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Fix prose typos and grammatical errors across 26 .qmd files

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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'
This commit is contained in:
Typo Fix Bot
2026-06-03 18:37:33 +00:00
committed by jverzani
parent b42c4492d1
commit d4b3be00a0
26 changed files with 45 additions and 45 deletions
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4
quarto/alternatives/makie_plotting.qmd
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@@ -207,7 +207,7 @@ lines(a..b, f)
As with `scatter`, `lines` returns an object that produces a graphic when displayed. As with `scatter`, `lines` returns an object that produces a graphic when displayed.
As with `scatter`, `lines` can can also be drawn using a vector of points: As with `scatter`, `lines` can also be drawn using a vector of points:
```{julia} ```{julia}
@@ -978,7 +978,7 @@ arrows(pts, dvs)
The rotational pattern becomes much clearer now. The rotational pattern becomes much clearer now.
The `streamplot` function also illustrates this phenomenon. This implements an "algorithm [that] puts an arrow somewhere and extends the streamline in both directions from there. Then, it chooses a new position (from the remaining ones), repeating the the exercise until the streamline gets blocked, from which on a new starting point, the process repeats." The `streamplot` function also illustrates this phenomenon. This implements an "algorithm [that] puts an arrow somewhere and extends the streamline in both directions from there. Then, it chooses a new position (from the remaining ones), repeating the exercise until the streamline gets blocked, from which on a new starting point, the process repeats."
The `streamplot` function expects a `Point` not a pair of values, so we adjust `f` slightly and call the function using the pattern `streamplot(g, xs, ys)`: The `streamplot` function expects a `Point` not a pair of values, so we adjust `f` slightly and call the function using the pattern `streamplot(g, xs, ys)`:

6
quarto/basics/calculator.qmd
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@@ -190,7 +190,7 @@ A right triangle has sides $a=11$ and $b=12$. Find the length of the hypotenus
##### Example ##### Example
A overview of a research paper published in [theconversation.com](https://theconversation.com/earth-harbours-20-000-000-000-000-000-ants-and-they-weigh-more-than-wild-birds-and-mammals-combined-190831) reviews six authors' work on estimating the number of ants currently on earth. This was covered in an An overview of a research paper published in [theconversation.com](https://theconversation.com/earth-harbours-20-000-000-000-000-000-ants-and-they-weigh-more-than-wild-birds-and-mammals-combined-190831) reviews six authors' work on estimating the number of ants currently on earth. This was covered in an
article in the [Washington Post](https://www.washingtonpost.com/climate-environment/2022/09/19/ants-population-20-quadrillion/). article in the [Washington Post](https://www.washingtonpost.com/climate-environment/2022/09/19/ants-population-20-quadrillion/).
@@ -374,7 +374,7 @@ The Google calculator has two built in constants, `e` and `π`. Julia provides t
pi pi
``` ```
Whereas, `e` is is not simply the character `e`, but *rather* a [Unicode](../unicode.html) character typed in as `\euler[tab]`. Whereas, `e` is not simply the character `e`, but *rather* a [Unicode](../unicode.html) character typed in as `\euler[tab]`.
```{julia} ```{julia}
@@ -797,7 +797,7 @@ numericq(val)
###### Question ###### Question
You are asked to cook chicken is an unfamiliar kitchen. Your recipe says to turn the oven to 200 Celsius, but you the oven is calibrated in Fahrenheit. Which value is closest? You are asked to cook chicken in an unfamiliar kitchen. Your recipe says to turn the oven to 200 Celsius, but the oven is calibrated in Fahrenheit. Which value is closest?
```{julia} ```{julia}
#| echo: false #| echo: false

8
quarto/basics/vectors.qmd
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@@ -102,7 +102,7 @@ Don't spend time thinking about the formulas if they are unfamiliar. The point e
Initially, our primary use of vectors will be as containers, but it is worthwhile to spend some time to discuss properties of vectors and their visualization. Initially, our primary use of vectors will be as containers, but it is worthwhile to spend some time to discuss properties of vectors and their visualization.
A line segment in the plane connects two points $(x_0, y_0)$ and $(x_1, y_1)$. The length of a line segment (its magnitude) is given by the distance formula $\sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}$. A line segment can be given a direction by assigning an initial point and a terminal point. A directed line segment has both a direction and a magnitude. A vector is an abstraction where just these two properties $-$ a **direction** and a **magnitude** $-$ are intrinsic. While a directed line segment can be represented by a vector, a single vector describes all such line segments found by translation. That is, how the the vector is located when visualized is for convenience, it is not a characteristic of the vector. In the figure above, all vectors are drawn with their tails at the position of the projectile over time. A line segment in the plane connects two points $(x_0, y_0)$ and $(x_1, y_1)$. The length of a line segment (its magnitude) is given by the distance formula $\sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2}$. A line segment can be given a direction by assigning an initial point and a terminal point. A directed line segment has both a direction and a magnitude. A vector is an abstraction where just these two properties $-$ a **direction** and a **magnitude** $-$ are intrinsic. While a directed line segment can be represented by a vector, a single vector describes all such line segments found by translation. That is, how the vector is located when visualized is for convenience, it is not a characteristic of the vector. In the figure above, all vectors are drawn with their tails at the position of the projectile over time.
We can visualize a (two-dimensional) vector as an arrow in space. This arrow has two components. We represent a vector mathematically as $\langle x,~ y \rangle$. For example, the vector connecting the point $(x_0, y_0)$ to $(x_1, y_1)$ is $\langle x_1 - x_0,~ y_1 - y_0 \rangle$. We can visualize a (two-dimensional) vector as an arrow in space. This arrow has two components. We represent a vector mathematically as $\langle x,~ y \rangle$. For example, the vector connecting the point $(x_0, y_0)$ to $(x_1, y_1)$ is $\langle x_1 - x_0,~ y_1 - y_0 \rangle$.
@@ -419,7 +419,7 @@ These properties may not all be desirable for one reason or the other and `Juli
### Arrays ### Arrays
Vectors are $1$-dimensional, but there are desires for other dimensions. Vectors are a implemented as a special case of a more general array type. Arrays are of dimension $N$ for various non-negative values of $N$. A common, and somewhat familiar, mathematical use of a $2$-dimensional array is a matrix. Vectors are $1$-dimensional, but there are desires for other dimensions. Vectors are implemented as a special case of a more general array type. Arrays are of dimension $N$ for various non-negative values of $N$. A common, and somewhat familiar, mathematical use of a $2$-dimensional array is a matrix.
Arrays can have their entries accessed by dimension and within that dimension their components. By default these are $1$-based, but other offsets are possible through the `OffsetArrays.jl` package. A matrix can refer to its values either by row and column indices or, as a matrix has linear indexing by a single index. Arrays can have their entries accessed by dimension and within that dimension their components. By default these are $1$-based, but other offsets are possible through the `OffsetArrays.jl` package. A matrix can refer to its values either by row and column indices or, as a matrix has linear indexing by a single index.
@@ -841,7 +841,7 @@ map(sin, xs)
The `map` function can be used with one or more iterators. The `map` function can be used with one or more iterators.
The `map` function can also be used in combination with `reduce`, a reduction. Reductions take a container with one or more dimensions and reduces the number of dimensions. A example might be: The `map` function can also be used in combination with `reduce`, a reduction. Reductions take a container with one or more dimensions and reduces the number of dimensions. An example might be:
```{julia} ```{julia}
sum(map(sin, xs)) sum(map(sin, xs))
@@ -861,7 +861,7 @@ There are other specialized reduction functions that reverse the order of the ma
sum(xs), prod(xs) sum(xs), prod(xs)
``` ```
These are reductions, which which fall back to a `mapreduce` call. They require a starting value (`init`) of `0` and `1` (which in this case can be determined from `xs`). The `sum` and `prod` function also allow as a first argument an initial function to map over the collection: These are reductions, which fall back to a `mapreduce` call. They require a starting value (`init`) of `0` and `1` (which in this case can be determined from `xs`). The `sum` and `prod` function also allow as a first argument an initial function to map over the collection:
```{julia} ```{julia}
sum(sin, xs) sum(sin, xs)

2
quarto/derivatives/condition.qmd
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@@ -64,7 +64,7 @@ $$
&= 0 + f'(r)\delta + \epsilon &= 0 + f'(r)\delta + \epsilon
\end{align*} \end{align*}
$$ $$
Rearranging gives $\lVert\delta/\epsilon\rVert \approx 1/\lVert f'(r)\rVert$. But the $|\delta|/|\epsilon|$ ratio is related to the the condition number: Rearranging gives $\lVert\delta/\epsilon\rVert \approx 1/\lVert f'(r)\rVert$. But the $|\delta|/|\epsilon|$ ratio is related to the condition number:
> The absolute condition number is $\hat{\kappa}_r = |f'(r)|^{-1}$. > The absolute condition number is $\hat{\kappa}_r = |f'(r)|^{-1}$.

2
quarto/derivatives/derivatives.qmd
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@@ -1574,7 +1574,7 @@ radioq(choices, answ)
###### Question ###### Question
The rate of change of volume with respect to height is $3h$. The rate of change of height with respect to time is $2t$. At at $t=3$ the height is $h=14$ what is the rate of change of volume with respect to time when $t=3$? The rate of change of volume with respect to height is $3h$. The rate of change of height with respect to time is $2t$. At $t=3$ the height is $h=14$ what is the rate of change of volume with respect to time when $t=3$?
```{julia} ```{julia}

2
quarto/derivatives/lhospitals_rule.qmd
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@@ -498,7 +498,7 @@ $$
\lim_{x \rightarrow \infty} \frac{1 - \cos(x)}{1}, \lim_{x \rightarrow \infty} \frac{1 - \cos(x)}{1},
$$ $$
as the function just oscillates. This shows that L'Hospital's rule does not apply when the limit of the the ratio of the derivatives does not exist. as the function just oscillates. This shows that L'Hospital's rule does not apply when the limit of the ratio of the derivatives does not exist.
##### Example: the assumptions matter ##### Example: the assumptions matter

2
quarto/derivatives/mean_value_theorem.qmd
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@@ -134,7 +134,7 @@ This insight holds for other types of functions:
When the derivative exists, this says the tangent line is flat. (If it had a slope, then the the function would increase by moving left or right, as appropriate, a point we pursue later.) When the derivative exists, this says the tangent line is flat. (If it had a slope, then the function would increase by moving left or right, as appropriate, a point we pursue later.)
For a continuous function $f(x)$, call a point $c$ in the domain of $f$ where either $f'(c)=0$ or the derivative does not exist a **critical** **point**. For a continuous function $f(x)$, call a point $c$ in the domain of $f$ where either $f'(c)=0$ or the derivative does not exist a **critical** **point**.

2
quarto/derivatives/more_zeros.qmd
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@@ -461,7 +461,7 @@ $$
&= 0 + f'(r)\delta + \epsilon &= 0 + f'(r)\delta + \epsilon
\end{align*} \end{align*}
$$ $$
Rearranging gives $\lVert\delta/\epsilon\rVert \approx 1/\lVert f'(r)\rVert$. But the $|\delta|/|\epsilon|$ ratio is related to the the condition number: Rearranging gives $\lVert\delta/\epsilon\rVert \approx 1/\lVert f'(r)\rVert$. But the $|\delta|/|\epsilon|$ ratio is related to the condition number:
> The absolute condition number is $\hat{\kappa}_r = |f'(r)|^{-1}$. > The absolute condition number is $\hat{\kappa}_r = |f'(r)|^{-1}$.

2
quarto/derivatives/optimization.qmd
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@@ -1361,7 +1361,7 @@ radioq(choices, answ)
###### Question ###### Question
The ladder problem has an trigonometry-free solution. We show one attributed to [Asma](http://www.mathematische-basteleien.de/ladder.htm). The ladder problem has a trigonometry-free solution. We show one attributed to [Asma](http://www.mathematische-basteleien.de/ladder.htm).
```{julia} ```{julia}

2
quarto/derivatives/taylor_series_polynomials.qmd
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@@ -1186,7 +1186,7 @@ answ = 1
radioq(choices, answ) radioq(choices, answ)
``` ```
Assuming the above is right, find the smallest value $k$ guaranteeing a error no more than $10^{-16}$. Assuming the above is right, find the smallest value $k$ guaranteeing an error no more than $10^{-16}$.
```{julia} ```{julia}

6
quarto/differentiable_vector_calculus/matrix_calculus_notes.qmd
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@@ -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 ## 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. 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 $\text{vec}(f(a))$. 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), \text{vec}(A dA) = \text{vec}(A dA I^T) = (I \otimes A) \text{vec}(dA),
@@ -861,7 +861,7 @@ $$
d(f')[dx] = f''(x)[d\tilde{x}][dx] = f''(x)[d\tilde{x}, dx]. d(f')[dx] = f''(x)[d\tilde{x}][dx] = f''(x)[d\tilde{x}, dx].
$$ $$
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*: Moreover, the following shows it is *symmetric*:

2
quarto/differentiable_vector_calculus/plots_plotting.qmd
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@@ -308,7 +308,7 @@ arrow!(p, v)
### The tangent plane ### 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} ```{julia}

2
quarto/differentiable_vector_calculus/scalar_functions_applications.qmd
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@@ -1544,7 +1544,7 @@ $$
\int \sqrt{1 + y'(x)^2} dx = L. \int \sqrt{1 + y'(x)^2} dx = L.
$$ $$
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 This is a variant of [Dido](http://www.ams.org/publications/journals/notices/201709/rnoti-p980.pdf)'s problem, described by Bandle as

4
quarto/differentiable_vector_calculus/vector_fields.qmd
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@@ -531,7 +531,7 @@ $$
J = [\nabla{f}']. J = [\nabla{f}'].
$$ $$
* For $f:R^2 \rightarrow R$, the Hessian matrix, was the matrix of $2$nd partial derivatives. This may be viewed as the total derivative of the the gradient function, $\nabla{f}$: * For $f:R^2 \rightarrow R$, the Hessian matrix, was the matrix of $2$nd partial derivatives. This may be viewed as the total derivative of the gradient function, $\nabla{f}$:
$$ $$
@@ -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 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 $(l=3, k=3)$. To visualize these four cases, we take $(l=2,k=1)$, $(l=3, k=2)$ (twice) and $(l=3, k=3)$.

2
quarto/differentiable_vector_calculus/vector_valued_functions.qmd
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@@ -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. 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} ```{julia}

6
quarto/differentiable_vector_calculus/vectors.qmd
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@@ -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. 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.) 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 ### 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: 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 @@ $$
\|\vec{u} \times \vec{v}\| \| \vec{w}\| \cos(\theta), \|\vec{u} \times \vec{v}\| \| \vec{w}\| \cos(\theta),
$$ $$
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} :::{.callout-note}

2
quarto/integral_vector_calculus/div_grad_curl.qmd
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@@ -1360,7 +1360,7 @@ $$
\nabla \cdot F = \lim \frac{1}{\Delta{V}}\oint_S F \cdot \hat{N} dS. \nabla \cdot F = \lim \frac{1}{\Delta{V}}\oint_S F \cdot \hat{N} dS.
$$ $$
Taking $V$ as a box in the curvilinear coordinates, with side lengths $h_udu$, $h_vdv$, and $h_wdw$ the surface integral is computed by projecting $F$ onto each normal area element and multiplying by the area. The task is similar to how the the divergence was derived above, only now the terms are like $\partial{(F_uh_vh_w)}/\partial{u}$ due to the scale factors ($F_u$ is the u component of $F$.) The result is: Taking $V$ as a box in the curvilinear coordinates, with side lengths $h_udu$, $h_vdv$, and $h_wdw$ the surface integral is computed by projecting $F$ onto each normal area element and multiplying by the area. The task is similar to how the divergence was derived above, only now the terms are like $\partial{(F_uh_vh_w)}/\partial{u}$ due to the scale factors ($F_u$ is the u component of $F$.) The result is:
$$ $$

6
quarto/integral_vector_calculus/double_triple_integrals.qmd
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@@ -193,7 +193,7 @@ a1, b1 = 0, 5 # R is area 20, so V = 60 = 3 ⋅ 20
hcubature(f, (a0, a1), (b0, b1)) 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} ```{julia}
@@ -795,7 +795,7 @@ $$
\iint_{(x,y): \phi(x,y) \leq 0} f(x,y) dx dy. \iint_{(x,y): \phi(x,y) \leq 0} f(x,y) dx dy.
$$ $$
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: 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 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 The exact formula to relate $(\rho, \theta, \phi)$ to $(x,y,z)$ is given by

8
quarto/integral_vector_calculus/line_integrals.qmd
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@@ -152,7 +152,7 @@ $$
which is $1$ on the path $C$. So $\int_C B\cdot\hat{T} ds = \int_C ds = 2\pi$. So the current satisfies $2\pi = \mu_0 I$, so $I = (2\pi)/\mu_0$. which is $1$ on the path $C$. So $\int_C B\cdot\hat{T} ds = \int_C ds = 2\pi$. So the current satisfies $2\pi = \mu_0 I$, so $I = (2\pi)/\mu_0$.
(Ampere's law is more typically used to find $B$ from an current, then $I$ from $B$, for special circumstances. The Biot-Savart does this more generally.) (Ampere's law is more typically used to find $B$ from a current, then $I$ from $B$, for special circumstances. The Biot-Savart does this more generally.)
### Line integrals and vector fields; work and flow ### Line integrals and vector fields; work and flow
@@ -355,7 +355,7 @@ The fact that work in a potential field is path independent is a consequence of
::: {.callout-note icon=false} ::: {.callout-note icon=false}
## The Fundamental Theorem of Line [Integrals](https://en.wikipedia.org/wiki/Gradient_theorem): ## The Fundamental Theorem of Line [Integrals](https://en.wikipedia.org/wiki/Gradient_theorem):
Let $U$ be an open subset of $R^n$, $f: U \rightarrow R$ a *differentiable* function and $\vec{r}: R \rightarrow R^n$ a differentiable function such that the the path $C = \vec{r}(t)$, $a\leq t\leq b$ is contained in $U$. Then Let $U$ be an open subset of $R^n$, $f: U \rightarrow R$ a *differentiable* function and $\vec{r}: R \rightarrow R^n$ a differentiable function such that the path $C = \vec{r}(t)$, $a\leq t\leq b$ is contained in $U$. Then
$$ $$
\int_C \nabla{f} \cdot d\vec{r} = \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. 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 ### Flow across a curve
@@ -521,7 +521,7 @@ quadgk(G, a, b)[1]
##### Example ##### 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} ```{julia}

4
quarto/integral_vector_calculus/review.qmd
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@@ -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 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: 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. 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. 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.

4
quarto/integral_vector_calculus/stokes_theorem.qmd
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@@ -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. 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: 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 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:
$$ $$

2
quarto/integrals/improper_integrals.qmd
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@@ -621,7 +621,7 @@ numericq(val)
###### Question ###### Question
Compute the the integral $\int_1^\infty \log(x)/x^2 dx$. Compute the integral $\int_1^\infty \log(x)/x^2 dx$.
```{julia} ```{julia}

2
quarto/integrals/substitution.qmd
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@@ -585,7 +585,7 @@ $$
$$ $$
Keeping in mind that that a circle with radius $a$ is an ellipse with $b=a$, we see that this gives the correct answer for a circle. Keeping in mind that a circle with radius $a$ is an ellipse with $b=a$, we see that this gives the correct answer for a circle.
## Questions ## Questions

2
quarto/integrals/surface_area.qmd
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@@ -837,7 +837,7 @@ radioq(choices, answ)
##### Questions ##### Questions
Find the surface area of the dome of sphere generated by rotating the the curve generated by $g(t) = \cos(t)$ and $f(t) = \sin(t)$ for $t$ in $0$ to $\pi/6$. Find the surface area of the dome of sphere generated by rotating the curve generated by $g(t) = \cos(t)$ and $f(t) = \sin(t)$ for $t$ in $0$ to $\pi/6$.
Numerically find the value. Numerically find the value.

4
quarto/limits/intermediate_value_theorem.qmd
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@@ -750,7 +750,7 @@ Related but different is the concept of a relative of *local extrema*:
::: {.callout-note icon=false} ::: {.callout-note icon=false}
## Local maximum, local minimum ## Local maximum, local minimum
A local maxima for $f$ is a value $f(c)$ where $c$ is in **some** *open* interval $I=(a,b)$, $I$ in the domain of $f$, and $f(c)$ is an absolute maxima for $f$ over $I$. Similarly, an local minima for $f$ is a value $f(c)$ where $c$ is in **some** *open* interval $I=(a,b)$, $I$ in the domain of $f$, and $f(x)$ is an absolute minima for $f$ over $I$. A local maxima for $f$ is a value $f(c)$ where $c$ is in **some** *open* interval $I=(a,b)$, $I$ in the domain of $f$, and $f(c)$ is an absolute maxima for $f$ over $I$. Similarly, a local minima for $f$ is a value $f(c)$ where $c$ is in **some** *open* interval $I=(a,b)$, $I$ in the domain of $f$, and $f(x)$ is an absolute minima for $f$ over $I$.
The term *local extrema* is used to describe either a local maximum or local minimum. The term *local extrema* is used to describe either a local maximum or local minimum.
@@ -883,7 +883,7 @@ plot(x -> x * exp(-x), 0, 5)
##### Example ##### Example
The tangent function does not have a *guarantee* of an absolute maximum or an minimum over $(-\pi/2, \pi/2),$ as it is not *continuous* at the endpoints. In fact, it doesn't have either extrema - it has vertical asymptotes at each endpoint of this interval. The tangent function does not have a *guarantee* of an absolute maximum or a minimum over $(-\pi/2, \pi/2),$ as it is not *continuous* at the endpoints. In fact, it doesn't have either extrema - it has vertical asymptotes at each endpoint of this interval.
##### Example ##### Example

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quarto/precalc/functions.qmd
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@@ -835,7 +835,7 @@ For example, jumping ahead a bit, the `plot` function of `Plots` expects functi
plot(Area, 0, 10) plot(Area, 0, 10)
``` ```
From the graph, we can see that that width for maximum area is $w=5$ and so $h=5$ as well. From the graph, we can see that the width for maximum area is $w=5$ and so $h=5$ as well.