daviddoji/CalculusWithJuliaNotes.jl
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some typos.
This commit is contained in:
Fang Liu 2023-05-31 08:54:07 +08:00
parent 0af9ca6ac4
commit f10e045d29
4 changed files with 46 additions and 44 deletions
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20
quarto/ODEs/euler.qmd
View File

@ -220,6 +220,7 @@ function euler(F, x0, xn, y0, n)
xs[i + 1] = xs[i] + h
ys[i + 1] = ys[i] + h * F(ys[i], xs[i])
end
xs[end] = xn
linterp(xs, ys)
end
```
@ -288,7 +289,6 @@ We graphically compare our approximate answer with the exact one:
```{julia}
plot(f, x0, xn)
𝒐ut = dsolve(D(u)(x) - F(u(x),x), u(x), ics = Dict(u(x0) => y0))
plot(rhs(𝒐ut), x0, xn)
plot!(f, x0, xn)
@ -348,7 +348,7 @@ The [Brachistochrone problem](http://www.unige.ch/~gander/Preprints/Ritz.pdf) wa
imgfile = "figures/bead-game.jpg"
caption = """
A child's bead game. What shape wire will produce the shortest time for a bed to slide from a top to the bottom?
A child's bead game. What shape wire will produce the shortest time for a bead to slide from a top to the bottom?
"""
#ImageFile(:ODEs, imgfile, caption)
@ -464,11 +464,11 @@ The race is on. An illustration of beads falling along a path, as can be seen, s
ImageFile(imgfile, caption)
```
Now, the natural question is which path is best? The solution can be [reduced](http://mathworld.wolfram.com/BrachistochroneProblem.html) to solving this equation for a positive $c$:
Now, the natural question is which path is best? The solution can be [reduced](http://mathworld.wolfram.com/BrachistochroneProblem.html) to solving this equation for a positive $C$:
$$
1 + (y'(x))^2 = \frac{c}{y}, \quad c > 0.
1 + (y'(x))^2 = \frac{C}{y}, \quad C > 0.
$$
Reexpressing, this becomes:
@ -482,7 +482,7 @@ This is a separable equation and can be solved, but even `SymPy` has trouble wit
$$
x(u) = c\cdot u - \frac{c}{2}\sin(2u), \quad y(u) = \frac{c}{2}( 1- \cos(2u)), \quad 0 \leq u \leq U.
x(u) = C\cdot u - \frac{C}{2}\sin(2u), \quad y(u) = \frac{C}{2}( 1- \cos(2u)), \quad 0 \leq u \leq U.
$$
The values of $U$ and $c$ must satisfy $(x(U), y(U)) = (B, A)$.
@ -495,7 +495,7 @@ The equation can be written in terms of $y'=F(y,x)$, where
$$
F(y,x) = \sqrt{\frac{c-y}{y}}.
F(y,x) = \sqrt{\frac{C-y}{y}}.
$$
But as $y_0 = 0$, we immediately would have a problem with the first step, as there would be division by $0$.
@ -609,7 +609,7 @@ u_{n+1} &= u_n + h v_n,\\
v_{n+1} &= v_n - h \cdot g/l \cdot \sin(u_n).
\end{align*}
Here we need *two* initial conditions: one for the initial value $u(t_0)$ and the initial value of $u'(t_0)$. We have seen if we start at an angle $a$ and release the bob from rest, so $u'(0)=0$ we get a sinusoidal answer to the linearized model. What happens here? We let $a=1$, $L=5$ and $g=9.8$:
Here we need *two* initial conditions: one for the initial value $u(t_0)$ and the initial value of $u'(t_0)$. We have seen if we start at an angle $a$ and release the bob from rest, so $u'(0)=0$ we get a sinusoidal answer to the linearized model. What happens here? We let $a=1$, $l=5$ and $g=9.8$:
We write a function to solve this starting from $(x_0, y_0)$ and ending at $x_n$:
@ -624,12 +624,12 @@ function euler2(x0, xn, y0, yp0, n; g=9.8, l = 5)
xs[i+1] = xs[i] + h
us[i+1] = us[i] + h * vs[i]
vs[i+1] = vs[i] + h * (-g / l) * sin(us[i])
end
linterp(xs, us)
end
linterp(xs, us)
end
```
Let's take $a = \pi/4$ as the initial angle, then the approximate solution should be $\pi/4\cos(\sqrt{g/l}x)$ with period $T = 2\pi\sqrt{l/g}$. We try first to plot then over 4 periods:
Let's take $a = \pi/4$ as the initial angle, then the approximate solution should be $\pi/4\cos(\sqrt{g/l}x)$ with period $T = 2\pi\sqrt{l/g}$. We try first to plot them over 4 periods:
```{julia}

22
quarto/ODEs/odes.qmd
View File

@ -88,7 +88,7 @@ Again, we can integrate to get an answer for any value $t$:
\begin{align*}
x(t) - x(t_0) &= \int_{t_0}^t \frac{dv}{dt} dt \\
x(t) - x(t_0) &= \int_{t_0}^t \frac{dx}{dt} dt \\
&= (v_0t + \frac{1}{2}a t^2 - at_0 t) |_{t_0}^t \\
&= (v_0 - at_0)(t - t_0) + \frac{1}{2} a (t^2 - t_0^2).
\end{align*}
@ -216,7 +216,7 @@ Let $F(y) = \int_{y_0}^y du/g(u)$, then a solution to the above is $F(y) = x - x
[Toricelli's Law](http://tinyurl.com/hxvf3qp) describes the speed a jet of water will leave a vessel through an opening below the surface of the water. The formula is $v=\sqrt{2gh}$, where $h$ is the height of the water above the hole and $g$ the gravitational constant. This arises from equating the kinetic energy gained, $1/2 mv^2$ and potential energy lost, $mgh$, for the exiting water.
An application of Torricelli's law is to describe the volume of water in a tank over time, $V(t)$. Imagine a cylinder of cross sectional area $A$ with a hole of cross sectional diameter $a$ at the bottom, Then $V(t) = A h(t)$, with $h$ giving the height. The change in volume over $\Delta t$ units of time must be given by the value $a v(t) \Delta t$, or
An application of Torricelli's law is to describe the volume of water in a tank over time, $V(t)$. Imagine a cylinder of cross sectional area $A$ with a hole of cross sectional area $a$ at the bottom, Then $V(t) = A h(t)$, with $h$ giving the height. The change in volume over $\Delta t$ units of time must be given by the value $a v(t) \Delta t$, or
$$
@ -511,7 +511,7 @@ $$
Here, the solution is in terms of sines and cosines, with period given by $T = 2\pi/\sqrt{k} = 2\pi\cdot\sqrt{l/g}$. The answer does not depend on the mass, $m$, of the bob nor the amplitude of the motion, provided the small-angle approximation is valid.
If we pull the bob back an angle $a$ and release it then the initial conditions are $\theta(0) = a$ and $\theta'(a) = 0$. This gives the solution:
If we pull the bob back an angle $a$ and release it then the initial conditions are $\theta(0) = a$ and $\theta'(0) = 0$. This gives the solution:
```{julia}
@ -652,10 +652,12 @@ Furthermore, we can solve for $t$ from $x(t)$, to get an equation describing $y(
```{julia}
#| hold: true
@syms x0::real y0::real v0::real alpha::real g::real
@syms t x u()
a1 = dsolve(D2(u)(x) ~ 0, u(x), ics=Dict(u(0) => x0, D(u)(0) => v0 * cos(alpha)))
a2 = dsolve(D2(u)(x) ~ -g, u(x), ics=Dict(u(0) => y0, D(u)(0) => v0 * sin(alpha)))
ts = solve(t - rhs(a1), x)[1]
@syms t::positive x u()
Dₜ = Differential(t)
D2ₜ = Dₜ ∘ Dₜ
a1 = dsolve(D2ₜ(u)(t) ~ 0, u(t), ics=Dict(u(0) => x0, Dₜ(u)(0) => v0 * cos(alpha)))
a2 = dsolve(D2ₜ(u)(t) ~ -g, u(t), ics=Dict(u(0) => y0, Dₜ(u)(0) => v0 * sin(alpha)))
ts = solve(x - rhs(a1), t)[1]
y = simplify(rhs(a2)(t => ts))
sympy.Poly(y, x).coeffs()
```
@ -676,9 +678,9 @@ We now attempt to solve these.
```{julia}
@syms alpha::real, γ::postive, t::positive, v()
@syms alpha::real, γ::postive, v()
@syms x_0::real y_0::real v_0::real
Dₜ = Differential(t)
eq₁ = Dₜ(Dₜ(u))(t) ~ - γ * Dₜ(u)(t)
eq₂ = Dₜ(Dₜ(v))(t) ~ -g - γ * Dₜ(v)(t)
@ -931,7 +933,7 @@ The differential equation with boundary values
$$
\frac{r^2 \frac{dc}{dr}}{dr} = 0, \quad c(1)=2, c(10)=1,
\frac{d(r^2 \frac{dc}{dr})}{dr} = 0, \quad c(1)=2, c(10)=1,
$$
can be solved with `SymPy`. What is the value of $c(5)$?

26
quarto/integrals/arc_length.qmd
View File

@ -148,9 +148,9 @@ $$
:::{.callout-note}
## Note
[Bressoud](http://www.math.harvard.edu/~knill/teaching/math1a_2011/exhibits/bressoud/) notes that Gregory (1668) proved this formula for arc length of the graph of a function by showing that the length of the curve $f(x)$ is defined by the area under $\sqrt{1 + f'(x)^2}$. (It is commented that this was also known a bit earlier by von Heurat.) Gregory went further though, as part of the fundamental theorem of calculus was contained in his work. Gregory then posed this inverse question: given a curve $y=g(x)$ find a function $u(x)$ so that the area under $g$ is equal to the length of the second curve. The answer given was $u(x) = (1/c)\int_a^x \sqrt{g^2(t) - c^2}$, which if $g(t) = \sqrt{1 + f'(t)^2}$ and $c=1$ says $u(x) = \int_a^x f(t)dt$.
[Bressoud](http://www.math.harvard.edu/~knill/teaching/math1a_2011/exhibits/bressoud/) notes that Gregory (1668) proved this formula for arc length of the graph of a function by showing that the length of the curve $f(x)$ is defined by the area under $\sqrt{1 + f'(x)^2}$. (It is commented that this was also known a bit earlier by von Heurat.) Gregory went further though, as part of the fundamental theorem of calculus was contained in his work. Gregory then posed this inverse question: given a curve $y=g(x)$ find a function $u(x)$ so that the area under $g$ is equal to the length of the second curve. The answer given was $u(x) = (1/c)\int_a^x \sqrt{g^2(t) - c^2}dt$, where $g(t) = \sqrt{1 + u'(t)^2}$ if $c=1$ says $\int_a^x\sqrt{1 + u'(t)^2}dt = \int_a^x g(t)dt$.
An analogy might be a sausage maker. These take a mass of ground-up sausage material and return a long length of sausage. The material going in would depend on time via an equation like $\int_0^t g(u) du$ and the length coming out would be a constant (accounting for the cross section) times $u(t) = \int_0^t \sqrt{1 + g'(s)} ds$.
An analogy might be a sausage maker. These take a mass of ground-up sausage material and return a long length of sausage. The material going in would depend on time via an equation like $\int_0^t g(u) du$ and the length coming out would be a constant (accounting for the cross section) times $\int_0^t \sqrt{1 + u'(s)^2} ds$.
:::
@ -483,7 +483,7 @@ What looks at first glance to be just a slightly more complicated equation is th
\begin{align*}
s(u) &= \int_0^u \sqrt{(-\sin(t))^2 + b\cos(t)^2} dt\\
&= \int_0^u \sqrt{\sin(t))^2 + \cos(t)^2 + c\cos(t)^2} dt \\
&= \int_0^u \sqrt{\sin(t)^2 + \cos(t)^2 + c\cos(t)^2} dt \\
&=\int_0^u \sqrt{1 + c\cos(t)^2} dt.
\end{align*}
@ -521,7 +521,7 @@ Here we see visually that the new parameterization yields the same curve:
g(t) = 𝒂 * cos(t)
f(t) = 𝒃 * sin(t)
plot(t -> g(𝒔(t)), t -> f(𝒔(t)), 0, 𝒔(2*pi))
plot(t -> g(𝒔(t)), t -> f(𝒔(t)), 0, sinv(2*pi))
```
#### Example: An implication of concavity
@ -704,7 +704,7 @@ For the latter claim, integrating in the $y$ variable gives
Now, the area under $h$ over $[u,c]$ is greater than that over $[c,v]$ as $(u+v)/2 < c$ or $v-c < c-u$. That means the area under $f$ over $[u,c]$ is greater than that over $[c,v]$.
> There is more arc length for $f$over $[a,u]$ than $[v,b]$; more arc length for $f$ over $[u,c]$ than $[c,v]$. In particular more arc length over $[a,c]$ than $[c,b]$.
> There is more arc length for $f$ over $[a,u]$ than $[v,b]$; more arc length for $f$ over $[u,c]$ than $[c,v]$. In particular more arc length over $[a,c]$ than $[c,b]$.
@ -712,10 +712,10 @@ let $\phi(z) = f_2^{-1}(f_1(z))$ be the function taking $u$ to $v$, and $a$ to $
$$
f'(\phi(z)) \cdot \phi'(z) = f'(z) < 0
f'(\phi(z)) \cdot \phi'(z) = f'(z) > 0
$$
or $\phi'(z) < 0$. Moreover, we have by the first assertion that $f'(z) < -f'(\phi(z))$ so $|\phi'(z)| = |f(z)/f'(\phi(z))| < 1$.
or $\phi'(z) < 0$. Moreover, we have by the first assertion that $f'(z) < -f'(\phi(z))$ so $|\phi'(z)| = |f'(z)/f'(\phi(z))| < 1$.
Using the substitution $x = \phi(z)$ gives:
@ -732,7 +732,7 @@ Using the substitution $x = \phi(z)$ gives:
\end{align*}
Letting $h=f(u) \rightarrow c$ we get the *inequality*
Letting $h=f(u \rightarrow c)$ we get the *inequality*
$$
@ -789,7 +789,7 @@ y''(t) &= -g - W(t,x(t), x'(t), y(t), y'(t)) \cdot y'(t)\\
with initial conditions: $x(0) = y(0) = 0$ and $x'(0) = v_0 \cos(\theta), y'(0) = v_0 \sin(\theta)$.
Only with certain drag forces, can this set of equations be be solved exactly, though it can be approximated numerically for admissible $W$, but if $W$ is strictly positive then it can be shown $x(t)$ is increasing on $[0, x_\infty)$ and so invertible, and $f(u) = y(x^{-1}(u))$ is three times differentiable with both $f$ and $f'$ being strictly concave, as it can be shown that (say $x(v) = u$ so $dv/du = 1/x'(v) > 0$):
Only with certain drag forces, can this set of equations be solved exactly, though it can be approximated numerically for admissible $W$, but if $W$ is strictly positive then it can be shown $x(t)$ is increasing on $[0, x_\infty)$ and so invertible, and $f(u) = y(x^{-1}(u))$ is three times differentiable with both $f$ and $f'$ being strictly concave, as it can be shown that (say $x(v) = u$ so $dv/du = 1/x'(v) > 0$):
@ -837,7 +837,7 @@ The length of the curve given by $f(x) = e^x$ between $0$ and $1$ is certainly l
#| hold: true
#| echo: false
f(x) = exp(x)
val = sqrt( (f(1) - f(0))^2 - (1 - 0)^2)
val = sqrt( (f(1) - f(0))^2 + (1 - 0)^2)
numericq(val)
```
@ -888,7 +888,7 @@ A [pursuit](http://www-history.mcs.st-and.ac.uk/Curves/Pursuit.html) curve is a
#| echo: false
f(x) = x^2 - log(x)
a, b= 1/10, 2
val, _ = quadgk( x -> sqrt(1 + (f)(x)^2), a, b)
val, _ = quadgk( x -> sqrt(1 + D(f)(x)^2), a, b)
numericq(val)
```
@ -918,7 +918,7 @@ sqrt((tan(pi/4) - tan(-pi/4))^2 + (pi/4 - -pi/4)^2)
###### Question
Find the length of the graph of the function $g(x) =\int_0^x \tan(x)dx$ between $0$ and $\pi/4$ by hand or numerically:
Find the length of the graph of the function $g(x) =\int_0^x \tan(t)dt$ between $0$ and $\pi/4$ by hand or numerically:
```{julia}
@ -981,7 +981,7 @@ f(x,a) = a * cosh(x/a)
inside = 1 + diff(f(x,a), x)^2
```
Just trying `integrate(sqrt(inside), x)` will fail, but if we try `integrate(sqrt(simplify(inside), x))` an antiderivative can be found. What is it?
Just trying `integrate(sqrt(inside), x)` will fail, but if we try `integrate(sqrt(simplify(inside)), x)` an antiderivative can be found. What is it?
```{julia}

22
quarto/integrals/surface_area.qmd
View File

@ -83,7 +83,7 @@ To see why this formula is as it is, we look at the parameterized case, the firs
Let a partition of $[a,b]$ be given by $a = t_0 < t_1 < t_2 < \cdots < t_n =b$. This breaks the curve into a collection of line segments. Consider the line segment connecting $(g(t_{i-1}), f(t_{i-1}))$ to $(g(t_i), f(t_i))$. Rotating this around the $x$ axis will generate something approximating a disc, but in reality will be the frustum of a cone. What will be the surface area?
Consider a right-circular cone parameterized by an angle $\theta$ and the largest radius $r$ (so that the height satisfies $r/h=\tan(\theta)$). If this cone were made of paper, cut up a side, and layed out flat, it would form a sector of a circle, whose area would be $R\gamma$ where $R$ is the radius of the circle (also the side length of our cone), and $\gamma$ an angle that we can figure out from $r$ and $\theta$. To do this, we note that the arc length of the circle's edge is $R\gamma$ and also the circumference of the bottom of the cone so $R\gamma = 2\pi r$. With all this, we can solve to get $A = \pi r^2/\sin(\theta)$. But we have a frustum of a cone with radii $r_0$ and $r_1$, so the surface area is a difference: $A = \pi (r_1^2 - r_0^2) /\sin(\theta)$.
Consider a right-circular cone parameterized by an angle $\theta$ and the largest radius $r$ (so that the height satisfies $r/h=\tan(\theta)$). If this cone were made of paper, cut up a side, and layed out flat, it would form a sector of a circle, whose area would be $R^2\gamma/2$ where $R$ is the radius of the circle (also the side length of our cone), and $\gamma$ an angle that we can figure out from $r$ and $\theta$. To do this, we note that the arc length of the circle's edge is $R\gamma$ and also the circumference of the bottom of the cone so $R\gamma = 2\pi r$. With all this, we can solve to get $A = \pi r^2/\sin(\theta)$. But we have a frustum of a cone with radii $r_0$ and $r_1$, so the surface area is a difference: $A = \pi (r_1^2 - r_0^2) /\sin(\theta)$.
Relating this to our values in terms of $f$ and $g$, we have $r_1=f(t_i)$, $r_0 = f(t_{i-1})$, and $\sin(\theta) = \Delta f / \sqrt{(\Delta g)^2 + (\Delta f)^2}$, where $\Delta f = f(t_i) - f(t_{i-1})$ and similarly for $\Delta g$.
@ -100,7 +100,7 @@ $$
(This is $2 \pi$ times the average radius times the slant height.)
As was done in the derivation of the formula for arc length, these pieces are multiplied both top and bottom by $\Delta t = t_{i} - t_{i-1}$. Carrying the bottom inside the square root and noting that by the mean value theorem $\Delta g/\Delta t = g(\xi)$ and $\Delta f/\Delta t = f(\psi)$ for some $\xi$ and $\psi$ in $[t_{i-1}, t_i]$, this becomes:
As was done in the derivation of the formula for arc length, these pieces are multiplied both top and bottom by $\Delta t = t_{i} - t_{i-1}$. Carrying the bottom inside the square root and noting that by the mean value theorem $\Delta g/\Delta t = g'(\xi)$ and $\Delta f/\Delta t = f'(\psi)$ for some $\xi$ and $\psi$ in $[t_{i-1}, t_i]$, this becomes:
$$
@ -160,14 +160,14 @@ ImageFile(imgfile, caption)
Lets see that the surface area of an open cone follows from this formula, even though we just saw how to get this value.
A cone be be envisioned as rotating the function $f(x) = x\tan(\theta)$ between $0$ and $h$ around the $x$ axis. This integral yields the surface area:
A cone can be envisioned as rotating the function $f(x) = x\tan(\theta)$ between $0$ and $h$ around the $x$ axis. This integral yields the surface area:
\begin{align*}
\int_0^h 2\pi f(x) \sqrt{1 + f'(x)^2}dx
&= \int_0^h 2\pi x \tan(\theta) \sqrt{1 + \tan(\theta)^2}dx \\
&= (2\pi\tan(\theta)\sqrt{1 + \tan(\theta)^2} x^2/2 \big|_0^h \\
&= (2\pi\tan(\theta)\sqrt{1 + \tan(\theta)^2}) x^2/2 \big|_0^h \\
&= \pi \tan(\theta) \sec(\theta) h^2 \\
&= \pi r^2 / \sin(\theta).
\end{align*}
@ -287,7 +287,7 @@ val
#| hold: true
g(u) = u
f(u) = u^u
S(u,v) = [g(u)*cos(v), g(u)*sin(v), f(u)]
S(u,v) = [g(u), f(u)*cos(v), f(u)*sin(v)]
us = range(0, 3/2, length=100)
vs = range(0, pi, length=100) # not 2pi (to see inside)
ws = unzip(S.(us,vs'))
@ -515,7 +515,7 @@ $$
A = \int_u^{u+h} 2\pi f(x) \sqrt{1 + f'(x)^2} dx.
$$
If we let $f(x) = y$ then $f'(x) = x/y$. With this, what does the integral above come down to after cancellations:
If we let $f(x) = y$ then $f'(x) = -x/y$. With this, what does the integral above come down to after cancellations:
```{julia}
@ -544,7 +544,7 @@ Numerically find the value.
#| echo: false
g(t) = cos(t)
f(t) = sin(t)
a, b = 0, pi/4
a, b = 0, pi/6
val, _ = quadgk(t -> 2pi* f(t) * sqrt(g'(t)^2 + f'(t)^2), a, b)
numericq(val)
```
@ -558,8 +558,8 @@ The [astroid](http://www-history.mcs.st-and.ac.uk/Curves/Astroid.html) is parame
```{julia}
#| hold: true
#| echo: false
g(t) = cos(t^3)
f(t) = sin(t^3)
g(t) = cos(t)^3
f(t) = sin(t)^3
a, b = 0, pi
val, _ = quadgk(t -> 2pi* f(t) * sqrt(g'(t)^2 + f'(t)^2), a, b)
numericq(val)
@ -574,8 +574,8 @@ For the curve parameterized by $g(t) = a\cos(t)^5$ and $f(t) = a \sin(t)^5$. L
```{julia}
#| hold: true
#| echo: false
g(t) = cos(t^5)
f(t) = sin(t^5)
g(t) = cos(t)^5
f(t) = sin(t)^5
a, b = 0, pi
val, _ = quadgk(t -> 2pi* f(t) * sqrt(g'(t)^2 + f'(t)^2), a, b)
numericq(val)