daviddoji/CalculusWithJuliaNotes.jl
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make pdf file generation work

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jverzani
2022-10-10 14:28:05 -04:00
parent a0b913eed8
commit a9ca131870
59 changed files with 884 additions and 1330 deletions
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12
quarto/integrals/area_between_curves.qmd
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@@ -688,9 +688,13 @@ imgfile="figures/cycloid-companion-curve.png"
caption = """
Figure from Martin showing the companion curve to the cycloid. As the generating circle rolls, from ``A`` to ``C``, the original point of contact, ``D``, traces out an arch of the cycloid. The companion curve is that found by congruent line segments. In the figure, when ``D`` was at point ``P`` the line segment ``PQ`` is congruent to ``EF`` (on the original position of the generating circle).
"""
ImageFile(:integrals, imgfile, caption)
# ImageFile(:integrals, imgfile, caption)
nothing
```
![Figure from Martin showing the companion curve to the cycloid. As the generating circle rolls, from ``A`` to ``C``, the original point of contact, ``D``, traces out an arch of the cycloid. The companion curve is that found by congruent line segments. In the figure, when ``D`` was at point ``P`` the line segment ``PQ`` is congruent to ``EF`` (on the original position of the generating circle).](./figures/cycloid-companion-curve.png)
In particular, it can be read that Roberval proved that the area between the cycloid and its companion curve is half the are of the generating circle. Roberval didn't know integration, so finding the area between two curves required other tricks. One is called "Cavalieri's principle." From the figure above, which of the following would you guess this principle to be:
@@ -738,5 +742,9 @@ 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``.
"""
ImageFile(:integrals, imgfile, caption)
# 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``.
](./figures/companion-curve-bisects-rectangle.png)