# 流体力学|Fluid mechanics II 3A3代写2023

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Assignment-daixieTM为您提供剑桥大学University of Cambridge Fluid mechanics II 3A3流体力学代写代考辅导服务！

## Instructions:

Fluids are generally considered to be those materials that have the ability to constantly change their shape by adapting to the container, which is why liquids, vapors and gases are considered to be fluids. Fluid mechanics consists of two main branches:

Fluid mechanics deals with fluids that are stationary in an inertial system, i.e. with constant velocity in time and homogeneity in space. Historically, it was the first step towards the study of mechanics.
Fluid dynamics or fluid mechanics (including specifically aerodynamics, hydrodynamics, and oil dynamics), deals with fluids in motion.
Fluids are characterized by having their own volume and a density very similar to that of solids, which means that at the microscopic level, the distances between molecules remain small and the interaction forces are high. This is a fundamental difference from gaseous substances, which have a low density and therefore low intermolecular interactions, allowing them to expand at any volume.

(a) Supersonic flow enters a straight pipe of constant cross-sectional area. Heat transfer is negligible, but the pipe wall is rough. Draw a labelled graph to show how the Mach number distribution along the pipe evolves as the skin-friction coefficient increases from zero. You may assume that the exit pressure is low enough to ensure that the inlet conditions are always the same.

(a) The graph below shows the variation of the Mach number with distance along the pipe as the skin friction coefficient increases from zero. As the skin friction coefficient increases, the velocity near the pipe wall decreases, causing the boundary layer to thicken. This results in a reduction in the effective cross-sectional area available for flow, which in turn reduces the mass flow rate and increases the Mach number. The Mach number gradually increases until it reaches the sonic condition at the throat of the pipe, after which it remains constant until the exit.

(b) Air flows in a pipe of length $5.9 \mathrm{~m}$ and inside diameter $0.2 \mathrm{~m}$. The inlet stagnation pressure is 2.7 bar, and the static pressure at the pipe exit is 1 bar. If the exit is choked, and there are no shocks in the pipe, find: (i) the two possible values of the Mach number at the inlet; (ii) the skin-friction coefficient $c_f$ corresponding to each.

(b) From the given data, we can use the choked flow condition to find the Mach number at the inlet. The choked flow condition occurs when the flow velocity at the throat of the pipe reaches the local speed of sound. At the throat, the Mach number is therefore 1.

Using the isentropic relations for a perfect gas, we can relate the Mach number to the pressure ratio across the throat:

$\frac{P_{02}}{P_{01}}=\left(\frac{1+\frac{\gamma-1}{2} M_1^2}{\frac{\gamma+1}{2}}\right)^{\frac{\gamma}{\gamma-1}}=\frac{P_{02}}{P_e}=\left(\frac{A_e}{A_{02}}\right)^2=1$,

where $P_{01}$ is the stagnation pressure at the inlet, $P_{02}$ is the pressure at the throat, $P_e$ is the static pressure at the exit, $A_{02}$ is the area of the throat, and $A_e$ is the area of the exit.

Solving for $M_1$ using the given values, we find that there are two possible values of the Mach number at the inlet:

$M_1=\sqrt{\frac{2}{\gamma-1}\left[\left(\frac{P_{02}}{P_{01}}\right)^{\frac{\gamma-1}{\gamma}}-1\right]}=0.747,2.11$.

Next, we can use the Prandtl-Meyer function to find the Mach number corresponding to a given skin-friction coefficient $c_f$. The Prandtl-Meyer function is a relation between the Mach number and the turning angle of a supersonic flow, and it depends only on the specific heat ratio $\gamma$ of the gas. The skin-friction coefficient $c_f$ can be related to the friction Reynolds number $Re_{\tau}$ using the law of the wall:

$c_f=\frac{\tau_w}{\frac{1}{2} \rho_1 V_1^2}=\frac{0.026}{R e_\tau^{0.2}}$,

where $\tau_w$ is the wall shear stress, $\rho_1$ is the density at the inlet, and $V_1$ is the velocity at the inlet.

For air at room temperature and pressure, $\gamma=1.4$. Using a table or a calculator, we can find that the Prandtl-Meyer function for $\gamma=1.4$ is approximately $\nu=30.5^{\circ}$ at $M=0.747$, and $\nu=66.1^{\circ}$ at $M=2.11$.

To find the turning angle corresponding to a given skin-friction coefficient, we can use the relation

$\theta=\frac{c_f}{2 \nu}$

# 市场营销|Marketing 3E2代写2023

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Assignment-daixieTM为您提供剑桥大学University of Cambridge Marketing 3E2市场营销代写代考辅导服务！

## Instructions:

Marketing refers to the process of identifying, anticipating, and satisfying the needs and wants of consumers through the creation, promotion, distribution, and pricing of goods and services. The ultimate goal of marketing is to create value for customers and build strong relationships with them, in order to generate profitable customer actions, such as sales, repeat purchases, and brand loyalty.

Marketing involves a variety of activities, including market research, product development, branding, advertising, sales promotion, public relations, and customer service. It is a vital function of any business or organization, as it helps to connect producers with consumers, and enables businesses to effectively communicate their offerings and value propositions to their target audiences. Successful marketing strategies require a deep understanding of consumer behavior, market trends, and competitive dynamics, as well as creativity and innovation in developing and executing marketing campaigns.

The Copper Kettle is a historic café and restaurant located on King’s Parade in Cambridge. Established more than 100 years ago, it serves drinks, breakfast, lunch and dinner every day and can also host special events. Currently they have no loyalty programme of any kind, but the General Manager is considering launching one. You have been approached to help them decide whether to go ahead with a loyalty programme or not, and how to increase customer retention. Outline your recommendations to the General Manager by addressing the following questions: (a) What are the advantage and disadvantages of loyalty programmes, and based on these, what would you recommend? Should the company launch a loyalty programme or not, and why?

(a) Loyalty programmes can offer various advantages, such as incentivizing repeat purchases, increasing customer retention and engagement, and providing valuable data and insights about customers’ preferences and behaviors. However, loyalty programmes can also be costly to implement and maintain, and there is a risk of customers signing up for the programme just for the rewards rather than being loyal to the business.

In the case of the Copper Kettle, I would recommend launching a loyalty programme. Given the highly competitive nature of the hospitality industry, offering incentives to customers can help differentiate the business and encourage them to return. Additionally, a loyalty programme can provide valuable data and insights that can inform business decisions and improve the overall customer experience. To mitigate the risk of customers joining the programme for the rewards only, the Copper Kettle can focus on creating a programme that reinforces their brand values and fosters a sense of community among customers.

(b) Regardless of whether you think the Copper Kettle should go ahead with a formal loyalty programme or not, what would you recommend the business should do to increase its customer loyalty?

(b) Regardless of whether the Copper Kettle decides to launch a formal loyalty programme or not, there are several ways the business can increase customer loyalty. One effective approach is to focus on enhancing the overall customer experience, including personalized service, attention to detail, and a welcoming atmosphere. This can help create an emotional connection with customers and foster loyalty beyond just the incentives of a loyalty programme.

Additionally, the Copper Kettle can leverage digital marketing channels to communicate with customers and keep them engaged. This can include email newsletters, social media engagement, and targeted promotions. Providing a seamless and convenient ordering and payment process can also contribute to customer satisfaction and loyalty.

Finally, the Copper Kettle can offer special events and experiences to create unique and memorable experiences for customers. This can include wine tastings, cooking classes, and other events that align with the business’s brand and appeal to their target audience. By creating a sense of exclusivity and providing value beyond just the products and services offered, the Copper Kettle can increase customer loyalty and drive repeat business.

You have been approached by an entrepreneur who would like to launch a new smartphone app in the UK. The app allows users to whistle or hum a song into their phone and identifies the song for them. The entrepreneur is at the stage where she has developed a basic working version of the app, and she’s ready to launch it. She has asked you to help her with the following questions: (a) According to the product life cycle model, what should her marketing priorities be?

According to the product life cycle model, the marketing priorities for a new product like this app would depend on its current stage in the life cycle. The product life cycle has four stages: introduction, growth, maturity, and decline.

Assuming that the entrepreneur is in the introduction stage, her marketing priorities should be focused on creating awareness of the app and generating interest among potential users. This could involve creating a strong brand identity, developing a compelling value proposition, and establishing a strong social media presence to reach and engage with potential users.

Other marketing priorities at the introduction stage could include identifying and targeting early adopters who are more likely to try out new products and share their experiences with others, as well as establishing partnerships with influencers, bloggers, and other media outlets to help spread the word about the app.

Overall, the entrepreneur should focus on building a strong foundation for the app in the introduction stage, so that it can grow and mature in the subsequent stages of the product life cycle.

# 固体力学|Mechanics of solids 3C7代写2023

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Assignment-daixieTM为您提供剑桥大学University of Cambridge Mechanics of solids 3C7固体力学代写代考辅导服务！

## Instructions:

Mechanics of solids is a branch of mechanics that deals with the behavior of solid materials subjected to various external forces. It is concerned with the study of deformation, stress, and failure of materials under various loading conditions.

Some of the fundamental concepts and principles of mechanics of solids are:

1. Stress: Stress is the force per unit area acting on a material. It is a measure of the internal forces that hold the material together.
2. Strain: Strain is the change in dimension of a material under the influence of stress. It is a measure of the deformation of the material.
3. Elasticity: Elasticity is the ability of a material to deform under stress and return to its original shape when the stress is removed. A material that exhibits this behavior is said to be elastic.
4. Plasticity: Plasticity is the ability of a material to undergo permanent deformation when subjected to stress. A material that exhibits this behavior is said to be plastic.
5. Yield strength: Yield strength is the stress at which a material begins to exhibit plastic deformation.
6. Failure: Failure is the point at which a material can no longer withstand the applied stress and ruptures.

Mechanics of solids is used in many engineering applications, such as the design of structures, machines, and materials. It is also used in fields like aerospace, civil, mechanical, and materials engineering, among others.

Derive the equilibrium equation in the $z$-direction in terms of the shear stress components $\sigma_{z y}$ and $\sigma_{z x}$ with all other stress components equal to zero.

The equilibrium equation in the $z$-direction can be derived from the principle of equilibrium, which states that the sum of forces and moments acting on a body must be zero for the body to be in static equilibrium. For a small element in a three-dimensional stress field, the equilibrium equation in the $z$-direction can be expressed as:

$\frac{\partial \sigma_{x z}}{\partial x}+\frac{\partial \sigma_{y z}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$,

where $\sigma_{xz}$ and $\sigma_{yz}$ are the shear stresses acting on the $z$-plane, $\sigma_{zz}$ is the normal stress acting on the $z$-plane, and $f_z$ is the external force acting in the $z$-direction.

Since all other stress components are zero, we have $\sigma_{xx}=\sigma_{yy}=\sigma_{xy}=0$. Thus, the equilibrium equation reduces to:

$\frac{\partial \sigma_{z x}}{\partial x}+\frac{\partial \sigma_{z y}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$.

Since $\sigma_{zy}=\sigma_{yz}$, we can write:

$\frac{\partial \sigma_{z x}}{\partial x}+\frac{\partial \sigma_{y z}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$.

Finally, substituting $\sigma_{yz}$ and $\sigma_{zx}$ with $\sigma_{zy}$ and $\sigma_{xz}$ respectively, we get:

$\frac{\partial \sigma_{x z}}{\partial x}+\frac{\partial \sigma_{z y}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$.

Therefore, the equilibrium equation in the $z$-direction in terms of the shear stress components $\sigma_{zy}$ and $\sigma_{xz}$ with all other stress components equal to zero is:

$\frac{\partial \sigma_{x z}}{\partial x}+\frac{\partial \sigma_{z y}}{\partial y}+\frac{\partial \sigma_{z z}}{\partial z}+f_z=0$

For the case of a torsion of a shaft made from an isotropic elastic material, the strain components are given in terms of the warping function $w(x, y)$ and the twist $\beta$ per unit length of the shaft by $$\gamma_{z x}=-\beta y+\frac{\partial w}{\partial x}, \text { and } \gamma_{z y}=\beta x+\frac{\partial w}{\partial y} .$$ Hence show that the warping function satisfies $\nabla^2 w=0$.

To show that the warping function $w(x, y)$ satisfies $\nabla^2 w=0$, we need to calculate the Laplacian of $w$ which is defined as the sum of the second partial derivatives of $w$ with respect to its independent variables $x$ and $y$.

$\nabla^2 w=\frac{\partial^2 w}{\partial x^2}+\frac{\partial^2 w}{\partial y^2}$

Let’s start by differentiating the first strain component $\gamma_{zx}$ with respect to $y$ and the second strain component $\gamma_{zy}$ with respect to $x$:

\begin{aligned} & \frac{\partial \gamma_{z x}}{\partial y}=-\beta \ & \frac{\partial \gamma_{z y}}{\partial x}=\beta\end{aligned}

Next, we can differentiate these expressions again, using the product rule for differentiation, to obtain:

\begin{aligned} & \frac{\partial^2 \gamma_{z x}}{\partial y^2}=0 \ & \frac{\partial^2 \gamma_{z y}}{\partial x^2}=0\end{aligned}

We can now use the strain-displacement equations for torsion, which relate the strains to the warping function and the twist per unit length of the shaft, to express the second partial derivatives of the strain components in terms of the warping function:

\begin{aligned} & \frac{\partial^2 \gamma_{z x}}{\partial y^2}=\frac{\partial}{\partial y}\left(\frac{\partial \gamma_{z x}}{\partial x}\right)=\frac{\partial}{\partial y}\left(\frac{\partial w}{\partial x}\right)=\frac{\partial^2 w}{\partial x \partial y} \ & \frac{\partial^2 \gamma_{z y}}{\partial x^2}=\frac{\partial}{\partial x}\left(\frac{\partial \gamma_{z y}}{\partial y}\right)=\frac{\partial}{\partial x}\left(\frac{\partial w}{\partial y}\right)=\frac{\partial^2 w}{\partial y \partial x}\end{aligned}

Using the symmetry of mixed partial derivatives, $\frac{\partial^2 w}{\partial x \partial y} = \frac{\partial^2 w}{\partial y \partial x}$, we can add these expressions to obtain:

$\frac{\partial^2 w}{\partial x^2}+\frac{\partial^2 w}{\partial y^2}=0$

which is the Laplace equation for $w$. Therefore, we have shown that the warping function satisfies $\nabla^2 w = 0$ for the case of torsion of a shaft made from an isotropic elastic material.

A turbine rotor has the form of a cylinder of outer radius $40 \mathrm{~mm}$ and length $60 \mathrm{~mm}$ with a central circular hole of radius $20 \mathrm{~mm}$. The rotor is shrink fitted onto a shaft of radius slightly greater than the circular hole in the rotor. The rotor and the shaft are made from the same alloy steel, with Young’s modulus $E=210 \mathrm{GPa}$, Poisson’s ratio $v=0.3$, and uniaxial tensile yield strength $Y=240 \mathrm{MPa}$. If the maximum shear stress in the assembly is limited to $Y / 3$, comment on whether the above design of the rotor assembly is suitable.

To determine whether the design of the rotor assembly is suitable, we need to calculate the maximum shear stress in the assembly and compare it to the limit of $Y/3$.

The assembly consists of the rotor and the shaft, which are shrink fitted together. This means that the rotor is heated to expand it, and then placed onto the shaft while it is still hot. As the rotor cools and contracts, it tightly grips the shaft. This creates a state of residual stress in both the rotor and the shaft.

To calculate the maximum shear stress in the assembly, we need to consider the stress state due to both the shrink fitting and any external loads. Let’s assume that the assembly is subjected to an external torque, causing the shaft to twist relative to the rotor. This creates a shear stress in the radial direction of the shaft.

The stress due to the shrink fitting can be calculated using the equation:

$\Delta \sigma = \frac{E \alpha \Delta T}{1-\nu}$

where $\Delta \sigma$ is the change in stress due to the shrink fitting, $E$ is the Young’s modulus, $\alpha$ is the coefficient of thermal expansion, $\Delta T$ is the temperature change during the fitting process, and $\nu$ is the Poisson’s ratio.

Assuming a temperature change of $100^{\circ}\mathrm{C}$ during the fitting process, and a coefficient of thermal expansion of $12\times10^{-6}\mathrm{/K}$ for steel, we can calculate the change in stress:

$\Delta \sigma = \frac{(210 \times 10^9 \mathrm{Pa})(12 \times 10^{-6} \mathrm{/K})(100^{\circ}\mathrm{C})}{1-0.3} \approx 2.52 \mathrm{MPa}$

This change in stress creates a compressive stress in the rotor and a tensile stress in the shaft.

Now let’s consider the external torque. The maximum shear stress in the assembly will occur at the outer surface of the shaft, where the diameter is the largest. The maximum shear stress due to the torque can be calculated using the formula:

$\tau = \frac{T r}{J}$

where $\tau$ is the shear stress, $T$ is the applied torque, $r$ is the radius at the location of interest, and $J$ is the polar moment of inertia of the shaft cross-section.

The polar moment of inertia of a solid cylinder is $J=\frac{\pi}{2}r^4$. Using $r=20\mathrm{~mm}$ for the shaft radius, and assuming a torque of $100\mathrm{~Nm}$, we can calculate the maximum shear stress due to the external torque:

$\tau = \frac{(100\mathrm{~Nm})(20\mathrm{~mm})}{\frac{\pi}{2}(20\mathrm{~mm})^4} \approx 8.03 \mathrm{MPa}$

The maximum shear stress in the assembly is the sum of the stress due to the shrink fitting and the stress due to the external torque:

$\tau_{max} = \Delta \sigma + \tau \approx 10.55 \mathrm{MPa}$

This is less than the limit of $Y/3 = 80\mathrm{~MPa}$, so the design of the rotor assembly is suitable.

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Assignment-daixieTM为您提供剑桥大学University of Cambridge Business economics 3E1商业经济学代写代考辅导服务！

## Instructions:

Business economics is the study of the principles and theories that underlie the decision-making process of businesses. It involves the application of economic concepts and theories to analyze business problems and make strategic decisions that maximize profits and efficiency.

Business economics encompasses a wide range of topics, including microeconomics (the study of individual markets and consumer behavior), macroeconomics (the study of the economy as a whole), accounting, finance, and management. It can be applied to a variety of industries, including manufacturing, finance, healthcare, retail, and technology.

Some of the key concepts in business economics include demand and supply analysis, market structure and competition, cost and production analysis, pricing strategies, risk management, and financial analysis. By applying these concepts, businesses can make informed decisions that optimize their resources and achieve their objectives.

(a) The demand for organic yogurt for a monopoly firm has been estimated according to the following linear regression model, where $\ln \mathrm{X}$ stands for the natural logarithm of variable $\mathrm{X}$ : $$\ln \mathrm{Q}=7-1.4 \ln \mathrm{P}+1.8 \ln \mathrm{A}+0.1 \ln \mathrm{M}$$ std errors: $(5.2)(0.45) \quad(2.2) \quad(0.04)$ Q is quantity demanded of yogurt, P is price of yogurt, A is advertising and M is income. The marginal cost of producing the yogurt is $£ 0.5$. The standard errors are given in parenthesis below each estimate. Interpret the results. What price would you advise the CEO to charge in order to maximize profits? Explain.

(a) The demand function for the organic yogurt can be expressed as:

$Q=e^{7-1.4 \ln P+1.8 \ln A+0.1 \ln M}$

Taking the derivative of the demand function with respect to P, we get:

$\frac{\partial Q}{\partial P}=-\frac{1.4 e^{7-1.4 \ln P+1.8 \ln A+0.1 \ln M}}{P}$

The marginal revenue function for the firm is given by:

$M R=\frac{\partial T R}{\partial Q}=\frac{\partial(P Q)}{\partial Q}+Q \frac{\partial P}{\partial Q}=P\left(1-\frac{1.4 e^{7-1.4 \ln P+1.8 \ln A+0.1 \ln M}}{Q}\right)$

The profit-maximizing condition for the monopoly firm is:

$M R=M C$

Substituting the values of MR and MC, we get:

$P\left(1-\frac{1.4 e^{7-1.4 \ln P+1.8 \ln A+0.1 \ln M}}{Q}\right)=0.5$

Solving for P, we get:

$P=e^{2.764+0.7 \ln A-0.05 \ln M}$

To determine the optimal price, we need to know the values of A and M. Assuming A = 100 and M = 10,000, we get:

$P=e^{2.764+0.7 \ln 100-0.05 \ln 10000}=£ 2.19$

Therefore, the CEO should charge £2.19 for the organic yogurt to maximize profits.

(b) A utility company faces economies of scale over all relevant levels of output. If the government wants to regulate the monopoly to produce the quantity that would be expected to hold under perfect competition, what price should the government stipulate? Include a diagram in your answer.

Under perfect competition, the firm produces where marginal cost (MC) equals marginal revenue (MR), which is also equal to price (P). This can be shown in the following diagram:

In this case, the government wants the monopoly to produce the quantity that would be expected to hold under perfect competition. This is shown as Qpc in the diagram. Since the monopoly faces economies of scale over all relevant levels of output, its average total cost (ATC) is decreasing over the range of output. This means that the price (P) must be equal to the minimum point of the ATC curve at Qpc in order for the firm to produce at the efficient scale. This is shown in the following diagram:

Therefore, the government should stipulate a price equal to the minimum point of the ATC curve at Qpc in order to regulate the monopoly to produce the efficient quantity.

(c) The ‘invisible hand’ is an old-fashioned idea of how markets should function. Discuss.

(c) The “invisible hand” is an old-fashioned idea that suggests that markets will allocate resources efficiently without the need for government intervention. According to this idea, individuals pursuing their own self-interest in a competitive market will create a balance between supply and demand that leads to the efficient allocation of resources. The concept was popularized by the economist Adam Smith in his book “The Wealth of Nations”.

However, the idea of the “invisible hand” has been subject to much criticism. Some argue that markets do not always function efficiently and can lead to market failures such as externalities, public goods

# 数理方法|Mathematical methods 1P4代写2023

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Assignment-daixieTM为您提供剑桥大学University of Cambridge Partial differential equations and variational methods 4M12偏微分方程和变分方法代写代考辅导服务！

## Instructions:

Mathematical methods are techniques used in mathematics to solve problems and analyze data. These methods are often used in scientific and engineering fields to model real-world phenomena and make predictions. Some examples of mathematical methods include:

1. Calculus: This branch of mathematics deals with the study of functions and their rates of change. It is used to solve problems in physics, engineering, and economics, among other fields.
2. Linear algebra: This is the study of systems of linear equations and their properties. It is used in fields such as computer graphics, physics, and economics.
3. Differential equations: These are mathematical equations that describe how a quantity changes over time. They are used in physics, engineering, and other sciences to model natural phenomena.
4. Statistics: This is the branch of mathematics that deals with the collection, analysis, interpretation, presentation, and organization of data. It is used in fields such as psychology, sociology, and economics.
5. Numerical analysis: This is the study of algorithms and computational methods used to solve mathematical problems. It is used in fields such as engineering, finance, and science to make numerical predictions.

Overall, mathematical methods are essential tools for solving problems and understanding the world around us.

Consider the following equation $$\text { (c.a) } \mathbf{a}+(\mathbf{c} . \mathbf{b}) \mathbf{b}=\mathbf{p}$$ where $\mathbf{a}$ and $\mathbf{b}$ are known vectors which are not parallel. (i) What is the value of $\mathbf{p} .(\mathbf{a} \times \mathbf{b})$ ?

(i) We have $\mathbf{p} .(\mathbf{a} \times \mathbf{b})=\mathbf{p} \cdot \mathbf{n}$, where $\mathbf{n}=\mathbf{a} \times \mathbf{b}$ is the normal vector to the plane spanned by $\mathbf{a}$ and $\mathbf{b}$. This dot product gives the scalar projection of $\mathbf{p}$ onto $\mathbf{n}$, which is the signed distance between $\mathbf{p}$ and the plane, multiplied by the magnitude of $\mathbf{n}$. Since $\mathbf{a}$ and $\mathbf{b}$ are not parallel, the cross product $\mathbf{a} \times \mathbf{b}$ is a nonzero vector orthogonal to both $\mathbf{a}$ and $\mathbf{b}$, and therefore defines a plane that contains both vectors. Thus, $\mathbf{p} .(\mathbf{a} \times \mathbf{b})$ gives the signed distance between $\mathbf{p}$ and the plane containing $\mathbf{a}$ and $\mathbf{b}$, multiplied by the area of the parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}$. In other words, $\mathbf{p} .(\mathbf{a} \times \mathbf{b})$ is twice the volume of the parallelepiped defined by $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{p}$.

The response $y(t)$ of a linear system with input $f(t)$ is given by the convolution integral $$y(t)=\int_0^t g(t-\tau) f(\tau) d \tau$$ where $g(t)$ is the impulse response. Both $f(t)$ and $g(t)$ are zero for $t<0$. Show that $$\int_0^t g(t-\tau) f(\tau) d \tau=\int_0^t f(t-\tau) g(\tau) d \tau$$

We start by making a change of variables $\tau = t-s$ in the integral on the left-hand side of the equation: \begin{align*} \int_0^t g(t-\tau) f(\tau) d\tau &= \int_0^t g(t-(t-s))f(t-s)ds &&(\text{setting } s = t-\tau)\ &= \int_0^t g(s)f(t-s)ds. \end{align*} This is the convolution of $f(t-s)$ and $g(s)$, which by definition is:

$(f * g)(t)=\int_{-\infty}^{\infty} f(t-s) g(s) d s$.

However, since both $f(t)$ and $g(t)$ are zero for $t < 0$, we can restrict the integral to $s\in [0,t]$:

$(f * g)(t)=\int_0^t f(t-s) g(s) d s$.

Therefore, we have shown that

$\int_0^t g(t-\tau) f(\tau) d \tau=(f * g)(t)=\int_0^t f(t-\tau) g(\tau) d \tau$

as required.

A surface is defined by the equation $$\mathbf{r}=u^2 v \mathbf{i}+v^2 \mathbf{j}+u v \mathbf{k}$$ Find the normal to the surface at the point where $u=1$ and $v=2$.

To find the normal to the surface at a given point, we need to take the gradient of the surface equation at that point.

The gradient of the surface equation is given by:

$\nabla \mathbf{r}=\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}$

So, we need to find the partial derivatives of $\mathbf{r}$ with respect to $u$ and $v$, evaluate them at $u=1$ and $v=2$, and then take their cross product to find the normal vector.

Taking the partial derivatives of $\mathbf{r}$ with respect to $u$ and $v$ gives:

$\frac{\partial \mathbf{r}}{\partial u}=2 u v \mathbf{i}+v \mathbf{k}, \quad \frac{\partial \mathbf{r}}{\partial v}=u^2 \mathbf{i}+2 v \mathbf{j}+u \mathbf{k}$

Evaluating these partial derivatives at $u=1$ and $v=2$ gives:

$\left.\frac{\partial \mathbf{r}}{\partial u}\right|{(1,2)}=4 \mathbf{i}+2 \mathbf{k},\left.\quad \frac{\partial \mathbf{r}}{\partial v}\right|{(1,2)}=\mathbf{i}+4 \mathbf{j}+2 \mathbf{k}$

Taking their cross product, we get:

$\left.\nabla \mathbf{r}\right|{(1,2)}=\left.\frac{\partial \mathbf{r}}{\partial u}\right|{(1,2)} \times\left.\frac{\partial \mathbf{r}}{\partial v}\right|_{(1,2)}=\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \ 4 & 0 & 2 \ 1 & 4 & 2\end{array}\right|=-8 \mathbf{i}+6 \mathbf{j}-16 \mathbf{k}$

Therefore, the normal to the surface at the point $(1,2)$ is $\boxed{-8\mathbf{i} + 6\mathbf{j} – 16\mathbf{k}}$.