数学代写|MATH1111 Calculus

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这是悉尼大学 微积分课程的代写成功案例。

MATH1111课程简介

This unit is an introduction to the calculus of one variable. Topics covered include elementary functions, differentiation, basic integration techniques and coordinate geometry in three dimensions. Applications in science and engineering are emphasised.

Prerequisites 

At the completion of this unit, you should be able to:

• LO1. Apply mathematical logic and rigour to solving problems, and express mathematical ideas coherently in written and oral form;
• LO2. Demonstrate fluency in manipulating real numbers, their symbolic representations, operations, and solve associated algebraic equations and inequalities;
• LO3. Develop fluency with lines, coordinate geometry in two dimensions, the notion of a function, its natural domain, range and graph;
• LO4. Become conversant with elementary functions, including trigonometric, exponential, logarithmic and hyperbolic functions and be able to apply them to real phenomena and to yield solutions of associated equations;
• LO5. Perform operations on functions and be able to invert functions where appropriate;
• LO6. Understand the definitions of a derivative, definite and indefinite integral and be able to apply the definitions to elementary functions;
• LO7. Develop fluency in rules of differentiation, such as the product, quotient and chain rules, and use them to differentiate complicated functions;
• LO8. Understand and apply the Fundamental Theorem of Calculus; and develop fluency in techniques of integration, such as integration by substitution, the method of partial fractions and integration by parts;
• LO9. Develop some fluency with coordinate geometry in three dimensions, planes, surfaces, ellipsoids, paraboloids, level curves and qualitative features such as peaks, troughs and saddle points.

MATH1111 Calculus HELP（EXAM HELP， ONLINE TUTOR）

To prove existence using Bolzano’s theorem, we note that $f$ is continuous with
$$
f(-1)=-1-4+1<0, \quad f(0)=1>0, \quad f(1)=1-4+1<0 .
$$
In view of Bolzano’s theorem, $f$ must then have a root in $(-1,0)$ and another root in $(0,1)$, so it has two roots in $(-1,1)$. Suppose that it has three roots in $(-1,1)$. Then $f^{\prime}$ must have two roots in this interval by Rolle’s theorem. On the other hand,
$$
f^{\prime}(x)=3 x^2-8 x=x(3 x-8)
$$
has only one root in $(-1,1)$. This implies that $f$ can only have two roots in $(-1,1)$.
To use Newton’s method to approximate the roots, we repeatedly apply the formula
$$
x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime}\left(x_n\right)}=x_n-\frac{x_n^3-4 x_n^2+1}{3 x_n^2-8 x_n} .
$$
Starting with the initial guess $x_1=-1$, one obtains the approximations
$$
x_2=-0.6364, \quad x_3=-0.4972, \quad x_4=-0.4735, \quad x_5=-0.4728 .
$$
Starting with the initial guess $x_1=1$, one obtains the approximations
$$
x_2=0.6, \quad x_3=0.5398, \quad x_4=0.5374, \quad x_5=0.5374 .
$$
This suggests that the two roots are roughly -0.47 and 0.53 within two decimal places.

Let $x, y$ be the two sides of the rectangle and assume that $x$ lies along the base of the triangle. Then one can relate the two sides $x, y$ by noting that
$$
\tan 60^{\circ}=\frac{y}{(a-x) / 2} \quad \Longrightarrow \quad \sqrt{3}=\frac{2 y}{a-x} \quad \Longrightarrow \quad y=\frac{\sqrt{3}}{2}(a-x) .
$$
We need to maximise the area $A$ of the rectangle and this is given by
$$
A(x)=x y=\frac{\sqrt{3}}{2} x(a-x)=\frac{\sqrt{3}}{2}\left(a x-x^2\right), \quad 0 \leq x \leq a .
$$
Since $A^{\prime}(x)=\frac{\sqrt{3}}{2}(a-2 x)$, the only points at which the maximum value may occur are the points $x=0, x=a$ and $x=\frac{a}{2}$. Since $A(0)=A(a)=0$, the maximum is $A\left(\frac{a}{2}\right)=\frac{a^2 \sqrt{3}}{8}$.

Let $x$ be the horizontal distance between the base of the ladder and the wall, and let $y$ be the vertical distance between the top of the ladder and the floor. We must then have
$$
x(t)^2+y(t)^2=5^2 \quad \Longrightarrow \quad 2 x(t) x^{\prime}(t)+2 y(t) y^{\prime}(t)=0 .
$$
At the given moment, $x^{\prime}(t)=0.2=1 / 5$ and also $x(t)=3$, so it easily follows that
$$
y^{\prime}(t)=-\frac{x(t) x^{\prime}(t)}{y(t)}=-\frac{x(t) x^{\prime}(t)}{\sqrt{5^2-x(t)^2}}=-\frac{3 / 5}{\sqrt{5^2-3^2}}=-\frac{3}{20} .
$$

We now need to determine $\theta^{\prime}$. Using the chain rule along with the quotient rule, we get
$$
\tan \theta=\frac{x}{y} \quad \Longrightarrow \quad \sec ^2 \theta \cdot \theta^{\prime}=\frac{x^{\prime} y-y^{\prime} x}{y^2} \quad \Longrightarrow \quad \theta^{\prime}=\frac{x^{\prime} y-y^{\prime} x}{y^2} \cdot \cos ^2 \theta .
$$
Since $\cos \theta=y / 5$ and the other variables are already known, we may conclude that
$$
\theta^{\prime}=\frac{x^{\prime} y-y^{\prime} x}{y^2} \cdot \cos ^2 \theta=\frac{4(1 / 5)-3(-3 / 20)}{4^2} \cdot\left(\frac{4}{5}\right)^2=\frac{1}{20} .
$$

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