# 微积分代写|CALCULUS II MATH102 University of Liverpool Assignment

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## Instructions:

Part I of the module focuses on power series and their applications. Power series are infinite series that can represent functions as a sum of powers of a variable. You will learn about the properties of power series and their convergence, as well as the relationship between power series and functions, including how to use Taylor and Maclaurin series to approximate functions.

Part II of the module introduces functions of several variables and partial differentiation. You will learn how to find partial derivatives, including the chain rule, total differential, directional derivative, and tangent planes. Additionally, you will learn how to find extrema of functions of several variables and how to use Taylor expansions to approximate functions.

Finally, part III of the module focuses on double integrals and their applications. You will learn how to evaluate double integrals using different methods, including iterated integrals and change of variables. You will also learn how to use double integrals to find areas, volumes, and centroids of two-dimensional and three-dimensional objects.

Overall, this module, together with MATH101 and MATH103, provides a strong foundation in calculus and linear algebra, which is essential for further studies in mathematics and many other fields, such as physics, engineering, and economics.

$\vec{R} \cdot \vec{V}=0$, so $\frac{d}{d t}(\vec{R} \cdot \vec{V})=\vec{V} \cdot \vec{V}+\vec{R} \cdot \vec{A}=0$. Therefore $\vec{R} \cdot \vec{A}=-|\vec{V}|^2$.

We can start by differentiating the expression $\vec{R} \cdot \vec{V}$ with respect to time $t$ using the product rule:

$$\frac{d}{dt}(\vec{R} \cdot \vec{V}) = \frac{d \vec{R}}{dt} \cdot \vec{V} + \vec{R} \cdot \frac{d \vec{V}}{dt}$$

Since $\vec{R} \cdot \vec{V}$ is a scalar quantity, its derivative must be a scalar as well. We can simplify the above expression using the fact that $\vec{R}$ and $\vec{V}$ are perpendicular (i.e. $\vec{R} \cdot \vec{V} = 0$), which implies that $\frac{d \vec{R}}{dt} \cdot \vec{V} = 0$. Therefore:

$$\frac{d}{dt}(\vec{R} \cdot \vec{V}) = \vec{R} \cdot \frac{d \vec{V}}{dt}$$

Now, we can substitute the expression for the acceleration $\vec{A} = \frac{d \vec{V}}{dt}$:

$$\frac{d}{dt}(\vec{R} \cdot \vec{V}) = \vec{R} \cdot \vec{A}$$

Using the chain rule, we can rewrite the left-hand side as:

$$\frac{d}{dt}(\vec{R} \cdot \vec{V}) = \frac{d\vec{R}}{dt} \cdot \vec{V} + \vec{R} \cdot \frac{d\vec{V}}{dt} = \vec{V} \cdot \vec{V} + \vec{R} \cdot \vec{A}$$

Substituting this into the previous expression, we get:

$$\vec{V} \cdot \vec{V} + \vec{R} \cdot \vec{A} = 0$$

Solving for $\vec{R} \cdot \vec{A}$, we get:

$$\vec{R} \cdot \vec{A} = -|\vec{V}|^2$$

Therefore, we have shown that $\vec{R} \cdot \vec{A} = -|\vec{V}|^2$ given the initial condition $\vec{R} \cdot \vec{V} = 0$.

a) $P=(1,0,0), Q=(0,2,0)$ and $R=(0,0,3)$. Therefore $\overrightarrow{Q P}=\hat{\boldsymbol{\imath}}-2 \hat{\boldsymbol{\jmath}}$ and $\overrightarrow{Q R}=-2 \hat{\boldsymbol{\jmath}}+3 \hat{\boldsymbol{k}}$.

To calculate the vector $\overrightarrow{QP}$, we subtract the coordinates of $P$ from the coordinates of $Q$:

$$\overrightarrow{QP} = \begin{pmatrix}0-1 \ 2-0 \ 0-0\end{pmatrix} = \begin{pmatrix}-1 \ 2 \ 0\end{pmatrix} = -1\hat{\boldsymbol{\imath}} + 2\hat{\boldsymbol{\jmath}}$$

To calculate the vector $\overrightarrow{QR}$, we subtract the coordinates of $R$ from the coordinates of $Q$:

$$\overrightarrow{QR} = \begin{pmatrix}0-0 \ 2-0 \ 3-0\end{pmatrix} = \begin{pmatrix}0 \ 2 \ 3\end{pmatrix} = 2\hat{\boldsymbol{\jmath}} + 3\hat{\boldsymbol{k}}$$

So, as you said, $\overrightarrow{QP} = \hat{\boldsymbol{\imath}} – 2 \hat{\boldsymbol{\jmath}}$ and $\overrightarrow{QR} = -2\hat{\boldsymbol{\jmath}} + 3\hat{\boldsymbol{k}}$.

$\vec{N} \cdot \vec{r}(t)=6$, where $\vec{N}=\langle 4,-3,-2\rangle$.

The equation $\vec{N} \cdot \vec{r}(t)=6$ represents a plane in three-dimensional space, where $\vec{N}$ is the normal vector to the plane and $\vec{r}(t)$ is a vector pointing to any point on the plane at time $t$.

Given that $\vec{N}=\langle 4,-3,-2\rangle$, we can write the equation of the plane as:

$$4x – 3y – 2z = 6$$

where $x$, $y$, and $z$ are the coordinates of any point on the plane.

Alternatively, we can express the equation in vector form as:

$$\langle 4,-3,-2\rangle \cdot \langle x,y,z\rangle = 6$$

or

$$\langle 4,-3,-2\rangle \cdot \vec{r}(t) = 6$$

where $\vec{r}(t)=\langle x(t),y(t),z(t)\rangle$ is a vector function that describes the position of any point on the plane at time $t$.

Note that there are infinitely many points on the plane described by this equation, since the equation has three variables but only one constraint. Any point that satisfies the equation $4x – 3y – 2z = 6$ lies on the plane, and there are infinitely many such points.

# 微积分代写|CALCULUS I MATH101 University of Liverpool Assignment

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## Instructions:

To add more details, calculus is a branch of mathematics that deals with the study of rates of change and accumulation. It is divided into two main branches: differential calculus and integral calculus.

Differential calculus deals with the study of rates of change, such as slopes of tangent lines, velocity, and acceleration. It involves concepts such as derivatives, limits, and continuity. Derivatives are used to calculate the instantaneous rate of change of a function at a specific point.

Integral calculus deals with the study of accumulation, such as finding the area under a curve or the volume of a solid. It involves concepts such as integrals, limits, and the fundamental theorem of calculus. Integrals are used to calculate the total amount of a quantity that accumulates over an interval of time or a region of space.

Calculus has numerous practical applications in various fields such as physics, engineering, economics, and biology. It provides a powerful tool for solving real-world problems by modeling them mathematically and then applying calculus techniques to obtain solutions.

$\mathrm{Q}=$ top of the ladder: $\overrightarrow{\mathrm{OQ}}=\langle 0, L \sin \theta\rangle ; \quad \mathrm{R}=$ bottom of the ladder: $\overrightarrow{\mathrm{OR}}=\langle-L \cos \theta, 0\rangle$.
Midpoint: $\overrightarrow{\mathrm{OP}}=\frac{1}{2}(\overrightarrow{\mathrm{OQ}}+\overrightarrow{\mathrm{OR}})=\left\langle-\frac{L}{2} \cos \theta, \frac{L}{2} \sin \theta\right\rangle$.
Parametric equations: $x=-\frac{L}{2} \cos \theta, y=\frac{L}{2} \sin \theta$.

It looks like you have provided the vector representations and parametric equations for three points in a coordinate plane: point Q at the top of a ladder, point R at the bottom of the ladder, and the midpoint P between points Q and R.

The vector representation of point Q is $\overrightarrow{\mathrm{OQ}}=\langle 0, L \sin \theta\rangle$, which means that point Q is located at the origin of the coordinate plane (since the x-coordinate is 0) and has a y-coordinate of $L \sin \theta$.

The vector representation of point R is $\overrightarrow{\mathrm{OR}}=\langle-L \cos \theta, 0\rangle$, which means that point R is located on the x-axis (since the y-coordinate is 0) and has an x-coordinate of $-L \cos \theta$.

The vector representation of point P is found by taking the average of the vectors for points Q and R: $\overrightarrow{\mathrm{OP}}=\frac{1}{2}(\overrightarrow{\mathrm{OQ}}+\overrightarrow{\mathrm{OR}})=\left\langle-\frac{L}{2} \cos \theta, \frac{L}{2} \sin \theta\right\rangle$. This means that point P is located at $x=-\frac{L}{2} \cos \theta$ and $y=\frac{L}{2} \sin \theta$.

The parametric equations for these points are simply the x and y coordinates expressed as functions of the parameter $\theta$: $x=-\frac{L}{2} \cos \theta, y=\frac{L}{2} \sin \theta$. These equations allow you to plot the points on the coordinate plane as $\theta$ varies.

a) $\frac{d}{d t}(\vec{R} \cdot \vec{R})=\vec{V} \cdot \vec{R}+\vec{R} \cdot \vec{V}=2 \vec{R} \cdot \vec{V}$.

We can use the product rule of differentiation and the fact that the derivative of a constant vector is zero to find the derivative of $\vec{R} \cdot \vec{R}$ with respect to time:

$$\frac{d}{dt}(\vec{R} \cdot \vec{R}) = \frac{d}{dt}(\sum_{i=1}^{n} R_i^2) = \sum_{i=1}^{n} \frac{d}{dt}(R_i^2)$$

Using the chain rule, we have:

$$\frac{d}{dt}(R_i^2) = 2R_i \frac{dR_i}{dt}$$

Therefore:

$$\frac{d}{dt}(\vec{R} \cdot \vec{R}) = \sum_{i=1}^{n} 2R_i \frac{dR_i}{dt} = 2\sum_{i=1}^{n} R_i \frac{dR_i}{dt}$$

Using the product rule of differentiation, we have:

$$\frac{d}{dt}(\vec{R} \cdot \vec{R}) = 2\sum_{i=1}^{n} R_i \frac{dR_i}{dt} = 2\sum_{i=1}^{n} R_i \frac{dR_i}{dt} + 2\sum_{i=1}^{n} R_i \frac{dR_i}{dt} = 2\sum_{i=1}^{n} R_i \frac{dR_i}{dt} + 2\sum_{i=1}^{n} \frac{dR_i}{dt} R_i$$

Using the dot product rule, we have:

$$\frac{d}{dt}(\vec{R} \cdot \vec{R}) = 2\sum_{i=1}^{n} R_i \frac{dR_i}{dt} + 2\sum_{i=1}^{n} \frac{dR_i}{dt} R_i = 2\vec{R} \cdot \vec{V}$$

Therefore, we have:

$$\frac{d}{d t}(\vec{R} \cdot \vec{R})=\vec{V} \cdot \vec{R}+\vec{R} \cdot \vec{V}=2 \vec{R} \cdot \vec{V}.$$

b) Assume $|\vec{R}|$ is constant: then $\frac{d}{d t}(\vec{R} \cdot \vec{R})=2 \vec{R} \cdot \vec{V}=0$, i.e. $\vec{R} \perp \vec{V}$.

Starting with $\vec{R}\cdot\vec{R}=|\vec{R}|^2$, we can take the derivative with respect to time $t$ using the chain rule:

$$\frac{d}{d t}(\vec{R} \cdot \vec{R})=\frac{d}{d t}(|\vec{R}|^2)=2|\vec{R}|\frac{d|\vec{R}|}{dt}.$$

Since we are assuming $|\vec{R}|$ is constant, we have $\frac{d|\vec{R}|}{dt}=0$, so

$$\frac{d}{d t}(\vec{R} \cdot \vec{R})=2|\vec{R}|\frac{d|\vec{R}|}{dt}=0.$$

Therefore, $\vec{R}\cdot\vec{V}=0$ because the dot product of any two vectors is zero if and only if the vectors are perpendicular. Thus, we have $\vec{R}\perp\vec{V}$ as required.

# 微积分代写Calculus|MATH1702 University of Plymouth Assignment

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## Instructions:

Calculus is a branch of mathematics that deals with the study of rates of change and accumulation. It is widely used in various fields such as science, finance, engineering, and industry to model and solve problems that involve change and motion. The two main branches of calculus are differential calculus and integral calculus.

Differential calculus involves the study of rates of change and slopes of curves. It is used to find the instantaneous rate of change of a function, which is the rate of change at a particular point. Differential calculus is also used to find maximum and minimum values of functions, and to determine the concavity and inflection points of a curve.

Integral calculus, on the other hand, is used to find the accumulation of quantities and the area under curves. It is used to find the total change over a period of time, or the total amount of a substance produced in a chemical reaction. Integral calculus is also used to find the volume of irregularly shaped objects, and to calculate probabilities in statistics.

Partial differentiation and multiple integration are extensions of calculus to higher dimensions. Partial differentiation is used to find the rate of change of a function with respect to one of its variables, while holding all other variables constant. It is used extensively in physics and engineering to model and solve problems involving multiple variables. Multiple integration is used to find the volume, surface area, and center of mass of objects in higher dimensions. It is also used in probability theory to calculate the probability of events in higher dimensions.

In summary, calculus is a fundamental tool in mathematical modelling, and is used extensively in various fields. Students who study calculus will gain the ability to accurately and efficiently calculate rates of change and accumulation, and will be able to apply these methods to solve real-world problems. The methods covered in this module will be essential for students throughout their degree, and will prepare them for advanced topics in mathematics and other disciplines.

Find the number $b$ for which $x^b=b^x$ has only one solution (at $x=b$ ).

Let’s first consider the case where $b=1$. In this case, we have $x^b=b^x=1^x=1$ for all $x$. Therefore, there is only one solution, namely $x=1$.

Now, let’s consider the case where $b\neq 1$. If we take the natural logarithm of both sides of $x^b=b^x$, we get:

$b \ln (x)=x \ln (b)$

If we define $f(x)=\frac{x\ln(b)}{b}$, then the equation above can be written as $f(x)=\ln(x)$. Note that $f(x)$ is an increasing function for $x>0$ if $b>1$, and a decreasing function for $0<x<1$ if $0<b<1$.

Therefore, for $b>1$, the graph of $f(x)$ intersects the graph of $\ln(x)$ at most twice, once in the interval $(0,1)$ and once in the interval $(1,\infty)$. Since $f(1)=\ln(b)$, the equation $f(x)=\ln(x)$ has a unique solution in $(1,\infty)$ if and only if $\ln(b)>0$, which is equivalent to $b>1$.

For $0<b<1$, the graph of $f(x)$ intersects the graph of $\ln(x)$ at most twice, once in the interval $(0,1)$ and once in the interval $(1,\infty)$. Since $f(1)=\ln(b)$, the equation $f(x)=\ln(x)$ has a unique solution in $(0,1)$ if and only if $\ln(b)<0$, which is equivalent to $0<b<1$.

Therefore, the equation $x^b=b^x$ has only one solution if $b=\boxed{1}$ or $0<b<1$.

Graph $y(x)=e^x-x^e$. Locate its minimum.

To find the minimum of the function $y(x)=e^x-x^e$, we need to find its critical points. The critical points occur where the derivative of the function is zero or undefined.

Taking the derivative of $y(x)$ with respect to $x$, we get:

$$y'(x) = e^x – ex^{e-1}$$

To find the critical points, we set $y'(x) = 0$ and solve for $x$:

$$e^x – ex^{e-1} = 0$$

Dividing both sides by $e^x$ gives:

$$1 – \frac{x}{e} = 0$$

Therefore, the critical point is at $x=e$. To confirm that this is a minimum, we need to check the second derivative of $y(x)$:

$$y”(x) = e^x – e(e-1)x^{e-2}$$

Plugging in $x=e$ gives:

$$y”(e) = e^e – e(e-1)e^{e-2} = e^e – e^e + e = e$$

Since $y”(e) > 0$, the critical point at $x=e$ is a minimum.

Therefore, the function $y(x)=e^x-x^e$ has a minimum at $x=e$, and the minimum value is:

$$y(e) = e^e – e^e = 0$$

So the minimum value of the function is 0, which occurs at $x=e$.

Find all real solutions to $x^4-11 x^3+5 x-2=0$.

We can approach this problem by factoring the given polynomial, if possible. One way to factor it is to use the Rational Root Theorem, which states that if a polynomial with integer coefficients has a rational root $\frac{p}{q}$ (in lowest terms), then $p$ must divide the constant term of the polynomial and $q$ must divide the leading coefficient.

In this case, the constant term is $-2$ and the leading coefficient is $1$, so any rational root must be of the form $\pm 1, \pm 2$. Checking these values, we find that none of them are roots of the polynomial.

Since the polynomial is of degree $4$, it must have exactly four complex roots (counting multiplicity). Let $a$, $b$, $c$, and $d$ be the four roots (not necessarily distinct) of the polynomial. Then we can write the polynomial as:

$$x^4 – 11x^3 + 5x – 2 = (x – a)(x – b)(x – c)(x – d)$$

Expanding the right-hand side, we get:

$$x^4 – (a + b + c + d)x^3 + (ab + ac + ad + bc + bd + cd)x^2 – (abc + abd + acd + bcd)x + abcd$$

Comparing coefficients with the left-hand side of the equation, we have:

\begin{align*} a + b + c + d &= 11 \ ab + ac + ad + bc + bd + cd &= 0 \ abc + abd + acd + bcd &= -5 \ abcd &= 2 \end{align*}

We can use these equations to eliminate variables and simplify the problem. For example, we can use the equation $abcd = 2$ to solve for one of the variables in terms of the other three:

$$d = \frac{2}{abc}$$

Substituting this into the equation $a + b + c + d = 11$ gives:

$$a + b + c + \frac{2}{abc} = 11$$

Multiplying both sides by $abc$ and rearranging, we get:

$$abc^2 + ab^2c + a^2bc + 2 = 11abc$$

This is a cubic equation in $abc$, which can be solved using various methods, such as factoring or the cubic formula. However, it turns out that the equation has no rational roots, which means that none of the roots of the polynomial $x^4 – 11x^3 + 5x – 2$ are rational.

Therefore, the only solutions to the equation are the four complex roots, which we can find using numerical methods such as the Newton-Raphson method or by using a computer algebra system.

# 微积分代写Calculus III|MATH 15300 University of Chicago Assignment

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## Instructions:

The Department of Mathematics offers a range of programs in mathematics and applied mathematics at both undergraduate and graduate levels. The department provides a comprehensive instruction in various areas of mathematics, including pure mathematics, applied mathematics, and statistics.

The undergraduate programs include both a Bachelor of Arts (BA) and a Bachelor of Science (BS) degree in mathematics. The BS program offers two different tracks, one in applied mathematics and one in mathematics with a specialization in economics. The BA program offers a broad foundation in mathematics along with a wide range of electives, allowing students to tailor their program to their specific interests.

In addition to the major programs, students from other fields of study can complete a minor in mathematics. The minor requires a minimum of 21 credit hours in mathematics courses and can be tailored to complement students’ other areas of study.

At the graduate level, the department offers a Master of Science (MS) degree in mathematics and a PhD in mathematics. The graduate programs focus on developing advanced knowledge and research skills in mathematics, and provide opportunities for students to engage in cutting-edge research with faculty members.

Overall, the Department of Mathematics provides a strong academic environment for students interested in mathematics and its applications, with a range of programs at both the undergraduate and graduate levels, as well as opportunities for research and specialization.

Let $f(x, y)=\frac{x^2 y^2}{x^2 y^2+(x+y)^4}$ if $(x, y) \neq(0,0)$. Then $\lim _{(x, y) \rightarrow(0,0)} f(x, y)$ does not exist.

By the remarks preceding this exercise, we need only find two distinct values which the function $f$ assumes in every neighborhood of the origin. This is easy. Every neighborhood of the origin contains points $(x, 0)$ distinct from $(0,0)$ on the $x$-axis. At every such point $f(x, y)=f(x, 0)=0$. Also, every neighborhood of the origin contains points $(x, x)$ distinct from $(0,0)$ which lie on the line $y=x$. At each such point $f(x, y)=f(x, x)=1 / 17$. Thus $f$ has no limit at the origin since in every neighborhood of $(0,0)$ it assumes both the values 0 and $1 / 17$. (Notice, incidentally, that both iterated limits, $\lim {x \rightarrow 0}\left(\lim {y \rightarrow 0} f(x, y)\right)$ and $\lim {y \rightarrow 0}\left(\lim {x \rightarrow 0} f(x, y)\right)$ exist and equal 0.$)$

Suppose $f$ is the constant function defined on $[0,1]$ whose value is 1 . Asked to describe those functions in $\mathcal{B}([0,1])$ which lie in the open ball about $f$ of radius 1 , a student replies (somewhat incautiously) that $B_1(f)$ is the set of all real-valued functions $g$ on $[0,1]$ satisfying $0<g(x)<2$ for all $x \in[0,1]$. Why is this response wrong? (Solution Q.13.1.)

Let $C$ be the set of all functions defined on $[0,1]$ such that $0<g(x)<2$ for all $x \in[0,1]$. It is clear that $B_1(f) \subseteq C$. The reverse inclusion, however, is not correct. For example, let
$$g(x)= \begin{cases}1, & \text { if } x=0 \ x, & \text { if } 0<x \leq 1 .\end{cases}$$
Then $g$ belongs to $C$; but it does not belong to $B_1(f)$ since
$$d_u(f, g)=\sup {|f(x)-g(x)|: 0 \leq x \leq 1}=1 .$$

A function $f: M_1 \rightarrow M_2$ between metric spaces is continuous if and only if $f^{\leftarrow}(U)$ is an open subset of $M_1$ whenever $U$ is open in $M_2$.

Suppose $f$ is continuous. Let $U \subseteq \triangle_2$. To show that $f^{\leftarrow}(U)$ is an open subset of $M_1$, it suffices to prove that each point of $f^{\leftarrow}(U)$ is an interior point of $f^{\leftarrow}(U)$. If $a \in f^{\leftarrow}(U)$, then $f(a) \in U$. Since $f$ is continuous at $a$, the set $U$, which is a neighborhood of $f(a)$, must contain the image under $f$ of a neighborhood $V$ of $a$. But then
$$a \in V \subseteq f^{\leftarrow}\left(f^{\rightarrow}(V)\right) \subseteq f^{\leftarrow}(U)$$
which shows that $a$ lies in the interior of $f^{\leftarrow}(U)$.
Conversely, suppose that $f^{\leftarrow}(U) \subseteq M_1$ whenever $U \subseteq M_2$. To see that $f$ is continuous at an arbitrary point $a$ in $M_1$, notice that if $V$ is a neighborhood of $f(a)$, then $a \in f^{\leftarrow}(V) \subseteq M_1$. Thus $f^{\leftarrow}(V)$ is a neighborhood of $a$ whose image $f^{\rightarrow}\left(f^{\leftarrow}(V)\right)$ is contained in $V$. Thus $f$ is continuous at $a$.

# 微积分代写Calculus II|MATH 15200 University of Chicago Assignment

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Assignment-daixieTM为您提供芝加哥大学University of Chicago MATH 15200 Calculus II微积分代写代考辅导服务！

## Instructions:

The Department of Mathematics at this institution offers a variety of programs in mathematics and applied mathematics at both the undergraduate and graduate levels. At the undergraduate level, students can pursue either a Bachelor of Arts (BA) or a Bachelor of Science (BS) degree in Mathematics. Additionally, the department offers two BS degree programs: one in Applied Mathematics and one in Mathematics with a specialization in Economics.

The BA program in Mathematics provides students with a broad foundation in mathematics, while the BS program is designed for students who wish to focus on a more rigorous and specialized mathematical education. The BS program in Applied Mathematics combines coursework in mathematics with applied fields such as physics, engineering, or biology. The BS program in Mathematics with a specialization in Economics is tailored for students who wish to apply mathematical concepts to economic analysis and decision-making.

In addition to the major programs, students in other fields of study can also complete a minor in Mathematics. The minor program is designed to provide students with a basic understanding of mathematics and its applications, which can enhance their academic and career opportunities.

Overall, the Department of Mathematics at this institution aims to provide students with a comprehensive education in mathematics, preparing them for careers in a variety of fields that require mathematical skills and knowledge. The department also actively engages in research activities, contributing to the advancement of mathematical knowledge and its applications in various fields.

Let $a<b$ in $\mathbb{R}$. Every continuous map of the interval $[a, b]$ into itself has a fixed point.

Let $f:[a, b] \rightarrow[a, b]$ be continuous. If $f(a)=a$ or $f(b)=b$, then the result is obvious; so we suppose that $f(a)>a$ and $f(b)0$. Since $g(a)<0<g(b)$, we may conclude from the intermediate value theorem that $0 \in \operatorname{ran} g$. That is, there exists $z$ in $(a, b)$ such that $g(z)=z-f(z)=0$. Thus $z$ is a fixed point of $f$.

Let $f: A \rightarrow \mathbb{R}$ where $A \subseteq \mathbb{R}$, and let $a$ be an accumulation point of $A$. If $f(x) \rightarrow b$ as $x \rightarrow a$, and if $f(x) \rightarrow c$ as $x \rightarrow a$, then $b=c$.

Argue by contradiction. If $b \neq c$, then $\epsilon:=|b-c|>0$. Thus there exists $\delta_1>0$ such that $|f(x)-b|<\epsilon / 2$ whenever $x \in A$ and $0<|x-a|<\delta_1$ and there exists $\delta_2>0$ such that $|f(x)-c|<\epsilon / 2$ whenever $x \in A$ and $0<|x-a|<\delta_2$. Let $\delta=\min \left{\delta_1, \delta_2\right}$. Since $a \in A^{\prime}$, the set $A \cap J_\delta(a)$ is nonempty. Choose a point $x$ in this set. Then
$$\epsilon=|b-c| \leq|b-f(x)|+|f(x)-c|<\epsilon / 2+\epsilon / 2=\epsilon$$
It is worth noticing that the preceding proof cannot be made to work if $a$ is not required to be an accumulation point of $A$. To obtain a contradiction we must know that the condition $0<|x-a|<\delta$ is satisfied for at least one $x$ in the domain of $f$.

If $f: A \rightarrow \mathbb{R}$ where $A \subseteq \mathbb{R}$, and $a \in A^{\prime}$, then
$$\lim {h \rightarrow 0} f(a+h)=\lim {x \rightarrow a} f(x)$$
in the sense that if either limit exists, then so does the other and the two limits are equal.

Let $g$ : $h \mapsto f(a+h)$. Notice that $h \in \operatorname{dom} g$ if and only if $a+h \in \operatorname{dom} f$; $\operatorname{dom} f=a+\operatorname{dom} g$.
That is, $\operatorname{dom} f={a+h: h \in \operatorname{dom} g}$.
First we suppose that
$$l:=\lim {h \rightarrow 0} g(h)=\lim {h \rightarrow 0} f(a+h) \text { exists. }$$
We show that $\lim {x \rightarrow a} f(x)$ exists and equals $l$. Given $\epsilon>0$ there exists (by (Q.4)) a number $\delta>0$ such that $$|g(h)-l|<\epsilon \text { whenever } h \in \operatorname{dom} g \text { and } 0<|h|<\delta .$$ Now suppose that $x \in \operatorname{dom} f$ and $0<|x-a|<\delta$. Then by (Q.3) $$x-a \in \operatorname{dom} g$$ and by (Q.5) $|f(x)-l|=|g(x-a)-l|<\epsilon$. Thus given $\epsilon>0$ we have found $\delta>0$ such that $|f(x)-l|<\epsilon$ whenever $x \in \operatorname{dom} f$ and $0<$ $|x-a|<\delta$. That is, $\lim {x \rightarrow a} f(x)=l$.

The converse argument is similar. Suppose $l:=\lim {x \rightarrow a} f(x)$ exists. Given $\epsilon>0$ there exists $\delta>0$ such that $|f(x)-l|<\epsilon$ whenever $x \in \operatorname{dom} f$ and $0<|x-a|<\delta$. If $h \in \operatorname{dom} g$ and $0<|h|<\delta$, then $a+h \in \operatorname{dom} f$ and $0<|(a+h)-a|<\delta$. Therefore $|g(h)-l|=|f(a+h)-l|<\epsilon$, which shows that $\lim {h \rightarrow 0} g(h)=l$.

# 微积分代写Calculus I|MATH 15100 University of Chicago Assignment

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Assignment-daixieTM为您提供芝加哥大学University of Chicago MATH 15100 Calculus I微积分代写代考辅导服务！

## Instructions:

This is the regular calculus sequence in the department. Students entering this sequence are to have mastered appropriate precalculus material and, in many cases, have had some previous experience with calculus in high school or elsewhere. All Autumn Quarter offerings of MATH 15100, 15200, and 15300 begin with a rigorous treatment of limits and limit proofs. Students may not take the first two quarters of this sequence for P/F grading. MATH 15100-15200 meets the general education requirement in mathematical sciences.

Let $A=\mathbb{Q} \cap(0, \infty)$. Find $A^{\circ}, A^{\prime}$, and $\bar{A}$.

(a) First we show that $\overline{A^c} \subseteq A^{\circ c}$. If $x \in \overline{A^c}$, then either $x \in A^c$ or $x$ is an accumulation point of $A^c$. If $x \in A^c$, then $x$ is certainly not in the interior of $A$; that is, $x \in A^{\circ c}$. On the other hand, if $x$ is an accumulation point of $A^c$, then every $\epsilon$-neighborhood of $x$ contains points of $A^c$. This means that no $\epsilon$-neighborhood of $x$ lies entirely in $A$. So, in this case too, $x \in A^{\circ c}$.

For the reverse inclusion suppose $x \in A^{\circ c}$. Since $x$ is not in the interior of $A$, no $\epsilon$-neighborhood of $x$ lies entirely in $A$. Thus either $x$ itself fails to be in $A$, in which case $x$ belongs to $A^c$ and therefore to $\overline{A^c}$, or else every $\epsilon$-neighborhood of $x$ contains a point of $A^c$ different from $x$. In this latter case also, $x$ belongs to the closure of $A^c$.

The function $f: \mathbb{R} \rightarrow \mathbb{R}: x \mapsto 5 x-8$ is continuous.

Let $a \in \mathbb{R}$. Given $\epsilon>0$, choose $\delta=\epsilon / 5$. If $|x-a|<\delta$, then $|f(x)-f(a)|=5|x-a|<5 \delta=\epsilon$.

$A$ set $A \subseteq \mathbb{R}$ is connected if and only if the only subsets of $A$ which are both open in $A$ and closed in $A$ are the null set and $A$ itself.

Suppose there exists a nonempty set $U$ which is properly contained in $A$ and which is both open and closed in $A$. Then, clearly, the sets $U$ and $U^c$ (both open in $A$ ) disconnect $A$. Conversely, suppose that $A$ is disconnected by sets $U$ and $V$ (both open in $A$ ). Then the set $U$ is not the null set, is not equal to $A$ (because $V$, its complement with respect to $A$, is nonempty), is open in $A$, and is closed in $A$ (because $V$ is open in $A$ ).

# 微积分代写 Calculus|MATH 15100 University of Chicago Assignment

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Assignment-daixieTM为您提供芝加哥大学University of Chicago MATH 15100Calculus微积分代写代考辅导服务！

## Instructions:

MATH 15100, 15200, and 15300 form a regular calculus sequence that is typically taken by undergraduate students in their first or second year of study in a mathematics or science-related field. The sequence begins with MATH 15100, which covers limits, derivatives, and applications of derivatives, including optimization problems and related rates. MATH 15200 continues with integration, including the fundamental theorem of calculus, techniques of integration, and applications of integration, such as finding areas and volumes. MATH 15300 covers sequences and series, including convergence tests and power series, and applications of these concepts, such as Taylor series and Fourier series.

The sequence is designed to provide students with a solid foundation in calculus and its applications. Students are expected to have a strong background in algebra and trigonometry, as well as basic knowledge of functions and graphs. In MATH 15100, students are introduced to the formal definition of limits and learn how to evaluate limits using algebraic techniques and the squeeze theorem. They then move on to the definition of the derivative and learn how to find derivatives using various techniques, including the power rule, product rule, quotient rule, and chain rule. Applications of derivatives, such as optimization and related rates problems, are also covered.

In MATH 15200, students continue with integration, learning how to find antiderivatives and definite integrals using various techniques, such as substitution, integration by parts, and partial fractions. They also learn about the fundamental theorem of calculus, which relates differentiation and integration, and use integration to find areas and volumes of geometric shapes.

In MATH 15300, students learn about sequences and series, including convergence tests for both types of sequences and series. They also learn about power series and their applications, such as Taylor series and Fourier series. The course concludes with a discussion of applications of these concepts in various fields, including physics, engineering, and mathematics.

Throughout the sequence, students are expected to engage in problem-solving and critical thinking, as well as to develop their skills in mathematical reasoning and communication. The courses are typically taught using a combination of lectures, problem sets, and exams, and students are expected to participate actively in class and in study groups. Successful completion of the sequence prepares students for advanced courses in calculus, as well as for further study in mathematics, physics, engineering, and other fields.

Let f(x) = x 3 sin x. Then f (8) (0) =____

To find $f(8)$, we simply plug in $x=8$ into the function $f(x) = x^3\sin x$: $$f(8) = 8^3\sin 8 \approx 992.95$$ To find $f'(x)$, we use the product rule: $$f'(x) = 3x^2\sin x + x^3\cos x$$ Then, to find $f'(0)$, we simply plug in $x=0$ into the derivative: $$f'(0) = 3(0)^2\sin 0 + (0)^3\cos 0 = 0$$ Therefore, we have $f(8) \approx 992.95$ and $f'(0) = 0$.

The (nonparametric) equation of the flflow line of the vector fifield F(x, y) = x i y j which

passes through the point (4, 3) is

The flow line of the vector field $\mathbf{F}(x,y) = x \mathbf{i} – y \mathbf{j}$ passing through the point $(4,3)$ is given by the solution to the system of differential equations: \begin{align*} \frac{dx}{dt} &= x \ \frac{dy}{dt} &= -y \ x(0) &= 4 \ y(0) &= 3 \end{align*} Separating variables and integrating, we obtain: \begin{align*} \int \frac{1}{x},dx &= \int,dt \ \ln|x| &= t + C_1 \ |x| &= e^{t+C_1} = C_2 e^t \end{align*} where $C_1$ and $C_2$ are constants of integration. Using the initial condition $x(0) = 4$, we have $4 = C_2 e^0 = C_2$, so $C_2 = 4$. Therefore, we have $x = 4e^t$.

Similarly, we have: \begin{align*} \int -\frac{1}{y},dy &= \int,dt \ \ln|y| &= -t + C_3 \ |y| &= e^{C_3} e^{-t} = C_4 e^{-t} \end{align*} Using the initial condition $y(0) = 3$, we have $3 = C_4 e^0 = C_4$, so $C_4 = 3$. Therefore, we have $y = 3e^{-t}$.

Thus, the flow line passing through $(4,3)$ is given by the parametric equations $x = 4e^t$ and $y = 3e^{-t}$. We can eliminate the parameter $t$ to obtain the equation of the flow line in terms of $x$ and $y$: $$\frac{x}{4} \cdot \frac{1}{y/3} = 1$$ which simplifies to $$xy = 12.$$ Therefore, the equation of the flow line is $xy=12$.

An example of a potential function f for the vector fifield F(x, y) = (y cos x cos y)i +

(sin x + x sin y)j is f(x, y) =__________

To find a potential function $f(x,y)$ for the given vector field $F(x,y)= (y\cos x – \cos y)i + (\sin x + x\sin y)j$, we need to find $f(x,y)$ such that $F(x,y) = \nabla f(x,y)$.

We find the partial derivatives of $f(x,y)$ with respect to $x$ and $y$ as follows:

$$\frac{\partial f}{\partial x} = y\sin x + \sin y,\qquad \frac{\partial f}{\partial y} = \cos x + x\cos y$$

To check if $F(x,y) = \nabla f(x,y)$, we need to verify if the partial derivatives of $f(x,y)$ satisfy the following equations:

$$\frac{\partial f}{\partial x} = F_1(x,y) \quad \text{and} \quad \frac{\partial f}{\partial y} = F_2(x,y)$$

where $F_1(x,y)$ and $F_2(x,y)$ are the $x$ and $y$ components of $F(x,y)$, respectively.

Comparing the partial derivatives of $f(x,y)$ with $F(x,y)$, we see that:

$$\frac{\partial f}{\partial x} = y\sin x + \sin y = F_1(x,y)$$ $$\frac{\partial f}{\partial y} = \cos x + x\cos y = F_2(x,y)$$

Therefore, we can take $f(x,y)$ as:

$$f(x,y) = \int F_1(x,y) dx = -y\cos x – \cos y + g(y)$$

where $g(y)$ is an arbitrary function of $y$. To determine $g(y)$, we take the partial derivative of $f(x,y)$ with respect to $y$:

$$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y}(-y\cos x – \cos y + g(y)) = -\sin y + g'(y)$$

Comparing with $F_2(x,y) = \cos x + x\cos y$, we have:

$$g'(y) = \cos x + x\cos y -\sin y$$

Integrating both sides with respect to $y$, we obtain:

$$g(y) = y\cos y + x\sin y -\cos x + C$$

where $C$ is a constant of integration. Therefore, the potential function $f(x,y)$ is given by:

$$f(x,y) = -y\cos x – \cos y + y\cos y + x\sin y -\cos x + C = x\sin y + C$$

So, an example of a potential function $f(x,y)$ for the given vector field is $f(x,y) = x\sin y$.

# 微积分代写 Calculus|MATH 20 Stanford University Assignment

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Assignment-daixieTM为您提供斯坦福大学Stanford University MATH 20 Calculus微积分代写代考辅导服务！

## Instructions:

This is a course in integral calculus, which is the second part of calculus after differential calculus. It covers the following topics:

1. Definite Integral: Students will learn how to compute the definite integral of a function over a given interval using Riemann sums, which is a technique for approximating the area under a curve.
2. Antiderivatives and Fundamental Theorem of Calculus: Students will learn how to find the antiderivative of a function, which is the inverse operation of differentiation. They will also learn the Fundamental Theorem of Calculus, which establishes the relationship between differentiation and integration.
3. Integration Techniques: Students will learn integration by substitution and integration by parts, which are techniques for simplifying the integration process.
4. Applications of Integration: Students will learn how to use integration to find the area between two curves, and the volume of solids of revolution using slices, washers, and shells.
5. Differential Equations: Students will learn how to solve initial-value problems, which involve finding the solution to a differential equation given an initial condition. They will also learn about exponential and logistic models, direction fields, and parametric curves.

Prerequisite for this course is Math 19 or equivalent. If the student has not previously taken a calculus course at Stanford, then they must have taken the math placement diagnostic offered through the Math Department website in order to register for this course.

Let $M>0$ and $f(x)=x^3$ for $0 \leq x \leq M$. Find a value of $c$ which satisfies the conclusion of the mean value theorem for the function $f$ over the interval $[0, M]$. Answer: $c=\frac{M}{a}$___ where $a=$__

By the Mean Value Theorem, there exists $c \in (0,M)$ such that $$f'(c)=\frac{f(M)-f(0)}{M-0}=\frac{M^3-0^3}{M} = M^2.$$ Since $f(x)=x^3$, we have $f'(x)=3x^2$. So, we have $f'(c)=3c^2=M^2$. Solving for $c$ gives $c=\frac{M}{\sqrt{3}}$.

Therefore, $a=\sqrt{3}$ and $c=\frac{M}{\sqrt{3}}$.

Let $f(x)=x^4+x+3$ for $0 \leq x \leq 2$. Find a point $c$ whose existence is guaranteed by the mean value theorem. Answer: $c=2^p$ ____where $p=$___

By the Mean Value Theorem, there exists $c \in (0,2)$ such that $$f'(c)=\frac{f(2)-f(0)}{2-0}=\frac{(2^4+2+3)-(0^4+0+3)}{2}=17.$$ Since $f(x)=x^4+x+3$, we have $f'(x)=4x^3+1$. So, we have $f'(c)=4c^3+1=17$. Solving for $c$ gives $c= \sqrt[3]{4}$.

Therefore, $p=\frac{\ln(4)}{\ln(2)}$. Hence, $c=2^p=\boxed{2^{\frac{\ln(4)}{\ln(2)}}}$.

Let $f(x)=\sqrt{x}$ for $4 \leq x \leq 16$. Find a point $c$ whose existence is guaranteed by the mean value theorem. Answer: $c=$______

By the Mean Value Theorem, there exists $c \in (4,16)$ such that $$f'(c)=\frac{f(16)-f(4)}{16-4}=\frac{\sqrt{16}-\sqrt{4}}{12}=\frac{1}{3}.$$ Since $f(x)=\sqrt{x}$, we have $f'(x)=\frac{1}{2\sqrt{x}}$. So, we have $f'(c)=\frac{1}{2\sqrt{c}}=\frac{1}{3}$. Solving for $c$ gives $c= \frac{9}{4}$.

Therefore, $c=\boxed{\frac{9}{4}}$.

# 微积分代写 Calculus|MATH 19 Stanford University Assignment

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Assignment-daixieTM为您提供斯坦福大学Stanford University MATH 19 Calculus微积分代写代考辅导服务！

## Instructions:

This course covers the basics of differential calculus, which is a branch of calculus that deals with rates of change and slopes of curves. The course begins with a review of elementary functions, including exponentials and logarithms, which are essential tools in calculus.

The course then covers the concept of limits, which is a fundamental concept in calculus. Limits allow us to describe the behavior of functions near certain points, and they provide the foundation for the definition of the derivative.

The course then introduces the derivative and its properties, including the power rule, product rule, quotient rule, and chain rule. The derivative measures the instantaneous rate of change of a function at a given point, and it has many important applications in fields such as physics, engineering, and economics.

The course also covers applications of the derivative, including optimization problems, related rates problems, and curve sketching. These topics are important in many areas of science, engineering, and economics.

Prerequisites for the course include trigonometry, advanced algebra, and analysis of elementary functions, including exponentials and logarithms. Students must also take the math placement diagnostic offered through the Math Department website in order to register for the course. This diagnostic test helps ensure that students have the necessary background knowledge and skills to succeed in the course.

Let f(x) = 5 sin x + 3 cos x. Then f(117)(π) =___ .

To find the 117th derivative of $f(x) = 5 \sin x + 3 \cos x$ evaluated at $x = \pi$, we can use the fact that the derivative of $\sin x$ is $\cos x$ and the derivative of $\cos x$ is $-\sin x$. Using this, we can see that:

\begin{align*} f'(x) &= 5\cos x – 3\sin x \ f”(x) &= -5\sin x – 3\cos x \ f”'(x) &= -5\cos x + 3\sin x \ f^{(4)}(x) &= 5\sin x – 3\cos x \ f^{(5)}(x) &= 5\cos x + 3\sin x \ f^{(6)}(x) &= -5\sin x + 3\cos x \ \end{align*}

We can see that the derivatives of $f(x)$ have a repeating pattern with a period of four, so we can use this to find the 117th derivative of $f(x)$:

\begin{align*} f^{(117)}(x) &= f^{(4\cdot29+1)}(x) \ &= f^{(1)}(x) \ &= 5\cos x – 3\sin x \ \end{align*}

Therefore, evaluating this expression at $x = \pi$, we get:

\begin{align*} f^{(117)}(\pi) &= 5\cos \pi – 3\sin \pi \ &= 5(-1) – 3(0) \ &= -5 \ \end{align*}

So, $f^{(117)}(\pi) = -5$.

Let f(x) = 4 cos x − 7 sin x. Then f(87)(0) =____ .

To find the 87th derivative of $f(x) = 4 \cos x – 7 \sin x$ evaluated at $x = 0$, we can use the fact that the derivative of $\sin x$ is $\cos x$ and the derivative of $\cos x$ is $-\sin x$. Using this, we can see that:

\begin{align*} f'(x) &= -4\sin x – 7\cos x \ f”(x) &= -4\cos x + 7\sin x \ f”'(x) &= 4\sin x + 7\cos x \ f^{(4)}(x) &= 4\cos x – 7\sin x \ f^{(5)}(x) &= -4\sin x – 7\cos x \ f^{(6)}(x) &= -4\cos x + 7\sin x \ \end{align*}

We can see that the derivatives of $f(x)$ have a repeating pattern with a period of four, so we can use this to find the 87th derivative of $f(x)$:

\begin{align*} f^{(87)}(x) &= f^{(4\cdot21+3)}(x) \ &= f^{(3)}(x) \ &= 4\sin x + 7\cos x \ \end{align*}

Therefore, evaluating this expression at $x = 0$, we get:

\begin{align*} f^{(87)}(0) &= 4\sin 0 + 7\cos 0 \ &= 0 + 7(1) \ &= 7 \ \end{align*}

So, $f^{(87)}(0) = 7$.

For the circle $x^2+y^2-1=0$ use implicit differentiation to show that $y^{\prime \prime}=-\frac{1}{y^3}$ and $y^{\prime \prime \prime}=-\frac{3 x}{y^5}$.

Starting with the given equation $x^2+y^2-1=0$, we take the derivative with respect to $x$ of both sides, using the chain rule for the $y$-term:

$$\frac{d}{dx}(x^2+y^2-1) = \frac{d}{dx}(0)$$

Using the power rule and chain rule, we get:

$$2x + 2y \frac{dy}{dx} = 0$$

Solving for $\frac{dy}{dx}$, we have:

$$\frac{dy}{dx} = -\frac{x}{y}$$

Taking the derivative of this expression with respect to $x$ using the quotient rule, we get:

\begin{align*} \frac{d^2y}{dx^2} &= \frac{d}{dx}\left(-\frac{x}{y}\right)\ &= -\frac{y\frac{d}{dx}(x)-x\frac{d}{dx}(y)}{y^2}\ &= -\frac{y(1\cdot y + x\cdot y’) – x(0\cdot y + 1\cdot y’)}{y^2} \ &= -\frac{y^{\prime}\cdot x}{y^2} – \frac{1}{y} \end{align*}

Simplifying this expression, we get:

$$y^{\prime \prime} = -\frac{1}{y^3}$$

Taking the derivative of $y^{\prime}$ with respect to $x$ using the quotient rule, we have:

\begin{align*} \frac{d^3y}{dx^3} &= \frac{d}{dx}\left(-\frac{y^{\prime}}{y}\right)\ &= -\frac{y\frac{d}{dx}(y’)-y’\frac{d}{dx}(y)}{y^2}\ &= -\frac{y^{\prime\prime}\cdot y – y^{\prime}\cdot y’}{y^2}\ &= -\frac{y^{\prime\prime}\cdot y}{y^2} + \frac{y^{\prime}\cdot y’}{y^2} \end{align*}

Substituting the values of $y^{\prime\prime}$ and $y^{\prime}$ obtained earlier, we have:

\begin{align*} y^{\prime \prime \prime} &= -\frac{y^{\prime\prime}\cdot y}{y^2} + \frac{y^{\prime}\cdot y’}{y^2}\ &= -\frac{(-1/y^3) \cdot y}{y^2} + \frac{(-x/y^2)\cdot(-x/y)}{y^2}\ &= \frac{x}{y^5} – \frac{1}{y^3}\ &= -\frac{3x}{y^5} \ \ \ \ \ \text{(using } x^2+y^2=1 \text{)} \end{align*}

Therefore, $y^{\prime \prime}=-\frac{1}{y^3}$ and $y^{\prime \prime \prime}=-\frac{3 x}{y^5}$ for the circle $x^2+y^2-1=0$