December 21, 2022 – Physicspupil

December 21, 2022December 21, 2022

202212211715 Solution to 1968-CE-AMATH-II-X

Find $\displaystyle{\int\frac{\mathrm{d}x}{\sqrt{9-4x^2}}}$

i. by using the substitution $\displaystyle{x=\frac{3}{2}\cos\theta}$ ;
ii. by using the substitution $\displaystyle{x=\frac{3}{2}\sin\theta}$ .

Explain why you appear to get two different answers.

Roughwork.

i.

$\begin{aligned} &\quad\enspace \int \frac{1}{\sqrt{9-4x^2}}\,\mathrm{d}x \\ \dots\, & \textrm{let }x=\frac{3}{2}\cos\theta \textrm{ s.t. }\mathrm{d}x=-\frac{3}{2}\sin\theta\,\mathrm{d}\theta \,\dots \\ & = -\int \frac{3\sin\theta}{2\sqrt{9-4(\frac{3}{2}\cos\theta )^2}}\,\mathrm{d}\theta \\ & = -\int \frac{3\sin\theta}{6\sin\theta}\,\mathrm{d}\theta \\ & = -\frac{\theta}{2} +C_1\textrm{ for some constant}\\ & = -\frac{1}{2}\arccos \bigg(\frac{2x}{3}\bigg) +C_1 \\ \end{aligned}$

ii.

$\begin{aligned} &\quad\enspace \int \frac{1}{\sqrt{9-4x^2}}\,\mathrm{d}x \\ \dots\, & \textrm{let }x=\frac{3}{2}\sin\theta \textrm{ s.t. }\mathrm{d}x=\frac{3}{2}\cos\theta\,\mathrm{d}\theta \,\dots \\ & = \int \frac{3\cos\theta}{2\sqrt{9-4(\frac{3}{2}\sin\theta )^2}}\,\mathrm{d}\theta \\ & = \int \frac{3\cos\theta}{6\cos\theta}\,\mathrm{d}\theta \\ & = \frac{\theta}{2} +C_2\textrm{ for some constant} \\ & = \frac{1}{2}\arcsin \bigg(\frac{2x}{3}\bigg) +C_2 \\ \end{aligned}$

Let’s see if

$\displaystyle{-\frac{1}{2}\arccos\bigg(\frac{2x}{3}\bigg) +C_1 \stackrel{\textrm{?}}{\equiv}\frac{1}{2}\arcsin\bigg(\frac{2x}{3}\bigg)} + C_2$

or, to be rephrased, whether

$\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}\bigg[\arcsin \bigg(\frac{2x}{3}\bigg) + \arccos\bigg(\frac{2x}{3}\bigg) \bigg]\stackrel{\textrm{?}}{\equiv} 0}$

This is left an exercise for the reader.

December 21, 2022December 21, 2022

202212211654 Solution to 2009-CE-AMATH-II-6

$C$ is a curve with equation $y^3+x^3y=10$ .

(a) Find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ .
(b) Find the equation of the tangent to the curve $C$ at the point $(1,2)$ .

Roughwork.

$\begin{aligned} y^3+x^3y & = 10 \\ 3y^2\,\mathrm{d}y+x^3\,\mathrm{d}y+3x^2y\,\mathrm{d}x & = 0 \\ -\frac{3x^2y}{x^3+3y^2} = \frac{\mathrm{d}y}{\mathrm{d}x} & = y'(x,y) \\ \end{aligned}$

$\displaystyle{y'(1,2)=-\frac{3(1)^2(2)}{(1)^3+3(2)^2}=-\frac{6}{13}}$

$\begin{aligned} &\quad\enspace \frac{y-2}{x-1} = -\frac{6}{13} \\ & \Rightarrow \big\{ L:\quad 6x+13y-32 = 0\big\} \\ \end{aligned}$

This problem is not to be attempted.

December 21, 2022December 21, 2022

202212211306 Solution to 1987-CE-AMATH-I-2

Let $x=y+\sin y$ . Find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ and $\displaystyle{\frac{\mathrm{d}^2y}{\mathrm{d}x^2}}$ in terms of $y$ .

Roughwork.

$\begin{aligned} x & = y+\sin y \\ \mathrm{d}x & = \mathrm{d}y+\cos y\,\mathrm{d}y \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \frac{1}{1+\cos y} \\ \bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 & = \frac{1}{(1+\cos y)^2} \\ \frac{\mathrm{d}}{\mathrm{d}x}\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)^2 & = 2\bigg(\frac{\mathrm{d}y}{\mathrm{d}x}\bigg)\bigg(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}\bigg) \\ \end{aligned}$

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

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

$\begin{aligned} \quad\enspace \frac{\mathrm{d}^2y}{\mathrm{d}x^2} & = \frac{1+\cos y}{2}\cdot \frac{2\sin y}{(1+\cos y)^4} \\ & = \frac{\sin y}{(1+\cos y)^3} \\ \end{aligned}$

Stretching the focus, compromise the better for it.

December 21, 2022December 21, 2022

202212211204 Solution to 1972-CE-AMATH-I-X

$F(x)\equiv A(x-1)^2+B(x-1)+C$ where $A$ , $B$ , and $C$ are constants. Find the values of $A$ , $B$ , and $C$ , given that $(x+2)$ is a factor of $F(x)$ and that the remainders when $F(x)$ is divided by $(x-1)$ and $(x+1)$ are $9$ and $-11$ respectively.

Roughwork.

$\begin{aligned} & \quad\enspace (x+2)\textrm{ is a factor of }F(x) \\ & \Leftrightarrow F(x)=(x+2)\cdot G(x) \\ & \Leftrightarrow F(-2)=0 \\ & \Leftrightarrow A((-2)-1)^2+B((-2)-1)+C=0 \\ \end{aligned}$

$\begin{aligned} & \quad\enspace \mathrm{9}\textrm{ remains when dividing }F(x)\textrm{ by }(x-1) \\ &\Leftrightarrow F(1)=C=9 \\ \end{aligned}$

$\begin{aligned} & \quad\enspace \mathrm{-11}\textrm{ remains when dividing }F(x)\textrm{ by }(x+1) \\ &\Leftrightarrow F(-1)=A((-1)-1)^2+B((-1)-1)+C=-11 \\ \end{aligned}$

This problem is not to be attempted.