# 多变量微积分代写| MULTIVARIABLE CALCULUS MATHS2001 University of Glasgow Assignment

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## Instructions:

Multivariate calculus is a fundamental course that builds on the concepts learned in single-variable calculus and extends them to functions of multiple variables. This course is important for students who intend to pursue further studies in mathematics, engineering, physics, or any field that requires a solid understanding of calculus.

The course covers the concepts of partial derivatives, multiple integrals, line integrals, surface integrals, and vector calculus. These concepts are essential in various fields such as physics, engineering, economics, and computer science.

The emphasis in this course is on methods and applications. Students will learn how to use calculus to solve problems in real-world scenarios. They will also learn how to apply calculus concepts to different areas of study, including geometry, physics, and statistics.

Overall, multivariate calculus is a challenging but rewarding course that is essential for students who want to pursue advanced studies in mathematics, science, and engineering.

a) Find the equation in the form $A x+B y+C z=D$ of the plane $\mathcal{P}$ which contains the line $L$ given by $x=1-t, \quad y=1+2 t, \quad z=2-3 t$ and the point $(-1,1,2)$

a) Line $L$ has direction vector $\mathbf{v}=\langle-1,2,-3\rangle$ which lies in $\mathcal{P}$.

To get a point $P_0$ on $L$ take $t=0 \Rightarrow P_0=(1,1,2)$.
$\Rightarrow \overrightarrow{\mathbf{P}{\mathbf{0}} \mathbf{Q}}=\langle-1,1,2\rangle-\langle 1,1,2\rangle=\langle-2,0,0\rangle$ also lies in $\mathcal{P}$. $\Rightarrow$ A normal to $\mathcal{P}$ is $$\mathbf{n}=\mathbf{v} \times \overrightarrow{\mathbf{P}{\mathbf{0}} \mathbf{Q}}=\left|\begin{array}{rrr} \mathbf{i} & \mathbf{j} & \mathbf{k} \ -1 & 2 & 3 \ -2 & 0 & 0 \end{array}\right|=\mathbf{i}(0)-\mathbf{j}(-6)+\mathbf{k}(4)=\langle 0,6,4\rangle .$$
So, the equation of $\mathcal{P}$ is
$$0(x-1)+6(y-1)+4(z-2)=0 \quad \text { or } \quad 6 y=4 z=14 \quad \text { or } \quad 3 y+2 z=7 .$$
b) $\mathbf{n}_Q=\langle 2,1,1\rangle \Rightarrow \widehat{\mathbf{n}}=\frac{1}{\sqrt{6}}\langle 2,1,1\rangle, \quad \mathbf{v}=\langle-1,2,-3\rangle \Rightarrow \widehat{\mathbf{v}}=\frac{1}{\sqrt{14}}\langle-1,2,-3\rangle$ Component of $\widehat{\mathbf{n}}$ on $\widehat{\mathbf{v}}$ is
$$\widehat{\mathbf{n}} \cdot \widehat{\mathbf{v}}=\frac{1}{\sqrt{6 \cdot 14}}(2+2-3)=-\frac{3}{\sqrt{84}}$$

Let $\mathbf{r}(t)=\left\langle\cos \left(e^t\right), \sin \left(e^t\right), e^t\right\rangle$.
a) (5 pts.) Compute and simplify the unit tangent vector $\mathbf{T}(t)=\frac{\mathbf{r}^{\prime}(t)}{\left|\mathbf{r}^{\prime}(t)\right|}$.
b) (5 pts.) Compute $\mathbf{T}^{\prime}(t)$

\begin{aligned} & \text { a) } \mathbf{r}^{\prime}(t)=\left\langle-\sin \left(\mathrm{e}^t\right) \mathrm{e}^t, \cos \left(\mathrm{e}^t\right) \mathrm{e}^t, \mathrm{e}^t\right\rangle \Rightarrow\left|\mathbf{r}^{\prime}(t)\right|=\mathrm{e}^t \sqrt{1+1}=\mathrm{e}^t \sqrt{2} \ & \Rightarrow \mathbf{T}(t)=\frac{\mathbf{r}^{\prime}(t)}{\left|\mathbf{r}^{\prime}(t)\right|}=\frac{1}{\sqrt{2}}\left\langle-\sin \left(\mathrm{e}^t\right), \cos \left(\mathrm{e}^t\right), 1\right\rangle \end{aligned}
b)
$$\mathbf{T}^{\prime}(t)=\frac{1}{\sqrt{2}}\left\langle-\cos \left(\mathrm{e}^t\right),-\sin \left(\mathrm{e}^t\right), 0\right\rangle=-\frac{\mathrm{e}^t}{\sqrt{2}}\left\langle\cos \left(\mathrm{e}^t\right), \sin \left(\mathrm{e}^t\right), 0\right\rangle .$$

Let $f(x, y)=x+4 y+\frac{2}{x y}$.
a) Find the critical point(s) of $f(x, y)$
b) Use the second-derivative test to test the critical point(s) found in part(a).

\begin{aligned} & \left.\left.\begin{array}{l} f_x=1-2 /\left(x^2 y\right)=0 \ f_y=4-2 /\left(x y^2\right)=0 \end{array}\right} \Rightarrow \begin{array}{l} x^2 y=2 \ x y^2=1 / 2 \end{array}\right} \Rightarrow x=4 y \ & \Rightarrow 4 y^3=\frac{1}{2} \Rightarrow y^3=\frac{1}{8} \Rightarrow y=\frac{1}{2} \Rightarrow x=2 . \ & \end{aligned}
There is one critical point at $(x, y)=(2,1 / 2)$.
b) $f_{x x}=4 /\left(x^2 y\right), \quad f_{y y}=4 /\left(x y^3\right), \quad f_{x y}=f_{y x}=2 /\left(x^2 y^2\right)$
$A=f_{x x}(2,1 / 2)=1, \quad C=f_{y y}(2,1 / 2)=16, \quad B=f_{x y}(2,1 / 2)=2$
$\Rightarrow A C-B^2=12>0, A>0 \Rightarrow f$ has a relative minimum at $(2,1 / 2)$.