数学代写|MATH141 Calculus

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这是惠灵顿维多利亚大学微积分课程的代写成功案例。

MATH141课程简介

This course provides a thorough development of the differential calculus and an introduction to the integral calculus. The course builds on the ideas of functions and limits to define derivatives and integrals, as well as rules for computing these and applications to physical modelling.
This course is designed for in-person study, and it is strongly recommended that students attend lectures and tutorials on campus. In particular some assessment items will have a requirement of in-person attendance, although exceptions can be made under special circumstances.

Queries about any such exceptions can be sent to [email protected].

Prerequisites 

This course provides a thorough introduction to the differential calculus and an introduction to the integral calculus. Following a review of coordinate geometry and the use of equations to represent straight lines and circles, we introduce the concept of a function and many important examples including polynomial, rational, trigonometric, exponential and logarithm functions. The idea of a limit is central and leads to considering continuity and differentiability of a function. The definition of a derivative and rules for computing derviatives are deduced. Derivatives are applied to model physical problems. The course concludes by introducing the idea of an integral of a function and we demonstrate the fundamental property that integrals can be calculated using the inverse of differentiation.As well as developing methods of computation and applications of the calculus, the course focuses on its underlying concepts, the correct use of symbolic representation in mathematics and the importance of providing logical justification for its results and methods.

MATH141 Calculus HELP（EXAM HELP， ONLINE TUTOR）

Let us write $x^2 e^y+3 x e^{2 y}=2$ for simplicity. Differentiating both sides, we get
$$
2 x e^y+x^2 e^y y^{\prime}+3 e^{2 y}+3 x e^{2 y} \cdot 2 y^{\prime}=0 .
$$
We now collect the terms that contain $y^{\prime}$ on the left hand side and we get
$$
\left(x^2 e^y+6 x e^{2 y}\right) y^{\prime}=-2 x e^y-3 e^{2 y} \quad \Longrightarrow \quad y^{\prime}=-\frac{2 x e^y+3 e^{2 y}}{x^2 e^y+6 x e^{2 y}}=-\frac{2 x+3 e^y}{x^2+6 x e^y} .
$$
To determine the value of $y$ that corresponds to $x=2$, we note that
$$
x^2 e^y+3 x e^{2 y}=2 \quad \Longrightarrow \quad 4 e^y+6 e^{2 y}=2 \quad \Longrightarrow \quad 3 e^{2 y}+2 e^y-1=0 .
$$
Let $z=e^y$ for convenience. Then $3 z^2+2 z-1=0$ and the quadratic formula gives
$$
z=\frac{-2 \pm \sqrt{4+4 \cdot 3}}{6}=\frac{-2 \pm 4}{6}=\frac{1}{3},-1 .
$$
Since $z=e^y$ must be positive, the only acceptable solution is $z=e^y=1 / 3$ and so
$$
y^{\prime}=-\frac{2 x+3 e^y}{x^2+6 x e^y}=-\frac{2 x+1}{x^2+2 x}=-\frac{5}{8} .
$$

Being a polynomial, $f$ is continuous on the interval $[0,1]$ and we also have
$$
f(0)=1, \quad f(1)=2-3-4+1=-4 .
$$
Since $f(0)$ and $f(1)$ have opposite signs, $f$ must have a root that lies in $(0,1)$. To show it is unique, suppose that $f$ has two roots in $(0,1)$. Then $f^{\prime}$ must have a root in this interval by Rolle’s theorem. On the other hand, it is easy to check that
$$
f^{\prime}(x)=0 \quad \Longrightarrow \quad 6 x^2-6 x-4=0 \quad \Longrightarrow \quad x=\frac{3 \pm \sqrt{33}}{6} .
$$
Since $f^{\prime}$ has no roots in $(0,1)$, we conclude that $f$ has exactly one root in $(0,1)$.

The first limit has the form $0 / 0$, so one may use L’Hôpital’s rule to find that
$$
L_1=\lim {x \rightarrow 2} \frac{6 x-5}{4 x-7}=\frac{12-5}{8-7}=7 . $$ The second limit has the form $\infty / \infty$ and one may apply L’Hôpital’s rule to get $$ L_2=\lim {x \rightarrow \infty} \frac{2(\ln x) \cdot 1 / x}{1}=\lim {x \rightarrow \infty} \frac{2 \ln x}{x} . $$ This is still a limit of the form $\infty / \infty$ and another application of L’Hôpital’s rule gives $$ L_2=\lim {x \rightarrow \infty} \frac{2 / x}{1}=\lim {x \rightarrow \infty} \frac{2}{x}=0 . $$ The third limit involves a non-constant exponent which can be eliminated by writing $$ \ln L_3=\ln \lim {x \rightarrow 0^{+}}\left(e^{3 x}+\sin x\right)^{2 / x}=\lim {x \rightarrow 0^{+}} \ln \left(e^{3 x}+\sin x\right)^{2 / x}=\lim {x \rightarrow 0^{+}} \frac{2 \ln \left(e^{3 x}+\sin x\right)}{x} .
$$
This gives a limit of the form $0 / 0$, so one may use L’Hôpital’s rule to find that
$$
\ln L_3=\lim _{x \rightarrow 0^{+}} \frac{2\left(e^{3 x}+\sin x\right)^{-1} \cdot\left(3 e^{3 x}+\cos x\right)}{1}=\frac{2(3+1)}{1+0}=8 .
$$
Since $\ln L_3=8$, the original limit $L_3$ is then equal to $L_3=e^{\ln L_3}=e^8$.

To say that $f(x)$ is increasing is to say that $f^{\prime}(x)>0$. Let us then compute
$$
f^{\prime}(x)=\frac{1}{4 x^2+1} \cdot\left(4 x^2+1\right)^{\prime}=\frac{8 x}{4 x^2+1} .
$$
Since the denominator is always positive, $f(x)$ is increasing if and only if $x>0$. Next, we look at concavity. To say that $f(x)$ is concave up is to say that $f^{\prime \prime}(x)>0$. In this case,
$$
f^{\prime \prime}(x)=\frac{8\left(4 x^2+1\right)-8 x \cdot 8 x}{\left(4 x^2+1\right)^2}=\frac{8\left(4 x^2+1-8 x^2\right)}{\left(4 x^2+1\right)^2}=\frac{8(1+2 x)(1-2 x)}{\left(4 x^2+1\right)^2} .
$$
To determine the sign of this expression, one needs to find the sign of each of the factors. According to the table below, $f(x)$ is concave up if and only if $x \in(-1 / 2,1 / 2)$.
\begin{tabular}{c|c|c|c}
& \multicolumn{2}{|c}{$-1 / 2$} & \multicolumn{2}{c}{$1 / 2$} \
\hline $8(1+2 x)$ & – & + & + \
\hline $1-2 x$ & + & + & – \
\hline$f^{\prime \prime}(x)$ & – & + & –
\end{tabular}

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