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Doing your best means never stop trying.
The best place to find a helping hand is at the end of your own arm, Swedish Proverb.
Definition. A vector $\vec{AB}$ is a geometric object or quantity that has both magnitude (or length) and direction. Vectors in an n-dimensional Euclidean space can be represented as coordinates vectors in a Cartesian coordinate system.
Definition. The magnitude of the vector $\vec{A}$, also known as length or norm, is given by the square root of the sum of its components squared, $|\vec{A}|~ or~ ||\vec{A}|| = \sqrt{a_1^2+a_2^2+a_3^2}$, e.g., $||< 3, 2, 1 >|| = \sqrt{3^2+2^2+1^2}=\sqrt{14}$, $||< 3, -4, 5 >|| = \sqrt{3^2+(-4)^2+5^2}=\sqrt{50}=5\sqrt{2}$, or $||< 1, 0, 0 >|| = \sqrt{1^2+0^2+0^2}=\sqrt{1}=1$.
The sum of two vectors is the sum of their components, $\vec{A}+ \vec{B} = (a_1+b_1)\vec{i}+(a_2+b_2)\vec{j}+(a_3+b_3)\vec{k}$ = < (a1+b1), (a2+b2), (a3+b3) >.
The subtraction of two vectors is similar to addition and is also done component-wise, it is given by simply subtracting their corresponding components (x, y, and z in 3D), $\vec{A} - \vec{B} = (a_1-b_1)\vec{i}+(a_2-b_2)\vec{j}+(a_3-b_3)\vec{k}$ = < (a1-b1), (a2-b2), (a3-b3) >.
Scalar multiplication is the multiplication of a vector by a scalar, a real number, changing its magnitude without altering its direction. It is effectively multiplying each component of the vector by the scalar value, $c\vec{A} = (ca_1)\vec{i}+(ca_2)\vec{j}+(ca_3)\vec{k} = < ca_1, ca_2, ca_3>$.
The dot or scalar product is a fundamental operation between two vectors. It produces a scalar quantity that represents the projection of one vector onto another. The dot product is the sum of the products of their corresponding components: $\vec{A}·\vec{B} = \sum a_ib_i = a_1b_1 + a_2b_2 + a_3b_3.$, e.g. $\vec{A}·\vec{B} = \sum a_ib_i = ⟨2, 2, -1⟩·⟨5, -3, 2⟩ = a_1b_1 + a_2b_2 + a_3b_3 = 2·5+2·(-3)+(-1)·2 = 10-6-2 = 2.$
It is the product of their magnitudes multiplied by the cosine of the angle between them, $\vec{A}·\vec{B}=||\vec{A}||·||\vec{B}||·cos(θ).$
The cross product, denoted by $\vec{A}x\vec{B}$, is a binary operation on two vectors in three-dimensional space. It results in a vector that is perpendicular to both of the input vectors and has a magnitude equal to the area of the parallelogram formed by the two input vectors.
The cross product $\vec{A}x\vec{B}$ can be computed using the following formula, $\vec{A}x\vec{B} = det(\begin{smallmatrix}i & j & k\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\end{smallmatrix}) =|\begin{smallmatrix}a_2 & a_3\\ b_2 & b_3\end{smallmatrix}|\vec{i}-|\begin{smallmatrix}a_1 & a_3\\ b_1 & b_3\end{smallmatrix}|\vec{j}+|\begin{smallmatrix}a_1 & a_2\\ b_1 & b_2\end{smallmatrix}|\vec{k}$
Matrices offers an efficient way for solving systems of linear equations. Utilizing matrices, we can represent these equations more concisely and conveniently.
For a system like,
$(\begin{smallmatrix}2 & 3 & 3\\ 2 & 4 & 4\\ 1 & 1 & 2\end{smallmatrix}) (\begin{smallmatrix}x_1\\ x_2\\x_3\end{smallmatrix}) = (\begin{smallmatrix}u_1\\ u_2\\u_3\end{smallmatrix})$, A · X = U provides a more convenient and concise notation and an efficient way to solve system of linear equations where A is a 3 x 3 matrix representing coefficients, X is a 3 x 1 column vector representing the results where we basically do dot products between the rows of A (a 3 x 3 matrix) and the column vector of X (a 3 x 1 matrix).
Given a square matrix A (it is a matrix that has an equal number of rows and columns), an inverse matrix A-1 exists if and only if A is non-singular, meaning its determinant is non-zero (det(A)≠ 0).
The inverse matrix of A, denoted as A-1, has the property that when multiplied by A, it results in the identity matrix, i.e., $A \times A^{-1} = A^{-1} \times A = I$. Essentially, multiplying a matrix by its inverse reverses (“undoes”) the effect of the original matrix.
Consider a system of linear equations represented in matrix form as: AX = B, where A is a n×n matrix (coefficient matrix), X is an n×1 matrix (column vector of variables), and B is an n×1 matrix (column vector of constants).
Solving a system of linear equations expressed as AX = B involves finding or isolating the matrix X when we are given both matrices A and B. AX = B ⇒[A should be non-singular, meaning its determinant is non-zero, det(A)≠0, then we can multiply both side by A-1]⇒ X = A-1B.. $A^{-1} = \frac{1}{\text{det}(A)} \times \text{adj}(A)$. adj(A) = CT where $C_{ij} = (-1)^{i+j} \times \text{minor}(A_{ij})$ and det(A) =[A is a 3x3 matrix] a(ei - fh) - b(di - fg) + c(dh - eg)
Matrices provide an efficient way to solve systems of linear equations,
$(\begin{smallmatrix}2 & 3 & 3\\ 2 & 4 & 4\\ 1 & 1 & 2\end{smallmatrix}) (\begin{smallmatrix}x_1\\ x_2\\ x_3\end{smallmatrix}) = (\begin{smallmatrix}u_1\\ u_2\\ u_3\end{smallmatrix}) ↭ A · X = U$ (more convenient and concise notation) where we do the dot products between the rows of A (a 3 x 3 matrix) and the column vector of X (a 3 x 1 matrix).
Given a square matrix A, an inverse matrix A-1 exists if and only if A is non-singular, i.e., its determinant is non-zero (det(A)≠ 0). The inverse matrix of A, denoted as A-1, is a matrix such that when multiplied by A yields the identity matrix, i.e., $A \times A^{-1} = A^{-1} \times A = I$
Definition. A position vector indicates the position or location of a point relative to an arbitrary reference point. For a point P in space, this is typically written as:
$\vec{OP} =\vec{r}(t) = ⟨x(t), y(t), z(t)⟩$.
For example, consider a cycloid traced by a point on a wheel rolling along the x-axis at unit speed. If the wheel’s radius is 1, and the time t corresponds to the angle the wheel has rotated, then: $\vec{r}(t) = ⟨t -sin(t), 1 -cos(t)⟩$
Velocity is a vector quantity that measures the rate of change of position. It is a vector in the direction of motion. It has both magnitude and direction, and is given by the derivate of the position vector:
$\vec{v}(t) = \frac{d\vec{r}}{dt} = ⟨\frac{dx}{dt}, \frac{dy}{dt}⟩$, e.g., for our cycloid example, $\vec{v}(t) = ⟨1-cos(t), sin(t)⟩$.
The magnitude of the velocity vector, or speed, is: $|\vec{v}| = \sqrt{(1-cos(t))^2 + sin^2(t)} = \sqrt{1 -2cos(t) + cos^2(t)+ sin^2(t)} = \sqrt{2 -2cos(t)} $
Acceleration is the derivative of the velocity vector with respect to time: $\vec{a}(t) = \frac{d\vec{v}}{dt}$. For our cycloid example, $\vec{a} = ⟨sin(t), cos(t)⟩$. At t = 0, we have $\vec{v}(0) = \vec{0}, \vec{a} = ⟨0, 1⟩$. This indicates that the point is momentarily at rest and then accelerates upwards.
The arc length is the distance traveled by an object from one point to another point along a curve. Since the speed of a moving object is the length of its velocity vector, the distance the object travels from t = a to t = b is the integral of |r’(t)| over the interval [a, b], e.g., length of an arch in our cycloid is $\int_{a}^{b} |\frac{d\vec{r}}{dt}|dt = \int_{a}^{b} |\vec{v}(t)|dt =$[For our cycloid example] $\int_{0}^{2π} \sqrt{2 -2cos(t)}dt$
Definition. The unit tangent vector is a vector that points in the direction of the tangent to a curve at a given point and has a magnitude or length of 1. The unit tangent vector is used to describe the direction of motion along a curve and can be used to calculate the derivative of a curve at a given point, $\vec{T} = \frac{\vec{v}}{|\vec{v}|}$.
Therefore, the velocity vector can be expressed as $\vec{v}=\frac{d\vec{r}}{dt}$ =[The Chain Rule] $\frac{d\vec{r}}{ds}·\frac{ds}{dt}$
Since s is the arc length, $|\vec{v}|=\frac{ds}{dt}⇒ \vec{v}= \vec{T}·\frac{ds}{dt}$. The velocity vector is the product of the unit tangent and the speed, and $\vec{T}(s)=\frac{d\vec{r}}{ds}$, that is, the derivative of the position vector with respect to the arc length.
To sum up, velocity, being a vector, has both a magnitude and a direction. Its direction is always directed tangent to the object’s motion or trajectory. Its length is the speed, $|\vec{v}|=\frac{ds}{dt}$. In words, the speed is the magnitude of the velocity vector and is also the derivative of the arc length with respect to time.
When we approximate the change in position $\Delta r$ as $\vec{T}·\Delta s$ (Figure i and ii), the speed $\frac{\Delta s}{\Delta t}$ becomes more accurate as $\Delta t → 0$. Dividing by $\Delta t, \frac{\Delta r}{\Delta t} ≈ \vec{T}·\frac{\Delta s}{\Delta t}$ and taking the limit as $\Delta t → 0 ⇒ \frac{d\vec{r}}{dt}=\vec{T}·\frac{ds}{dt}$ that is the previous formula where the approximation becomes an equality (the approximation gets better and better).
Kepler's first law states that all planets orbit the Sun in an elliptical path with the Sun located at one of the two foci of the ellipse. This means that the distance between a planet and the Sun changes as the planet moves along its orbit.
Kepler’s second law states that planets move in planes, and the imaginary line joining a planet and the Sun (or the radius vector) sweeps out equal areas of space in equal intervals of time as the planet orbits the sun.
This means that planets do not move with constant speed along their orbits. Rather, their speed varies so that the line joining the centers of the Sun and the planet sweeps out equal areas in equal intervals of times, (Figure iii):
Isaac Newton later developed a theory, using the principles of Calculus and his laws of motion and universal gravitation, to explain why planets follow Kepler’s laws.
The cross product of two vectors $\vec{u}$ and $\vec{v}$ results in a vector whose magnitude is equal to the area of the parallelogram formed by $\vec{u}$ and $\vec{v}$. Hence, we can use the vector product to compute the area of a triangle formed by three points A, B and C in space, Area of △ABC = $\frac{1}{2}|\vec{AB}x\vec{AC}|$ (Figure iv).
To apply this to planetary motion, consider the area swept by the radius vector $\vec{r}$ (from the Sun to the planet) in a small time interval Δt. This area can be approximated by the area of the parallelogram formed by $\vec{r}$ and the change in the position vector $Δ\vec{r}$: Area ≈ $\frac{1}{2}|\vec{r} x \vec{\Delta r}|$ (Area swept by the planet in a small interval of time $\Delta t$) ≈[For small Δt, $Δ\vec{r} = \vec{v}Δt$] $\frac{1}{2}|\vec{r} x \vec{v}\Delta t| = \frac{1}{2}|\vec{r} x \vec{v}|\Delta t$. Kepler’s second law implies that $|\vec{r} x \vec{v}|$ must be constant because the area swept out in equal time intervals is constant.
Futhermore, since planets move in planes, the plane of motion contains both $\vec{r}$ and $\vec{v}$. The cross product $\vec{r}x\vec{v}$ is perpendicular to this plane. Given that $|\vec{r} x \vec{v}|$ is constant, $\vec{r} x \vec{v}$ is a constant vector. This can be expressed mathematically as:
$\frac{d}{dt}(\vec{r} x \vec{v}) = 0$.
Using the product rule for differentiation: $\frac{d\vec{r}}{dt}x\vec{v} + \vec{r} x \frac{d\vec{v}}{dt} = 0$.
Since $\frac{d\vec{r}}{dt} = \vec{v}, \frac{d\vec{v}}{dt} = \vec{a}$: $\vec{v}x\vec{v}+\vec{r}x\vec{a} = 0$.
Note that $∀\vec{v}, \vec{v}x\vec{v}=0$ because the cross product of any vector with itself is zero (geometrically, the cross product is the area of the parallelogram formed by both vectors, but as we are talking about the same vector, there is no area),
Therefore, $\vec{r}x\vec{a} = 0$.
A cross product of two vectors generates a perpendicular vector to both vectors. The two vectors are parallel if the cross product of their cross products is zero.
This indicates that the acceleration vector is parallel to $\vec{r}$.
$\vec{a} = \frac{d\vec{v}}{dt} || \vec{r}$. Gravitational acceleration is the acceleration an object experiences when gravity is the only force acting on it. Gravitational acceleration is directed towards the Sun (along $\vec{r}$).
To analyze the motion of the particle, we first need to find its velocity v(t) and acceleration a(t).
Velocity v(t) is the derivative of the position function x(t) with respect to time t: $v(t) = \frac{dx}{dt} = 3t^2 -24t + 36.$
Acceleration a(t) is the derivative of the velocity function v(t) with respect to time t: $a(t) = \frac{dv}{dt} = 6t -24.$
A particle changes direction when its velocity v(t) -it describes the rate of change of the particle’s position with respect to time- is zero (this implies that the particle momentarily stops before reversing its direction), and its acceleration a(t) is non-zero at that point.
If the acceleration is non-zero when the velocity is zero, it means that there is a force acting on the particle, causing its velocity to change. This non-zero acceleration ensures that the particle doesn’t remain stationary but instead starts moving in the opposite direction.
Set the velocity equal to zero and solve for t: $3t^2 -24t + 36 = 0 ⇒[\text{Divide the equation by 3 to simplify}] 0 = t^2 -8t + 12⇒[\text{Factor the quadratic equation:}] 0 = (t -6)(t-2)$. So, the solutions are t = 6 or t = 2.
Next, check the acceleration at these points to ensure it is non-zero: a(t) = 0 ↭ 6t = 24 ↭ t = 4. Since the acceleration is non-zero at both t = 2 and t = 6, the particle changes direction at these points.
A particle is slowing down when its velocity and acceleration have opposite signs.
A particle is slowing down if its speed (the magnitude of its velocity) is decreasing over time. This occurs when the direction of acceleration is opposite to the direction of velocity. If velocity and acceleration are in the same direction, the particle speeds up because the acceleration adds to the velocity. If velocity and acceleration are in opposite directions, the acceleration subtracts from the velocity, reducing its magnitude (speed).
To determine these intervals, we need to analyze the signs of v(t) and a(t) in different intervals. v(t) = (t -2)(t -6), so velocity changes sign at t = 2 and t = 6. a(t) = 0 ↭ t = 4, so acceleration changes sign at t = 4.
Now, let’s summarize the signs of velocity and acceleration in different intervals:
| Time | Velocity | Acceleration |
|---|---|---|
| 0 < t < 2 | + | - |
| 2 < t < 4 | - | - |
| 4 < t < 6 | - | + |
| t > 6 | + | + |
From the table, we see that the particle is slowing down when 0 < t < 2 and 4 < t < 6, because in these intervals, the velocity and acceleration have opposite signs (Figure B).
The position vector $\vec{r}(t)$ represents the location of a point in space at time t: $\vec{r}(t)=(2t-2)\vec{i}+(t^2+t-1)\vec{j}$. Here, $\vec{i}~\text{and}~\vec{j}$ are the unit vectors in the x and y directions respectively.
For t = -1: $\vec{r}(-1)=(2·(-1)-2)\vec{i}+((-1)^2+(-1)-1)\vec{j} = -4\vec{i}-\vec{j} = ⟨-4, -1⟩.$
The velocity vector $\vec{v}(t)$ is the derivative of the position vector $\vec{r}(t)$ with respect to time t: $\vec{v}(t)=\frac{d\vec{r}}{dt} = \frac{d}{dt}[(2t-2)\vec{i}+(t^2+t-1)\vec{j}] = 2\vec{i} + (2t+1)\vec{j} = ⟨2, 2t+1⟩$. It describes how fast the position is changing and in which direction.
For t = -1; $\vec{v}(-1) = ⟨2, 2·(-1)+1⟩ = ⟨2, -1⟩.$
The acceleration vector $\vec{a}(t)$ is the derivative of the velocity vector $\vec{v}(t)$ with respect to time: $\vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d}{dt}[2\vec{i} + (2t+1)\vec{j}] = 2\vec{j} = ⟨0, 2⟩$. It describes how fast the velocity is changing and in which direction.
Notice that the acceleration vector is constant and does not depend on t. $\vec{a}(-1) = \vec{a}(-1) = ⟨0, 2⟩$.
Eliminate the parameter (t) to express the equation in terms of x and y: x = t2-4, y = t ⇒ x = y2 -4 is a parabola with its vertex at (0, -4).
The position vector $\vec{r}(t)$ indicates the position of a point in space at time t: $\vec{r}(t)=(t^2-4)\vec{i}+t\vec{j}$. Here, $\vec{i}~\text{and}~\vec{j}$ are the unit vectors in the x and y directions respectively.
For t = 0: $\vec{r}(0)=(0^2-4)\vec{i}+0\vec{j} = ⟨-4, 0⟩.$
For t = 2: $\vec{r}(2)=(2^2-4)\vec{i}+2\vec{j} = +2\vec{j} = ⟨0, 2⟩.$
The velocity vector $\vec{v}(t)$ is the derivative of the position vector $\vec{r}(t)$ with respect to time t: $\vec{v}(t)=\frac{d\vec{r}}{dt} = \frac{d}{dt}[(t^2-4)\vec{i}+t\vec{j}] = (2t)\vec{i} + \vec{j} = ⟨2t, 1⟩$. It describes how fast the position is changing and in which direction.
For t = 0; $\vec{v}(0) = ⟨2·0, 1⟩ = ⟨0, 1⟩.$
For t = 2; $\vec{v}(2) = ⟨2·2, 1⟩ = ⟨4, 1⟩.$
The acceleration vector $\vec{a}(t)$ is the derivative of the velocity vector $\vec{v}(t)$ with respect to time: $\vec{a}(t) = \frac{d\vec{v}}{dt} = \frac{d}{dt}[(2t)\vec{i} + \vec{j}] = 2\vec{i} = ⟨2, 0⟩$. It describes how fast the velocity is changing and in which direction.
Notice that the acceleration vector is constant and does not depend on t. $\vec{a}(0) = \vec{a}(2) = ⟨2, 0⟩$.
For the given position vector, at t = 0, the object is at position ⟨−4, 0⟩, moving vertically upwards with a velocity ⟨0, 1⟩, and has a constant acceleration ⟨2, 0⟩ in the horizontal direction.
At t = 2, the object is at position ⟨0, 2⟩, moving in the direction ⟨4, 1⟩ (mostly horizontally), and still has the same constant horizontal acceleration ⟨2,0⟩.
Solution:
To find the velocity function $\vec{v}(t)$, we need to integrate the acceleration function $\vec{a}(t)$:
$\vec{v}(t) = \int \vec{a}(t)dt =\int (\vec{i}+2\vec{j}+6t\vec{k})dt = t\vec{i}+2t\vec{j}+3t^2\vec{k} + C,$ where C is the constant of integration.
Using the initial conditions $\vec{v}(0)=\vec{j}-\vec{k}$: $\vec{v}(0) = 0·\vec{i}+2·0·\vec{j}+3·0^2\vec{k} + C$. Thus, $C = \vec{j}-\vec{k}$
Therefore, the velocity function is: $\vec{v}(t) = t\vec{i}+2t\vec{j}+3t^2\vec{k} + \vec{j}-\vec{k} = t\vec{i}+(2t+1)\vec{j}+(3t^2-1)\vec{k}$
Analogously, we can calculate the position function by integrating the velocity function: $\vec{r}(t) = \int \vec{v}(t)dt = \int (t\vec{i}+(2t+1)\vec{j}+(3t^2-1)\vec{k})dt = \frac{t^2}{2}\vec{i}+(t^2+t)\vec{j} + (t^3-t)\vec{k} + C,$ where C is the constant of integration.
Using the initial condition $\vec{r}(0)=\vec{i}-2\vec{j}+3\vec{k}: \vec{r}(0) = \frac{0^2}{2}\vec{i}+(0^2+0)\vec{j} + (0^3-0)\vec{k} + C = C$. Thus, C = $\vec{i}-2\vec{j}+3\vec{k}$.
Therefore, the position function is $\vec{r}(t)=(\frac{t^2}{2}+1)\vec{i}+(t^2+t-2)\vec{j} + (t^3-t+3)\vec{k}$
This process illustrates how to integrate acceleration to find velocity, and then integrate velocity to find position, using the problem’s given initial conditions to determine the constants of integration.
The velocity vector is obtained by integrating the acceleration vector with respect to time:
$\vec{v}(t) = 3t\hat{\mathbf{i}} + 3t\hat{\mathbf{k}} + \vec{C}$. Here, $\vec{C}$ is the constant of integration. To determine $\vec{C}$, use the initial condition for the velocity:
$\vec{v}(0) = 8\hat{\mathbf{j}}⇒ \vec{C} = 8\hat{\mathbf{j}}⇒ \vec{v}(t) = 3t\hat{\mathbf{i}} + 8\hat{\mathbf{j}} + 3t\hat{\mathbf{k}}.$
The position vector is obtained by integrating the velocity vector with respect to time t.
$\vec{r}(t) = \frac{3t^2}{2}\hat{\mathbf{i}} + 8t\hat{\mathbf{j}} + \frac{3t^2}{2}\hat{\mathbf{k}} + \vec{C}$. Here, $\vec{C}$ is the constant of integration. To determine $\vec{C}$, use the initial condition for the position:
$\vec{r}(0) = \vec{0}⇒ \vec{C} = \vec{0}⇒ \vec{r}(t) = \frac{3t^2}{2}\hat{\mathbf{i}} + 8t\hat{\mathbf{j}} + \frac{3t^2}{2}\hat{\mathbf{k}}.$
Since the position vector $\vec{r}(t)$, which describes the position of a point in space as a function of time t, is said to have a constant length, the dot product $\vec{r}·\vec{r}$ remain constant. Let’s denote this constant by c1: $\vec{r}·\vec{r} = c_1$.
$\frac{d}{dt}(\vec{r}·\vec{r}) =$[Using the product rule for differentiation, we get:] $\frac{d}{dt}(\vec{r})·\vec{r} + \vec{r}·\frac{d}{dt}(\vec{r}) = \vec{v}·\vec{r}+\vec{r}·\vec{v} = $[The dot product is commutative] $2\vec{r}·\vec{v} =$[$\vec{r}·\vec{r} = c_1$ where c1 is a constant, so $\frac{d}{dt}(\vec{r}·\vec{r}) = 0$] 0⇒$\vec{r}·\vec{v} = 0$. This implies that $\vec{r}$ and $\vec{v}$ are orthogonal (perpendicular).
To demonstrate that $\vec{r}x\vec{v}$ is constant, we will prove that $\frac{d}{dt}(\vec{r}x\vec{v}) = 0$
$\frac{d}{dt}(\vec{r}x\vec{v}) = \frac{d}{dt}(\vec{r})x\vec{v} + \vec{r}x\frac{d}{dt}(\vec{v}) = \vec{v}x\vec{v} + \vec{r}x\vec{a}$
Note that $∀\vec{v}, \vec{v}x\vec{v}=0$ because the cross product of any vector with itself is zero (geometrically, the cross product of two vectors is the area of the parallelogram formed by both vectors, but as we are talking about the same vector, there is no area).
$\frac{d}{dt}(\vec{r}x\vec{v}) = \vec{r}x\vec{a} =$[By assumption, $\vec{a(t)} = c\vec{r}(t)$] $\vec{r}xc\vec{r}(t) = c(\vec{r}x\vec{r}) = c·\vec{0} = \vec{0}$. This shows that $\vec{r}x\vec{v}$ is constant over time.
An example of a position vector that satisfies these conditions is: $\vec{r}(t) = ⟨cos(t), sin(t), 0⟩$ (Figure A).
Let’s check this example:
Therefore, the example $\vec{r}(t) = ⟨cos(t), sin(t), 0⟩$ satisfies both conditions $\vec{r}·\vec{v} = 0$ and $\vec{r}x\vec{v}$ is constant.
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