# 工程原理 Engineering Principles GENG0005W1-01

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$$m=m_{0} \sqrt{1-\frac{v^{2}}{c^{2}}}$$
where
$m$ – mass of a moving body (also called the variable mass)
$m_{0}$ – rest mass of a body (velocity is zero)
$v-$ velocity of a moving body
$c$ – speed of light
The ratio, $v^{2} / c^{2}$ is usually denoted as $\beta^{2}$. Thus
$$m=m_{0} / \sqrt{1-\beta^{2}}$$
Similarly, the relativistic energy of a body moving with velocity $v$ is no more expressed as $m v^{2} / 2$, but
$$E=m_{0} c^{2} / \sqrt{1-\beta^{2}}$$

## GENG0005W1-01COURSE NOTES ：

$$E^{2}=(p c)^{2}+\left(m_{0} c^{2}\right)^{2}$$
By definition, momentum can be described as a function of the mass and velocity of a moving body:
$$p=m v=m_{0} v / \sqrt{1-\beta^{2}}$$
Squaring Eq. (3-6)
\begin{aligned} &E^{2}=\left(m c^{2}\right)^{2}=\left(m_{0} c^{2}\right)^{2} / 1-\frac{v^{2}}{c^{2}} \rightarrow m^{2} c^{4}\left(1-\frac{v^{2}}{c^{2}}\right)=m_{0}^{2} c^{4} \ &m^{2} c^{4}=E^{2}=m_{0}^{2} c^{4}+m^{2} c^{2} v^{2} \end{aligned}
where $p=m v$, thus giving
$$E^{2}=(p c)^{2}+\left(m_{0} c^{2}\right)^{2}$$
For a massless particle (like a photon) it follows that the total energy depends on its momentum and the speed of light: $E=p c$. This aspect will be discussed in greater detail in later sections.

# 机械科学 Mechanical Science GENG0003W1-01

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If final pressure of the gas is $p_{3}$, for a constant volume process $3-1$,
$$p_{3}=\frac{T_{3}}{T_{1}} p_{1}=\frac{323}{423} \times 10=7.6 \mathrm{bar}$$
Let us find the mass of the gas $m$ and
$$m=\frac{p_{1} \forall \forall_{1}}{R T_{1}}=\frac{10 \times 10^{5} \times 0.336}{293 \times 423}=2.7 \mathrm{~kg}$$
Change in internal energy
$$d U=U_{3}-U_{1}=m C_{7}\left(T_{3}-T_{1}\right)=2.7 \times 0.703(323-423)=-189.8 \mathrm{~kJ}$$
The negative sign indicates that there is a decrease in internal energy.

## GENG0003W1-01COURSE NOTES ：

The path followed by the system, $p_{1} \forall_{1}^{\prime}=p_{2} \forall_{2}$.
So,
$$\left(\frac{\forall_{2}}{\forall_{1}}\right)^{\gamma}=\frac{p_{1}}{p_{2}}$$
and
$$\forall_{2}=\left(\frac{p_{1}}{p_{2}}\right)^{1 / \gamma} \forall_{1}=\left(\frac{500}{100}\right)^{1 / L 4} \times 0.2=0.6313 \mathrm{~m}^{3}$$
Hence work done,
$$W_{1-2}=\frac{p_{1} \forall_{1}-p_{2} \forall_{2}}{\gamma-1}=\frac{(500 \times 0.2-100 \times 0.6313) 10^{3}}{(1.4-1) \times 10^{3}}=92.175 \mathrm{~kJ}$$

# 数学 Mathematics B GENG0002W1-01

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from which we see that $f+(-f)=0$. The additive inverse law follows. For the distributive laws note that for real numbers $a, b$ and continuous functions $f, g \in V$, we have that for all $0 \leq x \leq 1$,
$$a(f+g)(x)=a(f(x)+g(x))=a f(x)+a g(x)=(a f+a g)(x),$$
which proves the first distributive law. For the second distributive law, note that for all $0 \leq x \leq 1$,
$$((a+b) g)(x)=(a+b) g(x)=a g(x)+b g(x)=(a g+b g)(x),$$
and the second distributive law follows. For the scalar associative law, observe that for all $0 \leq x \leq 1$,
$$((a b) f)(x)=(a b) f(x)=a(b f(x))=(a(b f))(x),$$
so that $(a b) f=a(b f)$, as required. Finally, we see that
$$(1 f)(x)=1 f(x)=f(x),$$

## GENG0002W1-01COURSE NOTES ：

First let $f(x), g(x) \in V$ and let $c$ be a scalar. By definition of the set $V$ we have that $f(1 / 2)=0$ and $g(1 / 2)=0$. Add these equations together and we obtain
$$(f+g)(1 / 2)=f(1 / 2)+g(1 / 2)=0+0=0 .$$
It follows that $V$ is closed under addition with these operations. Furthermore, if we multiply the identity $f(1 / 2)=0$ by the real number $c$ we obtain that
$$(c f)(1 / 2)=c \cdot f(1 / 2)=c \cdot 0=0 .$$
It follows that $V$ is closed under scalar multiplication. Now certainly the zero function belongs to $V$, since this function has value 0 at any argument. Therefore, $V$ contains an additive identity element. Finally, we observe that the negative of a function $f(x) \in V$ is also an element of $V$, since
$$(-f)(1 / 2)=-1 \cdot f(1 / 2)=-1 \cdot 0=0 .$$

# 材料和结构 Materials and Structures FEEG2005W1-01

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The order parameter satisfies the time dependent Ginsburg-Landau equation
$$\frac{\partial \eta}{\partial t}=-M_{\eta} \frac{\delta F}{\delta \eta}$$
where $\mathrm{M}_{\eta}$ is a kinetic coefficient. The temperature field satisfies the equation of heat conduction with a source term [1]
$$\frac{\partial T}{\partial t}=\nabla \cdot(D \nabla T)+\frac{q}{c} \frac{d \eta}{d t}$$
where $\mathrm{D}$ is the thermal diffusivity, $\mathrm{q}$ is the latent heat, and $\mathrm{c}$ is the heat capacity.

## FEEG2005W1-01COURSE NOTES ：

$\dot{U}{Z}=\frac{\partial \dot{U}{R}}{\partial Z}=\frac{\partial \dot{U}{\theta}}{\partial Z}=0$ at $Z=0$ and $Z=L$, where $L$ is the length of the CNT. The homogeneous governing equations (9) and boundary conditions (10) constitute an Eigenvalue problem for the displacement increment $\dot{U}$. The Eigenvalue is the axial strain $E{z Z}$. In other words, Eqs. (9) and (10) have only the trivial solution $\dot{U}=0$ until the axial strain $E_{z z}$ reaches a critical value $\left(E_{\text {ZZ }}\right)_{\text {critical }}$ for bifurcation.

# 流体力学 Fluid Mechanics FEEG2003W1-01

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Start with the following vector identity (see the table of vector identities at the back of the book)
$$\frac{1}{2} \nabla(\mathbf{u} \cdot \mathbf{u})=\mathbf{u} \times \operatorname{curl} \mathbf{u}+(\mathbf{u} \cdot \nabla) \mathbf{u} .$$
Substituting this into the equations of motion yields
$$\frac{\partial \mathbf{u}}{\partial t}+\frac{1}{2} \nabla(\mathbf{u} \cdot \mathbf{u})-\mathbf{u} \times \operatorname{cur} \mathbf{u}=-\nabla w$$
Taking the curl and using the identity $\nabla \times \nabla f=0$ gives
$$\frac{\partial \xi}{\partial t}-\operatorname{curl}(\mathbf{u} \times \xi)=0$$
Using the identity (also from the back of the book)
$$\operatorname{curl}(\mathbf{F} \times \mathbf{G})=\mathbf{F} \operatorname{div} \mathbf{G}-\mathbf{G} \operatorname{div} \mathbf{F}+(\mathbf{G} \cdot \nabla) \mathbf{F}-(\mathbf{F} \cdot \nabla) \mathbf{G}$$

## FEEG2003W1-01COURSE NOTES ：

vortex sheet. Let $V$ denote the region of the vortex tube between $C_{1}$ and $C_{2}$ and $\Sigma=S \cup S_{1} \cup S_{2}$ denote the boundary of $V$. By Gauss’ theorem,
$$0=\int_{V} \nabla \cdot \xi d x=\int_{\Sigma} \xi \cdot d \mathbf{A}=\int_{S_{1} \cup S_{2}} \xi \cdot d \mathbf{A}+\int_{S} \xi \cdot d \mathbf{A} .$$
By Stokes’ theorem
$$\int_{C_{1}} \mathbf{u} \cdot d \mathbf{s}=\int_{S_{1}} \boldsymbol{\xi} \cdot d \mathbf{A} \quad \text { and } \quad \int_{C_{2}} \mathbf{u} \cdot d \mathbf{s}=-\int_{S_{2}} \boldsymbol{\xi} \cdot d \mathbf{A}$$
so (a) holds. Part (b) now follows from Kelvin’s circulation theorem.

# 机械学、机器和振动Mechanics, Machines and Vibration FEEG2002W1-01

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Another characteristic of a random variable is the power spectral density $S(\lambda)$, defined as the Fourier transform of the autocorrelation function
$$S(\lambda)=\int_{-\infty}^{\infty} \Psi(\tau) e^{i \lambda \tau} d \tau$$
The integral of function $S(\lambda)$ is the variance of function $y(t)$, i.e., if the average value is equal to zero, the square of its r.m.s. value
$$y_{r m s}=\sqrt{\int_{-\infty}^{\infty} S(\lambda) d \lambda} .$$

## FFEEG2002W1-01COURSE NOTES ：

The resultant of the axial force $F_{z}$ exerted by the other parts of the bar is
$$F_{z}+\frac{1}{2} \frac{\partial F_{z}}{\partial z} d z-F_{z}+\frac{1}{2} \frac{\partial F_{z}}{\partial z} d z .$$
The dynamic equilibrium equation can then be written in the form
$$\rho A \ddot{u}{z}=\frac{\partial F{z}}{\partial z}+f_{z}(z, t) .$$
The axial force $F_{z}$ is easily linked with the displacement by the usual formula from the theory of elasticity
$$F_{z}=A \sigma_{z}=E A \epsilon_{z}=E A \partial u_{z} / \partial z .$$

# 工程问题的解决 Long Answer Exam: Engineering Problem Solving FEEG1050W1-01

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Let $\left(x_{1}, x_{2}\right)$ be a pair of vectors belonging to $E \times E$.
The following is obtained:
$$I\left(x_{1}+x_{2}\right) \equiv L_{x_{1}+x_{2}}$$
with:
$$L_{x_{1}+x_{2}}(v)=\left(x_{1}+x_{2}, v\right){E}=\left(x{1}, v\right){E}+\left(x{2}, v\right){E}=L{x_{1}}(v)+L_{x_{2}}(v), \forall v \in E .$$
In other words:
$$I\left(x_{1}+x_{2}\right)=I\left(x_{1}\right)+I\left(x_{2}\right) .$$
b) Let $x$ be any vector belonging to $E$ and let $\lambda$ be a given arbitrary real number, then:
$$I(\lambda x) \equiv L_{\lambda x}$$
with:
$$\forall v \in E: L_{\lambda x}(v)=(\lambda x, v){E}=\lambda(x, v){E}=\lambda L_{x}(v)$$
Therefore,
$$I(\lambda x)=\lambda I(x)$$

## FEEG1050W1-01COURSE NOTES ：

where $\sigma$ refers to the stress tensor and $\varepsilon\left(\mathbf{U}^{}\right)$ to the linear strain tensor associated with the virtual displacements field $\mathbf{U}^{}$ :
$$\varepsilon_{i j}\left(\mathbf{U}^{}\right)=\frac{1}{2}\left[\frac{\partial U_{i}^{}}{\partial x_{j}}+\frac{\partial U_{j}^{*}}{\partial x_{i}}\right] .$$
It would be necessary to introduce the normal force $N(x)$ and the force loading $f(x)$ defined by:
$$N(x)=\iint_{S(x)} \sigma_{11} \mathrm{~d} S(x), \quad f(x)=\iint_{S(x)} f_{1} \mathrm{~d} S(x)$$
$S(x)$ refers to the section having abscissa $x$.

# 工程基础知识 Multiple Choice Exam: Engineering Fundamentals FEEG1040W1-01

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where $V$ is total volume of the element. Substituting for the stress and strain
$$U_{e}=\frac{E}{2} \int_{V} y^{2}\left(\frac{\mathrm{d}^{2} v}{\mathrm{~d} x^{2}}\right)^{2} \mathrm{~d} V$$
which can be written as
$$U_{e}=\frac{E}{2} \int_{0}^{L}\left(\frac{\mathrm{d}^{2} v}{\mathrm{~d} x^{2}}\right)^{2}\left(\int_{A} y^{2} \mathrm{~d} A\right) \mathrm{d} x$$
A gain recognizing the area integral as the moment of inertia $I_{z}$ about the centroidal axis perpendicular to the plane of bending, we have
$$U_{e}=\frac{E I_{z}}{2} \int_{0}^{L}\left(\frac{\mathrm{d}^{2} v}{\mathrm{~d} x^{2}}\right)^{2} \mathrm{~d} x$$

## FEEG1040W1-01COURSE NOTES ：

Using the specified data, The cross-sectional area is
$$A=1(1)=1 \mathrm{in}^{2}$$
And the area moment of inertia about the $z$ axis is
$$I_{z}=b h^{3} / 12=1 / 12=0.083 \text { in. }{ }^{4}$$
The characteristic axial stiffness is
$$A E / L=1\left(10 \times 10^{6}\right) / 20=\left(5 \times 10^{5}\right) \mathrm{lb} / \mathrm{in} .$$
and the characteristic bending stiffness is
$$E I_{z} / L^{3}=10 \times 10^{6}(0.083) / 20^{3}=104.2 \mathrm{lb} / \mathrm{in} .$$

# 机器学习 Machine Learning MATH6168W1-01

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At this point, we can gain some insight into the role of the bias parameter $w_{0}$. If we make the bias parameter explicit, then the error becomes
Setting the derivative with respect to $w{0}$ equal to zero, and solving for $w_{0}$, we obtain
$$w_{0}=\bar{t}-\sum_{j=1}^{M-1} w_{j} \overline{\phi_{j}}$$
where we have defined
$$\bar{t}=\frac{1}{N} \sum_{n=1}^{N} t_{n}, \quad \overline{\phi_{j}}=\frac{1}{N} \sum_{n=1}^{N} \phi_{j}\left(\mathbf{x}_{n}\right)$$

## MATH6168W1-01COURSE NOTES ：

Setting the derivative with respect to $\mu_{1}$ to zero and rearranging, we obtain
$$\boldsymbol{\mu}{1}=\frac{1}{N{1}} \sum_{n=1}^{N} t_{n} \mathbf{x}{n}$$ which is simply the mean of all the input vectors $\mathbf{x}{n}$ assigned to class $\mathcal{C}{1}$. By a similar argument, the corresponding result for $\mu{2}$ is given by
$$\boldsymbol{\mu}{2}=\frac{1}{N{2}} \sum_{n=1}^{N}\left(1-t_{n}\right) \mathbf{x}_{n}$$

# 代数拓扑学 Algebraic Topology MATH3080W1-01

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To define the cup product we consider cohomology with coefficients in a ring $R$, the most common choices being $\mathbb{Z}, \mathbb{Z}_{n}$, and $\mathrm{Q}$. For cochains $\varphi \in C^{k}(X ; R)$ and $\psi \in C^{\ell}(X ; R)$, the cup product $\varphi \smile \Psi \in C^{k+\ell}(X ; R)$ is the cochain whose value on a singular simplex $\sigma: \Delta^{k+\ell} \rightarrow X$ is given by the formula

$$(\Phi \smile \Psi)(\sigma)=\Phi\left(\sigma \mid\left[v_{0}, \cdots, v_{k}\right]\right) \psi\left(\sigma \mid\left[v_{k}, \cdots, v_{k+\ell}\right]\right)$$
where the right-hand side is the product in $R$. To see that this cup product of cochains induces a cup product of cohomology classes we need a formula relating it to the coboundary map:

## MATH3080W1-01COURSE NOTES ：

The cup product formula $(\varphi \sim \psi)(\sigma)=\varphi\left(\sigma \mid\left[v_{0}, \cdots, v_{k}\right]\right) \psi\left(\sigma \mid\left[v_{k}, \cdots, v_{k+\ell}\right]\right)$ also gives relative cup products
$H^{k}(X ; R) \times H^{\ell}(X, A ; R) \longrightarrow H^{k+\ell}(X, A ; R)$
$H^{k}(X, A ; R) \times H^{\ell}(X ; R) \longrightarrow H^{k+\ell}(X, A ; R)$
$H^{k}(X, A ; R) \times H^{\ell}(X, A ; R) \stackrel{\smile}{\longrightarrow} H^{k+\ell}(X, A ; R)$
since if $\varphi$ or $\psi$ vanishes on chains in $A$ then so does $\varphi \cup \psi$. There is a more general relative cup product
$$H^{k}(X, A ; R) \times H^{\ell}(X, B ; R) \longrightarrow H^{k+\ell}(X, A \cup B ; R)$$