some typos.

quarto/differentiable_vector_calculus/polar_coordinates.qmd

@@ -199,7 +199,7 @@ r(theta) = a * sin(k * theta)
 plot_polar(0..pi, r)
 ```
 
-This graph has radius $0$ whenever $\sin(k\theta) = 0$ or $k\theta =n\pi$. Solving means that it is $0$ at integer multiples of $\pi/k$. In the above, with $k=5$, there will $5$ zeroes in $[0,\pi]$. The entire curve is traced out over this interval, the values from $\pi$ to $2\pi$ yield negative value of $r$, so are related to values within $0$ to $\pi$ via the relation $(r,\pi +\theta) = (-r, \theta)$.
+This graph has radius $0$ whenever $\sin(k\theta) = 0$ or $k\theta =n\pi$. Solving means that it is $0$ at integer multiples of $\pi/k$. In the above, with $k=5$, there will $6$ zeroes in $[0,\pi]$. The entire curve is traced out over this interval, the values from $\pi$ to $2\pi$ yield negative value of $r$, so are related to values within $0$ to $\pi$ via the relation $(r,\pi +\theta) = (-r, \theta)$.
 
 
 ##### Example

quarto/differentiable_vector_calculus/scalar_functions.qmd

@@ -125,7 +125,7 @@ Then we can define an alternative method with just a single variable and use spl
 j(v) = j(v...)
 ```
 
-The we can call `j` with a vector or point:
+Then we can call `j` with a vector or point:
 
 
 ```{julia}

quarto/differentiable_vector_calculus/vector_valued_functions.qmd

@@ -104,7 +104,7 @@ However,  we will use a different approach, as the component functions are not n
 In `Plots`, the command `plot(xs, ys)`, where, say, `xs=[x1, x2, ..., xn]` and `ys=[y1, y2, ..., yn]`, will make a connect-the-dot plot between corresponding pairs of points. As previously discussed, this can be used as an alternative to plotting a function through `plot(f, a, b)`: first make a set of $x$ values, say `xs=range(a, b, length=100)`; then the corresponding $y$ values, say `ys = f.(xs)`; and then plotting through `plot(xs, ys)`.
 
 
-Similarly, were a third vector, `zs`, for $z$ components used, `plot(xs, ys, zs)` will make a $3$-dimensional connect the dot plot
+Similarly, were a third vector, `zs`, for $z$ components used, `plot(xs, ys, zs)` will make a $3$-dimensional connect the dot plot.
 
 
 However, our representation of vector-valued functions naturally generates a vector of points: `[[x1,y1], [x2, y2], ..., [xn, yn]]`, as this comes from broadcasting `f` over some time values. That is, for a collection of time values, `ts` the command `f.(ts)` will produce a vector of points. (Technically a vector of vectors, but points if you identify the $2$-$d$ vectors as points.)
@@ -1247,7 +1247,7 @@ $$
 \kappa = (x'y''-x''y')/\|r'\|^3. \quad (r(t) =\langle x(t), y(t) \rangle)
 $$
 
-Curvature can also be defined as derivative of the tangent vector, $\hat{T}$, *when* the curve is parameterized by arc length, a topic still to be taken up.  The vector $\vec{r}'(t)$ is the direction of motion, whereas $\vec{r}''(t)$ indicates how fast and in what direction this is changing. For curves with little curve in them, the two will be nearly parallel and the cross product small (reflecting the presence of $\cos(\theta)$ in the definition). For "curvy" curves, $\vec{r}''$ will be in a direction opposite of $\vec{r}'$ to the $\cos(\theta)$ term in the cross product will be closer to $1$.
+Curvature can also be defined as derivative of the tangent vector, $\hat{T}$, *when* the curve is parameterized by arc length, a topic still to be taken up.  The vector $\vec{r}'(t)$ is the direction of motion, whereas $\vec{r}''(t)$ indicates how fast and in what direction this is changing. For curves with little curve in them, the two will be nearly parallel and the cross product small (reflecting the presence of $\sin(\theta)$ in the definition). For "curvy" curves, $\vec{r}''$ will be in a direction orthogonal of $\vec{r}'$ to the $\sin(\theta)$ term in the cross product will be closer to $1$.
 
 
 Let $\vec{r}(t) = k \cdot \langle \cos(t), \sin(t), 0 \rangle$. This will have curvature:
@@ -1728,7 +1728,7 @@ $$
 \begin{align*}
 \vec{0} &= (\vec{U}
 + a(\alpha' - \kappa) \sin(\alpha) \vec{U}
-+ a(\alpha' - \kappa) \cos(\alpha)\vec{V}) \times
+- a(\alpha' - \kappa) \cos(\alpha)\vec{V}) \times
 (\vec{U} \cos(\alpha) +  \vec{V} \sin(\alpha)) \\
 &= (\vec{U} \times \vec{V}) \sin(\alpha) +
 a(\alpha' - \kappa) \sin(\alpha) \vec{U} \times  \vec{V} \sin(\alpha) -
@@ -1762,7 +1762,7 @@ $$
 $$
 
 
-From this $\|\vec{B}(u)\| = |\cos(\alpha)\|$. But $1 = \|d\vec{B}/dv\| = \|d\vec{B}/du \| \cdot |du/dv|$ and $|dv/du|=|\cos(\alpha)|$ follows.
+From this $\|\vec{B}'(u)\| = |\cos(\alpha)|$. But $1 = \|d\vec{B}/dv\| = \|d\vec{B}/du \| \cdot |du/dv|$ and $|dv/du|=|\cos(\alpha)|$ follows.
 
 
 
@@ -2804,7 +2804,7 @@ What is the resulting curve?
 choices = [
 "An astroid of the form ``c \\langle \\cos^3(t), \\sin^3(t) \\rangle``",
 "An cubic parabola of the form ``\\langle ct^3, dt^2\\rangle``",
-"An ellipse of the form ``\\langle a\\cos(t), b\\sin(t)``",
+"An ellipse of the form ``\\langle a\\cos(t), b\\sin(t)\\rangle``",
 "A cyloid of the form ``c\\langle t + \\sin(t), 1 - \\cos(t)\\rangle``"
 ]
 answ = 1

quarto/precalc/vectors.qmd

@@ -144,7 +144,7 @@ Two operations on vectors are fundamental.
 
 
   * Vectors can be multiplied by a scalar (a real number): $c\vec{v} = \langle cx,~ cy \rangle$. Geometrically this scales the vector by a factor of $\lvert c \rvert$ and switches the direction of the vector by $180$ degrees (in the $2$-dimensional case) when $c < 0$. A *unit vector* is one with magnitude $1$, and, except for the $\vec{0}$ vector, can be formed from $\vec{v}$ by dividing $\vec{v}$ by its magnitude. A vector's two parts are summarized by its direction given by a unit vector **and** its magnitude given by the norm.
-  * Vectors can be added: $\vec{v} + \vec{w} = \langle v_x + w_x,~ v_y + w_y \rangle$. That is, each corresponding component adds to form a new vector. Similarly for subtraction. The $\vec{0}$ vector then would be just $\langle 0,~ 0 \rangle$ and would satisfy $\vec{0} + \vec{v} = \vec{v}$ for any vector $\vec{v}$. Vector addition, $\vec{v} + \vec{w}$, is visualized by placing the tail of $\vec{w}$ at the tip of $\vec{v}$ and then considering the new vector with tail coming from $\vec{v}$ and tip coming from the position of the tip of $\vec{w}$. Subtraction is different, place both the tails of $\vec{v}$ and $\vec{w}$ at the same place and the new vector has tail at the tip of $\vec{v}$ and tip at the tip of $\vec{w}$.
+  * Vectors can be added: $\vec{v} + \vec{w} = \langle v_x + w_x,~ v_y + w_y \rangle$. That is, each corresponding component adds to form a new vector. Similarly for subtraction. The $\vec{0}$ vector then would be just $\langle 0,~ 0 \rangle$ and would satisfy $\vec{0} + \vec{v} = \vec{v}$ for any vector $\vec{v}$. Vector addition, $\vec{v} + \vec{w}$, is visualized by placing the tail of $\vec{w}$ at the tip of $\vec{v}$ and then considering the new vector with tail coming from $\vec{v}$ and tip coming from the position of the tip of $\vec{w}$. Subtraction is different, place both the tails of $\vec{v}$ and $\vec{w}$ at the same place and the new vector has tail at the tip of $\vec{w}$ and tip at the tip of $\vec{v}$.
 
 
 ```{julia}