August 2024 – Physicspupil

August 22, 2024

202408221017 Exercise 5.68

A particle moves along the space curve

$\mathbf{r}=e^{-t}\cos t\,\hat{\mathbf{i}}+e^{-t}\sin t\,\hat{\mathbf{j}}+e^{-t}\,\hat{\mathbf{k}}$ .

Find the magnitude of the (a) velocity and (b) acceleration at any time $t$ .

_{Extracted from Spiegel, M. R. (1971). Schaum’s Outline of Theory and Problems of Advanced Mathematics for Engineers and Scientists.}

Roughwork.

Let displacement

$\mathbf{r}=r_x\,\hat{\mathbf{i}}+r_y\,\hat{\mathbf{j}}+r_z\,\hat{\mathbf{k}}$

where

$\begin{aligned} r_x(t) & = e^{-t}\cos t \\ r_y(t) & = e^{-t}\sin t \\ r_z(t) & = e^{-t} \\ \end{aligned}$

then velocity

$\mathbf{v}=v_x\,\hat{\mathbf{i}}+v_y\,\hat{\mathbf{j}}+v_z\,\hat{\mathbf{k}}$

where

$\begin{aligned} v_x & = \frac{\mathrm{d}r_x}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(e^{-t}\cos t) \\ v_x(t)& = -e^{-t}\sin t-e^{-t}\cos t \\ v_y & = \frac{\mathrm{d}r_y}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(e^{-t}\sin t) \\ v_y(t)& = e^{-t}\cos t-e^{-t}\sin t \\ v_z & = \frac{\mathrm{d}r_z}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(e^{-t}) \\ v_z(t)& = -e^{-t} \\ \end{aligned}$

then acceleration

$\mathbf{a}=a_x\,\hat{\mathbf{i}}+a_y\,\hat{\mathbf{j}}+a_z\,\hat{\mathbf{k}}$

where

$\begin{aligned} a_x & = \frac{\mathrm{d}v_x}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(-e^{-t}\sin t-e^{-t}\cos t) \\ a_x(t)& = 2e^{-t}\sin t\\ a_y & = \frac{\mathrm{d}v_y}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(e^{-t}\cos t-e^{-t}\sin t) \\ a_y(t)& = -2e^{-t}\cos t\\ a_z & = \frac{\mathrm{d}v_z}{\mathrm{d}t} \\ & =\frac{\mathrm{d}}{\mathrm{d}t}(-e^{-t}) \\ a_z(t)& = e^{-t} \\ \end{aligned}$

Cheat.

$\mathbf{r}(t) = \underbrace{e^{-t}}_{\textrm{(1)}}\underbrace{(\cos t,\sin t, 1)}_{\textrm{(2)}}$

where

$\begin{aligned} \textrm{(1):}&\enspace \textrm{radial centripetal} \\ \textrm{(2):}&\enspace \textrm{circumferential anticlockwise} \\ \end{aligned}$

_{By courtesy of WolframAlpha}

In a cylindrical coordinate system, the position of a particle can be written as

$\mathbf{r}=\rho\,\hat{\boldsymbol{\rho}}+z\,\hat{\mathbf{z}}$ .

The velocity of the particle is the time derivative of its position,

$\mathbf{v}=\displaystyle{\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}}=\dot{\rho}\,\hat{\boldsymbol{\rho}}+\rho\dot{\varphi}\,\hat{\boldsymbol{\varphi}}+\dot{z}\,\hat{\mathbf{z}}$ ,

where the term $\rho\dot{\varphi}\,\hat{\boldsymbol{\varphi}}$ comes from the Poisson formula

$\displaystyle{\frac{\mathrm{d}\hat{\boldsymbol{\varphi}}}{\mathrm{d}t}}=\dot{\varphi}\,\hat{\mathbf{z}}\times\hat{\boldsymbol{\rho}}$ .

Its acceleration is

$\mathbf{a}=\displaystyle{\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t}}=(\ddot{\rho}-\rho\dot{\varphi}^2)\,\hat{\boldsymbol{\rho}}+(2\dot{\rho}\dot{\varphi}+\rho\ddot{\varphi})\,\hat{\boldsymbol{\varphi}}+\ddot{z}\,\hat{\mathbf{z}}$ .

_{Kinematics in Wikipedia on Cylindrical coordinate system}

Stop wandering! You are facing up to

$\begin{aligned} v=|\mathbf{v}|=\sqrt{v_x^2+v_y^2+v_z^2} & = \cdots \\ a=|\mathbf{a}|=\sqrt{a_x^2+a_y^2+a_z^2} & = \cdots \\ \end{aligned}$

But then, I’m fond of opting out.

This problem is not to be attempted.

August 21, 2024August 22, 2024

202408211136 Exercise 14.C.10

A stone is thrown vertically upwards so that its height, $s$ metres, above the ground, after $t$ seconds is given by $s=40t-5t^2$ . Find

(a) its velocity after $2\,\mathrm{s}$ ;
(b) its height above the ground when it is momentarily at rest;
(c) its initial velocity;
(d) its velocity when it is $15\,\mathrm{m}$ above the ground, giving your answer correct to the nearest $\mathrm{m\,s^{-1}}$ .

_{Extracted from K. S. Teh & C. Y. Loh. (2007). New Syllabus Additional Mathematics (8e)}

Roughwork.

Initially (i.e., at $t=0$ ), the stone is

$s(t=0)=40(0)-5(0)^2=0\,\mathrm{m}$

above the ground, i.e., at ground level $s=0$ .

$\begin{aligned} & s = 40t-5t^2 = t(40-5t) \\ \Rightarrow & \begin{cases} s =0 \Leftrightarrow t = \{0\}\cup\{ 8\} \\ s > 0 \Leftrightarrow t\in (0,8) \\ s < 0 \Leftrightarrow t<0\enspace\textrm{(rej.)}\enspace\textrm{\scriptsize{OR}}\enspace t>8 \\ \end{cases} \\ \end{aligned}$

Take upward positive. The (vertical) displacement of the stone is

$\begin{aligned} \mathbf{s}(t) & =s(t)\,\hat{\mathbf{j}} \\ & = (40t-5t^2)\,\hat{\mathbf{j}} \\ \end{aligned}$

its (vertical) instantaneous velocity

$\begin{aligned} \mathbf{v}(t) & = v(t)\,\hat{\mathbf{j}} \\ & = \frac{\mathrm{d}s(t)}{\mathrm{d}t}\,\hat{\mathbf{j}} \\ & = (40t-5t^2)'\,\hat{\mathbf{j}} \\ & = (40-10t)\,\hat{\mathbf{j}} \\ \end{aligned}$

and its (vertical) instantaneous acceleration

$\begin{aligned} \mathbf{a}(t) & = a(t)\,\hat{\mathbf{j}} \\ & = \frac{\mathrm{d}v(t)}{\mathrm{d}t}\,\hat{\mathbf{j}} \\ & = (40-10t)'\,\hat{\mathbf{j}} \\ \mathbf{a} & = -10\,\hat{\mathbf{j}} \\ \end{aligned}$

so are the answers to

(a) $\mathbf{v}(2)$
(b) $t'\in\{ t:v(t)=0\}\enspace\rightsquigarrow\enspace s(t')$
(c) $\mathbf{v}(0)$
(d) $t'\in\{ t:\mathbf{s}(t)=+15\,\hat{\mathbf{j}}\} \enspace\rightsquigarrow\enspace \mathbf{v}(t')$

This problem is not to be attempted.

August 20, 2024August 20, 2024

202408201533 Example 13.8

Let $\displaystyle{x=\frac{y-2}{y+2}}$ .

(a) Find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ in terms of $y$ .
(b) Make $y$ the subject of the given function. Hence, find $\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}}$ in terms of $x$ .
(c) Are the answers obtained in (a) and (b) different? Explain your assertion.

_{Extracted from S. W. Li et al. (2002). New Progress in Additional Mathematics Book 3}

Roughwork.

$\begin{aligned} x & = \frac{y-2}{y+2} \\ x & = \frac{y+2-4}{y+2} \\ \textrm{Eq. (1):}\quad x=f(y) & = 1-\frac{4}{y+2} \\ \frac{4}{y+2} & = 1-x \\ \frac{4}{1-x} & = y+2 \\ \textrm{Eq. (2):}\quad y=g(x) & = \frac{4}{1-x} - 2 \\ \end{aligned}$

domain: what can go into a function
codomain: what may possibly come out of a function
range: what actually comes out of a function

_{Kevin Sookocheff on Range, Domain, and Codomain (2018)}

Write, for function $f$ ,

$\begin{aligned} \textrm{domain: }& \mathbb{R}\backslash\{ -2\} \\ \textrm{codomain: }& \mathbb{R} \\ \textrm{range: }& \mathbb{R} \\ \end{aligned}$

and for function $g$ ,

$\begin{aligned} \textrm{domain: }& \mathbb{R}\backslash\{ 1\} \\ \textrm{codomain: }& \mathbb{R} \\ \textrm{range: }& \mathbb{R} \\ \end{aligned}$

We see $f$ discontinuous at $-2$ ; and $g$ , at $1$ ; whereas non-differentiable.

(to be continued)

August 19, 2024August 20, 2024

202408191743 Geometry 0

$\forall$ : for all/any/each/every
$\exists$ : there exist(s)/is/are

If, in any event $(x,y,z,t)$ :

i. no two points $a$ and $b$ occupy the same position. I.e.,

$\forall\, t\, \nexists\, a(\neq b)\enspace\textrm{s.t. }(x_a,y_a,z_a)=(x_b,y_b,z_b)$

ii. none any point $a$ occupies more than one position. I.e.,

$\forall\,a,t\,\exists\, !\mathbf{r}\enspace\textrm{s.t. }a\stackrel{\mathbf{r}(t)}{\longmapsto}(x_a,y_a,z_a)$

such points are said to be mass particles; or else massless particles.

(to be continued)