数学代写|微积分代写Calculus代写|MAST10006

statistics-lab™ 为您的留学生涯保驾护航 在代写微积分Calculus方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写微积分Calculus代写方面经验极为丰富，各种代写微积分Calculus相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代写|微积分代写Calculus代写|reversing the chain rule

The derivative calculation
$$\frac{d}{d x} \cos \left(x^2+3\right)=-\sin \left(x^2+3\right) \cdot 2 x=-2 x \sin \left(x^2+3\right)$$
can be reversed to give the antiderivative result
$$\int-2 x \sin \left(x^2+3\right) d x=\cos \left(x^2+3\right)+C .$$
However, if we start from scratch given the integral
$$\int-2 x \sin \left(x^2+3\right) d x$$
how do we proceed to find the antiderivative if we do not already know it? The integrand contains both a product $\left(-2 x\right.$ times $\left.\sin \left(x^2+3\right)\right)$ and a composition (sine of “not just plain $x^{” \prime}$ ). We need a clear strategy for how to handle such integrands.
To this end, recall the chain rule formula:
$$\frac{d}{d x} f(g(x))=f^{\prime}(g(x)) \cdot g^{\prime}(x) .$$

Suppose $F$ is an antiderivative of $f$ that is, $F^{\prime}(x)=f(x)$. We can rewrite the chain rule using $F$ as
$$\frac{d}{d x} F(g(x))=f(g(x)) \cdot g^{\prime}(x) .$$
Reversing the derivative calculation yields the antiderivative result
$$\int f(g(x)) \cdot g^{\prime}(x) d x=F(g(x))+C .$$
Next we make the substitution
\begin{aligned} u & =g(x) \ d u & =g^{\prime}(x) d x \end{aligned}
into the antiderivative formula:
$$\int f(u) d u=F(u)+C$$
This integral makes sense given that $F$ is an antiderivative of $f$. It also points the way to our procedure, for it gives us an easier integral to evaluate. The method of substitution takes an integral $\int f(g(x)) g^{\prime}(x) d x$ involving “not just plain $x^”$ (the expression $g(x)$ ), 1 makes the previous substitution to obtain an integral $\int f(u) d u$ with “just plain $u$,” (2) finds its (most general) antiderivative $F(u)+C$ using previously available antiderivative rules and formulas, and then 3 reverses the substitution, giving the antiderivative we are seeking: $F(g(x))+C$.

数学代写|微积分代写Calculus代写|More substitution examples

The next example explores what we can do if we run into a snag when making the substitution.
Example 3 Calculate $\int s^3\left(s^4+7\right)^{12} d s$.
Solution Once again we have an integrand with “not just plain $s$ ” to a power, so we try substitution and let $u$ be the “not just plain $s$ ” part:
$$\mathbf{u}=\mathrm{s}^4+7$$
Next we calculate the differential $d u$ :
$$d u=4 s^3 d s$$
Now we have a problem. The substitution requires replacing $4 s^3 d s$ with $d u$, but the 4 is missing from the integrand in $\int s^3\left(s^4+7\right)^{12} d s$. However, we can always multiply any expression by one without changing its value, so we can multiply by $4 \cdot \frac{1}{4}$ to obtain
$$\int 4 \cdot \frac{1}{4} \cdot s^3\left(s^4+7\right)^{12} d s$$
Now everything we need for the substitution is present and we can move forward:
\begin{aligned} \int 4 \cdot \frac{1}{4} \cdot s^3\left(s^4+7\right)^{12} d s . & =\int \frac{1}{4}(u)^{12} d u \ & =\frac{1}{4} \cdot \frac{u^{13}}{13}+C \ & =\frac{\left(s^4+7\right)^{13}}{52}+C \end{aligned}

微积分代考

数学代写|微积分代写Calculus代写|reversing the chain rule

$$\frac{d}{d x} \cos \left(x^2+3\right)=-\sin \left(x^2+3\right) \cdot 2 x=-2 x \sin \left(x^2+3\right)$$

$$\int-2 x \sin \left(x^2+3\right) d x=\cos \left(x^2+3\right)+C$$

$$\int-2 x \sin \left(x^2+3\right) d x$$

$$\frac{d}{d x} f(g(x))=f^{\prime}(g(x)) \cdot g^{\prime}(x)$$

$$\frac{d}{d x} F(g(x))=f(g(x)) \cdot g^{\prime}(x)$$

$$\int f(g(x)) \cdot g^{\prime}(x) d x=F(g(x))+C$$

$$u=g(x) d u \quad=g^{\prime}(x) d x$$

$$\int f(u) d u=F(u)+C$$

数学代写|微积分代写Calculus代写|More substitution examples

$$\mathbf{u}=s^4+7$$

$$d u=4 s^3 d s$$

$$\int 4 \cdot \frac{1}{4} \cdot s^3\left(s^4+7\right)^{12} d s$$

$$\int 4 \cdot \frac{1}{4} \cdot s^3\left(s^4+7\right)^{12} d s .=\int \frac{1}{4}(u)^{12} d u \quad=\frac{1}{4} \cdot \frac{u^{13}}{13}+C=\frac{\left(s^4+7\right)^{13}}{52}+C$$

有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

数学代写|线性代数代写linear algebra代考|MTH2106

statistics-lab™ 为您的留学生涯保驾护航 在代写线性代数linear algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写线性代数linear algebra代写方面经验极为丰富，各种代写线性代数linear algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代写|线性代数代写linear algebra代考|The Dimension of a Subspace

It can be shown that if a subspace $H$ has a basis of $p$ vectors, then every basis of $H$ must consist of exactly $p$ vectors. (See Exercises 27 and 28 .) Thus the following definition makes sense.
The dimension of a nonzero subspace $H$, denoted by $\operatorname{dim} H$, is the number of vectors in any basis for $H$. The dimension of the zero subspace ${0}$ is defined to be zero. ${ }^2$
The space $\mathbb{R}^n$ has dimension $n$. Every basis for $\mathbb{R}^n$ consists of $n$ vectors. A plane through 0 in $\mathbb{R}^3$ is two-dimensional, and a line through $\mathbf{0}$ is one-dimensional.

EXAMPLE 2 Recall that the null space of the matrix $A$ in Example 6 in Section $2.8$ had a basis of 3 vectors. So the dimension of $\operatorname{Nul} A$ in this case is 3 . Observe how each basis vector corresponds to a free variable in the equation $A \mathbf{x}=\mathbf{0}$. Our construction always produces a basis in this way. So, to find the dimension of $\mathrm{Nul} A$, simply identify and count the number of free variables in $A \mathbf{x}=\mathbf{0}$.
The rank of a matrix $A$, denoted by rank $A$, is the dimension of the column space of $A$.
Since the pivot columns of $A$ form a basis for $\operatorname{Col} A$, the rank of $A$ is just the number of pivot columns in $A$.

The row reduction in Example 3 reveals that there are two free variables in $A \mathbf{x}=\mathbf{0}$, because two of the five columns of $A$ are not pivot columns. (The nonpivot columns correspond to the free variables in $A \mathbf{x}=\mathbf{0}$.) Since the number of pivot columns plus the number of nonpivot columns is exactly the number of columns, the dimensions of Col $A$ and $\mathrm{Nul} A$ have the following useful connection. (See the Rank Theorem in Section $4.6$ for additional details.)
The Rank Theorem
If a matrix $A$ has $n$ columns, then $\operatorname{rank} A+\operatorname{dim} \operatorname{Nul} A=n$.
The following theorem is important for applications and will be needed in Chapters 5 and 6. The theorem (proved in Section 4.5) is certainly plausible, if you think of a $p$-dimensional subspace as isomorphic to $\mathbb{R}^p$. The Invertible Matrix Theorem shows that $p$ vectors in $\mathbb{R}^p$ are linearly independent if and only if they also span $\mathbb{R}^p$.

数学代写|线性代数代写linear algebra代考|Column Space and Null Space of a Matrix

Subspaces of $\mathbb{R}^n$ usually occur in applications and theory in one of two ways. In both cases, the subspace can be related to a matrix.
The column space of a matrix $A$ is the set $\operatorname{Col} A$ of all linear combinations of the columns of $A$.
If $A=\left[\begin{array}{lll}\mathbf{a}_1 & \cdots & \mathbf{a}_n\end{array}\right]$, with the columns in $\mathbb{R}^m$, then $\operatorname{Col} A$ is the same as Span $\left{\mathbf{a}_1, \ldots, \mathbf{a}_n\right}$. Example 4 shows that the column space of an $\boldsymbol{m} \times \boldsymbol{n}$ matrix is a subspace of $\mathbb{R}^m$. Note that $\operatorname{Col} A$ equals $\mathbb{R}^m$ only when the columns of $A$ span $\mathbb{R}^m$. Otherwise, $\operatorname{Col} A$ is only part of $\mathbb{R}^m$.

EXAMPLE 4 Let $A=\left[\begin{array}{rrr}1 & -3 & -4 \ -4 & 6 & -2 \ -3 & 7 & 6\end{array}\right]$ and $\mathbf{b}=\left[\begin{array}{r}3 \ 3 \ -4\end{array}\right]$. Determine whether $\mathbf{b}$ is in the column space of $A$.

SOLUTION The vector $\mathbf{b}$ is a linear combination of the columns of $A$ if and only if $\mathbf{b}$ can be written as $A \mathbf{x}$ for some $\mathbf{x}$, that is, if and only if the equation $A \mathbf{x}=\mathbf{b}$ has a solution. Row reducing the augmented matrix $\left[A \begin{array}{ll}A & \mathbf{b}\end{array}\right]$,
$$\left[\begin{array}{rrrr} 1 & -3 & -4 & 3 \ -4 & 6 & -2 & 3 \ -3 & 7 & 6 & -4 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & -3 & -4 & 3 \ 0 & -6 & -18 & 15 \ 0 & -2 & -6 & 5 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & -3 & -4 & 3 \ 0 & -6 & -18 & 15 \ 0 & 0 & 0 & 0 \end{array}\right]$$
we conclude that $A \mathbf{x}=\mathbf{b}$ is consistent and $\mathbf{b}$ is in $\operatorname{Col} A$.

The solution of Example 4 shows that when a system of linear equations is written in the form $A \mathbf{x}=\mathbf{b}$, the column space of $A$ is the set of all $\mathbf{b}$ for which the system has a solution.
The null space of a matrix $A$ is the set $\mathrm{Nul} A$ of all solutions of the homogeneous equation $A \mathbf{x}=\mathbf{0}$
When $A$ has $n$ columns, the solutions of $A \mathbf{x}=\mathbf{0}$ belong to $\mathbb{R}^n$, and the null space of $A$ is a subset of $\mathbb{R}^n$. In fact, $\mathrm{Nul} A$ has the properties of a subspace of $\mathbb{R}^n$.
The null space of an $m \times n$ matrix $A$ is a subspace of $\mathbb{R}^n$. Equivalently, the set of all solutions of a system $A \mathbf{x}=\mathbf{0}$ of $m$ homogeneous linear equations in $n$ unknowns is a subspace of $\mathbb{R}^n$.
PROOF The zero vector is in $\operatorname{Nul} A$ (because $A 0=0$ ). To show that $\mathrm{Nul} A$ satisfies the other two properties required for a subspace, take any $\mathbf{u}$ and $\mathbf{v}$ in $\mathrm{Nul} A$. That is, suppose $A \mathbf{u}=\mathbf{0}$ and $A \mathbf{v}=\mathbf{0}$. Then, by a property of matrix multiplication,
$$A(\mathbf{u}+\mathbf{v})=A \mathbf{u}+A \mathbf{v}=\mathbf{0}+\mathbf{0}=\mathbf{0}$$
Thus $\mathbf{u}+\mathbf{v}$ satisfies $A \mathbf{x}=\mathbf{0}$, and so $\mathbf{u}+\mathbf{v}$ is in $\operatorname{Nul} A$. Also, for any scalar $c, A(c \mathbf{u})=$ $c(A \mathbf{u})=c(0)=\mathbf{0}$, which shows that $c \mathbf{u}$ is in $\mathrm{Nul} A$.

To test whether a given vector $\mathbf{v}$ is in $\operatorname{Nul} A$, just compute $A \mathbf{v}$ to see whether $A \mathbf{v}$ is the zero vector. Because $\mathrm{Nul} A$ is described by a condition that must be checked for each vector, we say that the null space is defined implicitly. In contrast, the column space is defined explicitly, because vectors in Col A can be constructed (by linear combinations) from the columns of $A$. To create an explicit description of $\mathrm{Nul} A$, solve the equation $A \mathbf{x}=\mathbf{0}$ and write the solution in parametric vector form. (See Example 6 , below.) ${ }^2$

线性代数代考

数学代写|线性代数代写linear algebra代考|Column Space and Null Space of a Matrix

Veft{ $\left.\backslash m a t h b f{a} _1, \backslash d o t s, \backslash m a t h b f{a} _n \backslash r i g h t\right}$. 示例 4 显示了一个列空间 $\boldsymbol{m} \times \boldsymbol{n}$ 矩阵是一个子空间 $\mathbb{R}^m$. 注意 $\operatorname{Col} A$ 等于 $\mathbb{R}^m$ 只有当列 $A$ 跨度 $\mathbb{R}^m$. 否则， $\operatorname{Col} A$ 只是一部分 $\mathbb{R}^m$. $A$.

$$A(\mathbf{u}+\mathbf{v})=A \mathbf{u}+A \mathbf{v}=\mathbf{0}+\mathbf{0}=\mathbf{0}$$

有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

数学代写|线性代数代写linear algebra代考|MATH1051

statistics-lab™ 为您的留学生涯保驾护航 在代写线性代数linear algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写线性代数linear algebra代写方面经验极为丰富，各种代写线性代数linear algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代写|线性代数代写linear algebra代考|Perspective Projections

A three-dimensional object is represented on the two-dimensional computer screen by projecting the object onto a viewing plane. (We ignore other important steps, such as selecting the portion of the viewing plane to display on the screen.) For simplicity, let the $x y$-plane represent the computer screen, and imagine that the eye of a viewer is along the positive $z$-axis, at a point $(0,0, d)$. A perspective projection maps each point $(x, y, z)$ onto an image point $\left(x^, y^, 0\right)$ so that the two points and the eye position, called the center of projection, are on a line. See Figure 6(a).

The triangle in the $x z$-plane in Figure 6(a) is redrawn in part (b) showing the lengths of line segments. Similar triangles show that
$$\frac{x^}{d}=\frac{x}{d-z} \quad \text { and } \quad x^=\frac{d x}{d-z}=\frac{x}{1-z / d}$$
Similarly,
$$y^*=\frac{y}{1-z / d}$$
Using homogeneous coordinates, we can represent the perspective projection by a matrix, say, $P$. We want $(x, y, z, 1)$ to map into $\left(\frac{x}{1-z / d}, \frac{y}{1-z / d}, 0,1\right)$. Scaling these coordinates by $1-z / d$, we can also use $(x, y, 0,1-z / d)$ as homogeneous coordinates for the image. Now it is easy to display $P$. In fact,
$$P\left[\begin{array}{l} x \ y \ z \ 1 \end{array}\right]=\left[\begin{array}{cccc} 1 & 0 & 0 & 0 \ 0 & 1 & 0 & 0 \ 0 & 0 & 0 & 0 \ 0 & 0 & -1 / d & 1 \end{array}\right]\left[\begin{array}{c} x \ y \ z \ 1 \end{array}\right]=\left[\begin{array}{c} x \ y \ 0 \ 1-z / d \end{array}\right]$$
EXAMPLE 8 Let $S$ be the box with vertices $(3,1,5),(5,1,5),(5,0,5),(3,0,5)$, $(3,1,4),(5,1,4),(5,0,4)$, and $(3,0,4)$. Find the image of $S$ under the perspective projection with center of projection at $(0,0,10)$.

数学代写|线性代数代写linear algebra代考|Column Space and Null Space of a Matrix

Subspaces of $\mathbb{R}^n$ usually occur in applications and theory in one of two ways. In both cases, the subspace can be related to a matrix.
The column space of a matrix $A$ is the set $\operatorname{Col} A$ of all linear combinations of the columns of $A$.
If $A=\left[\begin{array}{lll}\mathbf{a}_1 & \cdots & \mathbf{a}_n\end{array}\right]$, with the columns in $\mathbb{R}^m$, then $\operatorname{Col} A$ is the same as Span $\left{\mathbf{a}_1, \ldots, \mathbf{a}_n\right}$. Example 4 shows that the column space of an $\boldsymbol{m} \times \boldsymbol{n}$ matrix is a subspace of $\mathbb{R}^m$. Note that $\operatorname{Col} A$ equals $\mathbb{R}^m$ only when the columns of $A$ span $\mathbb{R}^m$. Otherwise, $\operatorname{Col} A$ is only part of $\mathbb{R}^m$.

EXAMPLE 4 Let $A=\left[\begin{array}{rrr}1 & -3 & -4 \ -4 & 6 & -2 \ -3 & 7 & 6\end{array}\right]$ and $\mathbf{b}=\left[\begin{array}{r}3 \ 3 \ -4\end{array}\right]$. Determine whether $\mathbf{b}$ is in the column space of $A$.

SOLUTION The vector $\mathbf{b}$ is a linear combination of the columns of $A$ if and only if $\mathbf{b}$ can be written as $A \mathbf{x}$ for some $\mathbf{x}$, that is, if and only if the equation $A \mathbf{x}=\mathbf{b}$ has a solution. Row reducing the augmented matrix $\left[A \begin{array}{ll}A & \mathbf{b}\end{array}\right]$,
$$\left[\begin{array}{rrrr} 1 & -3 & -4 & 3 \ -4 & 6 & -2 & 3 \ -3 & 7 & 6 & -4 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & -3 & -4 & 3 \ 0 & -6 & -18 & 15 \ 0 & -2 & -6 & 5 \end{array}\right] \sim\left[\begin{array}{rrrr} 1 & -3 & -4 & 3 \ 0 & -6 & -18 & 15 \ 0 & 0 & 0 & 0 \end{array}\right]$$
we conclude that $A \mathbf{x}=\mathbf{b}$ is consistent and $\mathbf{b}$ is in $\operatorname{Col} A$.

The solution of Example 4 shows that when a system of linear equations is written in the form $A \mathbf{x}=\mathbf{b}$, the column space of $A$ is the set of all $\mathbf{b}$ for which the system has a solution.
The null space of a matrix $A$ is the set $\mathrm{Nul} A$ of all solutions of the homogeneous equation $A \mathbf{x}=\mathbf{0}$
When $A$ has $n$ columns, the solutions of $A \mathbf{x}=\mathbf{0}$ belong to $\mathbb{R}^n$, and the null space of $A$ is a subset of $\mathbb{R}^n$. In fact, $\mathrm{Nul} A$ has the properties of a subspace of $\mathbb{R}^n$.
The null space of an $m \times n$ matrix $A$ is a subspace of $\mathbb{R}^n$. Equivalently, the set of all solutions of a system $A \mathbf{x}=\mathbf{0}$ of $m$ homogeneous linear equations in $n$ unknowns is a subspace of $\mathbb{R}^n$.
PROOF The zero vector is in $\operatorname{Nul} A$ (because $A 0=0$ ). To show that $\mathrm{Nul} A$ satisfies the other two properties required for a subspace, take any $\mathbf{u}$ and $\mathbf{v}$ in $\mathrm{Nul} A$. That is, suppose $A \mathbf{u}=\mathbf{0}$ and $A \mathbf{v}=\mathbf{0}$. Then, by a property of matrix multiplication,
$$A(\mathbf{u}+\mathbf{v})=A \mathbf{u}+A \mathbf{v}=\mathbf{0}+\mathbf{0}=\mathbf{0}$$
Thus $\mathbf{u}+\mathbf{v}$ satisfies $A \mathbf{x}=\mathbf{0}$, and so $\mathbf{u}+\mathbf{v}$ is in $\operatorname{Nul} A$. Also, for any scalar $c, A(c \mathbf{u})=$ $c(A \mathbf{u})=c(0)=\mathbf{0}$, which shows that $c \mathbf{u}$ is in $\mathrm{Nul} A$.

To test whether a given vector $\mathbf{v}$ is in $\operatorname{Nul} A$, just compute $A \mathbf{v}$ to see whether $A \mathbf{v}$ is the zero vector. Because $\mathrm{Nul} A$ is described by a condition that must be checked for each vector, we say that the null space is defined implicitly. In contrast, the column space is defined explicitly, because vectors in Col A can be constructed (by linear combinations) from the columns of $A$. To create an explicit description of $\mathrm{Nul} A$, solve the equation $A \mathbf{x}=\mathbf{0}$ and write the solution in parametric vector form. (See Example 6 , below.) ${ }^2$

线性代数代考

数学代写|线性代数代写linear algebra代考|Perspective Projections

y^*=\frac{y}{1-z / d}
$$使用齐次坐标，我们可以用矩阵表示透视投影，比如说， P. 我们想要 (x, y, z, 1) 映射到 \left(\frac{x}{1-z / d}, \frac{y}{1-z / d}, 0,1\right). 缩放这些坐标 1-z / d ，我们也可以使用 (x, y, 0,1-z / d) 作为图像的齐次坐 标。现在很容易显示 P. 实际上，$$
P[x y z 1]=\left[\begin{array}{llllllllllllllll}
1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 / d & 1
\end{array}\right]\left[\begin{array}{ll}
x y z & -1
\end{array}\right]=\left[\begin{array}{lll}
x & 0 & 1
\end{array}\right.
$$例 8 让 S 是有顶点的盒子 (3,1,5),(5,1,5),(5,0,5),(3,0,5) ，(3,1,4),(5,1,4),(5,0,4) ， 和 (3,0,4). 找到图像 S 在投影中心位于的透视投影下 (0,0,10). 数学代写|线性代数代写linear algebra代考|Column Space and Null Space of a Matrix 通过将物体投影到观察平面上，三维物体在二维计算机屏幕上呈现。（我们忽略其他重要步㡜，例如选择 要在屏幕上显示的视图平面部分。) 为简单起见，让 x y-plane 代表计算机屏幕，并想象观众的眼睛沿着 正面 z-轴，在一点 (0,0, d). 透视投影映射每个点 (x, y, z) 到图像点 \ V \mathrm{eft}\left(\mathrm{X}^{\wedge}, y^{\wedge} ， 0 \backslash r i g h t\right) \$$ 上，这样两 个点和眼睛位置 (称为投影中心) 在一条线上。见图 6(a)。

$\left\langle f r a c\left{x^{\wedge}\right}{d}=|f r a c{x}{d z} \backslash q u a d| t e x t{\right.$ 和 $\left.} \backslash q u a d x^{\wedge}=\right| f r a c{d x}{d z}=\backslash f r a c{x}{1-z / d}$

$$y^*=\frac{y}{1-z / d}$$

$$P[x y z 1]=\left[\begin{array}{llllllllllllllll} 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & -1 / d & 1 \end{array}\right]\left[\begin{array}{ll} x y z & -1 \end{array}\right]=\left[\begin{array}{lll} x & 0 & 1 \end{array}\right.$$

有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

数学代写|线性代数代写linear algebra代考|MATH1014

statistics-lab™ 为您的留学生涯保驾护航 在代写线性代数linear algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写线性代数linear algebra代写方面经验极为丰富，各种代写线性代数linear algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代写|线性代数代写linear algebra代考|APPLICATIONS TO COMPUTER GRAPHICS

Computer graphics are images displayed or animated on a computer screen. Applications of computer graphics are widespread and growing rapidly. For instance, computeraided design (CAD) is an integral part of many engineering processes, such as the aircraft design process described in the chapter introduction. The entertainment industry has made the most spectacular use of computer graphics – from the special effects in Amazing Spider-Man 2 to PlayStation 4 and Xbox One.

Most interactive computer software for business and industry makes use of computer graphics in the screen displays and for other functions, such as graphical display of data, desktop publishing, and slide production for commercial and educational presentations. Consequently, anyone studying a computer language invariably spends time learning how to use at least two-dimensional (2D) graphics.

This section examines some of the basic mathematics used to manipulate and display graphical images such as a wire-frame model of an airplane. Such an image (or picture) consists of a number of points, connecting lines or curves, and information about how to fill in closed regions bounded by the lines and curves. Often, curved lines are approximated by short straight-line segments, and a figure is defined mathematically by a list of points.

Among the simplest 2D graphics symbols are letters used for labels on the screen. Some letters are stored as wire-frame objects; others that have curved portions are stored with additional mathematical formulas for the curves.

EXAMPLE 1 The capital letter $\mathrm{N}$ in Figure 1 is determined by eight points, or vertices. The coordinates of the points can be stored in a data matrix, $D$.
$x$-coordinate $\left[\begin{array}{cccccccc}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \ 0 & .5 & .5 & 6 & 6 & 5.5 & 5.5 & 0 \ 0 & 0 & 6.42 & 0 & 8 & 8 & 1.58 & 8\end{array}\right]=D$
In addition to $D$, it is necessary to specify which vertices are connected by lines, but we omit this detail.

The main reason graphical objects are described by collections of straight-line segments is that the standard transformations in computer graphics map line segments onto other line segments. (For instance, see Exercise 27 in Section 1.8.) Once the vertices that describe an object have been transformed, their images can be connected with the appropriate straight lines to produce the complete image of the original object.

数学代写|线性代数代写linear algebra代考|Homogeneous 3D Coordinates

By analogy with the $2 \mathrm{D}$ case, we say that $(x, y, z, 1)$ are homogeneous coordinates for the point $(x, y, z)$ in $\mathbb{R}^3$. In general, $(X, Y, Z, H)$ are homogeneous coordinates for $(x, y, z)$ if $H \neq 0$ and
$$x=\frac{X}{H}, \quad y=\frac{Y}{H}, \quad \text { and } \quad z=\frac{Z}{H}$$
Each nonzero scalar multiple of $(x, y, z, 1)$ gives a set of homogeneous coordinates for $(x, y, z)$. For instance, both $(10,-6,14,2)$ and $(-15,9,-21,-3)$ are homogeneous coordinates for $(5,-3,7)$.

The next example illustrates the transformations used in molecular modeling to move a drug into a protein molecule.
EXAMPLE 7 Give $4 \times 4$ matrices for the following transformations:
a. Rotation about the $y$-axis through an angle of $30^{\circ}$. (By convention, a positive angle is the counterclockwise direction when looking toward the origin from the positive half of the axis of rotation-in this case, the $y$-axis.)
b. Translation by the vector $\mathbf{p}=(-6,4,5)$.
SOLUTION
a. First, construct the $3 \times 3$ matrix for the rotation. The vector $\mathbf{e}_1$ rotates down toward the negative $z$-axis, stopping at $\left(\cos 30^{\circ}, 0,-\sin 30^{\circ}\right)=(\sqrt{3} / 2,0,-.5)$. The vector $\mathbf{e}_2$ on the $y$-axis does not move, but $\mathbf{e}_3$ on the $z$-axis rotates down toward the positive $x$-axis, stopping at $\left(\sin 30^{\circ}, 0, \cos 30^{\circ}\right)=(.5,0, \sqrt{3} / 2)$. See Figure 5. From Section $1.9$, the standard matrix for this rotation is
$$\left[\begin{array}{ccc} \sqrt{3} / 2 & 0 & .5 \ 0 & 1 & 0 \ -.5 & 0 & \sqrt{3} / 2 \end{array}\right]$$
So the rotation matrix for homogeneous coordinates is
$$A=\left[\begin{array}{cccc} \sqrt{3} / 2 & 0 & .5 & 0 \ 0 & 1 & 0 & 0 \ -.5 & 0 & \sqrt{3} / 2 & 0 \ 0 & 0 & 0 & 1 \end{array}\right]$$
b. We want $(x, y, z, 1)$ to map to $(x-6, y+4, z+5,1)$. The matrix that does this is
$$\left[\begin{array}{rrrr} 1 & 0 & 0 & -6 \ 0 & 1 & 0 & 4 \ 0 & 0 & 1 & 5 \ 0 & 0 & 0 & 1 \end{array}\right]$$

线性代数代考

数学代写|线性代数代写linear algebra代考|APPLICATIONS TO COMPUTER GRAPHICS

$\left[\begin{array}{llllllllllllllllllllllll}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 0 & .5 & .5 & 6 & 6 & 5.5 & 5.5 & 0 & 0 & 0 & 6.42 & 0 & 8 & 8 & 1.58 & 8\end{array}\right]$ 此外 $D$ ，有必要指定哪些顶点由线连接，但我们省略了这个细节。

数学代写|线性代数代写linear algebra代考|Homogeneous 3D Coordinates

$$x=\frac{X}{H}, \quad y=\frac{Y}{H}, \quad \text { and } \quad z=\frac{Z}{H}$$

a。旋转关于 $y$-轴通过一个角度 $30^{\circ}$. (按照惯例，正角是从旋转轴的正半边看原点时的逆时针方向一一在 这种情况下， $y$-轴。)
$\mathrm{b}$ 。向量翻译 $\mathbf{p}=(-6,4,5)$.

$\left(\cos 30^{\circ}, 0,-\sin 30^{\circ}\right)=(\sqrt{3} / 2,0,-.5)$. 载体 $\mathbf{e}_2$ 在 $y$-轴不移动，但 $\mathbf{e}_3$ 在 $z$-axis 向下旋转到正 $x$ 轴，停在 $\left(\sin 30^{\circ}, 0, \cos 30^{\circ}\right)=(.5,0, \sqrt{3} / 2)$. 参见图 5。从部分 $1.9$ ，这个旋转的标准矩阵是

b. 我们想要 $(x, y, z, 1)$ 映射到 $(x-6, y+4, z+5,1)$. 这样做的矩阵是

有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

数学代写|线性代数代写linear algebra代考|MTH2106

statistics-lab™ 为您的留学生涯保驾护航 在代写线性代数linear algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写线性代数linear algebra代写方面经验极为丰富，各种代写线性代数linear algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代写|线性代数代写linear algebra代考|ROW REDUCTION AND ECHELON FORMS

This section refines the method of Section $1.1$ into a row reduction algorithm that will enable us to analyze any system of linear equations. ${ }^1$ By using only the first part of the algorithm, we will be able to answer the fundamental existence and uniqueness questions posed in Section 1.1.

The algorithm applies to any matrix, whether or not the matrix is viewed as an augmented matrix for a linear system. So the first part of this section concerns an arbitrary rectangular matrix and begins by introducing two important classes of matrices that include the “triangular” matrices of Section 1.1. In the definitions that follow, a nonzero row or column in a matrix means a row or column that contains at least one nonzero entry; a leading entry of a row refers to the leftmost nonzero entry (in a nonzero row).

An echelon matrix (respectively, reduced echelon matrix) is one that is in echelon form (respectively, reduced echelon form). Property 2 says that the leading entries form an echelon (“steplike”) pattern that moves down and to the right through the matrix. Property 3 is a simple consequence of property 2 , but we include it for emphasis.
The “triangular” matrices of Section 1.1, such as
$$\left[\begin{array}{rrrc} 2 & -3 & 2 & 1 \ 0 & 1 & -4 & 8 \ 0 & 0 & 0 & 5 / 2 \end{array}\right] \text { and }\left[\begin{array}{lllr} 1 & 0 & 0 & 29 \ 0 & 1 & 0 & 16 \ 0 & 0 & 1 & 3 \end{array}\right]$$
are in echelon form. In fact, the second matrix is in reduced echelon form. Here are additional examples.

EXAMPLE 1 The following matrices are in echelon form. The leading entries ( $\boldsymbol{)}$ ) may have any nonzero value; the starred entries $(*)$ may have any value (including zero).

数学代写|线性代数代写linear algebra代考|Solutions of Linear Systems

The row reduction algorithm leads directly to an explicit description of the solution set of a linear system when the algorithm is applied to the augmented matrix of the system.
Suppose, for example, that the augmented matrix of a linear system has been changed into the equivalent reduced echelon form
$$\left[\begin{array}{rrrr} 1 & 0 & -5 & 1 \ 0 & 1 & 1 & 4 \ 0 & 0 & 0 & 0 \end{array}\right]$$
There are three variables because the augmented matrix has four columns. The associated system of equations is
$$\begin{array}{r} x_1-5 x_3=1 \ x_2+x_3=4 \ 0=0 \end{array}$$
The variables $x_1$ and $x_2$ corresponding to pivot columns in the matrix are called basic variables. ${ }^2$ The other variable, $x_3$, is called a free variable.

Whenever a system is consistent, as in (4), the solution set can be described explicitly by solving the reduced system of equations for the basic variables in terms of the free variables. This operation is possible because the reduced echelon form places each basic variable in one and only one equation. In (4), solve the first equation for $x_1$ and the second for $x_2$. (Ignore the third equation; it offers no restriction on the variables.)
$$\left{\begin{array}{l} x_1=1+5 x_3 \ x_2=4-x_3 \ x_3 \text { is free } \end{array}\right.$$
The statement ” $x_3$ is free” means that you are free to choose any value for $x_3$. Once that is done, the formulas in (5) determine the values for $x_1$ and $x_2$. For instance, when $x_3=0$, the solution is $(1,4,0)$; when $x_3=1$, the solution is $(6,3,1)$. Each different choice of $x_3$ determines a (different) solution of the system, and every solution of the system is determined by a choice of $x_3$.

线性代数代考

数学代写|线性代数代写linear algebra代考|ROW REDUCTION AND ECHELON FORMS

$1.1$ 节的“三角”矩阵，如

数学代写|线性代数代写linear algebra代考|Solutions of Linear Systems

$$\left[\begin{array}{llllllllllll} 1 & 0 & -5 & 1 & 0 & 1 & 1 & 4 & 0 & 0 & 0 & 0 \end{array}\right]$$

$$x_1-5 x_3=1 x_2+x_3=40=0$$

x_1=1+5 x_3 x_2=4-x_3 x_3 \text { is free }
$$正确的。 \ \$$

有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

数学代写|线性代数代写linear algebra代考|MATH1051

statistics-lab™ 为您的留学生涯保驾护航 在代写线性代数linear algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写线性代数linear algebra代写方面经验极为丰富，各种代写线性代数linear algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代写|线性代数代写linear algebra代考|Solving a Linear System

This section and the next describe an algorithm, or a systematic procedure, for solving linear systems. The basic strategy is to replace one system with an equivalent system (i.e., one with the same solution set) that is easier to solve.

Roughly speaking, use the $x_1$ term in the first equation of a system to eliminate the $x_1$ terms in the other equations. Then use the $x_2$ term in the second equation to eliminate the $x_2$ terms in the other equations, and so on, until you finally obtain a very simple equivalent system of equations.

Three basic operations are used to simplify a linear system: Replace one equation by the sum of itself and a multiple of another equation, interchange two equations, and multiply all the terms in an equation by a nonzero constant. After the first example, you will see why these three operations do not change the solution set of the system.

Row operations can be applied to any matrix, not merely to one that arises as the augmented matrix of a linear system. Two matrices are called row equivalent if there is a sequence of elementary row operations that transforms one matrix into the other.
It is important to note that row operations are reversible. If two rows are interchanged, they can be returned to their original positions by another interchange. If a row is scaled by a nonzero constant $c$, then multiplying the new row by $1 / c$ produces the original row. Finally, consider a replacement operation involving two rows -say, rows 1 and 2 -and suppose that $c$ times row 1 is added to row 2 to produce a new row 2. To “reverse” this operation, add $-c$ times row 1 to (new) row 2 and obtain the original row 2. See Exercises $29-32$ at the end of this section.

At the moment, we are interested in row operations on the augmented matrix of a system of linear equations. Suppose a system is changed to a new one via row operations. By considering each type of row operation, you can see that any solution of the original system remains a solution of the new system. Conversely, since the original system can be produced via row operations on the new system, each solution of the new system is also a solution of the original system. This discussion justifies the following statement.

数学代写|线性代数代写linear algebra代考|Existence and Uniqueness Questions

Section $1.2$ will show why a solution set for a linear system contains either no solutions, one solution, or infinitely many solutions. Answers to the following two questions will determine the nature of the solution set for a linear system.

To determine which possibility is true for a particular system, we ask two questions.

These two questions will appear throughout the text, in many different guises. This section and the next will show how to answer these questions via row operations on the augmented matrix.
EXAMPLE 2 Determine if the following system is consistent:
\begin{aligned} x_1-2 x_2+x_3 & =0 \ 2 x_2-8 x_3 & =8 \ 5 x_1-5 x_3 & =10 \end{aligned}
SOLUTION This is the system from Example 1. Suppose that we have performed the row operations necessary to obtain the triangular form
\begin{aligned} x_1-2 x_2+x_3 & =0 \ x_2-4 x_3 & =4 \ x_3 & =-1 \end{aligned} \quad\left[\begin{array}{rrrr} 1 & -2 & 1 & 0 \ 0 & 1 & -4 & 4 \ 0 & 0 & 1 & -1 \end{array}\right] At this point, we know $x_3$. Were we to substitute the value of $x_3$ into equation 2 , we could compute $x_2$ and hence could determine $x_1$ from equation 1 . So a solution exists; the system is consistent. (In fact, $x_2$ is uniquely determined by equation 2 since $x_3$ has only one possible value, and $x_1$ is therefore uniquely determined by equation 1 . So the solution is unique.)

线性代数代考

数学代写|线性代数代写linear algebra代考|Existence and Uniqueness Questions

$$x_1-2 x_2+x_3=02 x_2-8 x_3 \quad=85 x_1-5 x_3=10$$

$$x_1-2 x_2+x_3=0 x_2-4 x_3 \quad=4 x_3=-1 \quad\left[\begin{array}{llllllllllll} 1 & -2 & 1 & 0 & 0 & 1 & -4 & 4 & 0 & 0 & 1 & -1 \end{array}\right]$$

有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

数学代写|线性代数代写linear algebra代考|MATH1014

statistics-lab™ 为您的留学生涯保驾护航 在代写线性代数linear algebra方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写线性代数linear algebra代写方面经验极为丰富，各种代写线性代数linear algebra相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代写|线性代数代写linear algebra代考|SYSTEMS OF LINEAR EQUATIONS

A linear equation in the variables $x_1, \ldots, x_n$ is an equation that can be written in the form
$$a_1 x_1+a_2 x_2+\cdots+a_n x_n=b$$
where $b$ and the coefficients $a_1, \ldots, a_n$ are real or complex numbers, usually known in advance. The subscript $n$ may be any positive integer. In textbook examples and exercises, $n$ is normally between 2 and 5 . In real-life problems, $n$ might be 50 or 5000 , or even larger.
The equations
$$4 x_1-5 x_2+2=x_1 \quad \text { and } \quad x_2=2\left(\sqrt{6}-x_1\right)+x_3$$
are both linear because they can be rearranged algebraically as in equation (1):
$$3 x_1-5 x_2=-2 \text { and } 2 x_1+x_2-x_3=2 \sqrt{6}$$
The equations
$$4 x_1-5 x_2=x_1 x_2 \quad \text { and } \quad x_2=2 \sqrt{x_1}-6$$
are not linear because of the presence of $x_1 x_2$ in the first equation and $\sqrt{x_1}$ in the second. A system of linear equations (or a linear system) is a collection of one or more linear equations involving the same variables-say, $x_1, \ldots, x_n$. An example is
$$\begin{array}{r} 2 x_1-x_2+1.5 x_3=8 \ x_1-4 x_3=-7 \end{array}$$ A solution of the system is a list $\left(s_1, s_2, \ldots, s_n\right)$ of numbers that makes each equation a true statement when the values $s_1, \ldots, s_n$ are substituted for $x_1, \ldots, x_n$, respectively. For instance, $(5,6.5,3)$ is a solution of system ( 2 ) because, when these values are substituted in (2) for $x_1, x_2, x_3$, respectively, the equations simplify to $8=8$ and $-7=-7$.

数学代写|线性代数代写linear algebra代考|Matrix Notation

The essential information of a linear system can be recorded compactly in a rectangular array called a matrix. Given the system
\begin{aligned} x_1-2 x_2+x_3 & =0 \ 2 x_2-8 x_3 & =8 \ 5 x_1-5 x_3 & =10 \end{aligned}
with the coefficients of each variable aligned in columns, the matrix
$$\left[\begin{array}{rrr} 1 & -2 & 1 \ 0 & 2 & -8 \ 5 & 0 & -5 \end{array}\right]$$
is called the coefficient matrix (or matrix of coefficients) of the system (3), and
$$\left[\begin{array}{rrrr} 1 & -2 & 1 & 0 \ 0 & 2 & -8 & 8 \ 5 & 0 & -5 & 10 \end{array}\right]$$
is called the augmented matrix of the system. (The second row here contains a zero because the second equation could be written as $0 \cdot x_1+2 x_2-8 x_3=8$.) An augmented matrix of a system consists of the coefficient matrix with an added column containing the constants from the right sides of the equations.

The size of a matrix tells how many rows and columns it has. The augmented matrix (4) above has 3 rows and 4 columns and is called a $3 \times 4$ (read “3 by 4 “) matrix. If $m$ and $n$ are positive integers, an $\boldsymbol{m} \times \boldsymbol{n}$ matrix is a rectangular array of numbers with $m$ rows and $n$ columns. (The number of rows always comes first.) Matrix notation will simplify the calculations in the examples that follow.

线性代数代考

数学代写|线性代数代写linear algebra代考|SYSTEMS OF LINEAR EQUATIONS

$$a_1 x_1+a_2 x_2+\cdots+a_n x_n=b$$

$$4 x_1-5 x_2+2=x_1 \quad \text { and } \quad x_2=2\left(\sqrt{6}-x_1\right)+x_3$$

$$3 x_1-5 x_2=-2 \text { and } 2 x_1+x_2-x_3=2 \sqrt{6}$$

$$4 x_1-5 x_2=x_1 x_2 \quad \text { and } \quad x_2=2 \sqrt{x_1}-6$$

$$2 x_1-x_2+1.5 x_3=8 x_1-4 x_3=-7$$

数学代写|线性代数代写linear algebra代考|Matrix Notation

$$x_1-2 x_2+x_3=02 x_2-8 x_3=85 x_1-5 x_3=10$$

$$\left[\begin{array}{llllllll} 1 & -2 & 1 & 0 & 2 & -85 & 0 & -5 \end{array}\right]$$

$$\left[\begin{array}{lllllllllll} 1 & -2 & 1 & 0 & 0 & 2 & -8 & 85 & 0 & -5 & 10 \end{array}\right]$$

有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

数学代写|微积分代写Calculus代写|MAST10006

statistics-lab™ 为您的留学生涯保驾护航 在代写微积分Calculus方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写微积分Calculus代写方面经验极为丰富，各种代写微积分Calculus相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代写|微积分代写Calculus代写|Optimization example: maximum profit

Recall that for the revenue, cost, and profit functions from economics, marginal means “derivative.” Recall also that profit is revenue minus cost: $$P(x)=R(x)-C(x) .$$
If we wish to maximize profit, then we need to find the critical numbers of the profit function-that is, where $P^{\prime}(x)=0$ :
\begin{aligned} P^{\prime}(x) & =R^{\prime}(x)-C^{\prime}(x) \ R^{\prime}(x)-C^{\prime}(x) & =0 \ R^{\prime}(x) & =C^{\prime}(x) \end{aligned}
In other words, the critical numbers of the profit function occur where marginal revenue equals marginal cost. The traditional solution method for profit maximization problems is to equate marginal revenue and marginal cost. Because we also wish to check to ensure that profit is maximized rather than minimized, we still form the profit function and determine its maximum.

Example 4 Each month we can sell as many widgets as we can make for $\$ 12$each. The cost, in dollars, of making$x$widgets is given by $$C(x)=10000+7 x-0.002 x^2+\frac{1}{3} \cdot 10^{-6} x^3 .$$ How many widgets should we make to maximize profit? Solution Because we wish to maximize profit, this is an optimization problem. 1)-2 Notice that there is nothing geometric about this problem. No picture seems applicable, so we don’t draw one. Furthermore, the relevant variable has already been introduced;$x$is the number of widgets made in 1 month. 数学代写|微积分代写Calculus代写|Optimization example: minimum material Solution First we recognize this as an optimization problem because we are asked to minimize the cost. (1) We draw a picture of a utility on the bank of a straight river, with a manufacturer on the opposite side of the river but downstream. We also label the width of the river$(900 \mathrm{~m})$and the downstream distance to the manufacturer$(3000 \mathrm{~m})$. See figure 9. Visualizing different possibilities, we see the pipeline could go straight to the opposite shore to have the least amount of pipe under water (figure 10 , top). The pipeline could also go directly to the manufacturer, remaining under water the entire route, to have the least total amount of pipe (figure 10 , bottom). But we are not asked to minimize the amount of pipe under water or minimize the total length of pipe; instead, we are asked to minimize cost. It seems as if the least cost prompts us to follow a route like that in figure 9. (2) The variable amount in figure 9 appears to be the spot at which the pipe emerges from the river, which is in fact what we are asked for. Let’s let$x$represent the distance downstream from the utility at which the pipe emerges, and label this distance in the diagram; see figure 11 . We can then determine and label other lengths as well. The length of the pipe along the shore is$(3000-x) \mathrm{m}$, whereas the “vertical leg” of the right triangle is the width of the river,$900 \mathrm{~m}$. The length of the hypotenuse can be found using the Pythagorean theorem:$c^2=900^2+x^2$, or$c=\sqrt{900^2+x^2}$. These are labeled in figure 12 . (3) The quantity we are asked to optimize (minimize in this case) is the cost of the pipeline. The cost of the pipeline includes the cost of running pipe under the water and the cost of running pipe along the shoreline. Under water, the pipeline cost is$\$200 / \mathrm{m}$, and from the diagram we see that the length of pipe under the water is $\sqrt{900^2+x^2} \mathrm{~m}$. Therefore, the cost of the pipe under the water is
$$200 \sqrt{900^2+x^2}$$

Example 5 A cylindrical can must have volume $100 \mathrm{~cm}^3$. What dimensions should be used to minimize the amount of material used?

Solution We notice the phrase “minimize the amount of material used” and conclude that this is an optimization problem.
(1)-2) We are told the can is cylindrical, so we draw a cylindrical can (figure 13). We are asked for the dimensions to use, which include the can’s height and radius, so we visualize various possible shapes, such as tall and thin or short and wide (figure 14).
(3) We wish to minimize the amount of material used to make the can. The material of the can includes the top, bottom, and side of the can. Assuming a uniform thickness of the material, the material used is proportional to the surface area of the can. The formula for the surface area (SA) of a cylinder is
$$S A=2 \pi r h+2 \pi r^2 .$$

微积分代考

数学代写|微积分代写Calculus代写|Optimization example: maximum profit

$$P(x)=R(x)-C(x) .$$

$$P^{\prime}(x)=R^{\prime}(x)-C^{\prime}(x) R^{\prime}(x)-C^{\prime}(x) \quad=0 R^{\prime}(x)=C^{\prime}(x)$$

$$200 \sqrt{900^2+x^2}$$

(1)-2) 我们被告知罐子是圆柱形的，所以我们画一个圆柱形罐头 (图13)。我们被要求提供要使用的尺 寸，其中包括罐头的高度和半径，因此我们想象出各种可能的形状，例如又高又薄或又短又宽（图
14)。
（3）我们莃望尽量减少制造罐头所用的材料量。罐的材料包括罐的顶部、底部和侧面。假设材料厚度均 匀，则所用材料与罐的表面积成正比。圆柱表面积 (SA) 的公式为
$$S A=2 \pi r h+2 \pi r^2 .$$

有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

数学代写|微积分代写Calculus代写|MATH141

statistics-lab™ 为您的留学生涯保驾护航 在代写微积分Calculus方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写微积分Calculus代写方面经验极为丰富，各种代写微积分Calculus相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代写|微积分代写Calculus代写|Optimization example: maximum volume

Our next classical example involves making a box.
Example 2 An open-top box is to be made from a sheet of cardboard measuring 11 inches by 17 inches (legal size) by cutting squares from the corners and folding up the sides. What size squares should be cut from the corners to maximize the volume of the box?

Solution First we recognize this as an optimization problem because we are asked to “maximize the volume.”
(1) We draw a picture of a rectangular sheet of cardboard with its corners cut out (figure 5.) Visualizing different possibilities, we realize that cutting out small squares results in a wide, long, short box, whereas cutting out large squares results in a tall box that is less wide and long. Although we do not need to draw these diagrams if we can picture them mentally, they are given in figure 6.

(2) The fundamental quantity that seems to vary is the size of square that we cut from the corners. A square has the same length as width, so let’s call that length and width $x$. We then label the (original) diagram using the variable $x$; see figure 7 .
(3) The quantity to be optimized (in this case, maximized) is the volume of the box. The volume of a rectangular box is length times width times height:
$$V=\ell \cdot w \cdot h .$$
Consulting the diagram, we see that after the sides are folded up, the base of the box is the rectangle inside the creases. Thus, the length and width of the box are the lengths of these creases, 17 inches minus $2 x$ inches and 11 inches minus $2 x$ inches. It may be helpful to write these dimensions in the diagram as well; see figure 8 .

数学代写|微积分代写Calculus代写|Optimization example: best path

Solution First we recognize this as an optimization problem because we are asked to minimize the cost.
(1) We draw a picture of a utility on the bank of a straight river, with a manufacturer on the opposite side of the river but downstream. We also label the width of the river $(900 \mathrm{~m})$ and the downstream distance to the manufacturer $(3000 \mathrm{~m})$. See figure 9. Visualizing different possibilities, we see the pipeline could go straight to the opposite shore to have the least amount of pipe under water (figure 10 , top). The pipeline could also go directly to the manufacturer, remaining under water the entire route, to have the least total amount of pipe (figure 10 , bottom). But we are not asked to minimize the amount of pipe under water or minimize the total length of pipe; instead, we are asked to minimize cost. It seems as if the least cost prompts us to follow a route like that in figure 9.

(2) The variable amount in figure 9 appears to be the spot at which the pipe emerges from the river, which is in fact what we are asked for. Let’s let $x$ represent the distance downstream from the utility at which the pipe emerges, and label this distance in the diagram; see figure 11 . We can then determine and label other lengths as well. The length of the pipe along the shore is $(3000-x) \mathrm{m}$, whereas the “vertical leg” of the right triangle is the width of the river, $900 \mathrm{~m}$. The length of the hypotenuse can be found using the Pythagorean theorem: $c^2=$ $900^2+x^2$, or $c=\sqrt{900^2+x^2}$. These are labeled in figure 12 .
(3) The quantity we are asked to optimize (minimize in this case) is the cost of the pipeline. The cost of the pipeline includes the cost of running pipe under the water and the cost of running pipe along the shoreline. Under water, the pipeline cost is $\$ 200 / \mathrm{m}$, and from the diagram we see that the length of pipe under the water is$\sqrt{900^2+x^2} \mathrm{~m}$. Therefore, the cost of the pipe under the water is $$200 \sqrt{900^2+x^2}$$ 微积分代考 数学代写|微积分代写Calculus代写|Optimization example: maximum volume 我们的下一个经典示例涉及制作一个盒子。 示例 2 一个开顶盒将由一张 11 英寸$x 17$英寸（法定尺寸) 的纸板制成，方法是从角上切下正方形并将 边折炟起来。应该从角上切出多大的正方形才能使盒子的体积最大化? 解决方案 首先，我们将此视为优化问题，因为我们被要求“最大化音量”。 (1) 我们画了一张切掉角的长方形纸板（图 5)。可视化不同的可能性，我们意识到切出小方块会产生 宽、长、短的盒子，而切出大方块会产生在一个不太宽和不太长的高盒子里。虽然我们不需要画这些图， 如果我们可以在脑海中描绘它们，但它们在图 6 中给出。 (2) 似乎变化的基本量是我们从角上切出的正方形的大小。正方形的长度和宽度相同，所以我们称其为长 度和宽度$x$. 然后我们使用变量标记 (原始) 图表$x$；见图 7。 (3) 要优化的数量（在本例中为最大化）是盒子的体积。长方体的体积是长乘以宽乘以高： $$V=\ell \cdot w \cdot h .$$ 看图，边折起来后，盒子的底部就是折痕里面的长方形。因此，盒子的长度和宽度就是这些折痕的长度， 减去 17 英寸$2 x$英寸和 11 英寸负$2 x$英寸。将这些维度写在图表中也可能会有所帮助；见图 8。 数学代写|微积分代写Calculus代写|Optimization example: best path 解决方案 首先我们认识到这是一个优化问题，因为我们被要求最小化成本。 (1) 我们在一条笔直的河岸上画了一个公用事业公司的图片，在河的对面下游有一个制造商。我们还标注 了河流的宽度$(900 \mathrm{~m})$以及到制造商的下游距离$(3000 \mathrm{~m})$. 参见图 9。可视化不同的可能性，我们看到 管道可以直接通向对岸，从而使水下管道数量最少（图 10，顶部) 。管道也可以直接通向制造商，在整 个路线中保持在水下，以获得最少的管道总量（图 10，底部）。但我们并没有要求我们尽量减少水下管 道的数量或尽量减少管道的总长度；相反，我们被要求最小化成本。似乎最低成本促使我们遵循图 9 中 的路线。 (2) 图 9 中的变量似乎是管道从河流中露出的位置，这实际上是我们所要求的。让我们让$x$代表公用设施 下游管道出现的距离，并在图中标出该距离；见图 11。然后我们也可以确定和标记其他长度。沿岸管道 的长度为$(3000-x) \mathrm{m}$，而直角三角形的“垂直边”是河流的宽度，$900 \mathrm{~m}$. 可以使用毕达哥拉斯定理找 到斜边的长度:$c^2=900^2+x^2$，要么$c=\sqrt{900^2+x^2}$. 这些在图 12 中进行了标记。 (3) 我们被要求优化的数量 (在这种情况下最小化) 是管道的成本。管道成本包括在水下铺设管道的成本 和沿海岸线铺设管道的成本。在水下，管道成本为$\$200 / \mathrm{m}$ ，从图中我们可以看出水下管道的长度是 $\sqrt{900^2+x^2} \mathrm{~m}$. 因此，水下管道的造价为
$$200 \sqrt{900^2+x^2}$$

有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。

数学代写|微积分代写Calculus代写|MATH1051

statistics-lab™ 为您的留学生涯保驾护航 在代写微积分Calculus方面已经树立了自己的口碑, 保证靠谱, 高质且原创的统计Statistics代写服务。我们的专家在代写微积分Calculus代写方面经验极为丰富，各种代写微积分Calculus相关的作业也就用不着说。

• Statistical Inference 统计推断
• Statistical Computing 统计计算
• (Generalized) Linear Models 广义线性模型
• Statistical Machine Learning 统计机器学习
• Longitudinal Data Analysis 纵向数据分析
• Foundations of Data Science 数据科学基础

数学代写|微积分代写Calculus代写|Complete curve-sketching examples

Example 1 provided the information needed to sketch the curve. What if we need to gather this information? The complete process is lengthy but highly informative. The next example illustrates the entire process for a relatively simple function.
Example 2 Sketch the graph of $f(x)=\sqrt[3]{x^2}$.
Solution For any function, a good place to start is to determine its domain. Because $f$ contains only an odd root and no denominator, it is defined on all real numbers.

Our earlier list of information that calculus can provide gives us a list of what to explore. We start with discontinuities. Because root functions are continuous where defined, the function is continuous everywhere; there are no discontinuities.

Next on the list is corners or vertical tangents, which may be identified as part of the process of finding intervals of increase/decrease and extreme points. We defer this item momentarily.

For intervals of increase/decrease, we need to find the derivative and identify critical numbers. Differentiating, we have
$$f^{\prime}(x)=\frac{2}{3} x^{-1 / 3}=\frac{2}{3 \sqrt[3]{x}} .$$
To find critical numbers, we set both the numerator and the denominator equal to zero and solve; $2=0$ has no solutions and $3 \sqrt[3]{x}=0$ has solution $x=0$. The only critical number is $x=0$, and our chart is completed easily:
\begin{tabular}{c|c|c|c}
interval & sign of $f^{\prime}$ & inc/dec & local extrema \
\hline$(-\infty, 0)$ & $-$ & decreasing & $\leftarrow$ local min at $x=0$ \
$(0, \infty)$ & $+$ & increasing &
\end{tabular}
For the purpose of graphing, we need the extreme points and not just their locations, so we find the $y$-coordinate as well:
$$f(0)=\sqrt[3]{0^2}=0 .$$
The local minimum point is $(0,0)$. Notice that because $f^{\prime}$ is undefined at $x=0$, there is no horizontal tangent line at this local min. In other words, the local min might be at a corner in the graph.

We continue by finding intervals of concavity and inflection points. The second derivative is
$$f^{\prime \prime}(x)=-\frac{2}{9} x^{-4 / 3}=-\frac{2}{9 x^{4 / 3}}=-\frac{2}{9 \sqrt[3]{x^4}}$$

数学代写|微积分代写Calculus代写|Optimization example: maximum enclosed area

The general solution strategy for an optimization problem is to determine the quantity to be optimized, make that quantity the value of a function, and then find the extreme values of that function. Let’s examine the details of this strategy using an example.

Example 1 A farmer’s child has purchased a piglet. The farmer has given the child 60 feet of fencing left over from another project. Using the side of the barn as one side of a rectangular pig pen, the child wishes to enclose the largest area possible. What dimensions should be used?

Solution Perhaps the first step in solving an optimization problem is to recognize that it is an optimization problem. This is accomplished by noticing that the stated task involves the largest, the smallest, the greatest, the best, the maximum, the minimum, or [insert optimum word here]. In this case, largest is the word used to indicate an optimum value.

As when working any word problem, we draw a picture, if possible. The pig pen is described as a rectangle, so we draw a rectangle. We are lonking at the ground from above (a top view, or aerial view). We depict a barn along one side of the rectangle. See figure 1.

Although nothing in this example is changing (this is not a related rates exercise), so our diagram is static (diagrams for related rates exercises are dynamic), it is still helpful to visualize various possibilities. We are told that there is 60 feet of fencing to make the rectangular pig pen.

We could make the pig pen very wide but not very long (figure 2 , left), very long but not very wide (figure 2, right), or something in between.

We cannot draw all possible configurations and check their areas, for there are infinitely many possibilities. For this reason, we introduce one or more variables to help. Let’s use $\ell$ for length and $w$ for width, as in figure 3 . Then area, which is the quantity we wish to maximize, is given by
$$A=\ell \cdot w .$$

微积分代考

数学代写|微积分代写Calculus代写|Complete curve-sketching examples

$$f^{\prime}(x)=\frac{2}{3} x^{-1 / 3}=\frac{2}{3 \sqrt[3]{x}} .$$

$$f(0)=\sqrt[3]{0^2}=0$$

$$f^{\prime \prime}(x)=-\frac{2}{9} x^{-4 / 3}=-\frac{2}{9 x^{4 / 3}}=-\frac{2}{9 \sqrt[3]{x^4}}$$

数学代写|微积分代写Calculus代写|Optimization example: maximum enclosed area

$$A=\ell \cdot w$$

有限元方法代写

tatistics-lab作为专业的留学生服务机构，多年来已为美国、英国、加拿大、澳洲等留学热门地的学生提供专业的学术服务，包括但不限于Essay代写，Assignment代写，Dissertation代写，Report代写，小组作业代写，Proposal代写，Paper代写，Presentation代写，计算机作业代写，论文修改和润色，网课代做，exam代考等等。写作范围涵盖高中，本科，研究生等海外留学全阶段，辐射金融，经济学，会计学，审计学，管理学等全球99%专业科目。写作团队既有专业英语母语作者，也有海外名校硕博留学生，每位写作老师都拥有过硬的语言能力，专业的学科背景和学术写作经验。我们承诺100%原创，100%专业，100%准时，100%满意。

MATLAB代写

MATLAB 是一种用于技术计算的高性能语言。它将计算、可视化和编程集成在一个易于使用的环境中，其中问题和解决方案以熟悉的数学符号表示。典型用途包括：数学和计算算法开发建模、仿真和原型制作数据分析、探索和可视化科学和工程图形应用程序开发，包括图形用户界面构建MATLAB 是一个交互式系统，其基本数据元素是一个不需要维度的数组。这使您可以解决许多技术计算问题，尤其是那些具有矩阵和向量公式的问题，而只需用 C 或 Fortran 等标量非交互式语言编写程序所需的时间的一小部分。MATLAB 名称代表矩阵实验室。MATLAB 最初的编写目的是提供对由 LINPACK 和 EISPACK 项目开发的矩阵软件的轻松访问，这两个项目共同代表了矩阵计算软件的最新技术。MATLAB 经过多年的发展，得到了许多用户的投入。在大学环境中，它是数学、工程和科学入门和高级课程的标准教学工具。在工业领域，MATLAB 是高效研究、开发和分析的首选工具。MATLAB 具有一系列称为工具箱的特定于应用程序的解决方案。对于大多数 MATLAB 用户来说非常重要，工具箱允许您学习应用专业技术。工具箱是 MATLAB 函数（M 文件）的综合集合，可扩展 MATLAB 环境以解决特定类别的问题。可用工具箱的领域包括信号处理、控制系统、神经网络、模糊逻辑、小波、仿真等。