Assignment-daixieTM为您提供剑桥大学University of Cambridge Partial differential equations and variational methods 4M12偏微分方程和变分方法代写代考和辅导服务！
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Mathematical methods are techniques used in mathematics to solve problems and analyze data. These methods are often used in scientific and engineering fields to model real-world phenomena and make predictions. Some examples of mathematical methods include:
- Calculus: This branch of mathematics deals with the study of functions and their rates of change. It is used to solve problems in physics, engineering, and economics, among other fields.
- Linear algebra: This is the study of systems of linear equations and their properties. It is used in fields such as computer graphics, physics, and economics.
- Differential equations: These are mathematical equations that describe how a quantity changes over time. They are used in physics, engineering, and other sciences to model natural phenomena.
- Statistics: This is the branch of mathematics that deals with the collection, analysis, interpretation, presentation, and organization of data. It is used in fields such as psychology, sociology, and economics.
- Numerical analysis: This is the study of algorithms and computational methods used to solve mathematical problems. It is used in fields such as engineering, finance, and science to make numerical predictions.
Overall, mathematical methods are essential tools for solving problems and understanding the world around us.
Consider the following equation $$ \text { (c.a) } \mathbf{a}+(\mathbf{c} . \mathbf{b}) \mathbf{b}=\mathbf{p} $$ where $\mathbf{a}$ and $\mathbf{b}$ are known vectors which are not parallel. (i) What is the value of $\mathbf{p} .(\mathbf{a} \times \mathbf{b})$ ?
(i) We have $\mathbf{p} .(\mathbf{a} \times \mathbf{b})=\mathbf{p} \cdot \mathbf{n}$, where $\mathbf{n}=\mathbf{a} \times \mathbf{b}$ is the normal vector to the plane spanned by $\mathbf{a}$ and $\mathbf{b}$. This dot product gives the scalar projection of $\mathbf{p}$ onto $\mathbf{n}$, which is the signed distance between $\mathbf{p}$ and the plane, multiplied by the magnitude of $\mathbf{n}$. Since $\mathbf{a}$ and $\mathbf{b}$ are not parallel, the cross product $\mathbf{a} \times \mathbf{b}$ is a nonzero vector orthogonal to both $\mathbf{a}$ and $\mathbf{b}$, and therefore defines a plane that contains both vectors. Thus, $\mathbf{p} .(\mathbf{a} \times \mathbf{b})$ gives the signed distance between $\mathbf{p}$ and the plane containing $\mathbf{a}$ and $\mathbf{b}$, multiplied by the area of the parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}$. In other words, $\mathbf{p} .(\mathbf{a} \times \mathbf{b})$ is twice the volume of the parallelepiped defined by $\mathbf{a}$, $\mathbf{b}$, and $\mathbf{p}$.
The response $y(t)$ of a linear system with input $f(t)$ is given by the convolution integral $$ y(t)=\int_0^t g(t-\tau) f(\tau) d \tau $$ where $g(t)$ is the impulse response. Both $f(t)$ and $g(t)$ are zero for $t<0$. Show that $$ \int_0^t g(t-\tau) f(\tau) d \tau=\int_0^t f(t-\tau) g(\tau) d \tau $$
We start by making a change of variables $\tau = t-s$ in the integral on the left-hand side of the equation: \begin{align*} \int_0^t g(t-\tau) f(\tau) d\tau &= \int_0^t g(t-(t-s))f(t-s)ds &&(\text{setting } s = t-\tau)\ &= \int_0^t g(s)f(t-s)ds. \end{align*} This is the convolution of $f(t-s)$ and $g(s)$, which by definition is:
$(f * g)(t)=\int_{-\infty}^{\infty} f(t-s) g(s) d s$.
However, since both $f(t)$ and $g(t)$ are zero for $t < 0$, we can restrict the integral to $s\in [0,t]$:
$(f * g)(t)=\int_0^t f(t-s) g(s) d s$.
Therefore, we have shown that
$\int_0^t g(t-\tau) f(\tau) d \tau=(f * g)(t)=\int_0^t f(t-\tau) g(\tau) d \tau$
as required.
A surface is defined by the equation $$ \mathbf{r}=u^2 v \mathbf{i}+v^2 \mathbf{j}+u v \mathbf{k} $$ Find the normal to the surface at the point where $u=1$ and $v=2$.
To find the normal to the surface at a given point, we need to take the gradient of the surface equation at that point.
The gradient of the surface equation is given by:
$\nabla \mathbf{r}=\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v}$
So, we need to find the partial derivatives of $\mathbf{r}$ with respect to $u$ and $v$, evaluate them at $u=1$ and $v=2$, and then take their cross product to find the normal vector.
Taking the partial derivatives of $\mathbf{r}$ with respect to $u$ and $v$ gives:
$\frac{\partial \mathbf{r}}{\partial u}=2 u v \mathbf{i}+v \mathbf{k}, \quad \frac{\partial \mathbf{r}}{\partial v}=u^2 \mathbf{i}+2 v \mathbf{j}+u \mathbf{k}$
Evaluating these partial derivatives at $u=1$ and $v=2$ gives:
$\left.\frac{\partial \mathbf{r}}{\partial u}\right|{(1,2)}=4 \mathbf{i}+2 \mathbf{k},\left.\quad \frac{\partial \mathbf{r}}{\partial v}\right|{(1,2)}=\mathbf{i}+4 \mathbf{j}+2 \mathbf{k}$
Taking their cross product, we get:
$\left.\nabla \mathbf{r}\right|{(1,2)}=\left.\frac{\partial \mathbf{r}}{\partial u}\right|{(1,2)} \times\left.\frac{\partial \mathbf{r}}{\partial v}\right|_{(1,2)}=\left|\begin{array}{ccc}\mathbf{i} & \mathbf{j} & \mathbf{k} \ 4 & 0 & 2 \ 1 & 4 & 2\end{array}\right|=-8 \mathbf{i}+6 \mathbf{j}-16 \mathbf{k}$
Therefore, the normal to the surface at the point $(1,2)$ is $\boxed{-8\mathbf{i} + 6\mathbf{j} – 16\mathbf{k}}$.
