September 2021 – Page 2

September 16, 2021October 24, 2022

202109160920 Exercises 5.3.B (Q21)

Show that for all $x>0$ ,

$\displaystyle{\int_{0}^{x}\frac{\mathrm{d}t}{1+t^2}+\int_{0}^{1/x}\frac{\mathrm{d}t}{1+t^2}}$

is independent of $x$ .

Solution.

$\begin{aligned} & \qquad\int_{0}^{x}\frac{\mathrm{d}t}{1+t^2}+\int_{0}^{1/x}\frac{\mathrm{d}t}{1+t^2} \\ \dots &\enspace\textrm{as }\sin^2u +\cos^2u =1 \enspace\dots\\ \dots &\enspace\textrm{so }\tan^2u +1 = \sec^2u\enspace\dots \\ \dots &\enspace\textrm{let }t=\tan u \textrm{ then }\mathrm{d}t=\sec^2u\,\mathrm{d}u\enspace\dots \\ & = \int_{0}^{\arctan x}\frac{\sec^2u\,\mathrm{d}u}{1+\tan^2u} + \int_{0}^{\arctan (\frac{1}{x})}\frac{\sec^2u\,\mathrm{d}u}{1+\tan^2u} \\ & = \int_{0}^{\arctan x}\frac{\sec^2u\,\mathrm{d}u}{\sec^2u} + \int_{0}^{\arctan (\frac{1}{x})}\frac{\sec^2u\,\mathrm{d}u}{\sec^2u} \\ & = [u]|_{0}^{\arctan x} + [u]|_{0}^{\arctan (1/x)}\\ & = \arctan x + \arctan\bigg(\frac{1}{x}\bigg) \\ & \stackrel{(*)}{=} \frac{\pi}{2} \end{aligned}$

Lemma (*)

Let $y=\arctan x$ and $w=\arctan \displaystyle{\bigg(\frac{1}{x}\bigg)}$ ,

then $x=\tan y$ and $\displaystyle{\frac{1}{x}}=\tan w$ .

Thus,

$\begin{aligned} \tan y & = \frac{1}{\tan w} \\ \tan y & = \cot w \\ \tan y & = \tan \bigg(\frac{\pi}{2}-w\bigg) \\ y & = \frac{\pi}{2} - w \\ y + w & = \frac{\pi}{2} \\ \end{aligned}$

$\therefore\enspace \displaystyle{\arctan x + \arctan\bigg(\frac{1}{x}\bigg) = \frac{\pi}{2}}$

(independent of $x$ )

September 16, 2021July 25, 2022

202109160908 Exercises 5.1.A (Q18)

Use the free fall motion equation for position to show that the maximum height reached by an object launched straight up from the ground with an initial velocity $v_0$ is $\frac{v_0^2}{2g}$ .

Proof.

$\begin{aligned} v^2 & = u^2+2as \\ \dots \textrm{ upward} & \textrm{ taken to be positive }\dots \\ 0 & = v_0^2 + 2(-g)s \\ s & = \frac{v_0^2}{2g} \end{aligned}$

September 16, 2021October 24, 2022

202109160857 Exercises 5.1.A (Q17)

Integrate both sides of the equation

$\displaystyle{\frac{\mathrm{d}P}{P}+\frac{\mathrm{d}M}{M}=\frac{\mathrm{d}T}{2T}}$

to obtain the ideal gas continuity relation:

$\displaystyle{\frac{PM}{\sqrt{T}}=\textrm{constant}}$ .

Attempts.

$\begin{aligned} \int\frac{\mathrm{d}P}{P} + \int\frac{\mathrm{d}M}{M} & = \int\frac{\mathrm{d}T}{2T} \\ \ln |P| + \ln |M| - \frac{1}{2}\ln |T| & = C \\ \ln \bigg|\frac{PM}{T^{1/2}}\bigg| & = C \\ \frac{PM}{\sqrt{T}} & = \textrm{Const.} \\ \end{aligned}$

September 15, 2021October 24, 2022

202109151516 CYU 7.1

If $a=3+4i$ , what is the product $a^*a$ ?

Background. (Complex conjugate)

The complex conjugate of $a+bi$ is equal to $a-bi$ .

$\begin{aligned} a^*a & = (3-4i)(3+4i) \\ & = (3)(3)+(3)(4i)+(-4i)(3)+(-4i)(4i) \\ & = 9 - 16i^2 \\ & = 9 - 16(-1)\\ & = 9+16 \\ & = 25 \\ \end{aligned}$

September 15, 2021October 24, 2022

202109151444 CYU 3.4

When $1.00\,\mathrm{g}$ of ammonia boils at atmospheric pressure and $-33.0\,^\circ\mathrm{C}$ , its volume changes from $1.47\,\mathrm{cm^3}$ to $1130\,\mathrm{cm^3}$ . Its heat of vaporization at this pressure is $1.37\times 10^6\,\mathrm{J/kg}$ . What is the change in the internal energy of the ammonia when it vaporizes?

Solution.

With $L_v$ representing the latent heat of vaporization, the heat required to vaporize ammonia is

$\begin{aligned} Q & =mL_v \\ & =(0.001\,\mathrm{kg})(1.37\times 10^6\,\mathrm{J/kg}) \\ & =1.37\times 10^3\,\mathrm{J} \\ \end{aligned}$

Since the pressure on the system is constant at $1.00\,\mathrm{atm}=1.01\times 10^5\,\mathrm{N/m^2}$ , the work done by ammonia as it is vaporized is

$\begin{aligned} W & =p\Delta V \\ & =(1.01\times 10^5\,\mathrm{N/m^2})\big( (1130-1.47)\times (0.01\,\mathrm{m})^3 \big) \\ & = (1.01\times 10^5\,\mathrm{N\, m})(0.00112853) \\ & = 113.98153\,\mathrm{J} \\ \end{aligned}$

By the first law of thermodynamics, the internal (/thermal) energy of ammonia during its vaporization changes by

$\begin{aligned} \Delta E_{\textrm{int}} & =Q-W \\ & =1.37\times 10^3\,\mathrm{J} - 113.98153\,\mathrm{J}\\ & = 1256.01847\,\mathrm{J} \\ & \approx 1.26\,\mathrm{kJ} \end{aligned}$

September 15, 2021October 24, 2022

202109151432 CYU 1.3

If $25\,\mathrm{kJ}$ is necessary to raise the temperature of a rock from $25^\circ\mathrm{C}$ to $30^\circ\mathrm{C}$ , how much heat is necessary to heat the rock from $45^\circ\mathrm{C}$ to $50^\circ\mathrm{C}$ ?

Eq. (1.5):

$Q=mc\Delta T$

Substituting $25\,\mathrm{kJ}=25000\,\mathrm{J}$ for heat transferred $Q$ and $30\,^\circ\textrm{C} - 25\,^\circ\textrm{C} = 5\,\mathrm{C}^\circ$ for temperature change $\Delta T$ , we have

$mc=\displaystyle{\frac{Q}{\Delta T}=\frac{25000}{5}}=5000\,\mathrm{unit}$

To heat the rock from $45\,^\circ\mathrm{C}$ to $50\,^\circ\mathrm{C}$ , the heat needed is

$\begin{aligned} Q & = mc\Delta T \\ & = (5000)(50-45) \\ & = 25000\,\mathrm{J} \\ & = 25\,\mathrm{kJ} \end{aligned}$

September 15, 2021October 24, 2022

202109151235 CYU 1.7

How does the rate of heat transfer by conduction change when all spatial dimensions are doubled?

Background. (Rate of conductive heat transfer)

The rate of conductive heat transfer through a slab of material is given by $P=\displaystyle{\frac{\mathrm{d}Q}{\mathrm{d}t}=\frac{kA(T_{\textrm{h}}-T_{\textrm{c}})}{d}}$ where $P$ is the power or rate of heat tranfer, $A$ and $d$ are its surface area and thickness, $T_{\textrm{h}}-T_{\textrm{c}}$ is the temperature difference across the slab, and $k$ is the thermal conductivity of the material. More generally, $P=-kA\displaystyle{\frac{\mathrm{d}T}{\mathrm{d}x}}$ where $x$ is the coordinate in the direction of heat flow.

$\begin{aligned} A_{\textrm{initial}} & = l^2 \\ A_{\textrm{final}} & = l'^2 \\ (\,\dots\enspace \textrm{as } l' & =2l\enspace \dots\, ) \\ A_{\textrm{final}} & = (2l)^2 \\ & = 4l^2 \\ & = 4A_{\textrm{initial}} \end{aligned}$

and

$d_{\textrm{final}}=2d_{\textrm{initial}}$ ,

so,

$\begin{aligned} P_{\textrm{final}} & = \frac{kA_{\textrm{final}}(T_{\textrm{h}}-T_{\textrm{c}})}{d_{\textrm{final}}} \\ & = \frac{k(4A_{\textrm{initial}})(T_{\textrm{h}}-T_{\textrm{c}})}{(2d_{\textrm{initial}})} \\ & = 2\cdot \frac{kA_{\textrm{initial}}(T_{\textrm{h}}-T_{\textrm{c}})}{d_{\textrm{initial}}} \\ & = 2P_{\textrm{initial}} \end{aligned}$

$\therefore$ The rate of heat transfer by conduction increases by a factor of two.

September 15, 2021October 24, 2022

202109151148 CYU 1.9

How much greater is the rate of heat radiation when a body is at the temperature $40\,^\circ\mathrm{C}$ than when it is at the temperature $20\,^\circ\mathrm{C}$ ?

Background. (Stefan-Boltzmann law of radiation)

$P=\sigma AeT^4$

where $\sigma = 5.67\times 10^{-8}\,\mathrm{J/s}\cdot\mathrm{m^2}\cdot\mathrm{K^4}$ is the Stefan-Boltzmann constant; $A$ is the surface area of the object; and $T$ is its temperature in kelvins.

By temperature conversion formula,

$T_{\mathrm{(K)}}=T_{\mathrm{(^\circ C)}}+273.15$ .

$\begin{aligned} & T_{\mathrm{(^\circ C)}} = 20 \\ \Rightarrow \qquad & T_{\mathrm{(K)}} = 20 + 273.15 = 293.15 \\ & T_{\mathrm{(^\circ C)}} = 40 \\ \Rightarrow \qquad & T_{\mathrm{(K)}} = 40 + 273.15 = 313.15 \\ \end{aligned}$

The rate of heat transfer by emitted radiation is described by the Stefan-Boltzmann law of radiation:

$P=\sigma AeT^4$ .

When the temperature increases from $20\,^\circ\mathrm{C}$ to $40\,^\circ\mathrm{C}$ , ceteris paribus, the percentage change of rate of heat radiation is

$\begin{aligned} & \qquad \frac{P_{\textrm{new}}-P_{\textrm{old}}}{P_{\textrm{old}}} \times 100 \% \\ & = \frac{\sigma AeT_{\textrm{new}}^4-\sigma AeT_{\textrm{old}}^4}{\sigma AeT_{\textrm{old}}^4} \times 100 \% \\ & = \frac{T_{\textrm{new}}^4-T_{\textrm{old}}^4}{T_{\textrm{old}}^4} \times 100 \% \\ & = \frac{313.15^4-293.15^4}{293.15^4}\times 100 \% \\ & \approx 30.2 \% \qquad \textrm{(3\, s.f.)} \\ \end{aligned}$

$\therefore$ The rate of heat transfer increases by about $30\%$ of the original rate.

September 10, 2021July 25, 2022

202109101713 Exercises 1.2.1

Let $T:\mathbb{R}^2\rightarrow \mathbb{R}^3$ be the function defined by $T(x_1,x_2)=(x_1,x_2,0)$ . Show that $T$ is a linear transformation.

Definition. A function $T:\mathbb{R}^n\rightarrow \mathbb{R}^m$ is called a linear transformation if:
(1) $T(\mathbf{x}+\mathbf{y})=T(\mathbf{x})+T(\mathbf{y})$ ; and
(2) $T(c\mathbf{x})=cT(\mathbf{x})$ .
The conditions must be satisfied for all $\mathbf{x},\mathbf{y}$ in $\mathbb{R}^n$ and all $c$ in $\mathbb{R}$ .

Proof.

Let $\mathbf{x}=(x_1,x_2)$ and $\mathbf{y}=(y_1,y_2)$ .

Then

$\begin{aligned} T(\mathbf{x})& =T(x_1,x_2)=(x_1,x_2,0) \\ T(\mathbf{y})&=T(y_1,y_2)=(y_1,y_2,0) \\ T(\mathbf{x})+T(\mathbf{y}) & = (x_1,x_2,0) + (y_1,y_2,0) \\ \Rightarrow\enspace\textrm{RHS} & = (x_1+y_1,x_2+y_2,0)\\ \mathbf{x}+\mathbf{y} & = (x_1,x_2) + (y_1,y_2) \\ & = (x_1+y_1,x_2+y_2) \\ T(\mathbf{x}+\mathbf{y}) & = T(x_1+y_1,x_2+y_2) \\ \Rightarrow\enspace\textrm{LHS} & = (x_1+y_1,x_2+y_2,0) \\ \therefore\enspace \textrm{LHS} & = \textrm{RHS} \\ \end{aligned}$

Condition (1) is thus satisfied.

$\forall\, c\in\mathbb{R}$ and $\forall\, \mathbf{x}=(x_1,x_2)\in\mathbb{R}^2$ ,

$\begin{aligned} cT(\mathbf{x}) & = cT(x_1,x_2) \\ & = c(x_1,x_2,0) \\ \Rightarrow\enspace\textrm{RHS} & = (cx_1,cx_2,0) \\ T(c\mathbf{x}) & = T(c(x_1,x_2)) \\ & = T(cx_1,cx_2) \\ \Rightarrow \enspace\textrm{LHS} & = (cx_1,cx_2,0) \\ & = \textrm{RHS} \\ \end{aligned}$

Condition (2) is also satisfied.

In conclusion, $T$ is a linear transformation.

September 10, 2021October 24, 2022

202109101556 Exercises 1.2.C (Q16)

For Exercises 16-21, assuming that $f'(x)$ exists, prove the given formula.

$f'(x)=\displaystyle{\lim_{h\to 0}\frac{f(x+2h)-f(x-2h)}{4h}}$

Proof.

Renaming by dummy variables.

Let $y=x-2h$ , then $x+2h=(x-2h)+4h=y+4h$ .

Rewrite it as

$f'(x)=\displaystyle{\lim_{h\to 0}\frac{f(y+4h)-f(y)}{4h}}$ .

Note that

$\displaystyle{\lim_{h\to 0}}[\,\cdots ]\Rightarrow \displaystyle{\lim_{4h\to 0}}[\,\cdots ]$ .

So,

$\begin{aligned} f'(x) & = \lim_{4h\to 0}\frac{f(y+4h)-f(y)}{4h} \\ & = \lim_{\Delta y\to 0}\frac{f(y+\Delta y)-f(y)}{\Delta y} \\ & = \lim_{\Delta y\to 0}\frac{\Delta f}{\Delta y}\\ & = \frac{\mathrm{d}f}{\mathrm{d}y}\\ & = \dots\enspace \textrm{(discontinued)}\enspace \dots \\ \end{aligned}$

Do you spot the flaw in the ~~Proof~~?

(revised)

As left-hand limit and right-hand limit are equivalent,

i.e., $f'(x)=\displaystyle{\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{f(x)-f(x-h)}{h}}$ ,

in our scenario, do write

$\begin{aligned} f'(x) & = \lim_{2h\to 0}\frac{f(x+2h)-f(x)}{2h}=\lim_{2h\to 0}\frac{f(x)-f(x-2h)}{2h} \\ \frac{1}{2}f'(x) & =\lim_{2h\to 0}\frac{f(x+2h)-f(x)}{4h}=\lim_{2h\to 0}\frac{f(x)-f(x-2h)}{4h}\\ \end{aligned}$

Then

$\begin{aligned} & \quad \lim_{h\to 0}\frac{f(x+2h)-f(x-2h)}{4h} \\ & = \lim_{h\to 0}\frac{\big( f(x+2h)-f(x)\big) + \big( f(x)-f(x-2h) \big) }{4h} \\ & = \lim_{h\to 0}\frac{f(x+2h)-f(x)}{4h} + \lim_{h\to 0}\frac{f(x)-f(x-2h)}{4h} \\ & = \lim_{2h\to 0}\frac{f(x+2h)-f(x)}{4h} + \lim_{2h\to 0}\frac{f(x)-f(x-2h)}{4h} \\ & = \frac{1}{2}\cdot f'(x)+\frac{1}{2}\cdot f'(x) \\ & = f'(x) \\ \end{aligned}$

QED