April 2021 – Physicspupil

April 24, 2021October 24, 2022

202104241713 Homework 1 (Q7)

The height of a mountain is given by $h(x,y)=3000-2x^2-y^2$ , where the $y$ -axis points east, the $x$ -axis points north, and all distances are measured in meters. Suppose a mountain climber is at the point $(30,\, -20,\, 800)$ , will he ascend or descend if he moves in the southwest direction?

Solution.

(The solution below is based on the manuscript of 2015-2016 PHYS2155 Methods of Physics II Homework Solutions.)

The altitude $h(x,y)$ is given by a function of $x$ and $y$ :

$h(x,y)=3000-2x^2-y^2$ .

Now that the climber moves in the southwest direction

$\begin{aligned} \mathbf{n} & =-1\,\hat{\mathbf{i}}-1\,\hat{\mathbf{j}} \\ \hat{\mathbf{n}} & = \frac{1}{\sqrt{2}}(-\hat{\mathbf{i}}-\hat{\mathbf{j}}) \end{aligned}$

$\begin{aligned} h_{\hat{\mathbf{n}}}'(x,y) & = \nabla h(x,y)\cdot \hat{\mathbf{n}} \\ & = (-4x\,\hat{\mathbf{i}}-2y\,\hat{\mathbf{j}}) \cdot \frac{1}{\sqrt{2}}(-\hat{\mathbf{i}}-\hat{\mathbf{j}}) \\ & = \frac{1}{\sqrt{2}} (4x+2y) \end{aligned}$

At point $(30,\, -20,\, 800)$ ,

$\begin{aligned} h_{\hat{\mathbf{n}}}'(30,-20) & = \frac{1}{\sqrt{2}}\big( 4(30)+2(-20)\big) \\ & = \frac{1}{\sqrt{2}}\cdot 80 \qquad (>0) \end{aligned}$

he will ascend southwesterly.

April 22, 2021October 24, 2022

202104221623 Homework 2 (Q1)

Find the gradients of the following functions:

(a) $\phi (x,y,z)=x^2+y^2+z^2$ ;

(b) $\phi (x,y,z)=\exp (x+y)\cos z$ .

Solution.

(a)

$\begin{aligned} & \quad \nabla \phi \\ & = \bigg(\hat{\mathbf{i}}\frac{\partial}{\partial x}+\hat{\mathbf{j}}\frac{\partial}{\partial y}+\hat{\mathbf{k}}\frac{\partial}{\partial z}\bigg)\phi (x,y,z) \\ & = \hat{\mathbf{i}} \frac{\partial}{\partial x}(x^2+y^2+z^2) + \hat{\mathbf{j}} \frac{\partial}{\partial y}(x^2+y^2+z^2) + \hat{\mathbf{k}} \frac{\partial}{\partial z}(x^2+y^2+z^2) \\ & = 2x\,\hat{\mathbf{i}} + 2y\,\hat{\mathbf{j}} + 2z\,\hat{\mathbf{k}} \end{aligned}$

(b) It is left as an exercise to the reader.

April 22, 2021November 25, 2024

202104221542 Homework 2 (Q4)

Evaluate the limit, or explain why the limit fails to exist.

(a) $\displaystyle{\lim_{(x,y)\to (0,0),\, x\neq y} \frac{x^2-xy}{\sqrt{x}-\sqrt{y}}}$ ;

(b) $\displaystyle{\lim_{(x,y)\to (2,0)}\frac{x^2-y^2-4x+4}{x^2+y^2-4x+4}}$

Solution.

(a)

$\begin{aligned} & \quad \lim_{(x,y)\to (0,0),\, x\neq y} \frac{x^2-xy}{\sqrt{x}-\sqrt{y}} \\ & = \lim_{(x,y)\to (0,0),\, x\neq y} \frac{x(x-y)}{\sqrt{x}-\sqrt{y}} \\ & = \lim_{(x,y)\to (0,0),\, x\neq y} \frac{x(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})}{\sqrt{x}-\sqrt{y}} \\ & = \lim_{(x,y)\to (0,0),\, x\neq y} x(\sqrt{x}+\sqrt{y}) \\ & = (0)\big( \sqrt{(0)} + \sqrt{(0)} \big) \\ & = 0 \end{aligned}$

(b)

Roughwork.

If we take limits along the path $(2,y)\to (2,0)$ ,

$\begin{aligned} & \quad \lim_{(x,y)\to (2,0)}\frac{x^2-y^2-4x+4}{x^2+y^2-4x+4} \\ & = \lim_{(x,y)\to (2,0)} \frac{(x-2)^2-y^2}{(x-2)^2+y^2} \\ & = \lim_{y\to 0\textrm{ along }x=2}\frac{(x-2)^2-y^2}{(x-2)^2+y^2} \\ & = \lim_{y\to 0\textrm{ along }x=2}\frac{\big((2)-2\big)^2-y^2}{\big((2)-2\big)^2+y^2} \\ & = \lim_{y\to 0\textrm{ along }x=2}\frac{-y^2}{y^2} \\ & = -1 \end{aligned}$

whereas if we take limits along the path $(x,0)\to (2,0)$ ,

$\begin{aligned} & \quad \lim_{(x,y)\to (2,0)} \frac{(x-2)^2-y^2}{(x-2)^2+y^2} \\ & = \lim_{x\to 2\textrm{ along }y=0}\frac{(x-2)^2-y^2}{(x-2)^2+y^2} \\ & = \lim_{x\to 2\textrm{ along }y=0}\frac{(x-2)^2-(0)^2}{(x-2)^2+(0)^2} \\ & = 1 \end{aligned}$

The limit fails to exist because the limiting values vary with the paths of taking the limit $\lim_{(x,y)\to (2,0)}$ .

April 18, 2021October 24, 2022

202104181519 Homework 1 (Q4)

i. Find the infinitesimal small vector $\mathbf{dr}$ in the cylindrical coordinate induced by an infinitesimal small changes of $\mathrm{d}\rho$ , $\mathrm{d}\theta$ , and $\mathrm{d}z$ in terms of $\rho$ , $\theta$ , $z$ , $\mathrm{d}\rho$ , $\mathrm{d}\theta$ , $\mathrm{d}z$ and the corresponding unit vector.

ii. $f(u_1, u_2, u_3)$ is defined in $\mathbf{r}=(u_1, u_2, u_3)$ coordinate. Its gradient is defined

$\displaystyle{\lim_{\Delta l_i\to 0}\sum_{i=1}^{3}\frac{\Delta f_i}{\Delta l_i}\hat{\mathbf{u}}_l}$

where $\Delta l_i$ and $\Delta f_i$ are respectively the changes in length and functional value induced purely by the infinitesimal change in $u_i$ . $\hat{\mathbf{u}}_l$ is the unit vector of $\mathbf{u}_i$ . Thus find the gradient of $f$ in cylindrical coordinate.

Solution.

(The solution below is based on the manuscript of 2015-2016 PHYS2155 Methods of Physics II Homework Solutions.)

i.

$\mathbf{dr}=\mathrm{d}\rho\,\hat{\boldsymbol{\rho}}+\rho\,\mathrm{d}\theta\,\hat{\boldsymbol{\theta}}+\mathrm{d}z\,\hat{\mathbf{z}}$

Compare to the figure below.

ii.

$\begin{aligned} \nabla f & = \lim_{\Delta l_i\to 0}\sum_{i=1}^{3}\frac{\Delta f_i}{\Delta l_i}\hat{\mathbf{u}_i} \\ & = \lim_{\Delta\rho\to 0} \frac{\Delta f_\rho}{\Delta \rho}\,\hat{\boldsymbol{\rho}} + \lim_{\Delta\theta\to 0} \frac{\Delta f_\theta}{\rho\Delta\theta}\,\hat{\boldsymbol{\theta}} + \lim_{\Delta z\to 0}\frac{\Delta f_z}{\Delta z}\,\hat{\mathbf{z}} \\ & = \frac{\partial f}{\partial \rho}\,\hat{\boldsymbol{\rho}} + \frac{1}{\rho}\frac{\partial f}{\partial \theta}\,\hat{\boldsymbol{\theta}} + \frac{\partial f}{\partial z}\,\hat{\mathbf{z}} \end{aligned}$

April 16, 2021October 24, 2022

202104162114 Homework 1 (Q3)

Suppose $z=f(x,y)$ is differentiable. Find the expression

$\displaystyle{\bigg( \frac{\partial z}{\partial x} \bigg)^2 + \bigg( \frac{\partial z}{\partial y} \bigg)^2}$

in terms of polar coordinates $r$ and $\theta$ .

Solution.

Setup.

$\begin{aligned} & \begin{cases} \displaystyle{\frac{\partial z}{\partial r}} & = \displaystyle{\frac{\partial z}{\partial x}\frac{\partial x}{\partial r} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial r}} \\ \displaystyle{\frac{\partial z}{\partial \theta}} & = \displaystyle{\frac{\partial z}{\partial x}\frac{\partial x}{\partial \theta} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial \theta}} \\ \end{cases} \\ \Rightarrow & \begin{cases} \displaystyle{\frac{\partial z}{\partial r}} & = \displaystyle{\cos\theta\frac{\partial z}{\partial x} + \sin\theta\frac{\partial z}{\partial y}} \\ \displaystyle{\frac{\partial z}{\partial \theta}} & = \displaystyle{-r\sin\theta\frac{\partial z}{\partial x} + r\cos\theta\frac{\partial z}{\partial y}} \\ \end{cases} \\ \Rightarrow & \begin{bmatrix} \displaystyle{\frac{\partial z}{\partial r}} \\ \displaystyle{\frac{\partial z}{\partial \theta}}\end{bmatrix} = \begin{bmatrix} \cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{bmatrix} \begin{bmatrix} \displaystyle{\frac{\partial z}{\partial x}} \\ \displaystyle{\frac{\partial z}{\partial y}} \end{bmatrix} \\ \end{aligned}$

(to be continued)

Lemma.

_{Wikipedia on Cramer’s rule}

(continue)

$\begin{aligned} \frac{\partial z}{\partial x} & = \frac{\begin{vmatrix}\displaystyle{\frac{\partial z}{\partial r}} & \sin\theta \\ \displaystyle{\frac{\partial z}{\partial \theta}} & r\cos\theta \end{vmatrix}}{\begin{vmatrix}\cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{vmatrix}} \\ \\ & = \frac{r\cos\theta\displaystyle{\frac{\partial z}{\partial r}}-\sin\theta\displaystyle{\frac{\partial z}{\partial \theta}}}{r\cos^2\theta + r\sin^2\theta} \\\\ & = \cos\theta\displaystyle{\frac{\partial z}{\partial r}} - \frac{1}{r}\sin\theta\displaystyle{\frac{\partial z}{\partial \theta}} \\\\ \frac{\partial z}{\partial y} & = \frac{\begin{vmatrix}\cos\theta & \displaystyle{\frac{\partial z}{\partial r}} \\ -r\sin\theta & \displaystyle{\frac{\partial z}{\partial \theta}} \end{vmatrix}}{\begin{vmatrix}\cos\theta & \sin\theta \\ -r\sin\theta & r\cos\theta \end{vmatrix}} \\\\ & = \frac{1}{r}\cos\theta\frac{\partial z}{\partial \theta} + \sin\theta\frac{\partial z}{\partial r} \end{aligned}$

Thus,

$\begin{aligned} & \quad\bigg( \frac{\partial z}{\partial x}\bigg)^2 + \bigg( \frac{\partial z}{\partial y}\bigg)^2 \\ \\ & = \bigg[ \cos\theta\frac{\partial z}{\partial r} - \frac{1}{r}\sin\theta\frac{\partial z}{\partial \theta} \bigg]^2 + \bigg[ \frac{1}{r}\cos\theta\frac{\partial z}{\partial \theta} + \sin\theta\frac{\partial z}{\partial r} \bigg]^2 \\ \\ & = \cos^2\theta \bigg( \frac{\partial z}{\partial r}\bigg)^2 - \frac{\sin 2\theta}{r}\bigg(\frac{\partial z}{\partial r}\bigg)\bigg(\frac{\partial z}{\partial \theta}\bigg) + \frac{1}{r^2}\sin^2\theta \bigg(\frac{\partial z}{\partial \theta}\bigg)^2 \\ & \qquad\qquad + \frac{1}{r^2}\cos^2\theta \bigg(\frac{\partial z}{\partial \theta}\bigg)^2 + \frac{\sin 2\theta}{r}\bigg(\frac{\partial z}{\partial r}\bigg)\bigg(\frac{\partial z}{\partial \theta}\bigg) + \sin^2\theta \bigg( \frac{\partial z}{\partial r}\bigg)^2 \\ \\ & = \bigg( \frac{\partial z}{\partial r} \bigg)^2 + \frac{1}{r^2} \bigg( \frac{\partial z}{\partial \theta} \bigg)^2 \end{aligned}$

April 16, 2021October 24, 2022

202104162147 Homework 1 (Q2)

The angle $a$ of a triangle $ABC$ is increasing at a rate of $3\,\mathrm{^\circ\, s^{-1}}$ , the side of $AB$ is increasing at a rate of $1\,\mathrm{cm\, s^{-1}}$ , and the side of $AC$ is decreasing at a rate of $2\,\mathrm{cm\, s^{-1}}$ . How fast is the side $BC$ changing when $a=30^\circ$ , $AB=10\,\mathrm{cm}$ , and $AC=24\,\mathrm{cm}$ ? Is the length of $BC$ increasing or decreasing?

Solution.

Draw a figure below:

Rephrase the problem.

Given that If , , and , then

By cosine law,

$z^2=x^2+y^2-2xy\cos a$ .

Taking ordinary derivatives w.r.t. time $t$ ,

$\displaystyle{2z\frac{\mathrm{d}z}{\mathrm{d}t} = 2x\frac{\mathrm{d}x}{\mathrm{d}t} + 2y\frac{\mathrm{d}y}{\mathrm{d}t} + 2xy\sin a\frac{\mathrm{d}a}{\mathrm{d}t} - 2x\cos a\frac{\mathrm{d}y}{\mathrm{d}t} - 2y\cos a\frac{\mathrm{d}x}{\mathrm{d}t}}$

$\begin{aligned} z & =\sqrt{x^2+y^2-2xy\cos a} \\ & = \sqrt{(10)^2+(24)^2-2(10)(24)\cos 30^\circ} \\ & = 16.1341\qquad (4\,\mathrm{d.p.}) \end{aligned}$

Plugging in the value of each,

$\displaystyle{2(\cdot\cdot )\frac{\mathrm{d}z}{\mathrm{d}t} = 2(\cdot\cdot )\big(\cdot\cdot \big)+2(\cdot\cdot )\big(\cdot\cdot \big)+2(\cdot\cdot )(\cdot\cdot )\sin (\cdot\cdot )\big(\cdot\cdot \big) - 2(\cdot\cdot )\cos (\cdot\cdot )\big(\cdot\cdot \big) - 2(\cdot\cdot )\cos (\cdot\cdot )\big(\cdot\cdot \big)}$

you will know what $\displaystyle{\frac{\mathrm{d}z}{\mathrm{d}t}}$ is.

But now, I intend to treat it with partial derivatives.

Let $f(x,y,a) = x^2+y^2-2xy\cos a = z^2$ .

$\begin{aligned} \frac{\mathrm{d}f}{\mathrm{d}t} & = \frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t} + \frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t} + \frac{\partial f}{\partial a}\frac{\mathrm{d}a}{\mathrm{d}t} \\ & = (2x-2y\cos a)\frac{\mathrm{d}x}{\mathrm{d}t} + (2y-2x\cos a)\frac{\mathrm{d}y}{\mathrm{d}t} + 2xy\sin a\frac{\mathrm{d}a}{\mathrm{d}t} \\ \end{aligned}$

After $\displaystyle{\frac{\mathrm{d}f}{\mathrm{d}t}}$ is sought, recognise that

$\begin{aligned} \frac{\mathrm{d}f}{\mathrm{d}t} & =2z\frac{\mathrm{d}z}{\mathrm{d}t} \\ \frac{\mathrm{d}z}{\mathrm{d}t} & = \bigg(\frac{1}{2z}\bigg)\frac{\mathrm{d}f}{\mathrm{d}t} \end{aligned}$

you could have it also.

(to be continued)

April 16, 2021October 24, 2022

202104160754 Homework 1 (Q3)

A ship $A$ , which can sail at a constant speed $60\,\mathrm{km/hr}$ to meet a second ship $B$ which is $100\,\mathrm{km}$ away in the direction of $\mathrm{S60^\circ W}$ and is sailing due east at constant speed $30\,\mathrm{km/hr}$ . Find the sailing direction of $A$ and the time required to meet $B$ .

Solution.

(The solution below is based on the manuscript of 2014-2015 PHYS1250 Fundamental Physics Homework Solutions.)

Draw a diagram as follows:

Setup.

By the law of sines,

$\begin{aligned} \frac{V_A}{\sin 30^\circ} & = \frac{V_B}{\sin\theta} \\ \frac{60}{\sin 30^\circ} & = \frac{30}{\sin\theta} \\ \sin\theta & = 0.25 \\ \theta & = 14.5^\circ \end{aligned}$

Direction of $\mathbf{v}_A$ : $\mathrm{S45.5^\circ W}$

$\because 180^\circ -30^\circ -14.5^\circ - 90^\circ = 45.5^\circ$

Calculating $v_{AB}$ :

$\begin{aligned} |\mathbf{v}_{AB}| & = |\mathbf{v}_{A}|\cos\theta + |\mathbf{v}_{B}|\cos 30^\circ \\ & = 60 \cos 14.5^\circ + 30\cos 30^\circ \\ & = 84.1\,\mathrm{km/hr} \end{aligned}$

The time needed to meet ship $B$ is

$\begin{aligned} t & = \frac{100\,\mathrm{km}}{|\mathbf{v}_{AB}|} \\ & = \frac{100\,\mathrm{km}}{84.1\,\mathrm{km/hr}} \\ & = 1.19\,\mathrm{hr} \end{aligned}$

April 16, 2021October 24, 2022

202104160620 Homework 1 (Q4)

A particle is projected from a point $O$ on the horizontal floor. The range of the projectile is $R$ and the maximum height that the particle can reach is $h$ . Show that the equation of trajectory of the particle is

$\displaystyle{\frac{y}{h}=\frac{4x}{R}\bigg( 1-\frac{x}{R}\bigg)}$ .

Solution.

(The solution below is based on the manuscript of 2014-2015 PHYS1250 Fundamental Physics Homework Solutions.)

The trajectory of projectile motion must be a parabola, which can be expressed in the form of a quadratic equation:

$y=ax^2+bx+c$ ;

And since the particle passes through the points $(0,0)$ and $(R,0)$ , the equation of trajectory can be expressed in the form:

$y=A(x-0)(x-R)$ .

When the particle has traveled a horizontal distance $x=\displaystyle{\frac{R}{2}}$ , it reaches the maximum height $y=h$ .

Inserting the point $(\frac{R}{2},h)$ into the trajectory equation, we solve for the unknown $A$ :

$\begin{aligned} h & = A\bigg(\frac{R}{2}-0\bigg)\bigg(\frac{R}{2}-R\bigg) \\ h & = -\frac{AR^2}{4} \\ \Rightarrow \qquad A & = -\frac{4h}{R^2} \end{aligned}$

Thus,

$y = -\displaystyle{\frac{4h}{R^2}}x(x-R)$ ,

or,

$\boxed{\frac{y}{h} = \frac{4x}{R}\bigg( 1-\frac{x}{R}\bigg)}$

April 15, 2021October 24, 2022

202104150814 Homework 1 (Q2)

A particle is thrown with speed $v_0$ and an elevated angle $\theta$ on the floor. The air resistance is negligible.

(a) During the flight, the following quantities are investigated. Determine whether the following items are constants.

i. $\displaystyle{\frac{\mathrm{d}v}{\mathrm{d}t}}$ , where $v$ is the speed of the particle.

ii. $\displaystyle{\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}}$ , where $\mathbf{v}$ is the velocity of the particle.

(b) What is the radius of curvature of the path when the particle reaches the highest point?

Answer.

(The solution below is based on the manuscript of 2014-2015 PHYS1250 Fundamental Physics Homework Solutions.)

(a)

i. $\displaystyle{\frac{\mathrm{d}v}{\mathrm{d}t}}$ varies with time.

ii. $\displaystyle{\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}}=\mathbf{a}=\mathbf{g}=\textrm{Const.}$

Explanation.

(b)

$\begin{aligned} a=\frac{v^2}{r} & \Rightarrow g=\frac{v_0^2\cos^2\theta}{r} \\ & \Rightarrow r=\frac{v_0^2\cos^2\theta}{g} \end{aligned}$

April 15, 2021October 24, 2022

202104150729 Homework 1 (Q1)

A man starts from the origin and walks $30\,\mathrm{m}$ due east in $25\,\mathrm{s}$ . Then, he walks $10\,\mathrm{m}$ due south in $10\,\mathrm{s}$ and $18\,\mathrm{m}$ due northwest in $15\,\mathrm{s}$ . Please take the paths due east and due north as the positive $x$ and $y$ directions respectively in the Cartesian plane.

(a) Sketch the man’s path on the Cartesian plane.

(b) What is the average velocity of the man?

(c) What is the average speed of the man?

Solution.

(The solution below is based on the manuscript of 2014-2015 PHYS1250 Fundamental Physics Homework Solutions.)

(a)

(b) The average velocity of the man is

$\begin{aligned} \mathbf{OC} & = \mathbf{OA} + \mathbf{AB} + \mathbf{BC} \\ & = 30\,\hat{\mathbf{i}} - 10\,\hat{\mathbf{j}} + 18\bigg( -\frac{1}{\sqrt{2}}\,\hat{\mathbf{i}} + \frac{1}{\sqrt{2}}\,\hat{\mathbf{j}} \bigg) \\ & = \bigg( 30-\frac{18}{\sqrt{2}} \bigg) \hat{\mathbf{i}} + \bigg( \frac{18}{\sqrt{2}}-10 \bigg) \,\hat{\mathbf{j}}\\ & = 17.27\,\hat{\mathbf{i}} + 2.73\,\hat{\mathbf{j}} \end{aligned}$

$\begin{aligned} \therefore \mathbf{v}_{\textrm{avg}}& =\frac{\mathbf{OC}}{\Delta t} \\ & = \frac{17.27\,\hat{\mathbf{i}} + 2.73\,\hat{\mathbf{j}}}{50} \\ & = 0.35\,\hat{\mathbf{i}} + 0.055\,\hat{\mathbf{j}} \\ \therefore\quad |\mathbf{v}_{\textrm{avg}}| & = \sqrt{0.35^2+0.055^2} = 0.35\,\mathrm{m\, s^{-1}} \end{aligned}$

Direction of $\mathbf{v}_{\textrm{avg}}$ :

$\begin{aligned} \tan\theta & = \frac{0.055}{0.35}=0.16 \\ \theta & = 8.9^\circ \end{aligned}$

(c) The average speed of the man is

$\begin{aligned} v_{\textrm{avg}} & = \frac{OA+AB+BC}{\Delta t} \\ & = \frac{30+10+18}{50} \\ & = 1.16\,\mathrm{m\,s^{-1}} \end{aligned}$