tweaks

quarto/README-quarto.md

@@ -92,6 +92,12 @@ DONE * modify sympy's show method
 
 ## TODO
 
+* Old and new astronomy; : Proctor, Richard Anthony, 1837-1888 : Free Download, Borrow, and Streaming : Internet Archive
+
+
+https://archive.org/details/oldnewastronomy00procuoft
+https://ia800908.us.archive.org/16/items/oldnewastronomy00procuoft/oldnewastronomy00procuoft.pdf has interesting epicycle figures on p 186/886. Could add these in.
+
 * ImageFile -> ![]() as much as possible
 * use pandoc references(?)
 * mermaid, ojs?
@@ -103,3 +109,19 @@ DONE * remove pinned header
 DONE * clean up directory
 DONE (?) * JSXGraph files
 WON'T DO * download links to Pluto .jl files (if we have .jmd, but we might deprecate...) For *now* .jmd is derprecated; though we keep the files around ....
+
+## Cone, general
+# https://discourse.julialang.org/t/general-plotting-code-for-cone-in-3d-with-glmakie-or-plots/92104/3
+
+basecurve(u) = [cos(u), sin(u) + sin(u/2), 0]
+Vertex = [0, 3/4, 3]
+function Cone(u, t)
+	x, y, z = basecurve(u)
+	t * ([x,y,z] - Vertex) + Vertex
+end
+
+us = range(0, 2pi, length=50)
+ts = range(0, 1, length=50)
+
+xyzs = Cone.(us, ts')
+surface(CalculusWithJulia.unzip(xyzs))

quarto/derivatives/first_second_derivatives.qmd

@@ -668,6 +668,16 @@ answ = 3
 radioq(choices, answ)
 ```
 
+###### Question
+
+Consider the function `f(x) = airyai(x)`. What is the value of $f''(x) - x\cdot f(x)$ at $x=\pi$?
+
+```{julia}
+#| echo: false
+numericq(0; hint=raw"Any value of $x$ will produce the same answer")
+```
+
+
 ###### Question
 
 

quarto/differentiable_vector_calculus/plots_plotting.qmd

@@ -373,7 +373,7 @@ surface(xs, ys, zs)
 :::{.callout-note}
 ## Note
 The above may not work with all backends for `Plots`, even if those that support 3D graphics.
-    :::
+:::
 
 For convenience, the `plot_parametric` function from `CalculusWithJulia` can produce these plots using interval notation, `a..b`,  and a function:
 
@@ -384,6 +384,31 @@ F(theta, phi) = [X(1, theta, phi), Y(1, theta, phi), Z(1, theta, phi)]
 plot_parametric(0..pi, 0..pi/2, F)
 ```
 
+##### Example, the general cone
+
+The general equation of cone consists of a vertex, $V$, and a base curve. The cone consists of all line segments connecting $V$ to the base curve, here parameterized and in the $x-y$ plane: $r(u) = \langle x(u), y(u), 0 \rangle$. The equations for the cone are:
+
+$$
+\frac{x - V_x}{x(u)-V_x} = \frac{y - V_y}{y(u)-V_y} = \frac{z - V_z}{z(u)-V_z} = t,
+$$
+
+where $t \in [0,1]$ This gives a vector-valued function
+
+$$
+F(u, t) = t * (r(u) - V) + V, \quad \alpha \leq u \leq \beta, 0 \leq t \leq 1.
+$$
+
+To illustrate, we have:
+
+```{julia}
+# cf. https://discourse.julialang.org/t/general-plotting-code-for-cone-in-3d-with-glmakie-or-plots/92104/3
+
+basecurve(u) = [cos(u), sin(u) + sin(u/2), 0]
+Vertex = [0, 3/4, 3]
+Cone(u, t) = t * (basecurve(u) - Vertex) + Vertex
+plot_parametric(0..2pi, 0..1, Cone)
+```
+
 ### Plotting  F(x,y, z) = c
 
 

quarto/differentiable_vector_calculus/vector_valued_functions.qmd

@@ -232,6 +232,7 @@ plot_parametric(0..2pi, f, legend=false)
 scatter!([0],[0], markersize=4)
 ```
 
+
 ##### Example
 
 

quarto/index.qmd

@@ -11,7 +11,7 @@ This is a set of notes for learning
 programming language with an easy to learn syntax that is well suited
 for this task.
 
-Read "[Getting started with Julia](./misc/getting_started_with_julia.html)" to learn how to install and customize  `Julia` for following along with these notes. Read "[Julia interfaces](.misc/julia_interfaces.html) to review different ways to interact with a `Julia` installation.
+Read "[Getting started with Julia](./misc/getting_started_with_julia.html)" to learn how to install and customize  `Julia` for following along with these notes. Read "[Julia interfaces](./misc/julia_interfaces.html) to review different ways to interact with a `Julia` installation.
 
 Since the mid 90s there has been a push to teach calculus using many
 different points of view. The

quarto/integral_vector_calculus/double_triple_integrals.qmd

@@ -716,6 +716,9 @@ endpoints(ys, x) = callf.(ys, Ref(x))
     xs)
 ```
 
+The one issue with the above is the tolerances should increase for each outer integration. Otherwise, a certain precision is expected for a noisier integrand. This isn't an issue for many integrands, but can become one.
+
+
 ##### Example
 
 

quarto/integrals/area.qmd

@@ -608,26 +608,23 @@ This value is about $10^{-5}$ off from the actual answer of $1/3$.
 We should expect that larger values of $n$ will produce better approximate values, as long as numeric issues don't get involved.
 
 
-Before continuing, we define a function to compute the Riemann sum for us with an extra argument to specifying one of four methods for computing $c_i$:
+Before continuing, we define a function to compute  Riemann sums for us with an extra argument to specifying one of four methods for computing $c_i$:
 
 
 ```{julia}
 #| eval: false
-function riemann(f::Function, a::Real, b::Real, n::Int; method="right")
-    if method == "right"
-        meth = f -> (lr -> begin l,r = lr; f(r) * (r-l) end)
-    elseif method == "left"
-        meth = f -> (lr -> begin l,r = lr; f(l) * (r-l) end)
-    elseif method == "trapezoid"
-        meth = f -> (lr -> begin l,r = lr; (1/2) * (f(l) + f(r)) * (r-l) end)
-    elseif method == "simpsons"
-        meth = f -> (lr -> begin l,r=lr; (1/6) * (f(l) + 4*(f((l+r)/2)) + f(r)) * (r-l) end)
-    end
-
-    xs = range(a, b, n+1)
-    pairs = zip(xs[begin:end-1], xs[begin+1:end]) # (x₀,x₁), …, (xₙ₋₁,xₙ)
-    sum(meth(f), pairs)
-
+riemann(f, a, b, n; method="right") = riemann(f, range(a,b,n+1); method=method)
+function riemann(f, xs; method="right")
+    Ms = (left = (f,a,b) -> f(a),
+          right= (f,a,b) -> f(b),
+          trapezoid = (f,a,b) -> (f(a) + f(b))/2,
+          simpsons  = (f,a,b) -> (c = a/2 + b/2; (1/6) * (f(a) + 4*f(c) + f(b))),
+          )
+    _riemann(Ms[Symbol(method)], f, xs)
+end
+function _riemann(M, f, xs)
+    xs′ = zip(Iterators.drop(xs, 1), Iterators.rest(xs, 1))
+    sum(M(f, a, b) * (b-a) for (a,b) ∈ xs′)
 end
 ```
 

quarto/integrals/area_between_curves.qmd

@@ -276,6 +276,20 @@ find_zero(m -> area(m) - 1, (0, 1))
 (Which is a nice combination of using `find_zeros`, `quadgk` and `find_zero` to answer a problem.)
 
 
+##### Example
+
+In an early 2023 article appearing in the [New York Times](https://www.nytimes.com/2023/02/02/opinion/covid-pandemic-deaths.html) a discussion on *excess* deaths was presented. @fig-excess-deaths shows two curves from which the number of excess deaths can be computed.
+
+::: {#fig-excess-deaths}
+![Excess deaths](./figures/excess-deaths.png)
+
+Illustration of excess deaths. Figure from a Feb. 2023 New York Times article
+:::
+
+
+Consider the curve marked *Actual deaths*. The number of deaths per year is the sum over each day of the number of deaths per each day. Approximating this number with a curve and setting 1 day equal to 1 unit, the number of deaths is basically $\int_0^{365} d(t) dt$. This curve is *usually*, say, $u(t)$, so the expected number of deaths would be $\int_0^{365} u(t) dt$. The difference, $\int_0^{365} (d(t) - u(t))dt$ is interpreted as the number of *excess deaths*. This methodology has been used to estimate the true number of deaths attributable to the COVID-19 pandemic.
+
+
 ##### Example
 
 
@@ -740,11 +754,11 @@ Doubling the answer above gives a value that Galileo had struggled with for many
 #| echo: false
 imgfile="figures/companion-curve-bisects-rectangle.png"
 caption = """
-Roberval, avoiding a trignometric integral, instead used symmetry to show that the area under the companion curve was half the area of the rectangle, which in this figure is ``2\\pi``.
+Roberval, avoiding a trignometric integral, instead used symmetry to show that the area under the companion curve was half the area of the rectangle, which in this figure is `2pi`.
 """
 # ImageFile(:integrals, imgfile, caption)
 nothing
 ```
 
-![Roberval, avoiding a trignometric integral, instead used symmetry to show that the area under the companion curve was half the area of the rectangle, which in this figure is ``2\\pi``.
+![Roberval, avoiding a trignometric integral, instead used symmetry to show that the area under the companion curve was half the area of the rectangle, which in this figure is $2\pi$.
 ](./figures/companion-curve-bisects-rectangle.png)

quarto/precalc/calculator.qmd

@@ -190,32 +190,46 @@ A right triangle has sides $a=11$ and $b=12$. Find the length of the   hypotenus
 
 ##### Example
 
-An article in the [Washington Post](https://www.washingtonpost.com/climate-environment/2022/09/19/ants-population-20-quadrillion/) describes estimates for the number of ants on earth.
+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
+article in the [Washington Post](https://www.washingtonpost.com/climate-environment/2022/09/19/ants-population-20-quadrillion/).
 
 
-They describe the number of ants two ways:
+The authors describe the number of ants two ways:
 
-* There are $20$ *trillion* ants.
-* There are $12$ megatons of ants
+* There are $20 \cdot 10^{15}$ (20 *quadrillion* or 20 thousand million millions) ants.
 
-With this how many ants make up a pound?
+* There is an estimated total biomass of $12$ megatons of *dry* carbon.
 
-Below we use the underscore as a separator, which is parsed as commas are to separate groups of numbers. The calculation is the number of ants divided by the number of pounds of ants (one megaton is $1$ million pounds):
+
+The authors note in a supplement to their paper that over 15,700 species and subspecies of ants have been named. To get a good estimate, 489 studies of ant populations were combined spanning all continents and major habitats. The studies identify the number of *foraging* ants and combined yield the estimates for the *epigaeic ant abundance* ($3.02 \cdot 10^{15}$) and the *arboreal ant abundance* ($1.34\cdot 10^{15}$). Using an estimate of 22% of ants in a colony are foraging, they get the following estimate for the number of ants:
 
 ```{julia}
-20_000_000_000_000_000 / (1_000_000 * 12 * 2000)
+(3.02 * 10^15 + 1.34*10^15) * 100 / 22
 ```
 
-Or not quite a million per pound.
-
-Using a pound is $2.2$ killograms or $2,200$ grams, we can this many ants per gram:
+Shifting the decimal point, this gives a value rounded to $20\cdot 10^{15}$ ants.
 
+The authors used a value for the *dry weight* of an average (and representative) single ant. What was that value? (Which they indicate is perhaps unreliable,
+as, for example, small-bodied ants may be much more abundant than large-bodied ants).  We assume below that one "megaton" is $1$ million *metric* tons; a metric ton is $1,000$ kilograms; and a kilogram $1,000$ grams:
 
 ```{julia}
-20_000_000_000_000_000 / (1_000_000 * 12 * 2000) / 2200
+(12 * 1_000_000 * 1_000 * 1_000) / 20_000_000_000_000_000
 ```
 
-Such combinations will be easier to check for correctness when variable names are assigned the respective values.
+Which translates to an *average* dry *carbon* weight of $0.6/1000$ grams, that is $0.6$ milligrams ($0.62$ mg C was actually used).
+
+ The authors write that insects are generally considered to have a dry weight of 30% wet weight, and a carbon weight of 50% dry weight, so the weight in grams of an *average* living ant would be multiplied by $2$ and then $10/3$:
+
+```{julia}
+(12 * 1_000_000 * 1_000 * 1_000) / 20_000_000_000_000_000 * (2 * 10/3)
+```
+
+That is 4 milligrams, or 250 ants per gram on average.
+
+Numeric combinations, as above, will be easier to check for correctness when variable names are assigned to the respective values.
+
+Using the underscore, as above, to separate groups of digits, is helpful, as an alternate to scientific notation, when working with large numbers.
+
 
 
 
@@ -1094,7 +1108,21 @@ answ=1
 radioq(choices, answ)
 ```
 
+##### Question
 
+[Sagan's](https://en.wiktionary.org/wiki/Sagan%27s_number) number is defined to be the total number of stars in the observable universe. How big is it? A *sextillion* is 7 groups of three 0's after a leading 1. One estimate is $10$ sextillion. How might this be entered into `Julia`? Select the one that *doesn't* work:
+
+```{julia}
+choices = [
+"`10*10^21`",
+"`10*10.0^21`",
+"`10e21`",
+"`1e22`",
+    "`10_000_000_000_000_000_000_000`"]
+explanation = "With an *integer* base, `10^21` overflows. For typical integers, onmly `10^18` is defined as expected. Once `10^19` is entered the mathematical value is larger the the `typemax` for `Int64` and so the value *wraps* around. The number written out with underscores to separate groups of 0s is parsed as an integer with 128 bits, not 64."
+buttonq(choices, 1; explanation=explanation)
+
+The estimate of 10 sextillion for Sagan's number was made in 1980, a more modern estimate is 30 times larger.
 
 ##### Question
 

quarto/precalc/polynomial.qmd

@@ -459,6 +459,13 @@ plot!(x -> -x^4, -3,3, legend=false, xticks=false, yticks=false, subplot=3, titl
 plot!(x -> -x^5, -3,3, legend=false, xticks=false, yticks=false, subplot=4, title="n > odd, aₙ < 0")
 ```
 
+##### Example
+
+This graphic shows some of the above:
+
+[![you tube](https://img.youtube.com/vi/OFzqDatEvCo/3.jpg)](https://m.youtube.com/watch?v=OFzqDatEvCo)
+
+
 ##### Example
 
 

quarto/precalc/variables.qmd

@@ -111,6 +111,54 @@ a = v0 * cosd(theta)
 By defining a new variable `a` to represent a value that is repeated a few times in the expression, the last command is greatly simplified. Doing so makes it much easier to check for accuracy against the expression to compute.
 
 
+##### Example
+
+A [grass swale](https://stormwater.pca.state.mn.us/index.php?title=Design_criteria_for_dry_swale_(grass_swale)) is a design to manage surface water flow resulting from a storm. Swales detain, filter, and infiltrate runoff limiting erosion in the process.
+
+![Swale cross section](precalc/swale.png)
+
+There are a few mathematical formula that describe the characteristics of swale:
+
+The area is given by:
+
+$$
+A = (b + d/\tan(\theta)) d
+$$
+
+The *wetted* perimeter is given by
+
+$$
+P = b + 2 d/\sin(\theta)
+$$
+
+The *hydraulic radius* is given by
+
+$$
+R = \frac{b\cdot d \sin(\theta) + d^2 \cos(\theta)}{b\sin(\theta) + 2d}.
+$$
+
+Finally, the *flow quantity* is given by *Manning's* formula:
+
+$$
+Q = vA = \frac{R^{2/3} S^{1/2}}{n}, \quad R = \frac{A}{P}.
+$$
+
+With $n$ being Manning's coefficient, $v$ the velocity in feet per second, and $S$ being the slope. Velocity and slope are correlated.
+
+Manning's coefficient depends on the height of the vegetation in the grass swale. It is $0.025$ when the depth of flow is similar to the vegetation height.
+
+Given all this, compute the flow quantity when $S = 2/90$, $n=0.025$ and $v=1/10$ for a swale with characteristics $b=1$, $\theta=\pi/4$, $d=1$.
+
+```{julia}
+b, theta, d = 1, pi/4, 1
+n, S = 0.025, 2/90
+A = (b + d/tan(theta)) *  d
+P = b + 2d/sin(theta)
+R = A / P
+Q = R^(2/3) * S^(1/2) / n
+```
+
+
 ##### Example