Public Examinations in Mathematics – Page 3

December 13, 2022December 13, 2022

202212131509 Solution to 1985-CE-AMATH-II-6

Find the equations of the two tangents drawn from the point $(-1,0)$ to the parabola $y^2=4x$ .

Roughwork.

$\begin{aligned} y^2 & = 4x \\ 2y\,\mathrm{d}y & = 4\,\mathrm{d}x \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \frac{2}{y} \\ \end{aligned}$

Let $(a_1,b_1)$ and $(a_2,b_2)$ be two points at which the two tangents touches the parabola. Then,

$\displaystyle{\frac{2}{b_i} =\frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{(a_i,b_i)} =\frac{b_i-(0)}{a_i-(-1)}}$

By solving two equations in two unknowns:

$\begin{cases} b_i^2 = 2a_i+2 \\ b_i^2 = 4a_i \\ \end{cases} \Longrightarrow\quad \begin{pmatrix} a_i\\b_i \end{pmatrix}_{i=1,2} = \Bigg\{ \begin{pmatrix} 1\\ 2 \end{pmatrix} , \begin{pmatrix} 1 \\ -2 \end{pmatrix} \Bigg\}$

This problem is not to be attempted.

December 13, 2022December 13, 2022

202212131316 Solution to 1983-CE-AMATH-II-3

Use the substitution

$u=1+3x^2$

to evaluate

$\displaystyle{\int_{0}^{1}x^3\sqrt{1+3x^2}\,\mathrm{d}x}$ .

Roughwork.

$\begin{aligned} & \quad\enspace \int_{0}^{1}x^3\sqrt{1+3x^2}\,\mathrm{d}x \\ \dots &\textrm{ let }x=\frac{\tan\theta}{\sqrt{3}}\textrm{ so that }\dots \\ \dots &\textrm{ }\mathrm{d}x=\frac{1}{\sqrt{3}}\sec^2\theta\,\mathrm{d}\theta\textrm{ }\dots \\ \dots &\textrm{ }x=0\Leftrightarrow \theta = 0\textrm{ and }x=1\Leftrightarrow \theta =\frac{\pi}{3}\textrm{ }\dots \\ & = \int_{0}^{\frac{\pi}{3}}\bigg(\frac{\tan\theta}{\sqrt{3}}\bigg)^3\sqrt{1+\tan^2\theta}\,\bigg(\frac{1}{\sqrt{3}}\sec^2\theta\,\mathrm{d}\theta\bigg) \\ & = \frac{1}{9}\int_{0}^{\frac{\pi}{3}}\sec^3\theta\tan^3\theta\,\mathrm{d}\theta \\ \dots & \textrm{ but }\frac{\mathrm{d}(\sec^3\theta )}{\mathrm{d}\theta}=3\sec^2\theta\cdot\sec\theta\tan\theta\textrm{ so }\dots \\ & = \frac{1}{27}\int_{0}^{\frac{\pi}{3}}\tan^2\theta\,\mathrm{d}(\sec^3\theta )\\ \end{aligned}$

Whoops! Read and follow the instructions.

$\begin{aligned} & \quad\enspace \int_{0}^{1}x^3\sqrt{1+3x^2}\,\mathrm{d}x \\ \dots &\textrm{ let }u=1+3x^2\textrm{ then }\mathrm{d}x=\frac{\mathrm{d}u}{6x}\textrm{ }\dots \\ \dots &\textrm{ then }x=0\Leftrightarrow u=1\textrm{ and }x=1\Leftrightarrow u=4\textrm{ }\dots \\ & = \int_{1}^{4}x^3\sqrt{u}\,\bigg(\frac{\mathrm{d}u}{6x}\bigg) \\ & = \frac{1}{6}\int_{1}^{4}\bigg(\frac{u-1}{3}\bigg)\sqrt{u}\,\mathrm{d}u \\ & = \frac{1}{18}\int_{1}^{4}\big( u^{\frac{3}{2}}-u^{\frac{1}{2}}\big)\,\mathrm{d}u \\ & = \frac{1}{18}\bigg[ \frac{2u^{\frac{5}{2}}}{5}-\frac{2u^{\frac{3}{2}}}{3}\bigg]\bigg|_{1}^{4} \\ & = \cdots \\ \end{aligned}$

This problem is not to be attempted.

December 12, 2022December 13, 2022

202212121714 Solution to 1974-CE-AMATH-II-X

A bridge $ABC$ is in the form of the parabola given by the equation

$8y=16-x^2$

as shown in the figure.

The slope of the bridge at the point $P$ is $0.5$ . Find the coordinate of $P$ .

Roughwork.

$\begin{aligned} \bigg|\frac{\mathrm{d}y}{\mathrm{d}x}\bigg| & = \bigg|-\frac{1}{4}x\bigg| \\ 0.5 & = \frac{1}{4}|x| \\ x & = \pm 2 \\ 8y & = 16-(\pm 2)^2 \\ y & = 1.5 \\ \end{aligned}$

Answer. $P(2,1.5)$ .

December 12, 2022December 21, 2022

202212121040 Solution to 1968-CE-AMATH-II-X

Find the maximum and minimum values of $y$ on the curve

$y^2=x(1-x)^2$ .

Sketch the curve.

Roughwork. (true-negative/false-positive example)

All computations in the enclosed section below involving higher-order derivatives than the first are wrong owing to the mistaken assumption

$\times :\quad\displaystyle{\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2=\frac{(\mathrm{d}y)^2}{(\mathrm{d}x)^2}}$

which was to be detested. Show

$\begin{aligned} y^2 & = x(1-x)^2 \\ & = x(1-2x+x^2) \\ y^2 & = x^3-2x^2+x \\ \end{aligned}$

$\begin{aligned} 2y\,\mathrm{d}y & = 3x^2\,\mathrm{d}x-4x\,\mathrm{d}x+\mathrm{d}x \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \frac{3x^2-4x+1}{2y} \\ 2(\mathrm{d}y)^2 & = 6x(\mathrm{d}x)^2 - 4(\mathrm{d}x)^2 \\ \bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 & = 3x-2 \\ \end{aligned}$

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 & = 2\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)\cdot \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \\ 3 & = 2\bigg(\frac{3x^2-4x+1}{2y}\bigg)\cdot \frac{\mathrm{d}^2y}{\mathrm{d}x^2} \\ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} & = \frac{3y}{3x^2-4x+1} \\ \end{aligned}$

$\begin{aligned} y'(x) & = 0 \\ 3x^2-4x+1 & = 0 \\ (3x-1)(x-1) & = 0 \\ x & = \frac{1}{3},1 \\ \end{aligned}$

$\begin{aligned} y^2 & = \bigg(\frac{1}{3}\bigg)\bigg(1-\bigg(\frac{1}{3}\bigg)\bigg)^2 \\ y & =\pm\frac{2\sqrt{3}}{9}\\ y^2 & = (1)(1-(1))^2 \\ y & = 0 \\ \end{aligned}$

$\begin{pmatrix}x\\y\end{pmatrix} = \Bigg\{ \begin{pmatrix}\frac{1}{3}\\ -\frac{2\sqrt{3}}{9} \end{pmatrix}, \begin{pmatrix}\frac{1}{3}\\ \frac{2\sqrt{3}}{9} \end{pmatrix}, \begin{pmatrix}1\\0\end{pmatrix}\Bigg\}$ are extrema.

$\begin{aligned} y''(x) & = 0 \\ 3y & = 0 \\ \pm\sqrt{x(1-x)^2} & = 0 \\ x(1-x)^2 & = 0 \\ x & = 0,1 \\ \end{aligned}$

$\begin{pmatrix}x\\y\end{pmatrix} = \bigg\{ \begin{pmatrix}0\\ 0 \end{pmatrix}, \begin{pmatrix}1\\ 0\end{pmatrix}\bigg\}$ are inflexions.

To answer whether

$\displaystyle{\frac{\mathrm{d}^2y}{\mathrm{d}x^2}}\bigg|_{\big( \frac{1}{3},\pm\frac{2\sqrt{3}}{9}\big)}$

is positive or negative is critical to yielding the maximum and the minimum values of the function.

The correction is left a $\textrm{\scriptsize{MUST}}$ for the author.

December 12, 2022December 12, 2022

202212121004 Solution to 1968-CE-AMATH-II-X

Find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ if

$ax^2+2hxy+by^2+2gx+2fy+c=0$ .

Roughwork.

$\begin{aligned} ax^2+2hxy+by^2+2gx+2fy+c & =0 \\ 2ax\,\mathrm{d}x+2h(x\,\mathrm{d}y+y\,\mathrm{d}x)+2by\,\mathrm{d}y+2g\,\mathrm{d}x+2f\,\mathrm{d}y & = 0 \\ 2ax + 2h\bigg( x\frac{\mathrm{d}y}{\mathrm{d}x}+y\bigg) +2by\frac{\mathrm{d}y}{\mathrm{d}x}+2g+2f\frac{\mathrm{d}y}{\mathrm{d}x} & = 0 \\ (ax+hy+g) + (hx+by+f)\frac{\mathrm{d}y}{\mathrm{d}x} & = 0 \\ -\frac{ax+hy+g}{hx+by+f} & = y' \\ \end{aligned}$

This problem is not to be attempted.

December 12, 2022December 12, 2022

202212120958 Solution to 1967-CE-AMATH-II-X

A spherical balloon is of diameter $x$ . By what approximate amount does the volume increases, if the diameter is increased by a very small amount $\Delta x$ ?

Roughwork.

$\begin{aligned} V & = \frac{4}{3}\pi x^3 \\ \Delta V & = 4\pi x^2\,\Delta x \\ \end{aligned}$

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 6, 2022December 6, 2022

202212061712 Solution to 1967-CE-AMATH-I-X

If the population of Hong Kong is $4$ million in $1967$ and it increases by $3\%$ each year, what will be the estimated population in $1987$ ? Give your answer correct to $4$ significant figures.

Roughwork.

$\begin{aligned} & \quad\enspace (4\times 10^6)(1+3\%)^{(1987-1967)} \\ & = (4\times 10^6)(1+0.03)^{20} \\ & = (4\times 10^6)\sum_{n=0}^{20}\begin{pmatrix}20\\ n\end{pmatrix}(1)^{20-n}(0.03)^n \\ & = (4\times 10^6)\sum_{n=0}^{20}\begin{pmatrix}20\\ n\end{pmatrix}(0.03)^n \\ & = \cdots \\ \end{aligned}$

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.