October 19, 2022 – Physicspupil

October 19, 2022October 19, 2022

202210191606 Solution to 1969-CE-AMATH-II-XX

Differentiate $2(3-4x)^{\frac{5}{6}}$ with respect to $x$ .

Roughwork.

$\begin{aligned} & \quad\, \frac{\mathrm{d}}{\mathrm{d}x}\Big( 2(3-4x)^{\frac{5}{6}}\Big) \\ & = 2\bigg( \frac{5}{6}\bigg) (3-4x)^{\frac{5}{6}-1}(-4) \\ & = -\frac{20}{3}(3-4x)^{-\frac{1}{6}} \\ & = -\frac{20}{3\sqrt[6]{3-4x}} \\ \end{aligned}$

October 19, 2022October 19, 2022

202210191554 Solution to 1969-CE-AMATH-II-XX

Find the $x$ -coordinate of the point on the curve $y=ax^2+bx+c$ where the tangent is horizontal.

Roughwork.

The gradient $y'(x)$ of the curve $y(x)$ is

$y'(x)=2ax+b$ .

When the tangent is horizontal,

$\begin{aligned} y'(x) & =0 \\ 2ax+b & =0 \\ x & = -\frac{b}{2a} \\ y & = a\bigg( \frac{-b}{2a}\bigg)^2+b\bigg( \frac{-b}{2a}\bigg)+c \\ & = \dots \\ \end{aligned}$

October 19, 2022October 19, 2022

202210191506 Solution to 1970-CE-AMATH-II-XX

Calculate the gradient of the curve $x^2+xy+y^2=1$ at the point $(1,-1)$ .

Roughwork.

Let the function $f$ of variables $x$ and $y$ be

$f(x,y):=x^2+xy+y^2-1=0$

Then,

$\begin{aligned} \Delta f=2x\Delta x+x\Delta y+y\Delta x+2y\Delta y & =0 \\ 2x + x\frac{\Delta y}{\Delta x}+y+2y\frac{\Delta y}{\Delta x} & = 0 \\ (x+2y)\frac{\Delta y}{\Delta x} & = -2x-y \\ \frac{\Delta y}{\Delta x} & = - \frac{2x+y}{x+2y} \\ \end{aligned}$

The gradient $\nabla$ at point $(x=1,y=-1)$ is

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

October 19, 2022October 19, 2022

202210191436 Solution to 1969-CE-AMATH-II-XX

Calculate the gradient of the curve $x=3y+y^2$ at the point $(0,-3)$ .

Roughwork.

Differentiating $x(y)$ wrt to $y$ and taking reciprocal,

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

the gradient $\nabla y(x)$ at point $(x,y) = (0,-3)$ is

$\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{(0,-3)} = \frac{1}{3+2(-3)} = -\frac{1}{3}}$ .

October 19, 2022October 19, 2022

202210191139 Solution to 1973-CE-AMATH-II-XX

Find the area bounded by the curve $y=4x-x^2$ and the straight line $x-y+2=0$ .

Roughwork.

Rewriting,

$\begin{cases} C:y=-x^2+4x \\ L:y=x+2 \\ \end{cases}$

Solving,

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

$\begin{aligned} x=1 \Rightarrow y=(1)+2=3 \\ x=2 \Rightarrow y=(2)+2=4 \\ \end{aligned}$

Now that the points of intersection are $(1,3)$ and $(2,4)$ , plot a graph below.

By either definite integral,

$\textrm{Area} & = \begin{cases} \displaystyle{\int_{1}^{2}\big(y_C(x)-y_L(x)\big)\,\mathrm{d}x }\\ \displaystyle{\int_{3}^{4}\big(x_L(y)-x_C(y)\big)\,\mathrm{d}y }\\ \end{cases}$

The upper integral is easier solvable than the lower one. Exercise.

October 19, 2022October 19, 2022

202210191050 Solution to 1977-CE-AMATH-II-XX

Prove that

$\displaystyle{\cos^4\theta = \frac{1}{8}(\cos 4\theta + 4\cos 2\theta + 3)}$ .

Hence solve the equation

$\cos 4\theta + 4\cos 2\theta +1=0$

for $0^\circ\leqslant\theta\leqslant 360^\circ$ .

Roughwork.

$\begin{aligned} \textrm{RHS} & = \frac{1}{8}(\cos 4\theta + 4\cos 2\theta + 3) \\ & = \frac{1}{8}\big(\cos 2(2\theta ) + 4\cos 2\theta + 3\big) \\ & = \frac{1}{8}\big( (2\cos^22\theta -1) + 4\cos 2\theta + 3\big) \\ & = \frac{1}{8}(2\cos^22\theta +4\cos 2\theta +2) \\ \dots\,\because\enspace & \cos2\theta = 2\cos^2\theta -1\,\dots \\ & = \frac{1}{8}\big( 2(2\cos^2\theta -1)^2+4(2\cos^2\theta -1)+2\big) \\ & = \dots \\ & = \cos^4\theta \\ & = \textrm{LHS} \\ \end{aligned}$

The missing steps are left the reader.

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