\tan\theta \enspace\dots \\ \dots & \textrm{ let } \theta = \tan^{-1}\bigg(\frac{v}{a}\bigg) \textrm{ and } \theta_0 = \tan^{-1}\bigg(\frac{v_0}{a}\bigg) \\ & = -\int_{\theta_0}^{\theta} \frac{a\tan\theta}{a^2(a^2\tan^2\theta +a^2)}\cdot (a\sec^2\theta)\,\mathrm{d}\theta \\ & = -\int_{\theta_0}^{\theta} \frac{\tan\theta}{a(a^2\sec^2\theta)}\cdot (a\sec^2\theta )\,\mathrm{d}\theta \\ & = -\int_{\theta_0}^{\theta} \frac{\tan\theta}{a^2}\,\mathrm{d}\theta \\ & = -\bigg[ -\frac{\ln |\cos\theta |}{a^2} \bigg]\bigg|_{\theta_0}^{\theta} \\ & = \frac{1}{a^2} \bigg[\ln \bigg| \frac{a}{\sqrt{\dot{x}^2+a^2}} \bigg|\bigg]\bigg|_{v_0}^{v} \\ & = \frac{1}{a^2} \bigg( \ln\bigg| \frac{a}{\sqrt{v^2+a^2}} \bigg| -\ln\bigg| \frac{a}{\sqrt{v_0^2+a^2}} \bigg| \bigg) \\ & = \frac{1}{a^2} \ln \bigg( \frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}} \bigg) \\ \end{aligned}$

Back to an earlier line

$\displaystyle{\int_{v_0}^{v} \frac{1}{\dot{x}(\dot{x}^2+a^2)}\,\mathrm{d}\dot{x} = \int_{0}^{t}-k\,\mathrm{d}t}$

we have

$\begin{aligned} \frac{1}{a^2}\ln\bigg( \frac{v}{v_0} \bigg) + \frac{1}{a^2}\ln \bigg( \frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}}\bigg) & = -kt \\ \ln\bigg( \frac{v}{v_0} \bigg) + \ln \bigg( \frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}}\bigg) & = -a^2kt \\ \ln\bigg( \frac{v}{v_0}\frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}} \bigg) & =-a^2kt \\ \frac{v}{v_0}\frac{\sqrt{v_0^2+a^2}}{\sqrt{v^2+a^2}} & = e^{-a^2kt} \\ \end{aligned}$

or,

$\boxed{\displaystyle{\frac{v}{\sqrt{v^2+a^2}}=\frac{v_0}{\sqrt{v_0^2+a^2}}e^{-a^2kt}}}$

as desired.

It remains to be shown that for any value of the initial velocity, no matter how large, the particle will never travel a distance greater than $\pi /(2ka)$ .

(to be continued)

April 6, 2021October 14, 2023

202104060103 Solution to 1973-AL-AMATH-I-2

When a pan of mass $m$ is put on top of a vertical spring scale of negligible weight, the downward displacement of the spring is observed to be $d$ . A ball of mass $M$ is now dropped from a height $h$ above the pan onto the pan. Suppose that the coefficient of restitution is equal to zero.

i. Show that the velocity of the pan immediately after the impact is

$\displaystyle{\frac{M\sqrt{2gh}}{M+m}}$ .

ii. From energy considerations, determine the maximum deflection of the pan after the ball has been dropped onto it.

Note: Force in spring scale is proportional to displacement.

Solution. (bad, if not wrong)

i.

Background.

1. If the coefficient of restitution is zero, the collision is perfectly inelastic.

2. Recall Hooke’s law:

$\begin{aligned} \mathbf{F_s} & =-k\mathbf{x} \\ F & = kx \\ U(x) & =\displaystyle{\frac{1}{2}}kx^2 \\ \end{aligned}$

Let $t$ be the time it takes for the ball to free-fall a vertical distance $h$ .

$\begin{aligned} s & =ut+\frac{1}{2}at^2 \\ \dots & \enspace\textrm{(take downward to be positive) }\dots \\ h & =(0)t+\frac{1}{2}(g)t^2 \\ t & =\sqrt{\frac{2h}{g}} \end{aligned}$

The velocity of the ball right before the impact is given by

$\begin{aligned} v & =u+at \\ \dots & \enspace\textrm{(take downward to be positive) }\dots \\ & = (0) + (g)\cdot\bigg(\sqrt{\frac{2h}{g}}\bigg) \\ & = \sqrt{2gh} \end{aligned}$

By conservation of linear momentum,

$\begin{aligned} M\cdot \sqrt{2gh} & = (M+m)v \\ v & = \frac{M\sqrt{2gh}}{M+m} \end{aligned}$

is the velocity of the ball and the pan immediately after collision.

ii.

Setup.

From Hooke’s law, the spring force is given by $\mathbf{F}_s=-k\mathbf{x}$ , or $mg=kd$ . The spring constant, characteristic of the spring, is $k=mg/d$ . Let $x$ be the magnitude of downward displacement of the spring from its relaxed position in reaching the final state of equilibrium. Then,
$\begin{aligned} F & = kx \\ (M+m)g & = \bigg( \frac{mg}{d} \bigg) x \\ x & = \frac{(M+m)d}{m} \end{aligned}$

By the law of conservation of energy,

$\begin{aligned} \Delta\textrm{Mechanical Energy} & = 0 \\ \Delta\textrm{KE} + \Delta\textrm{PE} & = 0 \\ \Big( \textrm{KE}_{\textrm{final}} - \textrm{KE}_{\textrm{initial}}\Big) + \Big( \textrm{PE}_{\textrm{final}} - \textrm{PE}_{\textrm{initial}}\Big) & = 0 \end{aligned}$

$\begin{aligned} \textrm{KE}_{\textrm{initial}} & = \frac{1}{2}(M+m)\bigg( \frac{M\sqrt{2gh}}{M+m} \bigg)^2 \\ & = \frac{M^2gh}{M+m} \\ \textrm{KE}_{\textrm{final}} & = 0 \\ \textrm{PE}_{\textrm{initial}} & = \frac{1}{2}kd^2\\ \textrm{PE}_{\textrm{final}} & = \frac{1}{2}k(x+d)^2 - (M+m)gx \end{aligned}$

$\begin{aligned} & \textrm{KE}_{\textrm{final}}-\textrm{KE}_{\textrm{initial}} \\ & = 0 - \frac{M^2gh}{M+m} \\ & = - \frac{M^2gh}{M+m} \\ & \\ & \textrm{PE}_{\textrm{final}}-\textrm{PE}_{\textrm{initial}} \\ = & \frac{1}{2}k(x+d)^2-(M+m)gx - \frac{1}{2}kd^2 \\ = & \frac{1}{2}kx^2+kxd - (M+m)gx \\ = & \frac{1}{2}\bigg( \frac{mg}{d} \bigg) x^2 + \bigg( \frac{mg}{d} \bigg) xd - (M+m)gx \\ = & \bigg( \frac{mg}{2d}\bigg) x^2 - Mgx \end{aligned}$

Thus,

$\begin{aligned} 0 & = -\frac{M^2gh}{M+m} + \frac{mgx^2}{2d} - Mgx \\ 0 & = \bigg( \frac{m}{2d} \bigg) x^2 + (-M)x + \bigg( -\frac{M^2h}{M+m} \bigg) \\ x & = \frac{-(-M)\pm\sqrt{(-M)^2-4(\frac{m}{2d})(-\frac{M^2h}{M+m})}}{2(\frac{m}{2d})} \\ x & = \frac{Md}{m} \pm \frac{Md}{m} \sqrt{1+\frac{2hm}{(M+m)d}} \end{aligned}$

$\therefore$ The maximum deflection of the pan after the ball has been dropped onto it is

$\boxed{x = \displaystyle{\frac{Md}{m} \pm \frac{Md}{m} \sqrt{1+\frac{2hm}{(M+m)d}}}}$ .

Done. Let’s overdo.

ii. (energy approach)

One can derive the equation of motion by considering energy conservation:

$\begin{aligned} \textrm{KE} & =\frac{1}{2}(M+m)\dot{x}^2 \\ \textrm{PE} & = \frac{1}{2}kx^2-(M+m)gx \\ 0 & = \frac{\mathrm{d}(\textrm{KE}+\textrm{PE})}{\mathrm{d}t} \\ 0 & = (M+m)\dot{x}\ddot{x} + kx\dot{x} - (m+M)g\dot{x} \\ 0 & = (M+m)\ddot{x} + kx - (m+M)g \\ \ddot{x} & = -\frac{k}{(M+m)}x + g \\ \end{aligned}$

or, by considering the Euler-Lagrange equation

where the Lagrangian $\mathcal{L}$ is defined

and then computing,

the Euler-Lagrange equation now reads

which agrees with the equation of motion derived previously.

ii. (force approach)

One can also derive the equation of motion by considering Newton’s second law:

$\begin{aligned} \dots\enspace \textrm{recall }& \textrm{Newton's second law }\dots \\ \textrm{Net }\mathbf{F} & = m\mathbf{a} \\ \dots\enspace \textrm{downward } & \textrm{taken to be positive \dots} \\ (M+m)\mathbf{g} - k\mathbf{x} & = (M+m)\ddot{\mathbf{x}} \\ \dots\enspace \textrm{consider } & \textrm{magnitude only }\dots \\ \ddot{x} & = -\frac{k}{M+m}x + g\\ \ddot{x} & = -\bigg( \sqrt{\frac{k}{M+m}} \bigg)^2x+g \\ \ddot{x} & = -\omega^2x+g\qquad \textrm{where }\omega =\sqrt{\frac{k}{M+m}} \\ \dots \textrm{ go back } & \textrm{to an earlier line }\dots \\ (M+m)g - kx & = (M+m)\ddot{x} \\ \dots \textrm{let }x_{\textrm{eff}}\textrm{ be} & \textrm{ the effective displacement} \\ x_{\textrm{eff}} & = x-\frac{(M+m)g}{k} \\ \textrm{then\qquad} (M+m)\ddot{x}_{\textrm{eff}} & = -kx_{\textrm{eff}} \\ \end{aligned}$

(to be continued)

Remark.

The effective displacement $x_{\textrm{eff}}$ is defined by

$x_{\textrm{eff}} = x-\displaystyle{\frac{(M+m)g}{k}}$ .

Substituting the spring constant $k$ with

$k=\displaystyle{\frac{mg}{d}}$ ,

$x_{\textrm{eff}}$ can be rewritten as

$x_{\textrm{eff}} = x-\displaystyle{\frac{(M+m)d}{m}}$ .

As seen in ii. Setup,

$\displaystyle{\frac{(M+m)d}{m}}$

is the equilibrium position.

$\therefore$ The effective displacement $x_{\textrm{eff}}$ is thus the distance extended or compressed with reference to the equilibrium position.

(continue)

$\begin{aligned} (M+m)\ddot{x}_{\textrm{eff}} & = -kx_{\textrm{eff}} \\ \ddot{x}_{\textrm{eff}} & = - \bigg( \sqrt{\frac{k}{M+m}} \bigg)^2x_{\textrm{eff}} \\ \ddot{x}_{\textrm{eff}} & =-\omega^2x_{\textrm{eff}} \qquad \textrm{where }\omega = \sqrt{\frac{k}{M+m}} \\ \dots\textrm{ Solving it, }& \dots\\ x_{\textrm{eff}}(t) & = c_1\cos (\omega t) + c_2\sin (\omega t) \end{aligned}$

The constant coefficients (i.e., $c_1$ , $c_2$ ) should be obtained from the initial and boundary conditions.

Let $t=0$ be the instant the ball hits on the pan.

When $t=0$ ,

$\begin{aligned} x_{\textrm{eff}}(0) & = d - \frac{(M+m)d}{m} \\ c_1(1) + c_2(0) & = \frac{md-(M+m)d}{m} \\ c_1 & = -\bigg( \frac{M}{m} \bigg) d \\ \end{aligned}$

As the velocity of the pan immediately after the impact
was found in part (i) to be

$\displaystyle{\frac{M\sqrt{2gh}}{M+m}}$ ,

$\begin{aligned} \dot{x}_{\textrm{eff}}(t) & = -c_1\sin (\omega t)+c_2\cos (\omega t) \\ \dot{x}_{\textrm{eff}}(0) & = -c_1(0) + c_2(1) = \frac{M\sqrt{2gh}}{M+m} \\ c_2 & = \frac{M\sqrt{2gh}}{M+m} \end{aligned}$

$\begin{aligned} \therefore x_{\textrm{eff}}(t) & = -\bigg( \frac{Md}{m} \bigg)\cos \Bigg( \bigg(\sqrt{\frac{k}{M+m}}\bigg) (t) \Bigg)\\ & \qquad + \bigg( \frac{M\sqrt{2gh}}{M+m}\bigg) \sin\Bigg( \bigg( \sqrt{\frac{k}{M+m}}\bigg) ( t)\Bigg) \\ \end{aligned}$

If there is no friction or none any other dissipation of energy, the spring will continue indefinitely and uninhibitedly its simple harmonic motion.

(to be continued)

Aside.

The angular frequency $\omega$ ( $=2\pi f$ ) is

$\omega =\displaystyle{\sqrt{\frac{k}{M+m}}}$ .

Hence the frequency $f$ is

$f=\displaystyle{\frac{\omega}{2\pi}}=\displaystyle{\frac{1}{2\pi}\sqrt{\frac{k}{M+m}}}$ ,

and the period $T$ is

$T=\displaystyle{\frac{1}{f}=2\pi\sqrt{\frac{M+m}{k}}}$ .

At time $t=\displaystyle{\frac{T}{2}}$ , the spring is at the equilibrium position (i.e., $x_{\textrm{eff}}(\frac{T}{2})$ ), whereas at time(s) $t=\displaystyle{\frac{T}{4},\, \frac{3T}{4}}$ , it reaches the extreme points (i.e., $x_{\textrm{eff}}(\frac{T}{4}),\, x_{\textrm{eff}}(\frac{3T}{4})$ ).

(continue)

Lemma.

The sinusoidal function

$x(t)=c_1\cos (\omega t)+c_2\sin (\omega t)$

can be written in the form

$x(t)=A\cos (\omega t-\varphi )$

where $A=\sqrt{c_1^2+c_2^2}$ and $\tan\varphi =\displaystyle{\frac{c_2}{c_1}}$ .

The amplitude (i.e., $A=\sqrt{c_1^2+c_2^2}$ above) is the maximum displacement of the spring from its equilibrium position.

$\begin{aligned} A & = \sqrt{c_1^2+c_2^2} \\ & = \sqrt{\bigg( -\frac{Md}{m} \bigg)^2 + \bigg( \frac{M\sqrt{2gh}}{M+m} \bigg)^2} \\ & = \sqrt{\frac{M^2d^2}{m^2} + \frac{2M^2gh}{(M+m)^2}} \\ \end{aligned}$

August 19, 2020October 24, 2022

202008190508 Solution to 1973-AL-AMATH-I-1

In calm weather an aeroplane has a speed $v$ and a flying range (out-and-back) $R$ . In a north wind of speed $n$ ( $n<v$ ) and in a line of flight that makes a constant angle $\phi$ with the north, I wish to know its new flying range.

Assuming that the maximum time of flight is independent of wind conditions, and that the effect of turning round at the end of the outgoing journey can be ignored.

Let $\mathbf{v}=v_x\, \hat{\mathbf{i}} + v_y\,\hat{\mathbf{j}}$ denote the velocity of the aeroplane in calm weather (i.e., zero wind speed).

Then $\mathbf{v}=(v_x,\, v_y) = (v\sin\phi , \, v\cos\phi )$ , where $v=|\mathbf{v}|$ is its speed, and $\phi$ measured anticlockwise from the north.

In a north wind of speed $n$ , the outgoing speed $v_{\textrm{out}}$ of the plane will decrease by $(v\sin\phi ,\, v\cos\phi -n)$ , and on its return the speed $v_{\textrm{back}}$ will increase by $(v\sin\phi ,\, v\cos\phi +n)$ .

The new range $R'$ is calculated in two parts:

$R'_{\textrm{total}}=R'_{\textrm{out}}+R'_{\textrm{back}}$ .

$\because \qquad R'_{\textrm{out}}=R'_{\textrm{back}}$

$\therefore\qquad R'_{\textrm{out}}=R'_{\textrm{back}}=\displaystyle{\frac{R'}{2}}$

If the plane is on schedule, the time of flight $t=\displaystyle{\frac{R}{v}}$ should be kept unchanged.

I.e.,

$\begin{aligned} t & = t' \\ t & = t'_{\textrm{out}} + t'_{\textrm{back}} \\ \frac{R}{v} & = \frac{R'_{\textrm{out}}}{v'_{\textrm{out}}} + \frac{R'_{\textrm{back}}}{v'_{\textrm{back}}} \\ \frac{R}{v} & = \frac{R'/2}{\sqrt{(v\sin\phi )^2+(v\cos\phi -n)^2}} + \frac{R'/2}{\sqrt{(v\sin\phi )^2+(v\cos\phi +n)^2}} \\ \frac{R}{v} & = \frac{R'/2}{\sqrt{v^2\sin^2\phi +v^2\cos^2\phi -2nv\cos\phi +n^2}} + \frac{R'/2}{\sqrt{v^2\sin^2\phi +v^2\cos^2\phi +2nv\cos\phi +n^2}} \\ \frac{R}{v} & = \frac{R'/2}{\sqrt{v^2 -2nv\cos\phi +n^2}} + \frac{R'/2}{\sqrt{v^2 +2nv\cos\phi +n^2}} \\ \frac{R}{v} & = \frac{R'}{2}\Bigg\{ \frac{\sqrt{(v^2+n^2)+(2nv\cos\phi )}-\sqrt{(v^2+n^2)-(2nv\cos\phi )}}{\sqrt{(v^2+n^2)^2-(2nv\cos\phi )^2}}\Bigg\} \\ R' & = \frac{2R}{v} \Bigg\{ \frac{\sqrt{(v^2+n^2)^2-(2nv\cos\phi )^2}}{\sqrt{(v^2+n^2)+(2nv\cos\phi )}-\sqrt{(v^2+n^2)-(2nv\cos\phi )}}\Bigg\} \\ \end{aligned}$

Let $a=v^2+n^2$ and $b=2nv\cos\phi$ .

$\begin{aligned} R' & = \frac{2R}{v} \Bigg\{ \frac{\sqrt{a^2-b^2}}{\sqrt{a+b}-\sqrt{a-b}} \Bigg\} \\ R' & = \frac{2R}{v} \Bigg\{ \frac{\sqrt{a^2-b^2}(\sqrt{a+b}+\sqrt{a-b})}{(\sqrt{a+b})^2-(\sqrt{a-b})^2} \Bigg\} \\ R' & = \frac{R}{v}\bigg[\frac{(\sqrt{a-b})(a+b)+(\sqrt{a+b})(a-b)}{b}\bigg] \\ R' & = \frac{R}{v}\bigg[ \sqrt{a-b} \bigg( \frac{a}{b} \bigg) +\sqrt{a-b} + \sqrt{a+b} \bigg( \frac{a}{b} \bigg) - \sqrt{a+b} \bigg] \\ \end{aligned}$

(The Law of Cosines)

I need a break. To be continued.

After dinner, things are much clearer. Go back to the earlier line:

$\displaystyle{\frac{R}{v} = \frac{R'/2}{\sqrt{v^2 -2nv\cos\phi +n^2}} + \frac{R'/2}{\sqrt{v^2 +2nv\cos\phi +n^2}}}$ .

Now invoke its physical meaning:

i. Let $\mathbf{n}= -n\,\hat{\mathbf{j}}$ ( $n>0$ ) be the north wind velocity vector pointing in the southerly direction.

ii. Let $\mathbf{v}= -v_x\,\hat{\mathbf{i}} + v_y\,\hat{\mathbf{j}}$ ( $v_x,\, v_y>0$ ) be the velocity vector of the airplane pointing in the northwesterly direction.

iii. Let $\mathbf{s}=\mathbf{n}+\mathbf{v}$ be the side subtending the angle $\phi$ enclosed by the two vector arrows $\mathbf{n}$ and $\mathbf{v}$ . That is, the bottom of vector $\mathbf{s}$ is touching the arrowhead of $\mathbf{v}$ , and the arrowhead of $\mathbf{s}$ is touching the tip of $\mathbf{n}$ .

iv. The sides $\mathbf{n}$ , $\mathbf{v}$ , and $\mathbf{s}$ form a triangle of perimeter given by $n+v+s$ .

v. $s^2=|\mathbf{s}|^2= v^2+n^2-2nv\cos\phi$ .

vi. $2nv\cos\phi =2\mathbf{n}\cdot\mathbf{v}$ .

(continue)

$\begin{aligned} \frac{R}{v} & = \frac{R'/2}{\sqrt{v^2 -2nv\cos\phi +n^2}} + \frac{R'/2}{\sqrt{v^2 +2nv\cos\phi +n^2}} \\ \dots & = \frac{R'}{\sqrt{4(v^2 -2nv\cos\phi +n^2)}} + \frac{R'}{\sqrt{4(v^2 +2nv\cos\phi +n^2)}} \\ \dots & = \frac{R'}{\sqrt{4(s^2)}} + \frac{R'}{\sqrt{4(s^2 + 4\mathbf{n}\cdot\mathbf{v})}} \\ \dots & = \bigg( \frac{\sqrt{4(s^2+4\mathbf{n}\cdot \mathbf{v})}+\sqrt{4(s^2)}}{\sqrt{4(s^2)(s^2+4\mathbf{n}\cdot\mathbf{v})}} \bigg) R' \\ \dots & = \bigg( \frac{\sqrt{s^2+4\mathbf{n}\cdot \mathbf{v}}+ s}{\sqrt{s^4+4s^2\mathbf{n}\cdot\mathbf{v}}} \bigg) R' \\ R' & = \bigg( \frac{R}{v} \bigg) \bigg( \frac{\sqrt{s^4+4s^2\mathbf{n}\cdot\mathbf{v}}}{\sqrt{s^2+4\mathbf{n}\cdot \mathbf{v}}+ s} \bigg) \\ \end{aligned}$

Roughwork.

Simplifying first the numerator,

$\begin{aligned} & \sqrt{s^4+4s^2\mathbf{n}\cdot\mathbf{v}} \\ & = \sqrt{(v^4+n^4+2n^2v^2-4nv^3\cos\phi -4n^3v\cos\phi + 4n^2v^2\cos^2\phi )+4(v^2 + n^2 -2nv\cos\phi )(nv\cos\phi )} \\ & = \sqrt{(v^2+n^2)^2-(2nv\cos\phi )^2} \\ \end{aligned}$

then the denominator

$\begin{aligned} & \sqrt{s^2+4\mathbf{n}\cdot\mathbf{v}} + s \\ & = \sqrt{v^2+n^2-2nv\cos\phi+4nv\cos\phi } + \sqrt{v^2 + n^2 -2nv\cos\phi } \\ & = \sqrt{(v^2+n^2)+(2nv\cos\phi )} + \sqrt{(v^2 + n^2) -(2nv\cos\phi )} \\ \end{aligned}$

Putting them together,

$\begin{aligned} R' & = \bigg( \frac{R}{v}\bigg) \bigg( \frac{\sqrt{(v^2+n^2)^2-(2nv\cos\phi )^2}}{ \sqrt{(v^2+n^2)+(2nv\cos\phi )} + \sqrt{(v^2 + n^2) -(2nv\cos\phi )} } \bigg) \\ & = \bigg( \frac{R}{v}\bigg) \bigg( \frac{\sqrt{(v^2-n^2)^2+(2nv)^2-(2nv\cos\phi )^2}}{\sqrt{(v^2+n^2)+(2nv\cos\phi )} + \sqrt{(v^2 + n^2) -(2nv\cos\phi )}} \bigg) \\ \end{aligned}$

Unsatisfactory still.

Trimming the numerator again,

$\begin{aligned} & \sqrt{(v^2-n^2)^2\bigg( 1 + \frac{(2nv)^2-(2nv\cos\phi )^2}{(v^2-n^2)^2}\bigg)} \\ & = (v^2-n^2)\sqrt{\frac{(v^2+n^2)^2-(2nv\cos\phi )^2}{(v^2-n^2)^2}} \\ \end{aligned}$

and then dividing the denominator by $\sqrt{\frac{(v^2+n^2)^2-(2nv\cos\phi )^2}{(v^2-n^2)^2}}$ , or, multiplying the denominator by a factor of

$\displaystyle{\frac{v^2-n^2}{\sqrt{(v^2+n^2)^2-(2nv\cos\phi )^2}}}$

i.e.,

$(\sqrt{a+b}+\sqrt{a-b})\frac{v^2-n^2}{\sqrt{a^2-b^2}}$ .

Note.

(to be continued)

Ans.

$\displaystyle{\frac{R(v^2-n^2)}{v\sqrt{v^2-n^2\sin^2\phi}}}$