Bernoulli's Principle and Lift

Bernoulli's Principle

• Principle states that an increase in fluid speed occurs simultaneously with a decrease in pressure or the fluid's potential energy.

How Wings Generate Lift

• Airfoil refers to a wing's cross-sectional shape and is designed to control airflow, generating lift with minimal drag.

Lift Generation

• Air flowing over the wing travels a longer distance than air flowing under it in the same time.
• Air flows faster over the wing.
• Faster-moving air exerts less pressure.
• The pressure difference creates lift.

Angle of Attack

• Angle of attack is the angle between the wing and the oncoming airflow.
• Increasing the angle can increase lift, but only to a point.
• Exceeding the critical angle causes a stall, resulting in a loss of lift.

Other Factors Affecting Lift

Wing Area

• Larger wings generate more lift because they have more surface area for air to act upon.

Air Speed

• Faster air increases lift due to the increased pressure difference between the upper and lower wing surfaces.

Air Density

• Denser air increases lift because the wing deflects a greater mass of air.

Applications of Bernoulli's Principle

Airplane Wings

• Wings are shaped to create faster airflow above, resulting in lower pressure and lift.

Spoilers on Race Cars

• Spoilers are inverted airfoils designed to create downforce, improving traction and handling.

Curveballs in Baseball

• A spinning ball creates a pressure difference, causing it to curve.

Algèbre Linéaire

• Linear Algebra (French)

Vecteurs

• Vectors (French)

Définitions de base

• Basic definitions (French)
• Vector defined by direction (a line), sense (a side of the line), and length (a norm).
• Vectors are noted as $\overrightarrow{AB}$ or $\vec{u}$

Opérations de base

• Basic operations (French)
• Operations based on 2 vectors, $\vec{u}$ et $\vec{v}$, and a scalar (real number) $\lambda$

Addition

• Addition of vectors
• $\vec{u} + \vec{v} = \vec{w}$

Multiplication par un scalaire

• Multiply by a scalar
• $\lambda\vec{u} = \vec{w}$

Représentation en coordonnées

• Representation in coordinates (French)
• In an n-dimensional space, a vector can be represented by a list of n numbers called coordinates.
• In a 2-dimensional space: $\vec{u} = (x, y)$

Produit scalaire

• Scalar product (French)
• The scalar product of two vectors $\vec{u}$ and $\vec{v}$ is defined by:
• $\vec{u} \cdot \vec{v} = ||\vec{u}|| \cdot ||\vec{v}|| \cdot \cos(\theta)$
• $||\vec{u}||$ and $||\vec{v}||$ are the norms of the vectors $\vec{u}$ and $\vec{v}$, and $\theta$ is the angle between the two vectors.
• A scalar product is a real number(scalaire).

Produit vectoriel

• Vector Product (French)
• The vector product of two vectors $\vec{u}$ and $\vec{v}$ is defined by:
• $\vec{u} \times \vec{v} = \vec{w}$
• $\vec{w}$ is a vector orthogonal to both $\vec{u}$ and $\vec{v}$, and its norm is $||\vec{u}|| \cdot ||\vec{v}|| \cdot \sin(\theta)$.
• The vector product is a vector and defined in 3 dimensions only.

Matrices

• Matricies (French)

Définitions de base

• Basic Definitions (French)
• Matrix is a table of numbers.
• $A = \begin{bmatrix} 1 & 2 \ 3 & 4 \end{bmatrix}$

Opérations de base

• Basic Operations (French)
• Matrix operations based on 2 matricies, A & B, and a scalar (real number) $\lambda$

Addition

$A + B = C$

• Addition happens if the matrices A and B have the same dimensions.

Multiplication par un scalaire

• Multiplication by a Scalar (French)
• $\lambda A = C$

Multiplication de matrices

• Matric multiplication (French)
• $A \cdot B = C$

Matrice identité

• Identity matrix (French)
• The identity matrix is a square matrix with 1s on the diagonal and 0s elsewhere.
• $I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$

Transposition

• Transposition (French)
• Transpose of matrix A is obtained by exchanging the rows and columns of A.
• Indicated by $A^T$

Inverse

• Inverse (French)
• The inverse of a matrix A is a matrix $A^{-1}$ such that $A \cdot A^{-1} = A^{-1} \cdot A = I$.
• Inverse exists if the matrix is square and its determinant is non-zero.

Déterminant

• Determinate (French)
• The determinant of a matrix is a scalar calculated from the elements of the matrix.
• For a 2x2 matrix, $A = \begin{bmatrix} a & b \ c & d \end{bmatrix}$ then $det(A) = ad - bc$

Valeurs propres et vecteurs propres

• Eigenvalues and eigenvectors (French)
• Let A be a square matrix. An eigenvector of A is a vector $\vec{v}$ such that $A\vec{v} = \lambda\vec{v}$, where $\lambda$ is an eigenvalue of A.

Systèmes d'équations linéaires

• Systems of linear equations (French)

Définitions de base

• Basic definitions (French)
• A system of linear equations is a set of linear equations.
• $x + y = 3$
• $x - y = 1$

Résolution

• Solving (French)
• A system of linear equations can be solved using different methods: substitution, elimination, matrix method (matrix inversion, LU decomposition, etc.).

Applications

• Applications (French)
• Systems of linear equations are used in many fields: engineering, economics and computer science.

Remarques

• Remarks (French)
• Summary of basic concepts of linear algebra

Lecture 2: Welfare Economics

Introduction

• Welfare Economics: Study of the desirable properties of an economy.
• Normative Question: What "should" be; differs from positive economics.
• Tools: Pareto efficiency, welfare theorems.

Pareto Efficiency

• An allocation is Pareto efficient (or Pareto optimal) if it is feasible and there is no other feasible allocation that makes everyone at least as well off and at least one person strictly better off.
• Allocation: Specifies what goods are produced, how they are produced, and who consumes them.
• Feasible: Production is possible and total consumption of each good equals total production.
• Pareto Improvement: An allocation A is a Pareto improvement over allocation B if everyone is at least as well off in A as in B and at least one person is strictly better off.
• Objection: Pareto efficiency is silent about distribution.
  • Allocation where one person has everything is Pareto efficient.

Example: Pareto Efficiency

Setup

• 2 agents, denoted by $i = 1, 2$
• 2 goods, denoted by $j = 1, 2$
• $x_{ij}$: agent i's consumption of good j
• Allocation: $(x_{11}, x_{12}, x_{21}, x_{22})$
• Feasible allocation: $x_{11} + x_{21} \leq \omega_1$ and $x_{12} + x_{22} \leq \omega_2$
  • $\omega_j$: total endowment of good j
• Utility function: $u_i(x_{i1}, x_{i2})$
• Pareto efficient allocation: feasible allocation $(x_{11}^, x_{12}^, x_{21}^, x_{22}^)$ such that there is no other feasible allocation $(x_{11}, x_{12}, x_{21}, x_{22})$ with $u_i(x_{i1}, x_{i2}) \geq u_i(x_{i1}^, x_{i2}^)$ for $i = 1, 2$ with strict inequality for at least one i.

Pareto Efficiency and Calculus

$(x_{11}^, x_{12}^, x_{21}^, x_{22}^)$ is Pareto efficient if it solves: $$ \max_{x_{11}, x_{12}, x_{21}, x_{22}} u_1(x_{11}, x_{12}) \quad \text{subject to} \quad u_2(x_{21}, x_{22}) \geq \bar{u}2 $$ $$ x{11} + x_{21} \leq \omega_1, \quad x_{12} + x_{22} \leq \omega_2 $$ for some $\bar{u}_2$.

• Rewriting this as: $$ \max_{x_{11}, x_{12}, x_{21}, x_{22}} u_1(x_{11}, x_{12}) \quad \text{subject to} \quad u_2(x_{21}, x_{22}) \geq \bar{u}2 $$ $$ x{11} + x_{21} = \omega_1, \quad x_{12} + x_{22} = \omega_2 $$

• With monotone preferences, inequalities can be replaced with equalities.

• Lagrangian: $$ \mathcal{L} = u_1(x_{11}, x_{12}) + \lambda [u_2(x_{21}, x_{22}) - \bar{u}2] + \mu_1 [\omega_1 - x{11} - x_{21}] + \mu_2 [\omega_2 - x_{12} - x_{22}] $$

• First order conditions: $$ \frac{\partial \mathcal{L}}{\partial x_{11}} = \frac{\partial u_1}{\partial x_{11}} - \mu_1 = 0 \quad (1) $$ $$ \frac{\partial \mathcal{L}}{\partial x_{12}} = \frac{\partial u_1}{\partial x_{12}} - \mu_2 = 0 \quad (2) $$ $$ \frac{\partial \mathcal{L}}{\partial x_{21}} = \lambda \frac{\partial u_2}{\partial x_{21}} - \mu_1 = 0 \quad (3) $$ $$ \frac{\partial \mathcal{L}}{\partial x_{22}} = \lambda \frac{\partial u_2}{\partial x_{22}} - \mu_2 = 0 \quad (4) $$

• Rearranging: $$ \frac{\partial u_1 / \partial x_{11}}{\partial u_1 / \partial x_{12}} = \frac{\mu_1}{\mu_2} = \frac{\partial u_2 / \partial x_{21}}{\partial u_2 / \partial x_{22}} $$

• MRS condition: $$ MRS_{1} = MRS_{2} $$ where $$ MRS_i = \frac{\partial u_i / \partial x_{i1}}{\partial u_i / \partial x_{i2}} $$

• At optimum, agents must have same willingness to trade.

• Contract Curve: set of Pareto efficient allocations.

First Welfare Theorem

• Assumptions:
  • No externalities
  • Perfect information
  • Agents are price takers
  • Markets are complete
• First Welfare Theorem: If the assumptions above hold, then any competitive equilibrium is Pareto efficient.
  • Competitive Equilibrium: An allocation and a price vector such that:
    • Given prices, consumers choose their most preferred bundle from their budget set.
    • Given prices, firms maximize profits.
    • Markets clear.
• Under these assumptions, there is no need for government intervention.
• Examples of violations include externalities and public goods.

Second Welfare Theorem

• Assumptions:

  • No externalities
  • Perfect information
  • Agents are price takers
  • Markets are complete
  • Convex preferences and production sets

• Second Welfare Theorem: If the assumptions above hold, then any Pareto efficient allocation can be supported as a competitive equilibrium, after a suitable redistribution of initial endowments.

  • Efficiency and equity can be separated.
    • Lump-sum transfers to achieve desired distribution.
    • Let market do its work.
  • Lump-sum transfers are hard to implement.
    • Requires knowing preferences of agents.
    • Governments rarely have enough information.
  • Alternative is to use distortionary taxes to achieve desired distribution through optimal taxation(how to design taxes to minimize inefficiencies.

• Rely on strong assumptions that are often violated

• There is a potential role for government intervention.

Chapitre 6. Applications linéaires

• Chapter 6 Linear Applications

Définitions et Exemples

• Definitions and Examples (French) - Application $f$ from a vector space E to a vector space F (over the same field 𝕂) is said to be linear if: - $\forall u, v \in E, \quad f(u+v) = f(u)+f(v)$ - $\forall \lambda \in \mathbb{K}, \forall u \in E, :f(\lambda u)=\lambda f(u)$ - $\mathcal L(E,F)$ is the set of linear applications from E to F. - A linear application from E to E is called the endomorphism of E. - $\mathcal L(E)$ is the set of endomorphisms of E.

Propriétés

• Properties (French) - For $f ϵ \mathcal L(E,F)$ - $f(0_E) = 0_F$ - $\forall u \in E, :f(-u) = -f(u)$ - $\forall u, v \in E, :f(u-v) = f(u) - f(v)$ - $\forall n \in \mathbb{N}^*, \forall u_1, \ldots, u_n \in E, \forall \lambda_1, \ldots, \lambda_n \in \mathbb{K}$, $f\left(\sum_{i=1}^{n} \lambda_i u_i\right)=\sum_{i=1}^{n} \lambda_i f(u_i)$

Exemples

• Exemples
  • The null application from E to F that assigns vector E to the zero vector of F is linear.
  • The identity application of E is linear.
  • Soit $E = \mathcal{C}^1(\mathbb{R}, \mathbb{R})$ the set of derivable functions over $\mathbb{R}$ whose derivative is continuous, and $F = \mathcal{C}(\mathbb{R}, \mathbb{R})$ the set of continuous functions over $\mathbb{R}$. The application $f: E \rightarrow F$ defined by $f(g) = g'$ is linear.
  • Let E be a vector space over 𝕂 and λ ϵ 𝕂. The function f: E → E defined by f(u) = λu is linear. If λ = 0, the null function is returned. If λ = 1, the identity application is returned.

Image et Noyau

• Image and Kernel (French)

Définitions

• Definitions (French) - For $f \in \mathcal L(E,F)$. - The image of f is the set of vectors of F whose image is at least one vector in E. We note Im(f) = {v ϵ F | ∃ u ϵ E, v=f(u)} = {f(u) | u ϵ E} = f(E). - The kernel of f is the set of vectors of E that have the image of the null vector in F. We note Ker(f) = {u ϵ E | f(u)=0_F} = f^(-1)({0_F}).

Propriétés

• Properties (French) - For $f \in \mathcal L(E,F)$ - Im(f) is a vector subspace of F. - Ker(f) is a vector subspace of E.

Remarque

• Remarks (French)
  • f is surjective if and only if Im(f) = F
  • f is injective if and only if Ker(f) = {0_E}

Exemples

• Examples (French) - For $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ defined by $f(x,y)=(x+y,x+y)$. So $\operatorname {Im}(f)={(x,x) \mid x \in \mathbb{R}}$ and $\operatorname {Ker}(f)={(x,-x) \mid x \in \mathbb{R}}$.

Rang d'une Application Linéaire

• Rank of a Linear Application

Définition

• Definition (French)
  • For $f ϵ \mathcal L(E,F)$. The rank of f is the dimension of Im(f). Written as $\operatorname {rg}(f)= \operatorname {dim}(\operatorname {Im}(f))$.

Théorème du rang

• Rank Theorem (French)
  • For $f ϵ \mathcal L(E,F)$, where E is a vector space of finite dimensions. So: $\operatorname {dim}(E)=\operatorname {dim}(\operatorname {Ker}(f))+\operatorname {rg}(f)$.

Remarque

• Remarks (French)
  • If E is of finite dimensions, so Im(f) is of finite dimensions. So $\operatorname {rg}(f)\leq \operatorname {dim}(F)$ and $\operatorname {rg}(f) {dim}(E)$ and $\operatorname {rg}(f) {dim}(F)$. $((1,0),(0,1))$. $f((1,0))=(1,1)$ and $f((0,1))=(1,-1)$. So $\operatorname{Mat}_{\mathcal B}(f)= \begin{pmatrix} 1&&1 \ 1&&-1 \end{pmatrix}$.