# 微积分2|Calculus 2 MAST10006代写

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This subject will extend knowledge of calculus from school. Students are introduced to hyperbolic functions and their inverses, the complex exponential and functions of two variables. Techniques of differentiation and integration will be extended to these cases. Students will be exposed to a wider class of differential equation models, both first and second order, to describe systems such as population models, electrical circuits and mechanical oscillators. The subject also introduces sequences and series including the concepts of convergence and divergence.

Find the mass of a ball 9 of radius $a$ whose density is numerically equal to the distance from a fixed diametral plane.

Let the ball be the inside of the sphere $x^{2}+y^{2}+z^{2}=a^{2}$, and let the fixed diametral plane be $z=0$. Then $M=\iiint|z| d V$. Use the upper hemisphere and double the result. In spherical coordinates.
\begin{aligned} M &=2 \int_{0}^{2 \pi} \int_{0}^{2 \pi} \int_{0}^{a} z \cdot \rho^{2} \sin \phi d \rho d \phi d \theta=2 \int_{0}^{2 \pi} \int_{0}^{\pi / 2} \int_{0}^{a} \rho \cos \phi \cdot \rho^{2} \sin \phi d \rho d \phi d \theta \ &\left.=2 \int_{0}^{2 \pi} \int_{0}^{\pi / 2} \frac{1}{4} \rho^{4} \cos \phi \sin \phi\right]{0}^{a} d \phi d \theta=\frac{1}{2} a^{4} \int{0}^{2 \pi} \frac{1}{2} \sin ^{2} \phi \int_{0}^{\pi / 2} d \theta=\frac{1}{4} a^{4} \int_{0}^{2 \pi} d \theta=\frac{1}{4} a^{4} \cdot 2 \pi=\frac{1}{2} \pi a^{4} \end{aligned}

## MAST10006 COURSE NOTES ：

Find the surface area $S$ of the part of the sphere $x^{2}+y^{2}+z^{2}=4 z$ inside the paraboloid $z=x^{2}+y^{2}$.
$\square$ The region $\mathscr{R}$ under the spherical cap (Fig. 44-33) is obtained by finding the intersection of $x^{2}+y^{2}+z^{2}=4 z$ and $z=x^{2}+y^{2}$. This gives $z(z-3)=0$. Hence, the paraboloid cuts the sphere when $z=3$, and $\mathscr{B}$ is the disk $x^{2}+y^{2} \leq 3 . \quad S=\iint_{a} \sqrt{1+\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}} d A . \quad 2 x+2 z \frac{\partial z}{\partial x}=4 \frac{\partial z}{\partial x}, \quad \frac{\partial z}{\partial x}=-\frac{x}{z-2} . \quad$ Similarly, $\frac{\partial z}{\partial y}=-\frac{y}{z-2}$. Hence,
$$1+\left(\frac{\partial z}{\partial x}\right)^{2}+\left(\frac{\partial z}{\partial y}\right)^{2}=1+\frac{x^{2}}{(z-2)^{2}}+\frac{y^{2}}{(z-2)^{2}}=\frac{(z-2)^{2}+x^{2}+y^{2}}{(z-2)^{2}}=\frac{\left(x^{2}+y^{2}+z^{2}\right)-4 z+4}{(z-2)^{2}}=\frac{4}{(z-2)^{2}}$$
Therefore,
$$\left.S=\iint_{\pi} \frac{2}{z-2} d A=\int_{0}^{2 w} \int_{0}^{\sqrt{3}} \frac{2}{\sqrt{4-r^{2}}} r d r d \theta=-\int_{0}^{2 \theta} 2 \sqrt{4-r^{2}}\right]{0}^{\sqrt{3}} d \theta=-2 \int{0}^{2 \pi}(1-2) d \theta=2 \cdot 2 \pi=4 \pi$$

# 量子物理学的基础 Foundations of Quantum Physics PHYS104

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$$\sum_{\lambda} F_{\lambda ‘ \lambda}^{j} C_{\lambda j}=\epsilon_{j} \sum_{\lambda} S_{\lambda^{\prime} \lambda} C_{\lambda j},$$
where
\begin{aligned} F_{\lambda^{\prime} \lambda}^{j}=&\left\langle\chi_{\lambda^{\prime}}|h| \chi_{\lambda}\right\rangle+\sum_{\delta \kappa}\left[\gamma_{\delta_{k}}\left\langle\chi_{\lambda^{\prime}} \chi_{\delta}|g| \chi_{\lambda} \chi_{\chi^{\prime}}\right\rangle\right.\ &\left.-\gamma_{\delta \kappa}^{\text {exch }}\left\langle\chi_{\lambda^{\prime} \cdot} \chi_{\delta}|g| \chi_{\kappa} \chi_{\lambda}\right\rangle\right] . \end{aligned}
Here, $h$ is the one-electron part of the full Hamiltonian, $g$ is an electron-electron repulsion potential energy, and
$$\begin{gathered} \gamma_{\delta x}=\sum_{i}^{\prime} C_{\delta i} C_{\kappa i}, \ \gamma_{\delta \kappa}^{\text {exch }}=\sum_{i}^{n \prime} C_{\delta i} C_{\kappa i}, \end{gathered}$$

## PHYS104COURSE NOTES ：

Show that
if $f(x)=x^{3}$, then $f^{\prime}(x)=3 x^{2}$.
Prove by induction that for each positive integer $n$, $f(x)=x^{n} \quad$ has derivative $\quad f^{\prime}(x)=n x^{n-1} .$
HINT:
$$(x+h)^{k+1}-x^{k+1}=x(x+h)^{k}-x \cdot x^{k}+h(x+h)^{k} .$$

# 微积分 Calculus MATH101/MATH102

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For $h \neq 0$ and $x+h$ in the domain of $f$,
$$f(x+h)-f(x)=\frac{f(x+h)-f(x)}{h} \cdot h$$
With $f$ differentiable at $x$,
$$\lim {h \rightarrow 0} \frac{f(x+h)-f(x)}{h}=f^{\prime}(x)$$ Since $\lim {h \rightarrow 0} h=0$, we have
$$\lim {h \rightarrow 0}[f(x+h)-f(x)]=\left[\lim {h \rightarrow 0} \frac{f(x+h)-f(x)}{h}\right] \cdot\left[\lim _{h \rightarrow 0} h\right]=f^{\prime}(x) \cdot 0=0 .$$

## MATH101/MATH102COURSE NOTES ：

Show that
if $f(x)=x^{3}$, then $f^{\prime}(x)=3 x^{2}$.
Prove by induction that for each positive integer $n$, $f(x)=x^{n} \quad$ has derivative $\quad f^{\prime}(x)=n x^{n-1} .$
HINT:
$$(x+h)^{k+1}-x^{k+1}=x(x+h)^{k}-x \cdot x^{k}+h(x+h)^{k} .$$

# 微积分 Calculus MATH1006

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Given a series $\Sigma_{m=1}^{m} a_{n}=a_{1}+a_{2}+a_{s}+\cdots$, let $s_{a}$ denote its rth partial sum:
$$s_{n}=\sum_{i=1}^{n} a_{i}=a_{1}+a_{2}+\cdots+a_{n}$$

If the sequence $\left{s_{\mathrm{n}}\right}$ is convergent and $\lim {\mathrm{a} \rightarrow \mathrm{m}} s{\mathrm{a}}=s$ exists as a real number, then the series $\Sigma a_{n}$ is called convergent and we write
$$a_{1}+a_{2}+\cdots+a_{n}+\cdots=s \quad \text { or } \quad \sum_{n=1}^{\infty} a_{n}=s$$
The number $s$ is called the sum of the series. Otherwise, the series is called divergent.

## MATH1006COURSE NOTES ：

If $r=1$, then $s_{n}=a+a+\cdots+a=n a \rightarrow \pm \infty$. Since $\lim {n \rightarrow-} s{n}$ doesn’t exist, the geometric series diverges in this case.
If $r \neq 1$, we have
\begin{aligned} &s_{\mathrm{a}}=a+a r+a r^{2}+\cdots+a r^{\mathrm{n}-1} \ &r s_{\mathrm{a}}=\quad a r+a r^{2}+\cdots+a r^{\mathrm{n}-1}+a r^{\mathrm{n}} \end{aligned}
Subtracting these equations, we get
$$\begin{array}{r} s_{\mathrm{a}}-r s_{\mathrm{a}}=a-a r^{n} \ s_{\mathrm{n}}=\frac{a\left(1-r^{\mathrm{n}}\right)}{1-r} \end{array}$$
If $-1<r<1$, we know from $(11.1 .9)$ that $r^{n} \rightarrow 0$ as $n \rightarrow \infty$, so
$$\lim {n \rightarrow \infty} s{n}=\lim {n \rightarrow \infty} \frac{a\left(1-r^{n}\right)}{1-r}=\frac{a}{1-r}-\frac{a}{1-r} \lim {n \rightarrow \infty} r^{n}=\frac{a}{1-r}$$

# 微积分 Calculus MAT00001C

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This is
$$\left[\left[\begin{array}{c} R \circ g \ -Q \circ g \ P \circ g \end{array}\right] \cdot\left[\left[\frac{\partial g}{\partial u}\right] \times\left[\frac{\partial g}{\partial v}\right]\right]\right] d u \wedge d v$$
(The permutation of the $P, Q, R$ (and the minus sign) come from the way the $d x \wedge d y$ acts on a piece of surface normal to the $(d) z$ direction.)

We can rewrite this as
$$\left|\frac{\partial g}{\partial u} \times \frac{\partial g}{\partial v}\right|\left[\left[\begin{array}{c} R \circ g \ -Q \circ g \ P \circ g \end{array}\right] \cdot \hat{\boldsymbol{n}}[u, v]\right] \quad d u \wedge d v$$
where $\hat{\boldsymbol{n}}[u, v]$ is the unit normal to the surface at $g\left[\begin{array}{l}u \ v\end{array}\right]$, and
$$\left|\frac{\partial g}{\partial u} \times \frac{\partial g}{\partial v}\right|$$
is the “area stretching factor”.
We have that
$$\int_{g\left(I^{2}\right)} \omega$$

## BMAT00001C COURSE NOTES ：

is the limit of the sums of values of $\omega$ on small elements of the surface $g\left(I^{2}\right)$. Suppose $g$ takes a rectangle $\triangle u \times \triangle v$ in $I^{2}$ to a (small) piece of the surface. $\omega$ at $g\left[\begin{array}{l}u \ v\end{array}\right]$ is, say,
$$P d x \wedge d y+Q d x \wedge d z+R d y \wedge d z$$
and the unit normal to the surface is $\hat{\boldsymbol{n}}[u, v]$ (located at $\left.g\left[\begin{array}{l}u \ v\end{array}\right]\right)$.
Write $\hat{\boldsymbol{n}}[u, v]$ as
$$\left[\begin{array}{l} \hat{n} x \ \hat{n} y \ \hat{n} z \end{array}\right]$$

# 微积分I|Calculus I代写   4CCM111A

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so
$e^{y}-2 x-e^{-y}=0$
or, multiplying by $e^{y}$,
$$e^{2 y}-2 x e^{y}-1=0$$
This is really a quadratic equation in $e^{y}$ :
$$\left(e^{y}\right)^{2}-2 x\left(e^{y}\right)-1=0$$

Solving by the quadratic formula, we get
$$e^{y}=\frac{2 x \pm \sqrt{4 x^{2}+4}}{2}=x \pm \sqrt{x^{2}+1}$$
Note that $e^{y}>0$, but $x-\sqrt{x^{2}+1}<0$ (because $x<\sqrt{x^{2}+1}$ ). Thus the minus sign is inadmissible and we have
$$e^{y}=x+\sqrt{x^{2}+1}$$
Therefore
$$y=\ln \left(e^{y}\right)=\ln \left(x+\sqrt{x^{2}+1}\right)$$

## 4CCM111ACOURSE NOTES ：

Using Table 6 and the Chain Rule, we have
\begin{aligned} \frac{d}{d x}\left[\tanh ^{-1}(\sin x)\right] &=\frac{1}{1-(\sin x)^{2}} \frac{d}{d x}(\sin x) \ &=\frac{1}{1-\sin ^{2} x} \cos x=\frac{\cos x}{\cos ^{2} x}=\sec x \end{aligned}

# 微积分作业代写calculus代考

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## 微分的记号Notation for differentiation代写

• 莱布尼茨的记号Leibniz’s notation
• 微积分基本定理Fundamental theorem
• 策梅洛-弗兰克尔集合论Zermelo–Fraenkel set theory
• 连续统的势 Cardinality of the continuum

## 微积分的历史

Johannes Kepler’s work Stereometrica Doliorum formed the basis of integral calculus.Kepler developed a method to calculate the area of an ellipse by adding up the lengths of many radii drawn from a focus of the ellipse.

## 微积分课后作业代写

The first equation yields $y=3 x^{2}$, substituting that into the second equation yields $x-6 x^{2}=0$, which has the solutions $x=0$ and $x=\frac{1}{6}$. So $x=0 \Rightarrow y=3(0)=0$ and $x=\frac{1}{6} \Rightarrow y=3\left(\frac{1}{6}\right)^{2}=\frac{1}{12}$.
So the critical points are $(x, y)=(0,0)$ and $(x, y)=\left(\frac{1}{6}, \frac{1}{12}\right)$.
To use Theorem 2.6, we need the second-order partial derivatives:
$$\frac{\partial^{2} f}{\partial x^{2}}=-6 x, \quad \frac{\partial^{2} f}{\partial y^{2}}=-2, \quad \frac{\partial^{2} f}{\partial y \partial x}=1$$
So
$$D=\frac{\partial^{2} f}{\partial x^{2}}(0,0) \frac{\partial^{2} f}{\partial y^{2}}(0,0)-\left(\frac{\partial^{2} f}{\partial y \partial x}(0,0)\right)^{2}=(-6(0))(-2)-1^{2}=-1<0$$ and thus $(0,0)$ is a saddle point. Also, $$D=\frac{\partial^{2} f}{\partial x^{2}}\left(\frac{1}{6}, \frac{1}{12}\right) \frac{\partial^{2} f}{\partial y^{2}}\left(\frac{1}{6}, \frac{1}{12}\right)-\left(\frac{\partial^{2} f}{\partial y \partial x}\left(\frac{1}{6}, \frac{1}{12}\right)\right)^{2}=\left(-6\left(\frac{1}{6}\right)\right)(-2)-1^{2}=1>0$$

# 微积分calculus|MAT‑012 Assignment

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Suppose $a_{n} \rightarrow a$ and $b_{n} \rightarrow$ b as $n \rightarrow \infty$. Then the following results hold.
(i) $a_{n}+b_{n} \rightarrow a+b$ as $n \rightarrow \infty$.
(ii) $c a_{n} \rightarrow c$ a as $n \rightarrow \infty$ for any real number $c$.
(iii) If $a_{n} \leq b_{n}$ for all $n \in \mathbb{N}$, then $a \leq b$.
(iv) (Sandwich theorem) If $a_{n} \leq c_{n} \leq b_{n}$ for all $n \in \mathbb{N}$, and if $a=b$, then $c_{n} \rightarrow a$ as $n \rightarrow \infty$.

Proof Let $\varepsilon>0$ be given.
(i) Note that, for every $n \in \mathbb{N}$,
\begin{aligned} \left|\left(a_{n}+b_{n}\right)-(a+b)\right| &=\left|\left(a_{n}-a\right)+\left(b_{n}-b\right)\right| \ & \leq\left|a_{n}-a\right|+\left|b_{n}-b\right| \end{aligned}
Since $a_{n} \rightarrow a$ and $b_{n} \rightarrow b$, the above inequality suggests that we may take $\varepsilon_{1}=\varepsilon / 2$, and consider $N_{1}, N_{2} \in \mathbb{N}$ such that
$$\left|a_{n}-a\right|<\varepsilon_{1} \quad \forall n \geq N_{1} \text { and }\left|b_{n}-b\right|<\varepsilon_{1} \quad \forall n \geq N_{2}$$
so that
$$\left|\left(a_{n}+b_{n}\right)-(a+b)\right| \leq\left|a_{n}-a\right|+\left|b_{n}-b\right|<2 \varepsilon_{1}=\varepsilon$$
for all $n \geq N:=\max \left{N_{1}, N_{2}\right}$.
(ii) Note that
$$\left|c a_{n}-c a\right|=|c|\left|a_{n}-a\right| \quad \forall n \in \mathbb{N} .$$

(Ratio test) Suppose $a_{n}>0$ for all $n \in \mathbb{N}$ such that $\lim {n \rightarrow \infty} \frac{a{n+1}}{a_{n}}=\ell$ for some $\ell \geq 0$. Then the following hold.
(i) If $\ell<1$, then $a_{n} \rightarrow 0$. (ii) If $\ell>1$, then $a_{n} \rightarrow \infty$.

Suppose $\ell<1$. Let $q$ be such that $\ell1$. Let $q$ be such that $1<q<\ell$. Then, taking for example the open interval $I$ containing $\ell$ as $I=(q, \ell+1)$, there exists $N \in \mathbb{N}$

## MAT 012: Precalculus

Instructor: Emily Meyer
Contact: Time & Place:
[email protected] MWF 8:00-9:40 am, Chemistry 166
Office: MSB 3127 Office hours MWR 3:00-5:00 pm
Course Website
Homework, worksheets, updates and supplementary material will be online at
http://math.ucdavis.edu/~emeyer/math12/~index.html, as well as on Canvas:
http://canvas.ucdavis.edu/courses/282963.
Textbook
• Precalculus (Seventh Edition, 2012) by Cohen, Lee, and Sklar.
Prerequisites/Placement Exam
• It is expected that you have taken two years of high school algebra, plane geometry, and
plane trigonometry.
• You must have received a qualifying score (25 or more) on the UC Davis Math Placement
Exam to take this course. If you have not yet taken the exam, you can take it during the
testing session from 1pm July 30 to 1pm August 7, 2018. See department website for
further details.
NOTE: after completing this course, you are also required to take the Math Placement Exam
again before you may enroll in calculus. There are two testing windows for fall quarter:
1pm September 5 to 1pm September 11, and 1pm September 26 to 1pm October 2.
Course Information
• 40% final exam
• 20% midterm exam
1
MAT 012 SSII18
• 20% quizzes
• 20% homework and worksheets
Class Expectations
Class time will be interactive. I will not take attendance, but you are expected to be in class.
Being present and engaged is crucial to your academic success.
Homework and Quizzes
• Homework will be in the form of worksheets given in class that you will have time to start
in class and will be due at the beginning of the following class. These will be graded out
of 2 points, and you can redo each one once for a regrade as long as your first version is
complete and your redo is turned in within one week of the original due date.
• Additional homework problems will be provided but will not be graded. It is not expected
that you do all of the additional problems, but that you work through enough problems
from each section to develop and solidify the skills and knowledge you have learned. This
is the best way to study for exams! You are welcome to ask about these problems in office
hours, or to ask for general qualitative feedback from the instructor on your work.
• Quizzes will be given every Friday on which there is no exam (see Exams below). These
are intended not to be especially difficult but to give you practice thinking about math in a
similar environment to an exam.
Exams
There will be two exams:
• Midterm on Friday, August 24
• Final on Friday, September 14 (last day of class)
Each will be 100 minutes.
Other
• Make-ups and Absences: Make-up exams and quizzes will be given only in the case of
documented emergencies. Your lowest quiz grade (out of four) will be dropped, so if you
need to miss a quiz for a non-emergency reason, plan on using that as your dropped quiz.
• Office hours: I will hold office hours (tentatively) 3-5pm Monday, Wednesday and Thursday. You are highly encouraged to attend! This is your opportunity to ask me to repeat
or rephrase anything that did not make sense in class, to get individualized help on your
homework, or to ask any lingering questions you might have.
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MAT 012 SSII18