%\documentclass[12pt]{article}
\documentclass[12pt]{scrartcl}
\title{AE498 SDO - HW1}
\nonstopmode
%\usepackage[utf-8]{inputenc}
\usepackage{graphicx} % Required for including pictures
\usepackage[figurename=Figure]{caption}
\usepackage{float} % For tables and other floats
\usepackage{verbatim} % For comments and other
\usepackage{amsmath}
\usepackage{physics}% For math
\usepackage{amssymb} % For more math
\usepackage{fullpage} % Set margins and place page numbers at bottom center
\usepackage{paralist} % paragraph spacing
\usepackage{listings} % For source code
\usepackage{subfig} % For subfigures
%\usepackage{physics} % for simplified dv, and
\usepackage{enumitem} % useful for itemization
\usepackage{siunitx} % standardization of si units
\begin{align*}
\nabla \theta &= \begin{pmatrix} \pdv{\theta}{R}\\[6pt] \pdv{\theta}{H}\end{pmatrix}=\begin{pmatrix} 4\pi R+2\pi H\\ 2\pi R\end{pmatrix}\\[6pt]
\nabla h &= \begin{pmatrix} \pdv{h}{R}\\[6pt] \pdv{h}{H}\end{pmatrix}=\begin{pmatrix} \frac{2(0.001)}{\pi R^3}\\ 1\end{pmatrix}
\end{align*}
Equation (1) yields 2 equations and equation (2) yields 1 equation. Therefore we have 3 equations to solve for 3 unknowns ($R, H,\lambda$).
\begin{align*}
4\pi R+2\pi H\ &= \lambda \frac{0.002}{\pi R^3} \\[6pt]
2\pi R&= \lambda,\qquad R=\frac{\lambda}{2\pi} \\[6pt]
h(R,H)&=H-\frac{0.001}{\pi R^2} =0, \qquad H=\frac{0.004\pi}{ \lambda^2} \\[6pt]
\end{align*}
答案1
为了进行比较,我只在第一个方程中增加了向量的大小。为此我\displaystyle
之前使用过pdv
:
\documentclass[12pt]{scrartcl}
\title{AE498 SDO - HW1}
\nonstopmode
\usepackage[utf8]{inputenc}% <-- corrected
\usepackage{graphicx} % Required for including pictures
\usepackage[figurename=Figure]{caption}
\usepackage{float} % For tables and other floats
\usepackage{verbatim} % For comments and other
\usepackage{amsmath}
\usepackage{physics}% For math
\usepackage{amssymb} % For more math
\usepackage{fullpage} % Set margins and place page numbers at bottom center
\usepackage{paralist} % paragraph spacing
\usepackage{listings} % For source code
\usepackage{subfig} % For subfigures
%\usepackage{physics} % for simplified dv, and <-- it is sufficient to load ones
\usepackage{enumitem} % useful for itemization
\usepackage{siunitx} % standardization of si units
\begin{document}
\begin{align*}
\nabla\theta & = \begin{pmatrix}
\displaystyle\pdv{\theta}{R}\\[6pt] % corrected size by \displaystyle
\displaystyle\pdv{\theta}{H} % corrected size
\end{pmatrix}
= \begin{pmatrix}
4\pi R+2\pi H\\
2\pi R\end{pmatrix} \\[6pt]
\nabla h & = \begin{pmatrix}
\pdv{h}{R}\\[6pt] % not corrected
\pdv{h}{H} % not corrected
\end{pmatrix}
= \begin{pmatrix}
\dfrac{2(0.001)}{\pi R^3}\\
1
\end{pmatrix}
\end{align*}
Equation (1) yields 2 equations and equation (2) yields 1 equation. Therefore we have 3 equations to solve for 3 unknowns ($R, H,\lambda$).
\begin{gather*}
4\pi R+2\pi H
= \lambda \frac{0.002}{\pi R^3} \\
2\pi R
= \lambda,\quad R=\frac{\lambda}{2\pi} \\
h(R,H)
= H-\frac{0.001}{\pi R^2} =0, \quad H=\frac{0.004\pi}{ \lambda^2}
\end{gather*}
\end{document}