December 13, 2022 – Physicspupil

December 13, 2022December 13, 2022

202212131615 Solution to 1989-CE-AMATH-I-6

$p$ and $q$ are real numbers such that

$(p+qi)^2=21-20i$ .

Find the values of $p$ and $q$ . Hence write down the two square roots of $21-20i$ .

Roughwork.

$\begin{aligned} \textrm{LHS} & = (p+qi)^2 \\ & = (p^2-q^2)+(2pq)i \\ \textrm{RHS} & = (21)+(-20)i \\ \end{aligned}$

$\begin{cases} p^2-q^2=21 \\ pq = -10 \end{cases}$

$\therefore$ $(p,q)=(\pm 5,\mp 2)$ .

December 13, 2022December 13, 2022

202212131544 Solution to 1975-CE-AMATH-I-X

A rod $PQ$ of length $4$ units slides with $P$ on the $x$ -axis and $Q$ on the $y$ -axis. $R$ is the point on $PQ$ such that $PR:RQ=1:3$ . Find the equation of the locus of $R$ .

Roughwork.

Let $P=(a,0)$ and $Q=(0,b)$ where $a,b\in [0,4]$ and $a^2+b^2=(4)^2$ . The locus of $R(x,y)$ is

$\begin{aligned} &\quad\enspace \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} \frac{3}{4}a \\ \frac{1}{4}b \end{pmatrix} \\ &\Longrightarrow \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} \frac{4}{3}x \\ 4y \end{pmatrix} \\ &\Longrightarrow \bigg(\frac{4}{3}x\bigg)^2+ (4y)^2 = 16 \\ &\Longrightarrow \enspace R:\big\{ x^2+9y^2-9=0\big\} \\ \end{aligned}$

This problem is not to be attempted.

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.