[Analysis] Various fixes

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/differentiation/06_critical_points.tex

@@ -66,4 +66,4 @@ Finally, evaluate if the points are minima or maxima. It is often easiest to com
 where the lowest value is the global minimum and the highest value the global maximum (obviously).
 Always consider the corners as possible maxima or minima (if some corners are critical points, all are highly likely to be).
 
-The tangent plane at a critical point of a function $f : \R^n \rightarrow \R$, is of the form $\{ (x, y, z) \dividees z = \text{const} \}$, with $z = f(x_0)$.
+The tangent plane at a critical point of a function $f : \R^n \rightarrow \R$, is of the form $\{ (x, y, z) \divides z = \text{const} \}$, with $z = f(x_0)$.

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/00_line_integrals.tex

@@ -32,7 +32,7 @@ we have $\displaystyle\int_{\gamma} f(s) \cdot \dx \vec{s} = \int_{\sigma} f(s)
 
 \mrmvspace
 \setLabelNumber{all}{8}
-\compactdef{Conservative Vector Field} If for any $x_1, x_2 \in X$ the line integral $\displaystyle\int_{\gamma} f(s) \dx \vec{s}$ is of the independent choice of $\gamma$ in $X$
+\compactdef{Conservative Vector Field} If for any $x_1, x_2 \in X$ the line integral $\displaystyle\int_{\gamma} f(s) \dx \vec{s}$ is independent of the choice of $\gamma$ in $X$
 
 \mrmvspace
 \shortremark $f$ conservative iff $\int_{\gamma} f(s) \dx \vec{s} = 0$ for a \textit{closed} ($\gamma(a) = \gamma(b)$) parametrized curve\\
@@ -65,7 +65,7 @@ $\text{curl}(f) = \begin{bmatrix}
     \end{bmatrix}$
 \dnrmvspace
 
-If $\text{curl}(f) = 0$, then $f$ is irrational.
+If $\text{curl}(f) = 0$, then $f$ is \bi{irrational}.
 Below a chart to figure out some properties:
 \begin{center}
     \begin{tikzpicture}[node distance = 0.5cm and 0.5cm, >={Classical TikZ Rightarrow[width=7pt]}]

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/01_int_in_rn.tex

@@ -68,6 +68,7 @@ Image of par. curve $\gamma: [a, b] \rightarrow \R^n$ is negligible, since $\gam
 \rmvspace
 \bi{How to compute the integral:} We compute each integral "inside out". For a definite integral, don't just find the anti-derivative, compute the actual integral!
 For an integral as seen in the harder example, we compute it as we normally would, simply using the $\pm 2x$ as the $a$ and $b$.
+For \bi{set unions} (i.e. $S = A \cup B$): $\int_{A} f(\vec{x}) \dx \vec{x} + \int_B f(\vec{x}) \dx \vec{x}$
 
 Using a change of variables into polar coordinates may come in handy, e.g. for a set like $\{ (x, y) \in \R^2 \divides 1 \leq x^2 + y^2 \leq 4 \}$,
 we can use polar coordinates and the integral is then $\int_{0}^{2\pi} \int_{1}^{2} f(x, y) \dx r \dx \varphi$ (or flipped of course)

semester3/analysis-ii/cheat-sheet-jh/parts/vectors/integration/04_green_formula.tex

@@ -53,7 +53,7 @@ That set is derived from the image that is given for the line.
 Be cognizant of what direction the integral goes, if the set is on the right hand side of the curve, the final result has to be negated to change the direction of the integral.
 If the curve doesn't fully enclose the set, then we can simply compute the line integrals of the missing sections and subtract them from the final result.
 
-We can also use known formulas to compute the area of discs, etc (like $r^2 \cdot \pi$ for a circle).
+We can also use \hl{known formulas} to compute the area of discs, etc (like $2 \cdot r \cdot \pi$ for the outline of a circle).
 To calculate the area enclosed by a curve using Green's formua, if not given a vector field, we can use the vector field $F(x, y) = (0, x)$.
 
 \shade{gray}{Center of mass}
@@ -65,3 +65,5 @@ The center of mass of an object $\cU$ is given by $\displaystyle \overline{x}_i
 \rmvspace
 \shade{gray}{Matrix-Vector product} Given vector $v\in \R^m$ and matrix $A \in \R^{n \times m}$,
 we have $A \cdot v = u$ where $ \displaystyle u_j = \sum_{i = 1}^{m} v_i \cdot A_{j, i}$.
+
+\hl{Important} If you can't see the set parametrization, it is likely faster to directly do line integrals.