116 lines
3.2 KiB
Markdown
116 lines
3.2 KiB
Markdown
###### Question (Ladder [questions](http://www.mathematische-basteleien.de/ladder.htm))
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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?
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```julia; hold=true; echo=false
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l = 7
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adj = 1.5
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opp = sqrt(l^2 - adj^2)
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numericq(opp, 1e-3)
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```
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----
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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?
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You might find this code of help:
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```julia; eval=false
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@syms x y
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l, b = 7, 1
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eq = (b+x)^2 + (b+y)^2
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eq = subs(eq, x=> b*(b/y)) # x/b = b/y
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solve(eq ~ l^2, y)
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```
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What is the value `b+y` in the above?
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```julia; echo=false
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radioq(("The height of the ladder",
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"The height of the box plus ladder",
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"The distance from the base of the ladder to the box,"
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"The distance from the base of the ladder to the base of the wall"
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),1)
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```
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What is the height of the ladder
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```julia; hold=true; echo=false
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numericq(6.90162289514212, 1e-3)
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```
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----
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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?
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Consider this picture
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```julia; hold=true; echo=false
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p = plot(; axis=nothing, legend=false, aspect_ratio=:equal)
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x,y=1,2
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b = sqrt(x*y)
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plot!(p, [0,0,b+x], [b+y,0,0], linestyle=:dot)
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plot!(p, [0,b+x],[b,b], color=:black, linestyle=:dash)
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plot!(p, [b,b],[0,b+y], color=:black, linestyle=:dash)
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plot!(p, [b+x,0], [0, b+y], color=:black)
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```
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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)``.
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```julia; hold=true; echo=false
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b = 5
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l = 4*b
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@syms x y
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eq = (b+x)^2 + (b+y)^2
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eq =subs(eq, y=> b^2/x)
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x₀ = N(maximum(filter(>(0), solve(eq ~ l^2, x))))
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y₀ = b^2/x₀
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θ₀ = Float64(atan((b+y₀)/(b+x₀)))
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numericq(θ₀, 1e-2)
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```
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-----
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Two ladders of length ``a`` and ``b`` criss-cross between two walls of width ``x``. They meet at a height of ``c``.
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```julia; hold=true; echo=false
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p = plot(; legend=false, axis=nothing, aspect_ratio=:equal)
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ya,yb,x = 2,3,1
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plot!(p, [0,x],[ya,0], color=:black)
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plot!(p, [0,x],[0, yb], color=:black)
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plot!(p, [0,0], [0,yb], color=:blue, linewidth=5)
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plot!(p, [x,x], [0,yb], color=:blue, linewidth=5)
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plot!(p, [0,x], [0,0], color=:blue, linewidth=5)
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xc = ya/(ya+yb)
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c = yb*xc
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plot!(p, [xc,xc],[0,c])
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p
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```
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Suppose ``c=1``, ``b=3``, and ``a=5``. Find ``x``.
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Introduce ``x = z + y``, and ``h`` and ``k`` the heights of the ladders along the left wall and the right wall.
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The ``z/c = x/k`` and ``y/c = x/h`` by similar triangles. As ``z + y`` is ``x`` we can solve to get
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```math
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x = z + y = \frac{xc}{k} + \frac{xc}{h}
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= \frac{xc}{\sqrt{b^2 - x^2}} + \frac{xc}{\sqrt{a^2 - x^2}}
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```
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With ``a,b,c`` as given, this can be solved with
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```julia; hold=true; echo=false
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a,b,c = 5, 3, 1
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f(x) = x*c/sqrt(b^2 - x^2) + x*c/sqrt(a^2 - x^2) - x
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find_zero(f, (0, b))
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```
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The answer is ``2.69\dots``.
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