The excitement of learning separates youth from old age. As long as you’re learning, you’re not old, Rosalyn S.Yalow.
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, $\vec{A}+ \vec{B} = (a_1+b_1)\vec{i}+(a_2+b_2)\vec{j}+(a_3+b_3)\vec{k}$ = < (a_{1}+b_{1}), (a_{2}+b_{2}), (a_{3}+b_{3}) >.
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), $\vec{A} - \vec{B} = (a_1-b_1)\vec{i}+(a_2-b_2)\vec{j}+(a_3-b_3)\vec{k}$ = < (a_{1}-b_{1}), (a_{2}-b_{2}), (a_{3}-b_{3}) >.
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, $\vec{A}x\vec{B} = det(\begin{smallmatrix}i & j & k\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\end{smallmatrix}\bigl) =|\begin{smallmatrix}a_2 & a_3\\ b_2 & b_3\end{smallmatrix}|\vec{i}-|\begin{smallmatrix}a_1 & a_3\\ b_1 & b_3\end{smallmatrix}|\vec{j}+|\begin{smallmatrix}a_1 & a_2\\ b_1 & b_2\end{smallmatrix}|\vec{k}$
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,
$\begin{cases} 2x_1 +3x_2 + 3x_3 = u_1 \\ 2x_1 +4x_2 + 4x_3 = u_2 \\ x_1 + x_2 + 2x_3 = u_3 \end{cases}$
$(\begin{smallmatrix}2 & 3 & 3\\ 2 & 4 & 4\\ 1 & 1 & 2\end{smallmatrix}\bigl) (\begin{smallmatrix}x_1\\ x_2\\x_3\end{smallmatrix}\bigl) = (\begin{smallmatrix}u_1\\ u_2\\u_3\end{smallmatrix}\bigl)$, 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).
Solve for t in the first equation (i) $2t = x + 4 ↭ t = \frac{x+4}{2}$ and then substitute t into the second equation, $y = 4(\frac{x+4}{2})^2 =[\text{Simplify the expression}] = [\text{The 4’s cancel out:}]4·\frac{(x+4)^2}{4} = (x+4)^2$. Resulting Cartesian equation: y = (x +4)^{2}.
Parametric curves have a direction of motion. The direction of motion is given by increasing t. So, when plotting parametric curves, we also include arrows that show the direction of motion.
This is the equation of a parabola that opens upwards, with its vertex at (−4, 0). As x increases, t also increases, indicating that the direction of the parabola is upwards.
Solve each equation for the trigonometric function. From (i): $2cos(t) = x -3 ↭ cos(t) = \frac{x-3}{2}$ and (ii) $2sin(t) = y + 1↭ sin(t) = \frac{y+1}{2}$.
Substitute cos(t) and sin(t) from the above equations and use the Pythagorean identity: $sin^2(θ)+cos^2(θ) = 1⇒ \frac{(x-3)^2}{2^2}+\frac{(y+1)^2}{2^2} = 1 ⇒ (x-3)^2 + (y+1)^2 = 4.$ This is the equation of a circle with center at (3, -1) and radius 2. This sequence of points shows that as t increases from 0 to 2π, the curve traces a circle in a counterclockwise direction.
t | x | y |
---|---|---|
0 | 5 | -1 |
π/2 | 3 | 1 |
π | 1 | -1 |
3π/2 | 3 | -3 |
2π | 5 | -1 |
Create a Table of Values:
t | x | y |
---|---|---|
-3 | -8 | -7 |
-2 | -5 | -2 |
-1 | -2 | 1 |
0 | 1 | 2 |
1 | 4 | 1 |
2 | 7 | -2 |
3 | 10 | -7 |
Solve for t in the first equation (i), $t = \frac{x-1}{3}$, and substitute into the second equation (ii) $y = 2 -t^2 = 2 -(\frac{(x-1)}{3})^2 ⇒y = \frac{-1}{9}(x-1)^2+2$. As t increases, x increases and y first increases to a maximum 2 (t = 0, (x, y) = (1, 2)) and then decreases.
The Cartesian equation is a parabola: y = a(x -h)^{2} + k, where the vertex is (h, k). In this example, the vertex is (1, 2). The coefficient $\frac{-1}{9}$ indicates the parabola opens downwards and is vertically compressed.
Solve each equation for the trigonometric function. From (i): $cos(θ) = \frac{x}{4}$ and (ii) $sin(θ) = \frac{y}{3}$.
Substitute cos(t) and sin(t) from the above equations and use the Pythagorean identity: $sin^2(θ)+cos^2(θ) = 1⇒ \frac{x}{4}^2+\frac{y}{3}^2 = 1.$ This is the equation of an ellipse with center at the origin (0, 0). Since $\frac{-π}{2}≤θ≤\frac{π}{2}$, we are only considering the right half of the ellipse.
The semi-major axis is along the x-axis with a length of 4. The semi-minor axis is along the y-axis with a length of 3. This movement is in the counter-clockwise direction, starting from the bottom point (0, −3), moving through (4, 0), and ending at (0, 3).
t | x | y |
---|---|---|
-π/2 | 0 | -3 |
0 | 4 | 0 |
π/2 | 0 | 3 |
First, we solve for t in the first equation (i): $e^x = e^{ln(t+2)}↭ e^x = t + 2 ↭ t = e^x -2.$ Substitute this result into the second equation: $y = \frac{1}{e^x-2+1} = \frac{1}{e^x -1}$
There are three sections:
Vertical asymptote at x = 0. Horizontal asymptote at y = 0. Direction of motion: From x→−∞, y→0, through the point t = 0, (ln(2), 1), to x→∞, y→0.
A cycloid is the path traced by a point P on the rim of a wheel with radius a, as it rolls along a straight line (the x axis) without slipping. P is a point on the rim of a wheel, and it starts at the origin (0, 0).
To describe the cycloid, we need to express the coordinates x and y of point P as functions of the angle θ by which the wheel has rotated. Refer to Figures ii and iii in the diagram for a visual aid.
As the wheel rolls, the center O of the wheel moves horizontally. $\vec{OP}$ represents the distance traveled by the wheel’s center O. $\vec{AB}$ represents the vertical distance from the center of the wheel to the point A, that is, ⟨0, a⟩. $\vec{BP}$ represents the position of point P relative to the center of the wheel after it has rotated by an angle θ.
$\vec{OP} = \vec{OA} + \vec{AB} + \vec{BP}$ where $\vec{OA} = ⟨θ·a, 0⟩$ is (provided the angle θ is expressed in radians and the wheel is not slipping) the amount by which the wheel has rolled or moved horizontally. $\vec{OA}$ is equal to the Arc length from A to P, $\vec{AB} = ⟨0, a⟩, \vec{BP} = ⟨-a·sin(θ), -a·cos(θ)⟩$
Therefore, $\vec{OP} = \vec{OA} + \vec{AB} + \vec{BP} = ⟨aθ -asin(θ), a - acos(θ)⟩.$
Thus, the parametric equations for the cycloid are:
$\begin{cases} x(θ) = aθ -asin(θ)\\ y(θ) = a - acos(θ) \end{cases}$
When analyzing the cycloid near the origin, we simplify by setting a = 1 (Figure iv) This helps us understand how the cycloid behaves as the angle θ approaches zero.
Parametric Equations Simplified:
$\begin{cases} x(θ) = θ -sin(θ)\\ y(θ) = 1 - cos(θ) \end{cases}$
Using Taylor series approximations for small t, $f(t) ≈ f(0) + tf’(0) + \frac{t^2f’’(0)}{2} + \frac{t^3f’’’(0)}{6}$ is a pretty good approximation. In our particular case, $cos(θ) ≈ 1 -\frac{θ^2}{2}, sin(θ)≈θ -\frac{θ^3}{6}$
Now, substitute these approximations into the parametric equations: $x(θ) = θ - sin(θ) ≈ θ -(θ -\frac{θ^3}{6}) = \frac{θ^3}{6}, y(θ) = 1 - cos(θ) ≈ 1 -(1 -\frac{θ^2}{2}) = \frac{θ^2}{2}$.
To find the slope near the origin, we compute the radio $\frac{y(θ)}{x(θ)}=\frac{\frac{θ^2}{2}}{\frac{θ^3}{6}} = \frac{6θ^2}{2θ^3} =[\text{Simplify the expression:}]\frac{3}{θ}$.
As θ approaches zero, the slope $\frac{3}{θ}$ approaches infinity. This indicates that the slope at the origin is vertical, confirming that the correct sketch of the cycloid near θ ≈ 0 is the one with a vertical tangent, as shown in Figure iv, sketch 4.