And yet despite the look on my face, you’re still talking and thinking that I care, Anonymous
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, $\vec{A}x\vec{B} = det(\begin{smallmatrix}i & j & k\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\end{smallmatrix}) =|\begin{smallmatrix}a_2 & a_3\\ b_2 & b_3\end{smallmatrix}|\vec{i}-|\begin{smallmatrix}a_1 & a_3\\ b_1 & b_3\end{smallmatrix}|\vec{j}+|\begin{smallmatrix}a_1 & a_2\\ b_1 & b_2\end{smallmatrix}|\vec{k}$
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^{-1}B.. $A^{-1} = \frac{1}{\text{det}(A)} \times \text{adj}(A)$. adj(A) = C^{T} 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)
A function of two variables f: ℝ x ℝ → ℝ assigns to each ordered pair in its domain a unique real number, e.g., Area = $\frac{1}{2}b·h$, z = f(x, y) = 2x + 3y, f(x, y) = x^{2} + y^{2}, e^{x+y}, etc.
Partial derivatives are an extension of derivatives to functions of multiple variables. They measure how a function changes as one of its variables changes, while keeping the other variables constant. This concept is fundamental in multivariable calculus and has various applications in fields like physics, engineering, and economics.
Consider a function f(x_{1}, x_{2}, ···, x_{n}) that depends on n variables. The partial derivative of f with respect to one of these variables, say x_{i}, is the derivative of f while holding the other variables constant.
For a function f(x,y), the partial derivative with respect to x at a point (x_{0}, y_{0}) is defined as:
$\frac{\partial f}{\partial x}(x_0, y_0) = \lim_{\Delta x \to 0} \frac{f(x_0+\Delta x, y_0)-f(x_0, y_0)}{\Delta x}$
Using a different notation, $\frac{\partial f}{\partial x}(x, y) = \lim_{h \to 0} \frac{f(x+h, y)-f(x, y)}{h}$
Similarly, the partial derivative with respect to y at a point (x_{0}, y_{0}) is defined as: $\frac{\partial f}{\partial y}(x_0, y_0) = \lim_{\Delta y \to 0} \frac{f(x_0, y_0+\Delta y)-f(x_0, y_0)}{\Delta y}$
Using a different notation, $\frac{\partial f}{\partial y}(x, y) = \lim_{h \to 0} \frac{f(x, y+h)-f(x, y)}{h}$
Geometrically, the partial derivative $\frac{\partial f}{\partial x}(x_0, y_0)$ at a point (x_{0}, y_{0}) can be interpreted as the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane y = y_{0}. This plane is parallel to the xz-plane and passes through the point (x_{0}, y_{0}).
Similarly, $\frac{\partial f}{\partial y}(x_0, y_0)$ represents the slope of the tangent line to the curve formed by the intersection of the surface defined by f(x,y) and the plane x = x_{0} This plane is parallel to the yz-plane and passes through the point (x_{0}, y_{0}).(Figure A).
Find all the 1st order partial derivatives of the following functions:
Let f(x, y) = 2x^{2}y^{3}, $\frac{\partial f}{\partial x} = f_x =[\text{Here, y is treated as a constant}] 4x·y^3, \frac{\partial f}{\partial y} = f_y =[\text{Here, x is treated as a constant}] 6x^2·y^2.$
Let f(x, y) = x^{3}y + y^{2}, $\frac{\partial f}{\partial x} = f_x = 3x^2·y, \frac{\partial f}{\partial y} = f_y = x^3+2y.$
z = $x\sqrt{y}-\frac{y}{\sqrt{x}}$. It easy when we can rewrite it as z = $xy^{\frac{1}{2}}-yx^{\frac{-1}{2}}$. Then, the partial derivatives are: $\frac{\partial f}{\partial x} = y^{\frac{1}{2}}-y(\frac{-1}{2})x^{-\frac{3}{2}} = \sqrt{y}+\frac{y}{2x^{\frac{3}{2}}}, \frac{\partial f}{\partial y} = \frac{1}{2}xy^{\frac{-1}{2}}-x^{\frac{-1}{2}} = \frac{x}{2\sqrt{y}}-\frac{1}{\sqrt{x}}$.
Let f(x, y) = sin(x^{2}y -2x + 4). $\frac{\partial f}{\partial x} = f_x =[\text{Chain Rule}] cos(x^2y-2x+4)·(2xy-2). \frac{\partial f}{\partial y} = f_y = cos(x^2y-2x+4)·x^2$
w = $cos(x^2+2y)-e^{4x-z^4y}+y^3$. $\frac{\partial w}{\partial x} =[\text{Chain Rule}]-2xsin(x^2+2y)-4e^{4x-z^4y}, \frac{\partial w}{\partial y} =-2sin(x^2+2y)+z^4e^{4x-z^4y}+3y^2, \frac{\partial w}{\partial z} =4z^3y4e^{4x-z^4y}$
Understanding local minimums and maximums in multivariable functions involves examining the behavior of the function’s partial derivatives.
A function f(x, y) has a relative minimum (maximum) at the point (a, b) if f(x, y) ≥ f(a, b) (f(x, y) ≤ f(a, b)) for all points (x, y) in some region around (a, b). ⚠ Outside of this region, it is completely possible for the function to be smaller (bigger) than f(a, b).
At a local minimum or maximum of a function f(x,y), the partial derivatives with respect to each variable are zero: $\frac{\partial f}{\partial x}(x_0, y_0) = 0,\text{ and } \frac{\partial f}{\partial y}(x_0, y_0) = 0$.
This condition indicates that the tangent plane to the graph z = f(x, y) is horizontal at (x_{0}), y_{0} ($Δz ≈ f_xΔx+f_yΔy = 0$, Figure C). A horizontal tangent plane means there is no instantaneous rate of change in any direction from this point.
However, it’s essential to understand that this condition is necessary for identifying local extrema (minimums and maximums), it is not sufficient on its own. Critical points can also be saddle points or points of inflection, not just local extrema.
Critical points in multivariable functions can indeed be local minimums, local maximums or saddle points where the slopes (derivatives) in orthogonal directions are all zero (a critical point), but which is not a local extremum of the function, e.g., a relative minimum along one axial direction and a relative maximum along the crossing axis, f(x, y) = x^{2} - y^{2}, $\frac{\partial f}{\partial x} = 2x, \frac{\partial f}{\partial y} = -2y$ has a critical point at (0, 0) that is a saddle point or point of inflection.
This point is a saddle point because f is a hyperbolic paraboloid. It curves upwards in one direction and downwards in another direction.
In single-variable calculus, critical points of a function y = f(x) are the x-values within the domain (D) of f where the first derivate is either zero (f’(x) = 0) or undefined. These points can indicate local minima, local maxima, or points of inflection.
Definition. Let f(x_{1}, x_{2}, ···, x_{n}) be a function of n variables. A point (x_{1}, x_{2}, ···, x_{n}) in the domain of f is considered a critical point if all its partial derivatives are zero or if the partial derivatives do not exist. This can be mathematically expressed as: $\frac{∂f}{∂x_1} = 0, \frac{∂f}{∂x_2} = 0, ···, \frac{∂f}{∂x_n} = 0$.
The gradient ∇f is a vector of the partial derivatives of f: ∇f = ($\frac{∂f}{∂x_1}, \frac{∂f}{∂x_2}, ···, \frac{∂f}{∂x_n}$).
At a critical point, the gradient of f at that point is the zero vector or does not exist.
For a function f(x, y) of two variables, a point (x_{0}, y_{0}) is a critical point if $\frac{\partial f}{\partial x}(x_0, y_0) = 0$ and $\frac{\partial f}{\partial y}(x_0, y_0) = 0$ or if these partial derivatives do not exist.
Theorem. If the point (x_{0}, y_{0}) is a local (relative) extremum of the function f(x, y) and the first order derivative of f(x, y) exist at (x_{0}, y_{0}), then (x_{0}, y_{0}) is also a critical point of f and $∇f(x_0, y_0) = \vec{0}$.
Proof.
Consider the new function g(x) = f(x, y_{0}). This is essentially a cross-section of f(x, y) where y is held constant at y_{0}.
Suppose that f(x, y) has a relative (local) extremum at the point (x_{0}, y_{0}). This means that the function g(x) also has a local extremum at x = x_{0}. This is because g(x) is just the function f along the line y = y_{0}.
By Fermat’s Theorem, if a function has a local extremum at a point and its derivative exists at that point, then the derivative at that point must be zero. Therefore, g’(x_{0}) = 0.
The derivative of g(x) with respect to x at x_{0} is the same as the partial derivative of f with respect to x at (x_{0}, y_{0}): g’(x_{0}) = f_{x}(x_{0}, y_{0}).
Notice that $g’(x) = \frac{∂f}{∂x}(x, y_0)$. When we evaluate this at x = x_{0}, we get $g’(x_0) = \frac{∂f}{∂x}(x_0, y_0)$
Since g’(x_{0}) = 0, it follows that f_{x}(x_{0}, y_{0}) = 0.
Similarly, consider the function h(y) = f(a, y), which is the cross-section of f where x is held constant at x_{0}. By the same reasoning as above, if f(x, y) has a local extremum at (x_{0}, y_{0}), then h(y) has a local extremum at y = y_{0}. By Fermat’s Theorem: h’(y_{0}) = 0.
The derivative of h(y) with respect to y at y_{0} is the same as the partial derivative of f with respect to y at (x_{0}, y_{0}): h’(y_{0}) = f_{y}(x_{0}, y_{0}).
Since h’(y_{0}) = 0, it follows that f_{y}(x_{0}, y_{0}) = 0.
Since both partial derivatives of f are zero at (x_{0}, y_{0}): $∇f(x_0, y_0) = (\frac{∂f}{∂x}(x_0, y_0), \frac{∂f}{∂y}(x_0, y_0)) = ⟨0, 0⟩ = \vec{0}$. This means that (x_{0}, y_{0}) is a critical point of f.