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/differentiable_vector_calculus/matrix_calculus_notes.qmd

@@ -415,7 +415,7 @@ In calculus, we typically have $n$ and $m$ are $1$, $2$,or $3$. But that need no
 
 ## Derivatives of matrix functions
 
-What is the the derivative of $f(A) = A^2$?
+What is the derivative of $f(A) = A^2$?
 
 The function $f$ takes a $n\times n$ matrix and returns a matrix of the same size.
 
@@ -564,7 +564,7 @@ all(l == r for (l, r) ∈ zip(L, R))
 
 Now to use this relationship to recognize $df = A dA + dA A$ with the Jacobian computed from $\text{vec}(f(a))$.
 
-We have $\text{vec}(A dA + dA A) = \text{vec}(A dA) + \text{vec}(dA A)$, by obvious linearity of $\text{vec}$. Now inserting an identity matrix, $I$, which is symmteric, in a useful spot we have:
+We have $\text{vec}(A dA + dA A) = \text{vec}(A dA) + \text{vec}(dA A)$, by obvious linearity of $\text{vec}$. Now inserting an identity matrix, $I$, which is symmetric, in a useful spot we have:
 
 $$
 \text{vec}(A dA) = \text{vec}(A dA I^T) = (I \otimes A) \text{vec}(dA),
@@ -861,7 +861,7 @@ $$
 d(f')[dx] = f''(x)[d\tilde{x}][dx] = f''(x)[d\tilde{x}, dx].
 $$
 
-The last equality a definition. As $f''$ is linear in the the application to $d\tilde{x}$ and also linear in application to $dx$, $f''(x)$ is a bilinear operator.
+The last equality a definition. As $f''$ is linear in the application to $d\tilde{x}$ and also linear in application to $dx$, $f''(x)$ is a bilinear operator.
 
 Moreover, the following shows it is *symmetric*:
 

quarto/differentiable_vector_calculus/plots_plotting.qmd

@@ -308,7 +308,7 @@ arrow!(p, v)
 ### The tangent plane
 
 
-Let $z = f(x,y)$ describe a surface, and $F(x,y,z) = f(x,y) - z$. The the gradient of $F$ at a point $p$ on the surface, $\nabla F(p)$, will be normal to the surface and for a function, $f(p) + \nabla f \cdot (x-p)$ describes the tangent plane. We can visualize each, as follows:
+Let $z = f(x,y)$ describe a surface, and $F(x,y,z) = f(x,y) - z$. The gradient of $F$ at a point $p$ on the surface, $\nabla F(p)$, will be normal to the surface and for a function, $f(p) + \nabla f \cdot (x-p)$ describes the tangent plane. We can visualize each, as follows:
 
 
 ```{julia}

quarto/differentiable_vector_calculus/scalar_functions_applications.qmd

@@ -1544,7 +1544,7 @@ $$
 \int \sqrt{1 + y'(x)^2} dx = L.
 $$
 
-The latter being the formula for arc length. This is very much like a optimization problem that Lagrange's method could help solve, but with one big difference: the answer is *not* a point but a *function*.
+The latter being the formula for arc length. This is very much like an optimization problem that Lagrange's method could help solve, but with one big difference: the answer is *not* a point but a *function*.
 
 
 This is a variant of  [Dido](http://www.ams.org/publications/journals/notices/201709/rnoti-p980.pdf)'s problem, described by Bandle as

quarto/differentiable_vector_calculus/vector_fields.qmd

@@ -531,7 +531,7 @@ $$
 J = [\nabla{f}'].
 $$
 
-  * For $f:R^2 \rightarrow R$, the Hessian matrix, was the matrix of $2$nd partial derivatives. This may be viewed as the total derivative of the the gradient function, $\nabla{f}$:
+  * For $f:R^2 \rightarrow R$, the Hessian matrix, was the matrix of $2$nd partial derivatives. This may be viewed as the total derivative of the gradient function, $\nabla{f}$:
 
 
 $$
@@ -1341,7 +1341,7 @@ With this, we get the following possibilities for $f$ with a zero of order $k$ a
 
 * If $l$ is odd and $k$ is even and $f^{(k)}(b_0)$ and $f^{(l)}(c_0)$ have  *opposite* signs, the $(b_0, c_0)$ is an isolated solution.
 
-* If $l$ is add and $k$ is odd, then there are two continuous solutions, but only defined in a a one-sided neighborhood of $b_0$ where $f^{(k)}(b_0) f^{(l)}(c_0) (b - b_0) > 0$.
+* If $l$ is add and $k$ is odd, then there are two continuous solutions, but only defined in a one-sided neighborhood of $b_0$ where $f^{(k)}(b_0) f^{(l)}(c_0) (b - b_0) > 0$.
 
 
 To visualize these four cases, we take $(l=2,k=1)$, $(l=3, k=2)$ (twice) and $(l=3, k=3)$.

quarto/differentiable_vector_calculus/vector_valued_functions.qmd

@@ -558,7 +558,7 @@ vvf = [cos(t), sin(t), t]
 We will see working with these expressions is not identical to working with a vector-valued function.
 
 
-To plot, we can avail ourselves of the the parametric plot syntax. The following expands to `plot(cos(t), sin(t), t, 0, 2pi)`:
+To plot, we can avail ourselves of the parametric plot syntax. The following expands to `plot(cos(t), sin(t), t, 0, 2pi)`:
 
 
 ```{julia}

quarto/differentiable_vector_calculus/vectors.qmd

@@ -247,7 +247,7 @@ quiver([0],[0], quiver=([1],[2]))
 The cumbersome syntax, `quiver(x, y, quiver=(u, v))`, is typical here. We naturally describe vectors and points using `[a,b,c]` to combine them, but the plotting functions want to plot many such at a time and expect vectors containing just the `x` values, just the `y` values, etc. The above usage looks a bit odd, as these vectors of `x` and `y` values have only one entry.
 
 
-Converting from the one representation to the other requires reshaping the data. We will use the `unzip` function from `CalculusWithJulia` which in turn just uses the the `invert` function of the `SplitApplyCombine` package ("return a new nested container by reversing the order of the nested container") for the bulk of its work.
+Converting from the one representation to the other requires reshaping the data. We will use the `unzip` function from `CalculusWithJulia` which in turn just uses the `invert` function of the `SplitApplyCombine` package ("return a new nested container by reversing the order of the nested container") for the bulk of its work.
 
 
 This function takes a vector of vectors, and returns a tuple containing the `x` values, the `y` values, etc. So if  `u=[1,2,3]` and `v=[4,5,6]`, then `unzip([u,v])` becomes `([1,4],[2,5],[3,6])`, etc. (The `zip` function in base does essentially the reverse operation, hence the name.) Notationally, `A = [u,v]` can have the third element of the first vector (`u`) accessed by `A[1][3]`, where as `unzip(A)[3][1]` will do the same. We use `unzip([u])` in the following, which for this `u` returns `([1],[2],[3])`. (Note the `[u]` to make a vector of a vector.)
@@ -679,7 +679,7 @@ But the associative property does not, as $(\vec{u} \cdot \vec{v}) \cdot \vec{w}
 ### Cross product
 
 
-In three dimensions, there is a another operation between vectors that is similar to multiplication, though we will see with many differences.
+In three dimensions, there is another operation between vectors that is similar to multiplication, though we will see with many differences.
 
 Let $\vec{u}$ and $\vec{v}$ be two $3$-dimensional vectors, then the *cross* product, $\vec{u} \times \vec{v}$, is defined as a vector with length:
 
@@ -875,7 +875,7 @@ $$
 \|\vec{u} \times \vec{v}\| \| \vec{w}\| \cos(\theta),
 $$
 
-that is, the area of the parallelepiped. Wait, what about $(\vec{v}\times\vec{u})\cdot\vec{w}$? That will have an opposite sign. Yes, in the above, there is an assumption that $\vec{n}$ and $\vec{w}$ have a an angle between them within $[0, \pi/2]$, otherwise an absolute value must be used, as volume is non-negative.
+that is, the area of the parallelepiped. Wait, what about $(\vec{v}\times\vec{u})\cdot\vec{w}$? That will have an opposite sign. Yes, in the above, there is an assumption that $\vec{n}$ and $\vec{w}$ have an angle between them within $[0, \pi/2]$, otherwise an absolute value must be used, as volume is non-negative.
 
 
 :::{.callout-note}