Fix prose typos and grammatical errors across 26 .qmd files

Scan of all .qmd files under quarto/ found 43 genuine prose errors
in 26 files. Changes by category:

Duplicate words removed:
- alternatives/makie_plotting.qmd: 'can can' -> 'can'; 'the the' -> 'the'
- basics/vectors.qmd: 'the the' -> 'the'; 'which which' -> 'which'
- derivatives/condition.qmd: 'the the' -> 'the'
- derivatives/derivatives.qmd: 'At at' -> 'At'
- derivatives/lhospitals_rule.qmd: 'the the' -> 'the'
- derivatives/mean_value_theorem.qmd: 'the the' -> 'the'
- derivatives/more_zeros.qmd: 'the the' -> 'the'
- differentiable_vector_calculus/matrix_calculus_notes.qmd: 3 instances
- differentiable_vector_calculus/plots_plotting.qmd: 'The the' -> 'The'
- differentiable_vector_calculus/vector_fields.qmd: 'the the'; 'a a' -> 'a'
- differentiable_vector_calculus/vector_valued_functions.qmd: 'the the'
- differentiable_vector_calculus/vectors.qmd: 'the the'
- integral_vector_calculus/div_grad_curl.qmd: 'the the'
- integral_vector_calculus/double_triple_integrals.qmd: 'The the'; 'over over'
- integral_vector_calculus/line_integrals.qmd: 'the the'
- integral_vector_calculus/review.qmd: 2x 'the the'
- integral_vector_calculus/stokes_theorem.qmd: 2x 'the the'
- integrals/improper_integrals.qmd: 'the the'
- integrals/substitution.qmd: 'that that' -> 'that'
- integrals/surface_area.qmd: 'the the'
- precalc/functions.qmd: 'that that width' -> 'that the width'

Article (a/an) corrections:
- basics/calculator.qmd: 'A overview' -> 'An overview'
- basics/vectors.qmd: 'A example' -> 'An example'; 'are a implemented' -> 'are implemented'
- derivatives/optimization.qmd: 'an trigonometry-free' -> 'a trigonometry-free'
- derivatives/taylor_series_polynomials.qmd: 'a error' -> 'an error'
- differentiable_vector_calculus/scalar_functions_applications.qmd: 'a optimization' -> 'an optimization'
- differentiable_vector_calculus/vectors.qmd: 'a another' -> 'another'; 'a an angle' -> 'an angle'
- integral_vector_calculus/double_triple_integrals.qmd: 'a azimuthal' -> 'an azimuthal'
- integral_vector_calculus/line_integrals.qmd: 'an current' -> 'a current'; 'an simply' -> 'a simply'; 'an rotational' -> 'a rotational'
- limits/intermediate_value_theorem.qmd: 'an local' -> 'a local'; 'an minimum' -> 'a minimum'

Other typos:
- basics/calculator.qmd: 'is is not' -> 'is not'; 'chicken is an unfamiliar' -> 'chicken in an unfamiliar'; 'but you the oven' -> 'but the oven'
- differentiable_vector_calculus/matrix_calculus_notes.qmd: 'symmteric' -> 'symmetric'

quarto/integral_vector_calculus/div_grad_curl.qmd

@@ -1360,7 +1360,7 @@ $$
 \nabla \cdot F = \lim \frac{1}{\Delta{V}}\oint_S F \cdot \hat{N} dS.
 $$
 
-Taking $V$ as a box in the curvilinear coordinates, with side lengths $h_udu$, $h_vdv$, and $h_wdw$ the surface integral is computed by projecting $F$ onto each normal area element and multiplying by the area. The task is similar to how the the divergence was derived above, only now the terms are like $\partial{(F_uh_vh_w)}/\partial{u}$ due to the scale factors ($F_u$ is the u component of $F$.) The result is:
+Taking $V$ as a box in the curvilinear coordinates, with side lengths $h_udu$, $h_vdv$, and $h_wdw$ the surface integral is computed by projecting $F$ onto each normal area element and multiplying by the area. The task is similar to how the divergence was derived above, only now the terms are like $\partial{(F_uh_vh_w)}/\partial{u}$ due to the scale factors ($F_u$ is the u component of $F$.) The result is:
 
 
 $$

quarto/integral_vector_calculus/double_triple_integrals.qmd

@@ -193,7 +193,7 @@ a1, b1 = 0, 5  # R is area 20, so V = 60 = 3 ⋅ 20
 hcubature(f, (a0, a1), (b0, b1))
 ```
 
-  * A wedge. Let $f(x,y) = x$ and $R= [0,1] \times [0,1]$. The the volume is a wedge, and should be half the value of the unit cube, or simply $1/2$:
+  * A wedge. Let $f(x,y) = x$ and $R= [0,1] \times [0,1]$. The volume is a wedge, and should be half the value of the unit cube, or simply $1/2$:
 
 
 ```{julia}
@@ -795,7 +795,7 @@ $$
 \iint_{(x,y): \phi(x,y) \leq 0} f(x,y) dx dy.
 $$
 
-It can also integrate over over boundaries of the form $\phi(x) = 0$. The latter can be visualized through `implicit_plot`.
+It can also integrate over boundaries of the form $\phi(x) = 0$. The latter can be visualized through `implicit_plot`.
 
 The main function from `ImplicitIntegration` is `integrate`. The package is imported below to avoid naming conflicts with `SymPy`'s `integrate` function:
 
@@ -1762,7 +1762,7 @@ integrate(1 * r, (z, -sqrt(4-r^2), sqrt(4-r^2)), (r, 0, a), (theta,0, 2PI))
 #### Spherical integrals
 
 
-Spherical coordinates describe a point in space by a radius from the origin, $r$ or $\rho$; a azimuthal angle $\theta$ in $[0, 2\pi]$ and an *inclination* angle $\phi$ (also called polar angle) in $[0, \pi]$. The $z$ axis is the direction of the zenith and gives a reference line to define the inclination angle. The $x$-$y$ plane is the reference plane, with the $x$ axis giving a reference direction for the azimuth measurement.
+Spherical coordinates describe a point in space by a radius from the origin, $r$ or $\rho$; an azimuthal angle $\theta$ in $[0, 2\pi]$ and an *inclination* angle $\phi$ (also called polar angle) in $[0, \pi]$. The $z$ axis is the direction of the zenith and gives a reference line to define the inclination angle. The $x$-$y$ plane is the reference plane, with the $x$ axis giving a reference direction for the azimuth measurement.
 
 
 The exact formula to relate $(\rho, \theta, \phi)$ to $(x,y,z)$ is given by

quarto/integral_vector_calculus/line_integrals.qmd

@@ -152,7 +152,7 @@ $$
 which is $1$ on the path $C$. So $\int_C B\cdot\hat{T} ds = \int_C ds = 2\pi$. So the current satisfies $2\pi = \mu_0 I$, so $I = (2\pi)/\mu_0$.
 
 
-(Ampere's law is more typically used to find $B$ from an current, then $I$ from $B$, for special circumstances. The Biot-Savart does this more generally.)
+(Ampere's law is more typically used to find $B$ from a current, then $I$ from $B$, for special circumstances. The Biot-Savart does this more generally.)
 
 
 ### Line integrals and vector fields; work and flow
@@ -355,7 +355,7 @@ The fact that work in a potential field is path independent is a consequence of
 ::: {.callout-note icon=false}
 ## The Fundamental Theorem of Line [Integrals](https://en.wikipedia.org/wiki/Gradient_theorem):
 
-Let $U$ be an open subset of $R^n$, $f: U \rightarrow R$ a *differentiable* function and $\vec{r}: R \rightarrow R^n$ a differentiable function such that the the path $C = \vec{r}(t)$, $a\leq t\leq b$ is contained in $U$. Then
+Let $U$ be an open subset of $R^n$, $f: U \rightarrow R$ a *differentiable* function and $\vec{r}: R \rightarrow R^n$ a differentiable function such that the path $C = \vec{r}(t)$, $a\leq t\leq b$ is contained in $U$. Then
 
 $$
 \int_C  \nabla{f} \cdot d\vec{r} =
@@ -430,7 +430,7 @@ quadgk(t -> Radial(r₂(t)) ⋅ r₂'(t), 0, pi)
 Not all vector fields are conservative.  How can a vector field in $U$ be identified as conservative? For now, this would require either finding a scalar potential *or* showing all line integrals are path independent.
 
 
-In dimension $2$ there is an easy to check method assuming $U$ is *simply connected*: If $F=\langle F_x, F_y\rangle$ is continuously differentiable in an simply connected region *and* $\partial{F_y}/\partial{x} - \partial{F_x}/\partial{y} = 0$ then $F$ is conservative. A similarly statement is available in dimension $3$. The reasoning behind this will come from the upcoming Green's theorem.
+In dimension $2$ there is an easy to check method assuming $U$ is *simply connected*: If $F=\langle F_x, F_y\rangle$ is continuously differentiable in a simply connected region *and* $\partial{F_y}/\partial{x} - \partial{F_x}/\partial{y} = 0$ then $F$ is conservative. A similarly statement is available in dimension $3$. The reasoning behind this will come from the upcoming Green's theorem.
 
 
 ### Flow across a curve
@@ -521,7 +521,7 @@ quadgk(G, a, b)[1]
 ##### Example
 
 
-Example, let $F(x,y) = \langle -y, x\rangle$ be a vector field. (It represents an rotational flow.) What is the flow across the unit circle?
+Example, let $F(x,y) = \langle -y, x\rangle$ be a vector field. (It represents a rotational flow.) What is the flow across the unit circle?
 
 
 ```{julia}

quarto/integral_vector_calculus/review.qmd

@@ -299,7 +299,7 @@ The gradient is then scalar "multiplication" on the left: $\nabla{f}$.
 The divergence is the dot product on the left: $\nabla\cdot{F}$.
 
 
-The curl is the the cross product on the left: $\nabla\times{F}$.
+The curl is the cross product on the left: $\nabla\times{F}$.
 
 
 These operations satisfy two vanishing properties:
@@ -393,7 +393,7 @@ To compute integrals over non-box-like regions, Fubini's theorem may be utilized
 For a parameterized curve, $\vec{r}(t)$, the **line integral** of a scalar function between $a \leq  t \leq b$ is defined by: $\int_a^b f(\vec{r}(t)) \| \vec{r}'(t)\| dt$. For a path parameterized by arc-length, the integral is expressed by $\int_C f(\vec{r}(s)) ds$ or simply $\int_C f ds$, as the norm is $1$ and $C$ expresses the path.
 
 
-A Jordan curve in two dimensions is a non-intersecting continuous loop in the plane. The Jordan curve theorem states that such a curve divides the plane into a bounded and unbounded region. The curve is *positively* parameterized if the the bounded region is kept on the left. A line integral over a Jordan curve is denoted $\oint_C f ds$.
+A Jordan curve in two dimensions is a non-intersecting continuous loop in the plane. The Jordan curve theorem states that such a curve divides the plane into a bounded and unbounded region. The curve is *positively* parameterized if the bounded region is kept on the left. A line integral over a Jordan curve is denoted $\oint_C f ds$.
 
 
 Some interpretations: $\int_a^b \| \vec{r}'(t)\| dt$ computes the *arc-length*. If the path represents a wire with density $\rho(\vec{x})$ then $\int_a^b \rho(\vec{r}(t)) \|\vec{r}'(t)\| dt$ computes the mass of the wire.

quarto/integral_vector_calculus/stokes_theorem.qmd

@@ -89,7 +89,7 @@ In an abstract setting, Stokes' theorem says exactly this with the relationship
 The related functions will involve the divergence and the curl, previously discussed.
 
 
-Many of the the examples in this section come from either [Strang](https://ocw.mit.edu/resources/res-18-001-calculus-online-textbook-spring-2005/) or [Schey](https://www.amazon.com/Div-Grad-Curl-All-That/dp/0393925161/).
+Many of the examples in this section come from either [Strang](https://ocw.mit.edu/resources/res-18-001-calculus-online-textbook-spring-2005/) or [Schey](https://www.amazon.com/Div-Grad-Curl-All-That/dp/0393925161/).
 
 
 To make the abstract concrete, consider the one dimensional case of finding the definite integral $\int_a^b F'(x) dx$. The Riemann sum picture at the *microscopic* level considers a figure like:
@@ -419,7 +419,7 @@ annotate!(p, [(a,del, "a"), (b,-del,"b")])
 p
 ```
 
-Let $A$ label the red line, $B$ the green curve, $C$ the blue line, and $D$ the black line. Then the area is given from Green's theorem by considering half of the the line integral of $F(x,y) = \langle -y, x\rangle$ or $\oint_C (xdy - ydx)$. To that matter we have:
+Let $A$ label the red line, $B$ the green curve, $C$ the blue line, and $D$ the black line. Then the area is given from Green's theorem by considering half of the line integral of $F(x,y) = \langle -y, x\rangle$ or $\oint_C (xdy - ydx)$. To that matter we have:
 
 
 $$