December 23, 2022 – Physicspupil

December 23, 2022December 29, 2022

202212231640 Solution to 2016-DSE-PHY-IA-2

$0.3\,\mathrm{kg}$ of water at temperature $50\,^\circ\mathrm{C}$ is mixed with $0.2\,\mathrm{kg}$ of ice at temperature $0\,^\circ\mathrm{C}$ in an insulated container of negligible heat capacity. What is the final temperature of the mixture?

Given: specific heat capacity of water $=4200\,\mathrm{J\,kg^{-1}\,^\circ C^{-1}}$ ; specific latent heat of fusion of ice $3.34\times 10^5\,\mathrm{J\, kg^{-1}}$ .

Roughwork.

For all $0.3\,\mathrm{kg}$ liquid water, a temperature drop of $50\,\mathrm{C^\circ}$ will release

$\begin{aligned} \textrm{Sensible heat} & = (0.3)(4200)(50) \\ & = \textrm{63,000}\,\mathrm{J} \\ \end{aligned}$

and a phase transition of freezing will release further latent heat (of fusion)

$\begin{aligned} \textrm{Latent heat}& = ml_f \\ & = (0.3)(3.34\times 10^5) \\ & = \textrm{100,200}\,\mathrm{J} \\ \end{aligned}$

whereas for $0.2\,\mathrm{kg}$ of solid ice to melt all at once, latent heat (of liquidization)

$\begin{aligned} \textrm{Latent heat} & = ml_f \\ & = (0.2)(3.34\times 10^5) \\ & = \textrm{66,800}\,\mathrm{J} \\ \end{aligned}$

is to be absorbed.

Lemma.

Heat always moves from hotter objects to colder objects, unless energy in some form is supplied to reverse the direction of heat flow.

_{Wikipedia on Second law of thermodynamics}

By the inequalities

$0<\textrm{66,800}-\textrm{63,000}<\textrm{100,200}$

one will know.

December 23, 2022December 23, 2022

202212231530 Solution to 1990-CE-AMATH-I-7

Given the curve $C: x^2+4xy+5y^2=1$ , find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ . Hence find the equations of the two tangents to $C$ which are parallel to the line $\displaystyle{y=-\frac{1}{2}x}$ .

Roughwork.

For the curve,

$\begin{aligned} &\quad\enspace x^2+4xy+5y^2 = 1 \\ &\Rightarrow 2x\,\mathrm{d}x + 4(x\,\mathrm{d}y + y\,\mathrm{d}x) + 10y\,\mathrm{d}y = 0 \\ &\Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x}= -\frac{x+2y}{2x+5y} \\ \end{aligned}$

and for the line,

$\begin{aligned} &\quad\enspace y = -\frac{1}{2}x \\ &\Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{1}{2} \\ \end{aligned}$

equating them,

$\begin{aligned} &\quad\enspace -\frac{x+2y}{2x+5y} = -\frac{1}{2} \\ &\Rightarrow \begin{pmatrix} x \\ y \end{pmatrix} = \bigg\{\begin{pmatrix} 1 \\ 0 \end{pmatrix} , \begin{pmatrix} -1 \\ 0 \end{pmatrix}\bigg\} \\ \end{aligned}$

are points of contact at which the tangents

$\begin{aligned} &\quad\enspace \frac{y-(\pm 1)}{x-(0)} = -\frac{1}{2} \\ &\Rightarrow \big\{ x + 2y \pm 2 = 0 \big\} \\ \end{aligned}$

are drawn.

This problem is not to be attempted.

December 23, 2022December 23, 2022

202212231507 Solution to 1990-CE-AMATH-II-1

Let $f(x)=\sqrt{x^2+k}\sin 2x$ , where $k$ is a constant. If $f'(0)=1$ , find the value of $k$ .

Roughwork.

$\begin{aligned} f(x) & = \sqrt{x^2+k}\sin 2x \\ f'(x) & = \sqrt{x^2+k}\cdot \frac{\mathrm{d}}{\mathrm{d}x}(\sin 2x) + \sin 2x\cdot \frac{\mathrm{d}}{\mathrm{d}x} \big( \sqrt{x^2+k}\big) \\ & = \sqrt{x^2+k}\cdot (2\cos 2x) + \sin 2x\cdot \bigg(\frac{x}{\sqrt{x^2+k}}\bigg)\\ f'(0) & = \sqrt{(0)^2+k}\cdot (2\cos 2(0)) + \sin 2(0)\cdot \bigg(\frac{(0)}{\sqrt{(0)^2+k}}\bigg)\\ 1 & =: 2\sqrt{k} \\ k & = \frac{1}{4} \\ \end{aligned}$

This problem is not to be attempted.

December 23, 2022December 23, 2022

202212231206 Solution to 2002-CE-AMATH-I-18

(a) Let $z=\cos\theta +i\sin\theta$ , where $-\pi<\theta \leqslant \pi$ . Show that

$|z^2+1|^2=2(1+\cos 2\theta )$ .

Hence, or otherwise, find the greatest value of $|z^2+1|$ .
(b) $w$ is a complex number such that $|w|=3$ .
i. Show that the greatest value of $|w^2+9|$ is $18$ .
ii. Explain why the equation

$w^4-81=100i(w^2-9)$

has only two roots.

Roughwork.

(a)

$\begin{aligned} \textrm{LHS} & = |z^2+1|^2 \\ & = |(\cos\theta +i\sin\theta )^2+1|^2 \\ & = |(\cos^2\theta -\sin^2\theta +2i\sin\theta\cos\theta )+1|^2 \\ & = |(2\cos^2\theta )+i(2\sin\theta\cos\theta )|^2 \\ & = \big(\sqrt{(2\cos^2\theta )^2+(2\sin\theta\cos\theta )^2}\big)^2 \\ & = 4\cos^4\theta +4\sin^2\theta\cos^2\theta \\ & = 4\cos^2\theta (\cos^2\theta +\sin^2\theta ) \\ & = 4\cos^2\theta \\ \end{aligned}$

From the double-angle formula

$\cos 2\theta = 2\cos^2\theta -1$

_{Wikipedia on List of trigonometric identities}

the equality follows, i.e., $\textrm{LHS}=\textrm{RHS}$ . As $\max (\cos 2\theta )=1$ , the greatest value of $|z^2+1|$ is

$\sqrt{2(1+(1))} = 2$ .

(b) Not to be attempted.