daviddoji/CalculusWithJuliaNotes.jl
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use quarto, not Pluto to render pages

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2022-07-24 16:38:24 -04:00
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CwJ/integral_vector_calculus/line_integrals.jmd
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@@ -242,11 +242,9 @@ quadgk(t -> (Fₘ ∘ rₒ)(t) ⋅ rₒ'(t), 0, 1)
Still $0$. We will see next that this is not surprising if something about $F$ is known.
```julia; echo=false
note("""
The [Washington Post](https://www.washingtonpost.com/outlook/everything-you-thought-you-knew-about-gravity-is-wrong/2019/08/01/627f3696-a723-11e9-a3a6-ab670962db05_story.html") had an article by Richard Panek with the quote "Well, yes — depending on what we mean by 'attraction.' Two bodies of mass don’t actually exert some mysterious tugging on each other. Newton himself tried to avoid the word 'attraction' for this very reason. All (!) he was trying to do was find the math to describe the motions both down here on Earth and up there among the planets (of which Earth, thanks to Copernicus and Kepler and Galileo, was one)." The point being the formula above is a mathematical description of the force, but not an explanation of how the force actually is transferred.
""")
```
!!! note
The [Washington Post](https://www.washingtonpost.com/outlook/everything-you-thought-you-knew-about-gravity-is-wrong/2019/08/01/627f3696-a723-11e9-a3a6-ab670962db05_story.html") had an article by Richard Panek with the quote "Well, yes — depending on what we mean by 'attraction.' Two bodies of mass don’t actually exert some mysterious tugging on each other. Newton himself tried to avoid the word 'attraction' for this very reason. All (!) he was trying to do was find the math to describe the motions both down here on Earth and up there among the planets (of which Earth, thanks to Copernicus and Kepler and Galileo, was one)." The point being the formula above is a mathematical description of the force, but not an explanation of how the force actually is transferred.
#### Work in a *conservative* vector field
@@ -432,11 +430,8 @@ p
The flow integral is typically computed for a closed (Jordan) curve, measuring the total flow out of a region. In this case, the integral is written $\oint_C (F\cdot\hat{N})ds$.
```julia; echo=false
note(L"""
For a Jordan curve, the positive orientation of the curve is such that the normal direction (proportional to $\hat{T}'$) points away from the bounded interior. For a non-closed path, the choice of parameterization will determine the normal and the integral for flow across a curve is dependent - up to its sign - on this choice.
""")
```
!!! note
For a Jordan curve, the positive orientation of the curve is such that the normal direction (proportional to $\hat{T}'$) points away from the bounded interior. For a non-closed path, the choice of parameterization will determine the normal and the integral for flow across a curve is dependent - up to its sign - on this choice.
##### Example
@@ -725,10 +720,10 @@ In [Surface area](../integrals/surface_area.mmd) the following formula for the s
Consider the transformation $(x, \theta) \rightarrow \langle x, f(x)\cos(\theta), f(x)\sin(\theta)$. This maps the region $[a,b] \times [0, 2\pi]$ *onto* the surface of revolution. As such, the surface element would be:
```julia; hold=true
@syms f()::positive x::real theta::real
```julia
@syms 𝒇()::positive x::real theta::real
Phi(x, theta) = [x, f(x)*cos(theta), f(x)*sin(theta)]
Phi(x, theta) = [x, 𝒇(x)*cos(theta), 𝒇(x)*sin(theta)]
Jac = Phi(x, theta).jacobian([x, theta])
v1, v2 = Jac[:,1], Jac[:,2]
se = norm(v1 × v2)
@@ -994,8 +989,8 @@ raw" ``\langle a, b, c\rangle / \| \langle a, b, c\rangle\|``",
raw" ``\langle a, b, c\rangle``",
raw" ``\langle d-a, d-b, d-c\rangle / \| \langle d-a, d-b, d-c\rangle\|``",
]
ans = 1
radioq(choices, ans)
answ = 1
radioq(choices, answ)
```
Does it depend on $d$?
@@ -1006,8 +1001,8 @@ L"No. Moving $d$ just shifts the plane up or down the $z$ axis, but won't change
L"Yes. Of course. Different values for $d$ mean different values for $x$, $y$, and $z$ are needed.",
L"Yes. The gradient of $F(x,y,z) = ax + by + cz$ will be normal to the level curve $F(x,y,z)=d$, and so this will depend on $d$."
]
ans = 1
radioq(choices, ans)
answ = 1
radioq(choices, answ)
```
@@ -1035,8 +1030,8 @@ raw" ``2\pi + 2\pi^2``",
raw" ``2\pi^2``",
raw" ``4\pi``"
]
ans = 1
radioq(choices, ans)
answ = 1
radioq(choices, answ)
```
###### Question
@@ -1062,8 +1057,8 @@ choices =[
L"It will be $0$, as $\nabla{f}$ is orthogonal to the level curve and $\vec{r}'$ is tangent to the level curve",
L"It will $f(b)-f(a)$ for any $b$ or $a$"
]
ans = 1
radioq(choices, ans)
answ = 1
radioq(choices, answ)
```
@@ -1103,8 +1098,8 @@ choices = [
L"The field is a potential field, but the path integral around $0$ is not path dependent.",
L"The value of $d/dt(f\circ\vec{r})=0$, so the integral should be $0$."
]
ans =1
radioq(choices, ans)
answ =1
radioq(choices, answ)
```
The function $F = \nabla{f}$ is
@@ -1114,8 +1109,8 @@ choices = [
"Not continuous everywhere",
"Continuous everywhere"
]
ans = 1
radioq(choices, ans)
answ = 1
radioq(choices, answ)
```
###### Question
@@ -1165,8 +1160,8 @@ raw" ``\int_0^{2\pi} (a\cos(t)) \cdot (b\cos(t)) dt``",
raw" ``\int_0^{2\pi} (-b\sin(t)) \cdot (b\cos(t)) dt``",
raw" ``\int_0^{2\pi} (a\cos(t)) \cdot (a\cos(t)) dt``"
]
ans=1
radioq(choices, ans)
answ=1
radioq(choices, answ)
```
@@ -1182,8 +1177,8 @@ raw" ``\langle \cos(v), \sin(v), 1\rangle``",
raw" ``\langle -u\sin(v), u\cos(v), 0\rangle``",
raw" ``u\langle -\cos(v), -\sin(v), 1\rangle``"
]
ans = 1
radioq(choices, ans, keep_order=true)
answ = 1
radioq(choices, answ, keep_order=true)
```
Compute $\vec{v}_2 = \partial{\Phi}/\partial{u}$
@@ -1194,8 +1189,8 @@ raw" ``\langle \cos(v), \sin(v), 1\rangle``",
raw" ``\langle -u\sin(v), u\cos(v), 0\rangle``",
raw" ``u\langle -\cos(v), -\sin(v), 1\rangle``"
]
ans = 2
radioq(choices, ans, keep_order=true)
answ = 2
radioq(choices, answ, keep_order=true)
```
Compute $\vec{v}_1 \times \vec{v}_2$
@@ -1207,8 +1202,8 @@ raw" ``\langle \cos(v), \sin(v), 1\rangle``",
raw" ``\langle -u\sin(v), u\cos(v), 0\rangle``",
raw" ``u\langle -\cos(v), -\sin(v), 1\rangle``"
]
ans = 3
radioq(choices, ans, keep_order=true)
answ = 3
radioq(choices, answ, keep_order=true)
```
@@ -1288,8 +1283,8 @@ raw" ``\sqrt{2}/24``",
raw" ``2/\sqrt{24}``",
raw" ``1/12``"
]
ans = 1
radioq(choices, ans)
answ = 1
radioq(choices, answ)
```
@@ -1315,6 +1310,6 @@ raw" ``0``",
raw" ``7/36``",
raw" ``1/60``"
]
ans = 1
radioq(choices, ans)
answ = 1
radioq(choices, answ)
```