### 分类： 微积分代写

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

The two triangles are similar. Both triangles have right angles (at $B$ and at $D$ ), they share the angle at $E$, and by subtraction (or by the corresponding angles theorem from geometry), the angles at $A$ and $C$ are also congruent. Similar triangles are helpful because the ratios of the sides are the same in similar triangles:
$$\frac{B E}{A B}=\frac{D E}{C D} .$$
Therefore,
$$\frac{x+z}{16}=\frac{z}{5} .$$
This relationship can be simplified by cross-multiplication:
\begin{aligned} 5(x+z) &=16 z \ 5 x+5 z &=16 z \ 5 x &=11 z . \end{aligned}
(3) Next we differentiate with respect to time $t$ :
\begin{aligned} \frac{d}{d t}(5 x) &=\frac{d}{d t}(11 z) \ 5 \frac{d x}{d t} &=11 \frac{d z}{d t} . \end{aligned}
4) We know that $\frac{d x}{d t}=4 \mathrm{ft} / \mathrm{s}$ :
$$\begin{gathered} 5 \cdot 4=11 \frac{d z}{d t} \ \frac{20}{11}=\frac{d z}{d t}, \end{gathered}$$
which completes step 6 (there is no need for step 5 because the variables $x$ and $z$ are not in the equation after step (3).
(7) Rereading the question, we see that part (b) has been answered: the length of the shadow $(z)$ is increasing at a rate of $\frac{20}{11} \mathrm{ft} / \mathrm{s}$.

We still need an answer to part (a). How is the tip of the shadow moving? This rate is the rate of change of the distance $B E=x+z$. In other words, we want
$$\frac{d}{d t}(x+z)=\frac{d}{d t} x+\frac{d}{d t} z=\frac{d x}{d t}+\frac{d z}{d t}=4+\frac{20}{11}=\frac{64}{11} .$$
The rate at which the tip of the shadow is moving is $\frac{64}{11} \mathrm{ft} / \mathrm{s}$.

## Calculus_微积分_Rising water

Example $5 A$ water tank has the shape of an inverted right circular cone of altitude 12 feet and a base radius of 6 feet. Given that water is pumped into the tank at a rate of 10 gallons per minute, at what rate is the water level rising when the water is 3 feet deep?

Solution (1) begin our dynamic diagram by a drawing the objects (tank, water) and labeling the quantities that are not changing (tank radius, tank height). See figure 12 .

(b) Water is being pumped into the tank, which affects the volume of water; let’s call this variable $V$. The requested rate of change involves the water level (the distance from the bottom of the tank to the top of the water); let’s call this $y$. C We need to add an arrow for the change in water level and dabel the arrow with its rate of change. Although we do not write $V$ in the figure, we can write its rate of change at the side of the diagram. The result is in figure 13 .

(2) Next we need a geometric relationship between the variables $V$ and $y$. The volume of a cone is given by
$$V=\frac{1}{3} \pi r^2 h,$$
where $h$ is the height of the cone and $r$ is its radius. Because $V$ represents the volume of water, and the water is in the shape of a cone (see figure 13), the formula applies with the height of the cone of water being $h=y$. What about the radius? Notice that the smaller cone (the cone of water) is proportional (similar) to the larger cone (the tank). The radius of the tank is half its height ( 6 feet vs. 12 feet), so the same is true of the cone of water; the radius of the water is $r=\frac{1}{2} y$. Using these values in the formula for the volume of a cone,
$$V=\frac{1}{3} \pi\left(\frac{1}{2} y\right)^2 y=\frac{\pi}{12} y^3 .$$
(3) Differentiating both sides with respect to time $t$ gives
\begin{aligned} \frac{d}{d t}(V) &=\frac{d}{d t}\left(\frac{\pi}{12} y^3\right) \ \frac{d V}{d t} &=\frac{\pi}{12} \cdot 3 y^2 \cdot \frac{d y}{d t}=\frac{\pi}{4} y^2 \frac{d y}{d t} . \end{aligned}

## 微积分代考

$$\frac{B E}{A B}=\frac{D E}{C D} .$$

$$\frac{x+z}{16}=\frac{z}{5} .$$

$$5(x+z)=16 z 5 x+5 z \quad=16 z 5 x=11 z .$$
(3) 接下来我们对时间进行微分 $t$ :
$$\frac{d}{d t}(5 x)=\frac{d}{d t}(11 z) 5 \frac{d x}{d t}=11 \frac{d z}{d t} .$$
4) 我们知道 $\frac{d x}{d t}=4 \mathrm{ft} / \mathrm{s}$ :
$$5 \cdot 4=11 \frac{d z}{d t} \frac{20}{11}=\frac{d z}{d t},$$

(7) 重读问题，我们看到(b)部分已经回答: 阴影的长度 $(z)$ 正在以 $\frac{20}{11} \mathrm{ft} / \mathrm{s}$.

$$\frac{d}{d t}(x+z)=\frac{d}{d t} x+\frac{d}{d t} z=\frac{d x}{d t}+\frac{d z}{d t}=4+\frac{20}{11}=\frac{64}{11} .$$

## Calculus_微积分_Rising water

(b) 水被泵入水箱，影响水量；让我们称这个变量 $V$. 要求的变化率涉及水位 (从水箱底部到水顶的距离)；让我 们称之为 $y$. C我们需要为水位变化添加一个箭头，并用它的变化率标记箭头。虽然我们不写 $V$ 在图中，我们可以 将它的变化率写在图表的一侧。结果如图 13 所示。
(2) 接下来我们需要变量之间的几何关系 $V$ 和 $y$. 圆雉的体积由下式给出
$$V=\frac{1}{3} \pi r^2 h,$$

$$V=\frac{1}{3} \pi\left(\frac{1}{2} y\right)^2 y=\frac{\pi}{12} y^3 .$$
(3) 双方在时间上的微分 $t$ 给
$$\frac{d}{d t}(V)=\frac{d}{d t}\left(\frac{\pi}{12} y^3\right) \frac{d V}{d t} \quad=\frac{\pi}{12} \cdot 3 y^2 \cdot \frac{d y}{d t}=\frac{\pi}{4} y^2 \frac{d y}{d t}$$

## 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。

## Calculus_微积分_Non-Pythagorean Relationships

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

## Calculus_微积分_Melting ice

Example 2 The top of a silo has the shape of a hemisphere of diameter 20 feet. If it is coated uniformly with a layer of ice, and if the thickness of the $i c e$ is decreasing at a rate of $\frac{1}{4} \mathrm{in} / \mathrm{h}$, how fast is the volume of ice changing when the ice is 2 inches thick? See figure 3.Solution (1) We begin our dynamic diagram by a drawing the objects (top of silo, ice) and labeling the quantities that are not changing (diameter of top of silo), as in figure 4. A cross-section view is used to help visualize the ice on the outside of the structure.

(b) A quantity that is changing is the thickness of the ice. Label the thickness $k$. The quantity we are asked to find is the rate of change of the volume of the ice; call the volume $V$. C The thickness is decreasing, so we can draw arrows inward. (d) Finally, we label an arrow with the rate of change of thickness $\frac{d k}{d t}$. Because the thickness is decreasing, $\frac{d k}{d t}$ is negative. See figure 5 . Also, we need the radius rather than the diameter, and because the rate is expressed in terms of inches we convert the size of the radius, 10 feet, to inches, 120 inches.
(2) Next we need a geometric formula that relates the variables of step (1) b, $V$ and $k$. The formula for the volume of a sphere is $V=$ $\frac{4}{3} \pi r^3$, and the volume of a hemisphere is half that of the sphere:
$$V_{\text {hemisphere }}=\frac{2}{3} \pi r^3 \text {. }$$

The problem is that we need to relate the variables $V$ and $k$, not the variables $V$ and $r$. The formula gives the volume of a hemisphere, and the ice is not a hemisphere. On the other hand, the ice is the difference of two hemispheres. We can find the volume of the ice by taking the volume of the hemisphere that includes the silo top and the ice and then subtracting the volume of the silo top, which is also a hemisphere:
\begin{aligned} &V_{\text {ice }}=V_{\text {silo plus ice }}-V_{\text {silo }} \ &V_{\text {ice }}=\frac{2}{3} \pi(120+k)^3-\frac{2}{3} \pi(120)^3 . \end{aligned}
(3) Next we differentiate both sides with respect to time $t$ :
\begin{aligned} \frac{d}{d t}(V) &=\frac{d}{d t}\left(\frac{2}{3} \pi(120+k)^3-\frac{2}{3} \pi(120)^3\right) \ \frac{d V}{d t} &=\frac{2}{3} \pi \cdot 3(120+k)^2 \cdot \frac{d k}{d t}-0 . \end{aligned}

1. We know that $\frac{d k}{d t}=-\frac{1}{4} \mathrm{in} / \mathrm{h}$, and 5 we want to find $\frac{d V}{d t}$ when $k=2$ inches:
$$\frac{d V}{d t}=2 \pi(120+2)^2\left(-\frac{1}{4}\right)=-23380 \mathrm{in}^3 / \mathrm{h} .$$
Step 6 is complete.
(7) The volume of the ice is decreasing at a rate of $23380 \mathrm{in}^3 / \mathrm{h}$.

## Calculus_微积分_Expanding sphere

Example 1 Gas is being pumped into a spherical balloon at a rate of $5 \mathrm{ft}^3 / \mathrm{min}$. Find the rate at which the radius is changing when the diameter is 18 inches.

Solution As in section 2.8, we use the steps of the related rates solution method.
(1) We begin by drawing a dynamic diagram. (a) In figure 1 , we draw the objects (balloon) and label the quantities that are not changing (none): (b) Which variables are changing? Sometimes clues come from the rates that are given and requested. The rate that is given is the rate at which gas is being pumped into the balloon, $5 \mathrm{ft}^3 / \mathrm{min}$. Gas fills volume, so perhaps the volume is changing. Look also at the units on that rate: the numerator is cubic feet, which represents a volume. The denominator is minutes, which is a unit of time. Therefore, the rate is change in volume over time, so volume is a variable that is changing. Let’s label the volume $V$. The rate that is requested is the rate at which the radius is changing, so we label the radius $r$. It is not difficult to draw and label the radius of the balloon in the diagram, but drawing and labeling the volume is more challenging; we omit $V$ from the diagram.

C We place an arrow in the diagram to indicate how the radius is changing. Adding gas to the balloon causes the balloon to expand, so we draw the arrow outward, indicating an increasing radius. We can also visualize how the balloon’s volume is expanding. Id We label the arrow with its rate of change. Although we did not draw an arrow to represent the change in volume, we can write the rate of change $\frac{d V}{d t}$, along with its value, at the side of the diagram. See figure 2 .

(2) Now we need a geometric relationship between the variables of step (1) b, $V$ and $r$. There is a formula from geometry that relates the volume and radius of a sphere:
$$V=\frac{4}{3} \pi r^3 .$$
(3) Next we differentiate both sides of the equation (the geometric relationship) implicitly, with respect to time $t$ :
\begin{aligned} \frac{d}{d t}(V) &=\frac{d}{d t}\left(\frac{4}{3} \pi r^3\right) \ \frac{d V}{d t} &=\frac{4}{3} \pi \cdot 3 r^2 \cdot \frac{d r}{d t}=4 \pi r^2 \frac{d r}{d t} . \end{aligned}

1. We know $\frac{d V}{d t}=5 \mathrm{ft}^3 /$ min:
$$5=4 \pi r^2 \frac{d r}{d t} .$$

## Calculus_微积分_Melting ice

(b) 正在变化的量是冰的厚度。标注厚度 $k$. 我们被要求找到的量是冰体积的变化率; 调音量 $V . C$ 厚度在减小，所 以我们可以向内画箭头。(d) 最后，我们用厚度变化率标记一个箭头 $\frac{d k}{d t}$. 因为厚度在减少， $\frac{d k}{d t}$ 是负的。见图 5。 此外，我们需要的是半径而不是直径，因为速率是以英寸表示的，所以我们将半径的大小 (10 英尺) 转换为英 寸 (120 英寸)。
(2) 接下来我们需要一个几何公式来关联步骙 (1) b 的变量， $V$ 和 $k$. 球体的体积公式为 $V=\frac{4}{3} \pi r^3$ ，半球的体积 是球体的一半:
$$V_{\text {hemisphere }}=\frac{2}{3} \pi r^3 \text {. }$$

$$V_{\text {ice }}=V_{\text {silo plus ice }}-V_{\text {silo }} \quad V_{\text {ice }}=\frac{2}{3} \pi(120+k)^3-\frac{2}{3} \pi(120)^3 .$$
(3) 接下来我们根据时间区分双方 $t$ :
$$\frac{d}{d t}(V)=\frac{d}{d t}\left(\frac{2}{3} \pi(120+k)^3-\frac{2}{3} \pi(120)^3\right) \frac{d V}{d t} \quad=\frac{2}{3} \pi \cdot 3(120+k)^2 \cdot \frac{d k}{d t}-0 .$$

1. 我们知道 $\frac{d k}{d t}=-\frac{1}{4} \mathrm{in} / \mathrm{h}$ ，而 5 我们想找到 $\frac{d V}{d t}$ 什么时候 $k=2$ 英寸:
$$\frac{d V}{d t}=2 \pi(120+2)^2\left(-\frac{1}{4}\right)=-23380 \mathrm{in}^3 / \mathrm{h} .$$
步骙 6 完成。
(7) 冰的体积以 $23380 \mathrm{in}^3 / \mathrm{h}$.

## Calculus_微积分_Expanding sphere

(1) 我们先画一个动态图。(a) 在图 1 中，我们绘制对象 (气球) 并标记末变化的数量 (无)： (b) 哪些变量正在 变化? 有时线索来自提供和要求的费率。给出的速率是气体被䂿入气球的速率， $5 \mathrm{ft}^3 / \mathrm{min}$. 气体充满体积，所 以体积可能正在变化。还要查看该比率的单位：分子是立方英尺，代表体积。分母是分钟，是时间单位。因此， 速率是体积随时间的变化，因此体积是一个不断变化的变量。让我们标记音量 $V$. 请求的速率是半径变化的速 率，所以我们标记半径 $r$. 图中气球的半径的绘制和标注并不难，但是体积的绘制和标注比较有挑战性；我们省 略 $V$ 从图中。

C 我们在图中放了一个箭头来表示半径是如何变化的。向气球中加气会导致气球膨胀，因此我们将箭头向外画， 表示半径增加。我们还可以想象气球的体积是如何膨胀的。Id 我们用它的变化率来标记箭头。虽然我们没有画 箭头表示体积的变化，但是我们可以写出变化率 $\frac{d V}{d t}$ ，连同它的值，在图表的一侧。见图 2。
(2)现在我们需要步骤(1)b的变量之间的几何关系， $V$ 和 $r$. 有一个几何公式将球体的体积和半径联系起来:
$$V=\frac{4}{3} \pi r^3 .$$
(3) 接下来我们就时间隐含地微分方程的两边 (几何关系) $t$ :
$$\frac{d}{d t}(V)=\frac{d}{d t}\left(\frac{4}{3} \pi r^3\right) \frac{d V}{d t} \quad=\frac{4}{3} \pi \cdot 3 r^2 \cdot \frac{d r}{d t}=4 \pi r^2 \frac{d r}{d t} .$$

1. 我们知道 $\frac{d V}{d t}=5 \mathrm{ft}^3 /$ 分钟:
$$5=4 \pi r^2 \frac{d r}{d t} .$$

## 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。

## Calculus_微积分_Pythagorean Relationships

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

The other arrow in the diagram also represents change; it is the change in the variable $x$ with respect to time, $\frac{d x}{d x}$. We are told the value of that rate: the bottom of the ladder is sliding away from the building at a rate of $2 \mathrm{ft} / \mathrm{s}$. This can also be placed in the diagram next to its arrow. Our diagram, in figure 3, is finally complete. A diagram that indicates changing quantities, such as figure 3, is called a dynamic diagram. A diagram that does not indicate any changing quantity, such as in figure 4, is called a static diagram. Related rates exercises require dynamic diagrams.

Now that the diagram is drawn, how do we find the desired rate of change, $\frac{d y}{d t}$ ? The next step in the solution process is to determine a geometric relationship between the variables in the diagram. Because vertical and horizontal meet in a right angle, we recognize a right triangle in the diagram. The variables $x$ and $y$ are therefore related through the Pythagorean theorem:
$$x^2+y^2=20^2 .$$
However, this still does not tell us the relationship we really want to know, which is the relationship between the rates of change $\frac{d x}{d x}$ and $\frac{d y}{d t}$. This relationship can be found by differentiating both sides of the equation implicitly with respect to time $t$ :
\begin{aligned} \frac{d}{d t}\left(x^2+y^2\right) &=\frac{d}{d t}\left(20^2\right) \ 2 x \cdot \frac{d x}{d t}+2 y \cdot \frac{d y}{d t} &=0 . \end{aligned}
We know that $\frac{d x}{d t}=2$; thus,
$$2 x \cdot 2+2 y \cdot \frac{d y}{d t}=0 .$$
If we know values of $x$ and $y$, then we can easily solve for $\frac{d y}{d t}$. Rereading the exercise, we see that we want to know $\frac{d y}{d t}$ when the top of the ladder is 12 feet above the ground-that is, when $y=12$. Now we need only find $x$. Drawing a second diagram for the instant in which $y=12$ (figure 5), we recognize that the Pythagorean theorem can help us find $x$ at that instant in time:
\begin{aligned} x^2+12^2 &=20^2 \ x^2+144 &=400 \ x^2 &=256 \ x &=\pm 16 . \end{aligned}

## Calculus_微积分_Expanding sphere

Example 1 Gas is being pumped into a spherical balloon at a rate of $5 \mathrm{ft}^3 / \mathrm{min}$. Find the rate at which the radius is changing when the diameter is 18 inches.

Solution As in section 2.8, we use the steps of the related rates solution method.
(1) We begin by drawing a dynamic diagram. (a) In figure 1 , we draw the objects (balloon) and label the quantities that are not changing (none): (b) Which variables are changing? Sometimes clues come from the rates that are given and requested. The rate that is given is the rate at which gas is being pumped into the balloon, $5 \mathrm{ft}^3 / \mathrm{min}$. Gas fills volume, so perhaps the volume is changing. Look also at the units on that rate: the numerator is cubic feet, which represents a volume. The denominator is minutes, which is a unit of time. Therefore, the rate is change in volume over time, so volume is a variable that is changing. Let’s label the volume $V$. The rate that is requested is the rate at which the radius is changing, so we label the radius $r$. It is not difficult to draw and label the radius of the balloon in the diagram, but drawing and labeling the volume is more challenging; we omit $V$ from the diagram.

C We place an arrow in the diagram to indicate how the radius is changing. Adding gas to the balloon causes the balloon to expand, so we draw the arrow outward, indicating an increasing radius. We can also visualize how the balloon’s volume is expanding. Id We label the arrow with its rate of change. Although we did not draw an arrow to represent the change in volume, we can write the rate of change $\frac{d V}{d t}$, along with its value, at the side of the diagram. See figure 2 .

(2) Now we need a geometric relationship between the variables of step (1) b, $V$ and $r$. There is a formula from geometry that relates the volume and radius of a sphere:
$$V=\frac{4}{3} \pi r^3 .$$
(3) Next we differentiate both sides of the equation (the geometric relationship) implicitly, with respect to time $t$ :
\begin{aligned} \frac{d}{d t}(V) &=\frac{d}{d t}\left(\frac{4}{3} \pi r^3\right) \ \frac{d V}{d t} &=\frac{4}{3} \pi \cdot 3 r^2 \cdot \frac{d r}{d t}=4 \pi r^2 \frac{d r}{d t} . \end{aligned}

1. We know $\frac{d V}{d t}=5 \mathrm{ft}^3 /$ min:
$$5=4 \pi r^2 \frac{d r}{d t} .$$

## 微积分代考

$$x^2+y^2=20^2 .$$

$$\frac{d}{d t}\left(x^2+y^2\right)=\frac{d}{d t}\left(20^2\right) 2 x \cdot \frac{d x}{d t}+2 y \cdot \frac{d y}{d t}=0$$

$$2 x \cdot 2+2 y \cdot \frac{d y}{d t}=0 .$$

$$x^2+12^2=20^2 x^2+144 \quad=400 x^2=256 x \quad=\pm 16 .$$

## Calculus_微积分_Expanding sphere

(1) 我们从绘制动态图开始。(a) 在图 1 中，我们绘制了对象 (球) 并标记了没有变化的数量 (无)： (b) 哪些 变量正在变化? 有时线索来自给出和要求的费率。给出的速率是气体被䂿入气球的速率， $5 \mathrm{ft}^3 / \mathrm{min}$. 气体充满 体积，因此体积可能正在变化。还要查看该比率的单位：分子是立方英尺，代表体积。分母是分钟，是时间单 位。因此，速率是体积随时间的变化，因此体积是一个不断变化的变量。让我们标记音量 $V$. 请求的速率是半径 变化的速率，所以我们标记半径 $r$. 在图中绘制和标注气球的半径并不难，但绘制和标注体积更具挑战性; 我们 省略 $V$ 从图中。

C 我们在图中放置一个箭头来指示半径是如何变化的。给气球加气会使气球憉胀，所以我们向外画箭头，表示半 径增加。我们还可以想象气球的体积是如何膨胀的。Id 我们用它的变化率标记箭头。虽然我们没有画箭头来表 示体积的变化，但是我们可以写出变化率 $\frac{d V}{d t}$ ，连同它的值，在图表的一侧。见图 2。
(2) 现在我们需要步際 (1) b 的变量之间的几何关系， $V$ 和 $r$. 几何中有一个公式将球体的体积和半径联系起来:
$$V=\frac{4}{3} \pi r^3 .$$
(3) 接下来我们隐式微分方程的两边 (几何关系)，关于时间 $t$ :
$$\frac{d}{d t}(V)=\frac{d}{d t}\left(\frac{4}{3} \pi r^3\right) \frac{d V}{d t} \quad=\frac{4}{3} \pi \cdot 3 r^2 \cdot \frac{d r}{d t}=4 \pi r^2 \frac{d r}{d t} .$$

1. 我们知道 $\frac{d V}{d t}=5 \mathrm{ft}^3 /$ 分钟:
$$5=4 \pi r^2 \frac{d r}{d t} .$$

## 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|MATH1111

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|Power rule for negative integer exponents

The power rule was proved earlier for positive integer exponents only. We can use it for $\frac{d}{d x} x^6=6 x^5$, but we do not yet know that it works for $\frac{d}{d x} x^{-3}$. Since $x^{-3}=\frac{1}{x^3}$, the quotient rule can help:
$$\frac{d}{d x} x^{-3}=\frac{d}{d x}\left(\frac{1}{x^3}\right)=\frac{x^3 \cdot 0-1 \cdot 3 x^2}{\left(x^3\right)^2}=\frac{-3 x^2}{x^6}=-3 x^{-4}$$
This still conforms to the pattern $\frac{d}{d x} x^{-3}=-3 x^{-3-1}$. We can turn this calculation into a proof of the power rule for negative integers.
POWER RULE (VERSION 2)
Let $f(x)=x^n$ for a negative integer $n$. Then, $f^{\prime}(x)=n x^{n-1}$.
For the proof, let $n$ be a negative integer. Then $-n$ is a positive integer (if $n=-3$, then $-n=3$ ), and the power rule can be applied to $x^{-n}$. Then,
\begin{aligned} \frac{d}{d x} x^n &=\frac{d}{d x}\left(\frac{1}{x^{-n}}\right)=\frac{x^{-n} \cdot 0-1 \cdot-n x^{-n-1}}{\left(x^{-n}\right)^2} \ &=\frac{n x^{-n-1}}{x^{-2 n}}=n x^{-n-1-(-2 n)}=n x^{n-1} . \end{aligned}
Example 13 Find $f^{\prime \prime}(x)$ for $f(x)=\frac{1}{4 x}$.
Solution We could use the quotient rule to find the derivative, but we can also use the power rule if we rewrite the function using a negative exponent:

\begin{aligned} f(x) &=\frac{1}{4} x^{-1} \ f^{\prime}(x) &=\frac{1}{4}\left(-1 \cdot x^{-2}\right)=-\frac{1}{4} x^{-2}=-\frac{1}{4 x^2} \end{aligned}
The last step, expressing the derivative in terms of a fraction instead of a negative exponent, is not necessary but can be helpful for interpreting the result. On the other hand, thinking of $f^{\prime}(x)=-\frac{1}{4} x^{-2}$ is helpful for taking the second derivative using the power rule:
$$f^{\prime \prime}(x)=-\frac{1}{4}\left(-2 x^{-3}\right)=\frac{1}{2} x^{-3}=\frac{1}{2 x^3} .$$
Notice that as we take derivatives, the exponent in the denominator is getting larger, not smaller. A polynomial’s derivatives eventually reach zero, but this function’s derivatives never reach zero.

## 数学代写|微积分代写Calculus代写|Equations of tangent lines revisited

The procedure we used in section $1.8$ to find the equation of the tangent line to the curve $y=f(x)$ at $x=a$ (figure 1 ) is summarized in three steps:
(1) Find the slope $f^{\prime}(a)$.
(2) If necessary, calculate the $y$-coordinate $f(a)$ of the point of tangency.
(3) Use the point-slope form of the equation of a line with point $(a, f(a))$ and slope $f^{\prime}(a)$.

Example 1 Find the equation of the tangent line to the curve $y=7 x^3-$ $4 x^2+1$ at $x=1$.

Solution 1 To find the slope, we find the derivative using the derivative rules:
$$y^{\prime}=21 x^2-8 x .$$
Next we evaluate the derivative at $x=1$ :
$$\text { slope }=y^{\prime}(1)=21(1)^2-8(1)=13 .$$
(2) We have not been given the $y$-coordinate of the point of tangency, so it must be calculated:
$$y(1)=7(1)^3-4(1)^2+1=4 .$$

The point of tangency is $(1,4)$.
(3) We finish by using the point-slope equation of the line with point $(1,4)$ and slope $13:$
\begin{aligned} y-y_1 &=m\left(x-x_1\right) \ y-4 &=13(x-1) \ y-4 &=13 x-13 \ y &=13 x-9 \end{aligned}
The equation of the tangent line is $y=13 x-9$.
Reading Exercise 13 Find the equation of the tangent line to the curve $y=x^2+5$ at $x=3$.

## 数学代写|微积分代写Calculus代写|Power rule for negative integer exponents

$$\frac{d}{d x} x^{-3}=\frac{d}{d x}\left(\frac{1}{x^3}\right)=\frac{x^3 \cdot 0-1 \cdot 3 x^2}{\left(x^3\right)^2}=\frac{-3 x^2}{x^6}=-3 x^{-4}$$

$$\frac{d}{d x} x^n=\frac{d}{d x}\left(\frac{1}{x^{-n}}\right)=\frac{x^{-n} \cdot 0-1 \cdot-n x^{-n-1}}{\left(x^{-n}\right)^2} \quad=\frac{n x^{-n-1}}{x^{-2 n}}=n x^{-n-1-(-2 n)}=n x^{n-1} .$$

$$f(x)=\frac{1}{4} x^{-1} f^{\prime}(x) \quad=\frac{1}{4}\left(-1 \cdot x^{-2}\right)=-\frac{1}{4} x^{-2}=-\frac{1}{4 x^2}$$

$$f^{\prime \prime}(x)=-\frac{1}{4}\left(-2 x^{-3}\right)=\frac{1}{2} x^{-3}=\frac{1}{2 x^3} .$$

## 数学代写|微积分代写Calculus代写|Equations of tangent lines revisited

(1) 找到斜率 $f^{\prime}(a)$.
(2) 如有必要，计算 $y$-协调 $f(a)$ 的切点。
(3) 使用带点的直线方程的点斜形式 $(a, f(a))$ 和坡度 $f^{\prime}(a)$.

$$y^{\prime}=21 x^2-8 x .$$

$$\text { slope }=y^{\prime}(1)=21(1)^2-8(1)=13 .$$
(2) 我们没有得到 $y$-切点的坐标，因此必须计算:
$$y(1)=7(1)^3-4(1)^2+1=4$$

(3) 我们用带点的直线的点斜率方程来完成 $(1,4)$ 和坡度 13 :
$$y-y_1=m\left(x-x_1\right) y-4=13(x-1) y-4=13 x-13 y=13 x-9$$

## 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|Derivatives of polynomials

The derivative of any polynomial may be found using the derivative rules. Let’s look at the next example, which illustrates the procedure step-by-step.
Example 9 Determine $\frac{d}{d x}\left(5 x^7+x^4-8 x^2+7 x-3\right)$.
Solution First we use the sum and difference rules:
\begin{aligned} &\frac{d}{d x}\left(5 x^7+x^4-8 x^2+7 x-3\right) \ &\quad=\frac{d}{d x} 5 x^7+\frac{d}{d x} x^4-\frac{d}{d x} 8 x^2+\frac{d}{d x} 7 x-\frac{d}{d x} 3 . \end{aligned}
Next we use the constant multiple rule:
$$=5 \frac{d}{d x} x^7+\frac{d}{d x} x^4-8 \frac{d}{d x} x^2+\frac{d}{d x} 7 x-\frac{d}{d x} 3 .$$
Last we use the power, linear function, and constant rules:
$$=5 \cdot 7 x^6+4 x^3-8 \cdot 2 x+7-0=35 x^6+4 x^3-16 x+7 .$$
With practice, most of these rules can be applied rather quickly and the solution written down directly. It looks like this:
Example 10 Find $y^{\prime}$ for $y=4 x^3-2 x^2+5$.
Solution $y^{\prime}=12 x^2-4 x$.
Reading Exercise 8 Find $f^{\prime}(x)$ for $f(x)=x^2+2 x-7$
We see that the derivative of a polynomial function is also a polynomial function, which is defined everywhere. This means that polynomial functions are differentiable everywhere, and therefore their graphs are smooth and have no corners, as advertised previously.

## 数学代写|微积分代写Calculus代写|Higher-order derivatives

Consider the function $f(x)=3 x^3+7 x-5$. It has a derivative:
$$f^{\prime}(x)=3 \cdot 3 x^2+7-0=9 x^2+7 .$$
The derivative $f^{\prime}$ is a function in its own right. Therefore, it also has a derivative, $\left(f^{\prime}\right)^{\prime}$, customarily written as $f^{\prime \prime}$ :
$$f^{\prime \prime}(x)=9 \cdot 2 x+0=18 x .$$
We call $f^{\prime \prime}(x)$ the second derivative of $f$ and read it ” $f$ double-prime of $x$.” Why stop there? Since $f^{\prime \prime}$ is also a function, let’s take its derivative:
$$\left(f^{\prime \prime}\right)^{\prime}(x)=f^{\prime \prime \prime}(x)=18 .$$
This is the third derivative of $f$ and is read ” $f$ triple-prime of $x$.” For the sake of readability, beginning with the fourth derivative it is customary to write the number of the derivative in parentheses:
$$\left(f^{\prime \prime \prime}\right)^{\prime}(x)=f^{(4)}(x)=0 .$$
Because the degree of a polynomial decreases by one with each derivative taken, the derivatives of a polynomial eventually reach zero. This is not the case with other types of functions, as we shall see in section $2.4$ and chapter 5.

Leibniz notation for the second derivative is $\frac{d^2 y}{d x^2}$, which can be read “the second derivative of $y$ with respect to $x$ both times.” Third, fourth, and higher derivatives are similar.
Reading Exercise 9 Find $f^{\prime \prime}(x)$ for $f(x)=x^4$.

## 数学代写|微积分代写Calculus代写|Derivatives of polynomials

$$\frac{d}{d x}\left(5 x^7+x^4-8 x^2+7 x-3\right) \quad=\frac{d}{d x} 5 x^7+\frac{d}{d x} x^4-\frac{d}{d x} 8 x^2+\frac{d}{d x} 7 x-\frac{d}{d x} 3 .$$

$$=5 \frac{d}{d x} x^7+\frac{d}{d x} x^4-8 \frac{d}{d x} x^2+\frac{d}{d x} 7 x-\frac{d}{d x} 3 .$$

$$=5 \cdot 7 x^6+4 x^3-8 \cdot 2 x+7-0=35 x^6+4 x^3-16 x+7 .$$

## 数学代写|微积分代写Calculus代写|Higher-order derivatives

$$f^{\prime}(x)=3 \cdot 3 x^2+7-0=9 x^2+7 .$$

$$f^{\prime \prime}(x)=9 \cdot 2 x+0=18 x .$$

$$\left(f^{\prime \prime}\right)^{\prime}(x)=f^{\prime \prime \prime}(x)=18 .$$

$$\left(f^{\prime \prime \prime}\right)^{\prime}(x)=f^{(4)}(x)=0 .$$

## 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|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|Derivative of a constant function

The first rule is for the derivative of a constant function. Let’s begin with an example.
Example 1 Let $f(x)=7$. Find $f^{\prime}(x)$.
Solution We use the derivative formula (function version) as before:
$$f^{\prime}(x)=\frac{f(x+\alpha)-f(x)}{\alpha}=\frac{7-7}{\alpha}=\frac{0}{\alpha}=0 .$$
Look at the graph of $f(x)=7$ (figure 1) and this result should make sense. The graph of a constant function is a horizontal line, the derivative gives the slope of the tangent line, and the slope of a horizontal line is zero.
Figure 1 The graph of the constant function $f(x)=7$
The number 7 is not special. Any constant function (a function of the form $f(x)=c$ ) has a derivative of zero.

To prove a derivative rule, we use the definition of derivative by applying the derivative formula (function version). Here, we want the derivative of $f(x)=c$, where $c$ is a real number:
$$f^{\prime}(x)=\frac{f(x+\alpha)-f(x)}{\alpha}=\frac{c-c}{\alpha}=\frac{0}{\alpha}=0 .$$
Now that we know the derivative of any constant function is zero, we need not use the definition of derivative; we can apply the rule for the derivative of a constant function instead.
Example 2 Let $y=\frac{347.9 \pi \sqrt{14}}{\sin \frac{\pi}{9}}$. Find $y^{\prime}$.
Solution By the rule for the derivative of a constant function, $y^{\prime}=0$
Reading Exercise 4 Let $f(x)=48$. Find $f^{\prime}(x)$.

## 数学代写|微积分代写Calculus代写|Derivatives of linear functions

Example 3 For $f(x)=x$, find $f^{\prime}(x)$.
Solution Using the derivative formula (function version), we have
$$f^{\prime}(x)=\frac{f(x+\alpha)-f(x)}{\alpha}=\frac{x+\alpha-x}{\alpha}=\frac{\alpha}{\alpha}=1 .$$
This example is also easy to explain: the graph of $y=x$ is a line with slope 1 , and the derivative gives the slope. In fact, this should be true of any line $y=m x+b$ (figure 2). Because the slope of $y=m x+b$ is $m$ and the derivative gives the slope, the derivative is $m$.

Once again, a proof uses the derivative formula (function version):
\begin{aligned} f^{\prime}(x) &=\frac{f(x+\alpha)-f(x)}{\alpha}=\frac{m(x+\alpha)+b-(m x+b)}{\alpha} \ &=\frac{m x+m \alpha+b-m x-b}{\alpha}=\frac{m \alpha}{\alpha}=m . \end{aligned}
When the rule has been proved, it can be used instead of the definition.
Example 4 Find $\frac{d}{d x}(37 x-14)$.
Solution Using the rule for the derivative of a linear function, $\frac{d}{d x}(37 x-14)=37$
Reading Exercise 5 Find $y^{\prime}$ for $y=75 x+1$.

## 数学代写|微积分代写Calculus代写|Derivative of a constant function

$$f^{\prime}(x)=\frac{f(x+\alpha)-f(x)}{\alpha}=\frac{7-7}{\alpha}=\frac{0}{\alpha}=0 .$$

$$f^{\prime}(x)=\frac{f(x+\alpha)-f(x)}{\alpha}=\frac{c-c}{\alpha}=\frac{0}{\alpha}=0 .$$

## 数学代写|微积分代写Calculus代写|Derivatives of linear functions

$$f^{\prime}(x)=\frac{f(x+\alpha)-f(x)}{\alpha}=\frac{x+\alpha-x}{\alpha}=\frac{\alpha}{\alpha}=1 .$$

$$f^{\prime}(x)=\frac{f(x+\alpha)-f(x)}{\alpha}=\frac{m(x+\alpha)+b-(m x+b)}{\alpha} \quad=\frac{m x+m \alpha+b-m x-b}{\alpha}=\frac{m \alpha}{\alpha}$$

## 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|Oscillatory discontinuities

There are other types of discontinuity besides those already mentioned. One such type is called an oscillatory discontinuity. Consider the function $f(x)=\sin \frac{1}{x}$, which is undefined at $x=0$ and therefore not continuous at $x=0$. The function is graphed in figure 13 .

To see what is going on with this function, recall that $y=\sin x$ between $x=2 \pi$ and $x=4 \pi$ goes through one complete cycle of the graph (one period of the function). But, since $\frac{1}{1 /(2 \pi)}=2 \pi$ and $\frac{1}{1 /(4 \pi)}=4 \pi$, the function $f$ goes through one complete cycle of the $\sin$ function between $1 /(2 \pi)$ and $1 /(4 \pi)$, as seen in figure 14 . It goes through another cycle between $1 /(4 \pi)$ and $1 /(6 \pi)$, another between $1 /(6 \pi)$ and $1 /(8 \pi)$, and so on. This continues ad infinitum, and therefore the function oscillates infinitely many times as we near $x=0$. It is for this reason that we call this discontinuity an oscillatory discontinuity.
This type of discontinuity can be detected algebraically using the same procedures as demonstrated in examples 3-5.

Example 7 Is $f(x)=\sin \frac{1}{x}$ continuous at $x=0$ ? If not, what type of discontinuity does it have?

Solution To determine whether the function is continuous at $x=0$, we check to see if $\lim _{x \rightarrow 0} f(x)=f(0)$. We start with $f(0)$ :
$$f(0)=\sin \frac{1}{0},$$
which is undefined because of division by zero. The function $f$ is not continuous at $x=0$.

## 数学代写|微积分代写Calculus代写|Linear functions are continuous

A linear function is a function with a graph that is a line. Such functions are of the form $f(x)=m x+b$. There are no breaks in the graph of a line, so linear functions should be continuous everywhere.

Theorem 6 CONTINUITY OF LINEAR FUNCTIONS If $f(x)=$ $m x+b$ for real numbers $m$ and $b$, then $f$ is contimuous at every real number $k$.

Proof. We proceed as before, but use generic constants $m, b$, and $k$ in place of specific real numbers.

To determine whether $f$ is continuous at $x=k$, we check to see if $\lim {x \rightarrow k} f(x)=f(k)$. We start by determining $f(k)$ : $$f(k)=m k+b .$$ Next we check the limit: $$\lim {x \rightarrow k}(m x+b)=m(k+\alpha)+b=m k+m \alpha+b \approx m k+b .$$
Because $\lim _{x \rightarrow k} f(x)=f(k), f$ is continuous at $x=k$.
Now transport yourself back to algebra and graph $f(x)=2 x-1$ by plotting points. We use the following table of values: and plot the points (figure 1):and because we now know that every linear function is continuous, we know that we can safely connect the dots (figure 2). There are no holes, jumps, vertical asymptotes, oscillations, or any other discontinuities in the graph of the function!

There are, of course, some functions that have discontinuities. But, a quick perusal of examples 3-8 in section 1.6 reveals that whenever we had an algebraic formula for a function that was not piecewisedefined, the discontinuities of the function occurred only where the function was undefined. If we know that for all other values of $x$ that the function is continuous, we can play the connect-the-dot game on each piece. This helps motivate the following definition:

Definition 10 CONTINUOUS FUNCTION A function $f$ is continuous if it is continuous at $x=k$ for every real number $k$ in its domain.

## 数学代写|微积分代写Calculus代写|Oscillatory discontinuities

$$f(0)=\sin \frac{1}{0},$$

## 数学代写|微积分代写Calculus代写|Linear functions are continuous

$$f(k)=m k+b .$$

$$\lim x \rightarrow k(m x+b)=m(k+\alpha)+b=m k+m \alpha+b \approx m k+b$$

## 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|MATH1111

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|Alternate definition of continuity

The definition of continuity can be rephrased to give additional insight into the meaning of continuity. We know that $\lim _{x \rightarrow b} f(x)=L$ means that $f(b+\alpha)$ renders $L$ for every infinitesimal $\alpha$. The definition of continuity simply puts $f(b)$ in the place of $L$

Definition 9 CONTINUTTY (VERSION 2) Let $f$ be a function and let $b$ be a real number in the domain of $f$. Then, $f$ is continuous at $x=b$ if $f(b+\alpha)$ renders the real result $f(b)$ for every infinitesimal $\alpha$.
This means that a small change in the value of $x$ (from $x=b$ to $x=b+\alpha)$ produces only a small change in the value of $y=f(x)$ (because $f(b+\alpha)$ renders $f(b)$ and therefore must differ from $f(b)$ by, at most, an infinitesimal amount). In other words, small changes in the $x$-coordinate on the graph produce small changes in the $y$ coordinate on the graph. This is what keeps the points close together and prevents the graph from jumping or heading off to infinity. Small changes in $x$ producing small changes in $y$, rather than the ability to draw a graph in one piece, is the traditional intuitive concept of continuity.

## 数学代写|微积分代写Calculus代写|Identifying discontinuities graphically

Certain types of discontinuities in a function may be recognized easily from a graph. If we see what we have called a hole in the graph, then the function has a removable discontinuity; if we see a vertical asymptote, then the function has an infinite discontinuity; and if we see a jump in the graph, the function has a jump discontinuity. The locations of these discontinuities should be recorded using their $x$-coordinates, because the use of $y$-coordinates does not make sense for some types of discontinuities.

Example 2 Identify by type the discontinuities in the function $f$ graphed in figure 8.

Solution We recognize two “holes” in the graph, which represent removable discontinuities. The locations of these removable discontinuities are $x=-2$ and $x=5$ (figure 9).

Next we recognize a jump in the graph, which represents a jump discontinuity, at $x=1$ (figure 10 ).

Last, we recognize a vertical asymptote in the graph at $x=3$, which represents an infinite discontinuity (figure 11).

The final answer can be written as follows:
removable discontinuities at $x=-2$ and $x=5$
jump discontinuity at $x=1$
infinite discontinuity at $x=3$
Reading Exercise 24 Identify by type the discontinuities in the function fgraphed in figure 12.

## 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|Numerical estimation of limits

Suppose we wish to find $\lim {x \rightarrow 0} \frac{1-\cos x}{x}$. This is a two-sided limit, so we use $x=0+\alpha=\alpha$ : $$\lim {x \rightarrow 0} \frac{1-\cos x}{x}=\frac{1-\cos \alpha}{\alpha}=?$$
Although we will eventually learn the value of $\cos \alpha$, at this point we have not studied it. And even if we had, there will always be other functions for which an algebraic approach to finding limits hits some roadblock. It is not possible to determine algebraically the exact value of every limit we might want.

We have also studied the graphical approach. A graph of $f(x)=$ $\frac{1-\cos x}{x}$ is presented in figure 18 . Because the point $(0,0)$ appears to be on the graph (even though it can’t be there because of division by zero), it appears that the value of the limit is 0 , but we have no assurance that it is exactly 0 . We can zoom in repeatedly to smaller scales and look again, but at no point can we be exactly sure; we just have a very strong suspicion that the limit is 0.

When generating the graph, a calculator or computer must calculate a large number of values of the function. An alternative is to calculate a few of these values ourselves. Then, instead of checking the graph to see which $y$-coordinate the curve seems to be approaching, we can check the numbers for the same thing. Because numbers exhibit greater precision much more easily than points in a picture, we have greater assurance that the limit has been calculated relatively precisely. We call this process estimating a limit numerically.
Example 7 Estimate $\lim _{x \rightarrow 0} \frac{1-\cos x}{x}$ numerically.
Solution Our general strategy is to estimate the limit from the right and from the left by choosing values closer and closer to the target value of $x$. Choosing values that are $0.1,0.01,0.001$, and so on, to the right and to the left of the target value is generally effective.

First we check the limit from the right by checking values larger than the target value of 0 :

## 数学代写|微积分代写Calculus代写|Squeeze theorem

The squeeze theorem, sometimes called the sandwich theorem, is illustrated in figure 21. Two functions, $f$ and $h$, have the same limit $L$ as $x \rightarrow b$. Another function, $g$, is “squeezed” between $f$ and $h$. Under these constraints, it appears that $g$ has no choice but to take the same limit at $x=b$.

Theorem 5 SQUEEZE THEOREM If $f(x) \leq g(x) \leq h(x)$ for all val ues of $x$ in an open interval containing $b$, except possibly at $b$, and $\lim {x \rightarrow b} f(x)=L=\lim {x \rightarrow b} h(x)$, then $\lim _{x \rightarrow b} g(x)=L$ also.

Proof. Let’s prove the case that $L$ is a real number. Let $\alpha$ be an infinitesimal. Since $\lim _{x \rightarrow b} f(x)=L_3 f(b+\alpha)$ must render $L$. Therefore, $f(b+\alpha)=L+\gamma_1$ for some infinitesimal $\gamma_1$. Similarly, $h(b+\alpha)=L+\gamma_2$ for some infinitesimal $\gamma_2$. Since $f(x) \leq g(x) \leq h(x)$ on an open interval around $x=b$, then for any infinitesimal $\alpha$,
$$f(b+\alpha) \leq g(b+\alpha) \leq h(b+\alpha)$$
and thus
$$L+\gamma_1 \leq g(b+\alpha) \leq L+\gamma_2$$
and
$$\gamma_1 \leq g(b+\alpha)-L \leq \gamma_2 .$$

Then, $g(b+\alpha)-L$ must be an infinitesimal because it is between two infinitesimals, so that $g(b+\alpha)=L+$ infinitesimal $\doteq L$ Since $g(b+\alpha)$ renders $L$ for any infinitesimal $\alpha$, then $\lim {x \rightarrow b} g(x)=L$ The cases $L=\infty$ and $L=-\infty$ are similar. Example 10 Suppose $2 x \leq g(x) \leq x^2+1$ for all $x$. Find $\lim {x \rightarrow 1} g(x)$.
Solution We do not have a formula for $g$, so we cannot find its limit directly or estimate it numerically. But, it appears to be set up well for the squeeze theorem. To apply a theorem, all the hypotheses of the theorem must be met. In this case, we need to know that the functions on either side of $g$ have the same limit:
$$\lim {x \rightarrow 1} 2 x=2(1+\alpha)=2+2 \alpha \approx 2$$ and $$\lim {x \rightarrow 1}\left(x^2+1\right)=(1+\alpha)^2+1=1+2 \alpha+\alpha^2+1=2+2 \alpha+\alpha^2 \approx 2$$
Because the functions on either side of $g$ have the same limit, then $g$ is squeezed to that same limit and we conclude $\lim _{x \rightarrow 1} g(x)=2$.

## 数学代写|微积分代写Calculus代写|Numerical estimation of limits

$$\lim x \rightarrow 0 \frac{1-\cos x}{x}=\frac{1-\cos \alpha}{\alpha}=?$$

## 数学代写|微积分代写Calculus代写|Squeeze theorem

$$f(b+\alpha) \leq g(b+\alpha) \leq h(b+\alpha)$$

$$L+\gamma_1 \leq g(b+\alpha) \leq L+\gamma_2$$

$$\gamma_1 \leq g(b+\alpha)-L \leq \gamma_2 .$$
$g(b+\alpha)$ 渲染 $L$ 对于任何无穷小 $\alpha$ ，然后 $\lim x \rightarrow b g(x)=L$ 案例 $L=\infty$ 和 $L=-\infty$ 是相似的。示例 10 假设 $2 x \leq g(x) \leq x^2+1$ 对所有人 $x$. 寻找 $\lim x \rightarrow 1 g(x)$.

$$\lim x \rightarrow 12 x=2(1+\alpha)=2+2 \alpha \approx 2$$

$$\lim x \rightarrow 1\left(x^2+1\right)=(1+\alpha)^2+1=1+2 \alpha+\alpha^2+1=2+2 \alpha+\alpha^2 \approx 2$$

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## 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|MATH1111

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• 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|Infinite numbers

You may recall that reciprocals of small numbers are large numbers. For instance,
$$\frac{1}{0.1}=10, \quad \frac{1}{0.01}=100 \text {, and } \frac{1}{0.0000001}=1000000 \text {. }$$

The smaller the denominator, the larger the resulting number. Then what about
$\frac{1}{\omega}$ ?
The denominator is infinitely small, so the result must be infinitely large! This can be proved in the following manner. If we can show that $\frac{1}{\omega}$ is larger than any positive real number $r$, then it must be infinite. To this end, let $r$ be any positive real number. Then, $\frac{1}{r}$ is also a positive real number, and because $\omega$ is infinitesimal, we know that $\omega<\frac{1}{r}$. Taking reciprocals of numbers reverses the direction of the inequality (such as $\frac{1}{2}>\frac{1}{3}$ ); hence,
$$\frac{1}{\omega}>r,$$
as desired. The opposite is also true; the reciprocal of an infinite number is infinitesimal.

Because we will work with these infinite numbers often, it is convenient to write $\Omega$ in place of $\frac{1}{\omega}$. We do so throughout this text, and follow the convention that infinitesimals are represented by lowercase Greek letters whereas infinite numbers are represented by uppercase Greek letters-“little” letters for little numbers, “big” letters for big numbers.

Arithmetic with the infinite number $\Omega$ works just like arithmetic with $\omega$.
Example 3 Simplify $\Omega+4-(5 \Omega)$
Solution We treat $\Omega$ like any other algebraic quantity and collect like terms:
$$\Omega+4-5 \Omega=4-4 \Omega$$
When working with both $\Omega$ and $\omega$ it is helpful to remember the reciprocal relationships.

## 数学代写|微积分代写Calculus代写|Transfer principle

We asserted earlier that algebra in the hyperreal numbers works the same as it does for real numbers. This is part of what is called the transfer principle.
TRANSFER PRINCIPLE (CALCULUS VERSION)
Algebraic formulas are true in the real numbers if and only if they are true in the hyperreal numbers.

The full version of the transfer principle is more complicated, and its technical details are beyond calculus. As long as a statement can be written using only certain types of symbols and quantifiers, then it is true in the reals if and only if it is true in the hyperreals. Although there are types of statements that cannot be written in a form for use with the transfer principle, these types of statements are easily avoided in calculus. Furthermore, all the algebraic statements transfer, so they can be used worry-free.

Applications of the transfer principle go as follows. First we start with an algebraic statement that we know is true for the real numbers:
$-1 \leq \sin x \leq 1$ for every real number $x$.

Notice that the statement, as written, is true for every real number. Then, by the transfer principle, the statement is also true for every hyperreal number:
$-1 \leq \sin x \leq 1$ for every hyperreal number $x$.
Therefore, we can conclude, for instance, that
$$-1 \leq \sin \left(47 \Omega-500+2 \omega^2\right) \leq 1 .$$
We can put any hyperreal number we want inside $\sin$ and the result is still between $-1$ and 1 .

The transfer principle is almost like a magic wand that we can wave over algebraic statements and say, “Be true in the hyperreals!” It is perhaps the most powerful tool used by nonstandard analysts, mathematicians who study the hyperreal numbers and their applications to calculus and beyond.
Reading Exercise 5 True or false: $-1 \leq \cos \omega \leq 1$.
The transfer principle is part of a larger theorem known as \&os’ theorem, named after its discoverer Jerzy Loś, a Polish mathematician. This discovery led to the work of Abraham Robinson, who gave the first rigorous proof, in the $1960 \mathrm{~s}$, of the existence of a number system that includes infinitesimals and can be used to develop calculus.

The work of Robinson is beyond the scope of a course in calculus. Fortunately, everything we need for calculus can be developed from the assumptions of this section: that one infinitesimal exists and that the transfer principle applies.

## 数学代写|微积分代写Calculus代写|Infinite numbers

$$\frac{1}{0.1}=10, \quad \frac{1}{0.01}=100 \text {, and } \frac{1}{0.0000001}=1000000 \text {. }$$

$\frac{1}{\omega} ?$

$$\frac{1}{\omega}>r,$$

$$\Omega+4-5 \Omega=4-4 \Omega$$

## 数学代写|微积分代写Calculus代写|Transfer principle

$$-1 \leq \sin \left(47 \Omega-500+2 \omega^2\right) \leq 1 .$$

## 有限元方法代写

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## 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。