# 工程数学|ENG1005 Engineering mathematics代写 monash代写

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For $f(x)=3 x^{3}-x$ we have
\begin{aligned} f(-x) &=3(-x)^{3}-(-x)=-3 x^{3}+x \ &=-\left(3 x^{3}-x\right)=-f(x) \end{aligned}
So $3 x^{3}-x$ is an odd function.
For $f(x)=\frac{x^{2}}{1+x^{2}}$
$$f(-x)=\frac{(-x)^{2}}{1+(-x)^{2}}=\frac{x^{2}}{1+x^{2}}=f(x)$$
so this is even.

If $f(x)=\frac{2 x}{x^{2}-1}$ then
$$f(-x)=\frac{2(-x)}{(-x)^{2}-1}=-\frac{2 x}{x^{2}-1}=-f(x)$$
so $f(x)$ is odd.
If $f(x)=\frac{x^{2}}{x+1}$ we have
$$f(-x)=\frac{(-x)^{2}}{-x+1}=\frac{x^{2}}{1-x}$$
This is not equal to $f(x)$ or $-f(x)$ and so this function is neither odd nor even.

## ENG1005 COURSE NOTES ：

$$y=f(x)=\frac{1}{x-2} \quad(x \neq 2)$$
If $y=4$, find $x$.
We have
$$4=\frac{1}{x-2}$$
so
$$4(x-2)=1=4 x-8$$
and therefore
$$4 x=9$$
and
$$x=\frac{9}{4}$$

# 微分方程与建模|MTH2032 Differential equations with modelling代写 monash代写

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Let us denote the steady-state number of infectives by $\hat{I}$ and the steadystate number of susceptibles by $\hat{S}$, where $\hat{I}$ and $\hat{S}$ are constants. Then,
$$I_{k+1}=I_{k}=\hat{I} \text { and } S_{k+1}=S_{k}=\hat{S}$$
are the steady-state solutions. Substituting (4) into (3) we obtain
$$\hat{I}=f \hat{S} \hat{I}$$

$$\hat{S}=\hat{S}-f \hat{S} \hat{I}+B,$$
or
\begin{aligned} \hat{I}(1-f \hat{S}) &=0 \ f \hat{S} \hat{I}-B &=0 . \end{aligned}
Our aim is to solve for $\hat{I}$ and $\hat{S}$. From equation (5a) there are two cases to be considered: $\hat{I}=0$ or $\hat{S}=1 / f$.

## MTH2032 COURSE NOTES ：

$$Y_{k+1}=(1-r) Y_{k}$$
The solutions of the linear differenceare only approximate solutions of but they have the advantage that they can be found in closed form.
The closed-form solution of is
$$Y_{k}=(1-r)^{k} Y_{0}$$
where $Y_{0}$ can be calculated from the initial population $N_{0}$ using. Hence an approximate solution to the original is given by
$$N_{k} \simeq K+K(1-r)^{k} Y_{0}$$

# 多变量微积分|MTH2010 – Multivariable calculus代写 monash代写

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As the picture illustrates, a point $X=(x, y)$ lies on $\ell$ if and only if the vector $\mathbf{X}$ is of the form $\mathbf{P}{\mathbf{0}}+t \mathbf{v}$, for some (positive or negative) scalar $t$. (For the point $X$ shown, $t=2$.) Therefore, points on the line are those that satisfy the vector equation $$(x, y)=\left(x{0}, y_{0}\right)+t(a, b)$$

for some real number $t$. This means that we can think of the line as the image of the vectorvalued function $\mathbf{X}: \mathbb{R} \rightarrow \mathbb{R}^{2}$ defined for all real numbers $t$ by $\mathbf{X}(t)=\left(x_{0}, y_{0}\right)+t(a, b)$. In the language of parametric equations, $\ell$ can be parametrized by
$$x(t)=x_{0}+a t ; \quad y(t)=y_{0}+b t ; \quad-\infty<t<\infty .$$
As $t$ ranges through all real numbers, $\mathbf{X}(t)$ traces out the entire line, which is infinite in both directions.

## MTH2010 COURSE NOTES ：

The line is determined either by the vector-valued function
$$\mathbf{X}(t)=(0.28,-0.96,1.67)+t(.96,0.28,0.33)$$
or, equivalently, by the parametric equations
$$x=0.28+0.96 t ; \quad y=-0.96+0.28 t ; \quad z=1.67+0.33 t,$$
whichever we please. For good measure, here is yet another equivalent description:
$$\mathbf{X}(t)=(0.28+0.96 t) \mathbf{i}+(-0.96+0.28 t) \mathbf{j}+(1.67+0.33 t) \mathbf{k}$$
Notice that time $t=0$ corresponds to the point $(0.28,-0.96,1.67)$-no matter the form.

# 多变量微积分Multivariable Calculus MATH1011

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If $g: I \longrightarrow U \subseteq \mathbb{R}^{2}$ is a smooth embedding and if $\omega$ is a differential 1-form on $U, g^{} \omega$ is the differential 1-form on $I$ defined by $$g^{} \omega \triangleq P \circ g \frac{d x}{d t} d t+Q \circ g \frac{d y}{d t} d t$$
where
$$\omega \triangleq P d x+Q d y$$
and
$$g(t)=\left[\begin{array}{l} x(t) \ y(t) \end{array}\right]$$

## MATH1011COURSE NOTES ：

\begin{aligned} \omega & \triangleq P d x \wedge d y \ g^{} \omega & \triangleq(P \circ g) d x \wedge d y \end{aligned} and $$\left[\begin{array}{l} d x \ d y \end{array}\right]=\left[\begin{array}{ll} \frac{\partial x}{\partial s} & \frac{\partial x}{\partial t} \ \frac{\partial y}{\partial s} & \frac{\partial y}{\partial t} \end{array}\right]\left[\begin{array}{l} d s \ d t \end{array}\right]$$ where allows us to calculate $d x \wedge d y$ in terms of $d s \wedge d t$. This gives: \begin{aligned} d x \wedge d y &=\left(\frac{\partial x}{\partial s} d s+\frac{\partial x}{\partial t} d t\right) \wedge\left(\frac{\partial y}{\partial s} d s+\frac{\partial y}{\partial t} d t\right) \ &=\left(\frac{\partial x}{\partial s} \frac{\partial y}{\partial t}-\frac{\partial x}{\partial t} \frac{\partial y}{\partial s}\right) d s \wedge d t \end{aligned} So $$g^{}(\omega)=P \circ g\left(\frac{\partial x}{\partial s} \frac{\partial y}{\partial t}-\frac{\partial x}{\partial t} \frac{\partial y}{\partial s}\right) d s \wedge d t$$

# 多变量微积分 Calculus of Several Variables MATH20132

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$$g(x)=\sqrt{2 x+4}, \quad x \in[0,6] .$$
The domain of $g$ is given as the closed interval $[0,6]$. At $x=0, g$ takes on the value 2 :
$$g(0)=\sqrt{2 \cdot 0+4}=\sqrt{4}=2 \text {; }$$
at $x=6, g$ has the value 4 :
$$g(6)=\sqrt{2 \cdot 6+4}=\sqrt{16}=4$$

The most elementary way to sketch the graph of a function is to plot points. We plot enough points so that we can “see” what the graph may look like and then connect the points with a “curve.” Of course, if we can identify the curve in advance (for example, if we know that the graph is a straight line, a parabola, or some other familiar curve), then it is much easier to draw the graph.
The graph of the squaring function
$$f(x)=x^{2}, \quad x \in(-\infty, \infty)$$
The points that we plotted are indicated in the table and marked on the graph. The graph of the function
$$g(x)=\sqrt{2 x+4}, \quad x \in[0,6]$$

## MATH20132 COURSE NOTES ：

For even integers $n,(-x)^{n}=x^{n}$; for odd integers $n,(-x)^{n}=-x^{n}$. These simple observations prompt the following definitions:
A function $f$ is said to be even if
$$f(-x)=f(x) \quad \text { for all } \quad x \in \operatorname{dom}(f) ;$$
a function $f$ is said to be odd if
$$f(-x)=-f(x) \quad \text { for all } \quad x \in \operatorname{dom}(f) .$$
The graph of an even function is symmetric about the $y$-axis, and the graph of an odd function is symme tric about the origin. (Figures $1.5 .7$ and 1.5.8.)
The absolute value function is even:
$$f(-x)=|-x|=|x|=f(x) .$$

# 多变量微积分|Multivariate Calculus代写 MT2503

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This tells us that provided the form is defined on all of $\mathbb{R}^{3}$, which it is, then it must be exact, and the corresponding vector field is conservative. So there must be a 0 -form $f$. We must have
$$\frac{\partial f}{\partial x}=2 x y^{3} z^{4}$$
so integrating with respect to $x$ we get
$$x^{2} y^{3} z^{4}+u(y, z)$$

for some unknown function of $y$ and $z$ only. Similar integration for the other two functions leads us to the conclusion that $u=\mathbf{0}$ and
$$f(x, y, z)=x^{2} y^{3} z^{4}$$

## MT2503 COURSE NOTES ：

Find an expression for the arc length of the graph of $y=x^{2}$ from the origin to the point $[1,1]^{T}$. Using Mathematica or otherwise, find the length of the curve.

Solution We can write the problem as $\int_{c} d \ell$ where $d \ell$ is an ‘infinitesimal’ bit of the curve. It is reasonable to write
$$d \ell=\sqrt{(d x)^{2}+(d y)^{2}}$$
Parametrise the curve by
$$\left[\begin{array}{l} x \ y \end{array}\right]=\left[\begin{array}{c} t \ t^{2} \end{array}\right]$$
Then note that the quantity $d \ell$ becomes just the norm of the differential term so we get:
$\int_{0}^{1}\left|\begin{array}{c}\dot{x}(t) \ \dot{y}(t)\end{array}\right| d t$
$=\int_{0}^{1} \sqrt{1+4 t^{2}} d t$