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CwJ/TODO/AD.md

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+Good paper recommended here (https://discourse.julialang.org/t/learning-automatic-differentiation/56158/3)
+
+https://www.jmlr.org/papers/volume18/17-468/17-468.pdf

CwJ/TODO/arrows.md

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+This is really just
+
+plot!([0,cos(θ)],[0,sin(θ)], arrow=true)
+
+
+https://stackoverflow.com/questions/58219191/drawing-an-arrow-with-specified-direction-on-a-point-in-scatter-plot-in-julia
+
+https://github.com/m3g/CKP/blob/master/disciplina/codes/velocities.jl
+
+
+
+
+using Plots
+using LaTeXStrings
+
+function arch(θ₁,θ₂;radius=1.,Δθ=1.)
+  θ₁ = π*θ₁/180
+  θ₂ = π*θ₂/180
+  Δθ = π*Δθ/180
+  l = round(Int,(θ₂-θ₁)/Δθ)
+  x = zeros(l)
+  y = zeros(l)
+  for i in 1:l
+    θ = θ₁ + i*Δθ
+    x[i] = radius*cos(θ) 
+    y[i] = radius*sin(θ)
+  end
+  return x, y
+end
+
+plot()
+
+x, y = arch(0,360)
+plot(x,y,seriestype=:shape,label="",alpha=0.5)
+
+x, y = arch(0,360,radius=0.95)
+plot!(x,y,seriestype=:shape,label="",fillcolor=:white)
+
+x, y = arch(0,360,radius=0.7)
+plot!(x,y,seriestype=:shape,label="",alpha=0.5,fillcolor=:red)
+
+x, y = arch(0,360,radius=0.65)
+plot!(x,y,seriestype=:shape,label="",fillcolor=:white)
+
+plot!([0,0],[0,1.1],arrow=true,color=:black,linewidth=2,label="")
+plot!([0,1.1],[0,0],arrow=true,color=:black,linewidth=2,label="")
+
+x, y = arch(15,16,radius=0.65)
+plot!([0,x[1]],[0,y[1]],arrow=true,color=:black,linewidth=1,label="")
+
+x, y = arch(35,36,radius=0.95)
+plot!([0,x[1]],[0,y[1]],arrow=true,color=:black,linewidth=1,label="")
+
+plot!(draw_arrow=true)
+plot!(showaxis=:no,ticks=nothing,xlim=[-0.1,1.1],ylim=[-0.1,1.1],)
+plot!(xlabel="x",ylabel="y",size=(400,400))
+
+annotate!(0.58,-0.07,text(L"\Delta v_1",10))
+annotate!(0.88,-0.07,text(L"\Delta v_2",10))
+
+savefig("./velocities.pdf")

CwJ/TODO/earth.jl

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+# Calculate the temperature of the earth using the simplest model
+# @jake
+# https://discourse.julialang.org/t/seven-lines-of-julia-examples-sought/50416/121
+
+using Unitful, Plots
+p_sun = 386e24u"W"      # power output of the sun
+radius_a = 6378u"km"    # semi-major axis of the earth
+radius_b = 6357u"km"    # semi-minor axis of the earth
+orbit_a = 149.6e6u"km"  # distance from the sun to earth
+orbit_e = 0.017         # eccentricity of r = a(1-e^2)/(1+ecos(θ))  & time ≈ 365.25 * θ / 360 where θ is in degrees
+a = 0.75                # absorptivity of the sun's radiation
+e = 0.6                 # emmissivity of the earth (very dependent on cloud cover)
+σ = 5.6703e-8u"W*m^-2*K^-4"  # Stefan-Boltzman constant
+temp_sky = 3u"K"        # sky temperature
+
+
+
+t = (0:0.25:365.25)u"d"       # day of year in 1/4 day increments
+θ = 2*π/365.25u"d" .* t       # approximate angle around the sun
+r = orbit_a * (1-orbit_e^2) ./ (1 .+ orbit_e .* cos.(θ))    # distance from sun to earth
+area_projected = π * radius_a * radius_b    # area of earth facing the sun
+ec = sqrt(1-radius_b^2/radius_a^2)  # eccentricity of earth
+
+area_surface = 2*π*radius_a^2*(1 + radius_b^2/(ec*radius_b^2)*atanh(ec)) # surface area of the earth
+
+q_in = p_sun * a * area_projected ./ (4 * π .* r.^2) # total heat impacting the earth
+
+temp_earth = (q_in ./ (e*σ*area_surface) .+ temp_sky^4).^0.25 # temperature of the earth
+
+plot(t*u"d^-1", temp_earth*u"K^-1" .- 273.15, label = false, title = "Temperature of Earth", xlabel = "Day", ylabel = "Temperature [C]")

CwJ/TODO/ladder-questions.md

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+###### Question (Ladder [questions](http://www.mathematische-basteleien.de/ladder.htm))
+
+
+A ``7``meter ladder leans against wall with the base ``1.5``meters from wall at its base. At which height does the ladder touch the wall?
+
+```julia; hold=true; echo=false
+l = 7
+adj = 1.5
+opp = sqrt(l^2 - adj^2)
+numericq(opp, 1e-3)
+```
+
+
+----
+
+A ``7``meter ladder leans against the wall. Between the ladder and the wall is a ``1``m cube box. The ladder touches the wall, the box and the ground. There are two such positions, what is the height of the ladder of the more upright position?
+
+You might find this code of help:
+
+```julia; eval=false
+@syms x y
+l, b = 7, 1
+eq = (b+x)^2 + (b+y)^2
+eq = subs(eq, x=> b*(b/y)) # x/b = b/y
+solve(eq ~ l^2, y)
+```
+
+What is the value `b+y` in the above?
+
+```julia; echo=false
+radioq(("The height of the ladder",
+        "The height of the box plus ladder",
+        "The distance from the base of the ladder to the box,"
+        "The distance from the base of the ladder to the base of the wall"
+        ),1)
+```
+
+
+What is the height of the ladder
+
+```julia; hold=true; echo=false
+numericq(6.90162289514212, 1e-3)
+```
+
+
+----
+
+A ladder of length ``c`` is to moved through a 2-dimensional hallway of width ``b`` which has a right angled bend. If ``4b=c``, when will the ladder get stuck?
+
+Consider this picture
+
+```julia; hold=true; echo=false
+p = plot(; axis=nothing, legend=false, aspect_ratio=:equal)
+x,y=1,2
+b = sqrt(x*y)
+plot!(p, [0,0,b+x], [b+y,0,0], linestyle=:dot)
+plot!(p, [0,b+x],[b,b], color=:black, linestyle=:dash)
+plot!(p, [b,b],[0,b+y], color=:black, linestyle=:dash)
+plot!(p, [b+x,0], [0, b+y], color=:black)
+```
+
+
+Suppose ``b=5``, then with ``b+x`` and ``b+y``  being the lengths on the walls where it is stuck *and* by similar triangles ``b/x = y/b`` we can solve for ``x``. (In the case take the largest positive value. The answer would be the angle ``\theta`` with ``\tan(\theta) = (b+y)/(b+x)``.
+
+```julia; hold=true; echo=false
+b = 5
+l = 4*b
+@syms x y
+eq = (b+x)^2 + (b+y)^2
+eq =subs(eq, y=> b^2/x)
+x₀ = N(maximum(filter(>(0), solve(eq ~ l^2, x))))
+y₀ = b^2/x₀
+θ₀ = Float64(atan((b+y₀)/(b+x₀)))
+numericq(θ₀, 1e-2)
+```
+
+
+-----
+
+Two ladders of length ``a`` and ``b`` criss-cross between two walls of width ``x``. They meet at a height of ``c``.
+
+```julia; hold=true; echo=false
+p = plot(; legend=false, axis=nothing, aspect_ratio=:equal)
+ya,yb,x = 2,3,1
+plot!(p, [0,x],[ya,0], color=:black)
+plot!(p, [0,x],[0, yb], color=:black)
+plot!(p, [0,0], [0,yb], color=:blue, linewidth=5)
+plot!(p, [x,x], [0,yb], color=:blue, linewidth=5)
+plot!(p, [0,x], [0,0], color=:blue, linewidth=5)
+xc = ya/(ya+yb)
+c = yb*xc
+plot!(p, [xc,xc],[0,c])
+p
+```
+
+Suppose ``c=1``, ``b=3``, and ``a=5``. Find ``x``.
+
+Introduce ``x = z + y``, and ``h`` and ``k`` the heights of the ladders along the left wall and the right wall.
+
+The ``z/c = x/k`` and ``y/c = x/h`` by similar triangles. As ``z + y`` is ``x`` we can solve to get
+
+```math
+x = z + y = \frac{xc}{k} + \frac{xc}{h}
+ = \frac{xc}{\sqrt{b^2 - x^2}} + \frac{xc}{\sqrt{a^2 - x^2}}
+```
+
+With ``a,b,c`` as given, this can be solved with
+
+```julia; hold=true; echo=false
+a,b,c =  5, 3, 1
+f(x) = x*c/sqrt(b^2 - x^2) + x*c/sqrt(a^2 - x^2) - x
+find_zero(f, (0, b))
+```
+
+The answer is ``2.69\dots``.