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）

To simplify the first expression, let $\theta=\sin ^{-1} x$ and note that $\sin \theta=x$. When $x \geq 0$, the angle $\theta$ arises in a right triangle with an opposite side of length $x$ and a hypotenuse of length 1. It follows by Pythagoras’ theorem that the adjacent side has length $\sqrt{1-x^2}$, so the definition of cosine gives
$$
\cos \left(\sin ^{-1} x\right)=\cos \theta=\frac{\text { adjacent side }}{\text { hypotenuse }}=\sqrt{1-x^2} .
$$
When $x \leq 0$, the last equation holds with $-x$ instead of $x$. This changes the term $\sin ^{-1} x$ by a minus sign, but the cosine remains unchanged, so the equation is still valid.

To simplify the second expression, let $\theta=\tan ^{-1} x$ and note that $\tan \theta=x$. When $x \geq 0$, the angle $\theta$ arises in a right triangle with an opposite side of length $x$ and an adjacent side of length 1. It follows by Pythagoras’ theorem that the hypotenuse has length $\sqrt{1+x^2}$, so the definition of cosine gives
$$
\cos \left(\tan ^{-1} x\right)=\cos \theta=\frac{\text { adjacent side }}{\text { hypotenuse }}=\frac{1}{\sqrt{1+x^2}} .
$$
When $x \leq 0$, the last equation holds with $-x$ instead of $x$. This changes the term $\tan ^{-1} x$ by a minus sign, but the cosine remains unchanged, so the equation is still valid.
As for the third expression, one may use the properties of logarithms to get
$$
\log _3(54)-3 \log _3(18)+\log _3(36)=\log _3 \frac{54 \cdot 36}{18^3}=\log _3 \frac{3 \cdot 2}{18}=\log _3 3^{-1}=-1 .
$$

In this case, $x$ is approaching 3 and $f(x)$ is either $3 x-4$ or $4 x-7$. We thus expect the limit to be $L=5$. To prove this formally, we let $\varepsilon>0$ and estimate the expression
$$
|f(x)-5|=\left{\begin{array}{ll}
|3 x-9| & \text { if } x \leq 3 \
|4 x-12| & \text { if } x>3
\end{array}\right}=\left{\begin{array}{ll}
3|x-3| & \text { if } x \leq 3 \
4|x-3| & \text { if } x>3
\end{array}\right} .
$$
If we assume that $0 \neq|x-3|<\delta$, then we may use the last equation to get
$$
|f(x)-5| \leq 4|x-3|<4 \delta \text {. }
$$
Since our goal is to show that $|f(x)-5|<\varepsilon$, an appropriate choice of $\delta$ is thus $\delta=\varepsilon / 4$.

When it comes to the first limit, division of polynomials gives
$$
L=\lim {x \rightarrow 1} \frac{3 x^3-7 x^2+6 x-2}{x-1}=\lim {x \rightarrow 1}\left(3 x^2-4 x+2\right)=3-4+2=1 .
$$
When it comes to the second limit, division of polynomials gives
$$
M=\lim {x \rightarrow 2} \frac{2 x^3-7 x^2+4 x+4}{x^2-4 x+4}=\lim {x \rightarrow 2}(2 x+1)=4+1=5
$$

Let $f(x)=3 x^2-4 x+7$ for convenience. Then $f(2)=11$ and one has
$$
|f(x)-f(2)|=\left|3 x^2-4 x-4\right|=|x-2| \cdot|3 x+2| .
$$
The factor $|x-2|$ is related to our usual assumption that $0 \neq|x-2|<\delta$. To estimate the remaining factor $|3 x+2|$, we assume that $\delta \leq 1$ for simplicity and we note that
$$
\begin{aligned}
|x-2|<\delta \leq 1 & \Longrightarrow-1<x-2<1 \
& \Longrightarrow \quad 1<x<3 \quad \Longrightarrow \quad 5<3 x+2<11 .
\end{aligned}
$$
Combining the estimates $|x-2|<\delta$ and $|3 x+2|<11$, one may then conclude that
$$
|f(x)-f(2)|=|x-2| \cdot|3 x+2|<11 \delta \leq \varepsilon,
$$
as long as $\delta \leq \varepsilon / 11$ and $\delta \leq 1$. An appropriate choice of $\delta$ is thus $\delta=\min (\varepsilon / 11,1)$.

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