# 微积分代写|CALCULUS II MATH102 University of Liverpool Assignment

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## Instructions:

Part I of the module focuses on power series and their applications. Power series are infinite series that can represent functions as a sum of powers of a variable. You will learn about the properties of power series and their convergence, as well as the relationship between power series and functions, including how to use Taylor and Maclaurin series to approximate functions.

Part II of the module introduces functions of several variables and partial differentiation. You will learn how to find partial derivatives, including the chain rule, total differential, directional derivative, and tangent planes. Additionally, you will learn how to find extrema of functions of several variables and how to use Taylor expansions to approximate functions.

Finally, part III of the module focuses on double integrals and their applications. You will learn how to evaluate double integrals using different methods, including iterated integrals and change of variables. You will also learn how to use double integrals to find areas, volumes, and centroids of two-dimensional and three-dimensional objects.

Overall, this module, together with MATH101 and MATH103, provides a strong foundation in calculus and linear algebra, which is essential for further studies in mathematics and many other fields, such as physics, engineering, and economics.

$\vec{R} \cdot \vec{V}=0$, so $\frac{d}{d t}(\vec{R} \cdot \vec{V})=\vec{V} \cdot \vec{V}+\vec{R} \cdot \vec{A}=0$. Therefore $\vec{R} \cdot \vec{A}=-|\vec{V}|^2$.

We can start by differentiating the expression $\vec{R} \cdot \vec{V}$ with respect to time $t$ using the product rule:

$$\frac{d}{dt}(\vec{R} \cdot \vec{V}) = \frac{d \vec{R}}{dt} \cdot \vec{V} + \vec{R} \cdot \frac{d \vec{V}}{dt}$$

Since $\vec{R} \cdot \vec{V}$ is a scalar quantity, its derivative must be a scalar as well. We can simplify the above expression using the fact that $\vec{R}$ and $\vec{V}$ are perpendicular (i.e. $\vec{R} \cdot \vec{V} = 0$), which implies that $\frac{d \vec{R}}{dt} \cdot \vec{V} = 0$. Therefore:

$$\frac{d}{dt}(\vec{R} \cdot \vec{V}) = \vec{R} \cdot \frac{d \vec{V}}{dt}$$

Now, we can substitute the expression for the acceleration $\vec{A} = \frac{d \vec{V}}{dt}$:

$$\frac{d}{dt}(\vec{R} \cdot \vec{V}) = \vec{R} \cdot \vec{A}$$

Using the chain rule, we can rewrite the left-hand side as:

$$\frac{d}{dt}(\vec{R} \cdot \vec{V}) = \frac{d\vec{R}}{dt} \cdot \vec{V} + \vec{R} \cdot \frac{d\vec{V}}{dt} = \vec{V} \cdot \vec{V} + \vec{R} \cdot \vec{A}$$

Substituting this into the previous expression, we get:

$$\vec{V} \cdot \vec{V} + \vec{R} \cdot \vec{A} = 0$$

Solving for $\vec{R} \cdot \vec{A}$, we get:

$$\vec{R} \cdot \vec{A} = -|\vec{V}|^2$$

Therefore, we have shown that $\vec{R} \cdot \vec{A} = -|\vec{V}|^2$ given the initial condition $\vec{R} \cdot \vec{V} = 0$.

a) $P=(1,0,0), Q=(0,2,0)$ and $R=(0,0,3)$. Therefore $\overrightarrow{Q P}=\hat{\boldsymbol{\imath}}-2 \hat{\boldsymbol{\jmath}}$ and $\overrightarrow{Q R}=-2 \hat{\boldsymbol{\jmath}}+3 \hat{\boldsymbol{k}}$.

To calculate the vector $\overrightarrow{QP}$, we subtract the coordinates of $P$ from the coordinates of $Q$:

$$\overrightarrow{QP} = \begin{pmatrix}0-1 \ 2-0 \ 0-0\end{pmatrix} = \begin{pmatrix}-1 \ 2 \ 0\end{pmatrix} = -1\hat{\boldsymbol{\imath}} + 2\hat{\boldsymbol{\jmath}}$$

To calculate the vector $\overrightarrow{QR}$, we subtract the coordinates of $R$ from the coordinates of $Q$:

$$\overrightarrow{QR} = \begin{pmatrix}0-0 \ 2-0 \ 3-0\end{pmatrix} = \begin{pmatrix}0 \ 2 \ 3\end{pmatrix} = 2\hat{\boldsymbol{\jmath}} + 3\hat{\boldsymbol{k}}$$

So, as you said, $\overrightarrow{QP} = \hat{\boldsymbol{\imath}} – 2 \hat{\boldsymbol{\jmath}}$ and $\overrightarrow{QR} = -2\hat{\boldsymbol{\jmath}} + 3\hat{\boldsymbol{k}}$.

$\vec{N} \cdot \vec{r}(t)=6$, where $\vec{N}=\langle 4,-3,-2\rangle$.

The equation $\vec{N} \cdot \vec{r}(t)=6$ represents a plane in three-dimensional space, where $\vec{N}$ is the normal vector to the plane and $\vec{r}(t)$ is a vector pointing to any point on the plane at time $t$.

Given that $\vec{N}=\langle 4,-3,-2\rangle$, we can write the equation of the plane as:

$$4x – 3y – 2z = 6$$

where $x$, $y$, and $z$ are the coordinates of any point on the plane.

Alternatively, we can express the equation in vector form as:

$$\langle 4,-3,-2\rangle \cdot \langle x,y,z\rangle = 6$$

or

$$\langle 4,-3,-2\rangle \cdot \vec{r}(t) = 6$$

where $\vec{r}(t)=\langle x(t),y(t),z(t)\rangle$ is a vector function that describes the position of any point on the plane at time $t$.

Note that there are infinitely many points on the plane described by this equation, since the equation has three variables but only one constraint. Any point that satisfies the equation $4x – 3y – 2z = 6$ lies on the plane, and there are infinitely many such points.