align fix; theorem style; condition number

quarto/differentiable_vector_calculus/vector_valued_functions.qmd

@@ -800,7 +800,7 @@ Vector-valued functions do not have multiplication or division defined for them,
 For the dot product, the combination $\vec{f}(t) \cdot \vec{g}(t)$ we have a univariate function of $t$, so we know a derivative is well defined. Can it be represented in terms of the vector-valued functions? In terms of the component functions, we have this calculation specific to $n=2$, but that which can be generalized:
 
 
-
+$$
 \begin{align*}
 \frac{d}{dt}(\vec{f}(t) \cdot \vec{g}(t)) &=
 \frac{d}{dt}(f_1(t) g_1(t) + f_2(t) g_2(t))\\
@@ -808,6 +808,7 @@ For the dot product, the combination $\vec{f}(t) \cdot \vec{g}(t)$ we have a uni
 &= f_1'(t) g_1(t) + f_2'(t) g_2(t) + f_1(t) g_1'(t)  + f_2(t) g_2'(t)\\
 &= \vec{f}'(t)\cdot \vec{g}(t) + \vec{f}(t) \cdot \vec{g}'(t).
 \end{align*}
+$$
 
 
 Suggesting that a product rule like formula applies for dot products.
@@ -839,11 +840,12 @@ diff.(uₛ × vₛ, tₛ) - (diff.(uₛ, tₛ) × vₛ + uₛ × diff.(vₛ, t
 In summary, these two derivative formulas hold for vector-valued functions $R \rightarrow R^n$:
 
 
-
+$$
 \begin{align*}
 (\vec{u} \cdot \vec{v})' &= \vec{u}' \cdot \vec{v} + \vec{u} \cdot \vec{v}',\\
 (\vec{u} \times \vec{v})' &= \vec{u}' \times \vec{v} + \vec{u} \times \vec{v}'.
 \end{align*}
+$$
 
 
 ##### Application. Circular motion and the tangent vector.
@@ -896,11 +898,12 @@ Combining, Newton states $\vec{a} = -(GM/r^2) \hat{x}$.
 Now to show the first law. Consider $\vec{x} \times \vec{v}$. It is constant, as:
 
 
-
+$$
 \begin{align*}
 (\vec{x} \times \vec{v})' &= \vec{x}' \times \vec{v} + \vec{x} \times \vec{v}'\\
 &= \vec{v} \times \vec{v} + \vec{x} \times \vec{a}.
 \end{align*}
+$$
 
 
 Both terms are $\vec{0}$, as $\vec{a}$ is parallel to $\vec{x}$ by the above, and clearly $\vec{v}$ is parallel to itself.
@@ -912,34 +915,37 @@ This says, $\vec{x} \times \vec{v} = \vec{c}$ is a constant vector, meaning, the
 Now, by differentiating $\vec{x} = r \hat{x}$ we have:
 
 
-
+$$
 \begin{align*}
 \vec{v} &= \vec{x}'\\
 &= (r\hat{x})'\\
 &= r' \hat{x} + r \hat{x}',
 \end{align*}
+$$
 
 
 and so
 
 
-
+$$
 \begin{align*}
 \vec{c} &= \vec{x} \times \vec{v}\\
 &= (r\hat{x}) \times (r'\hat{x} + r \hat{x}')\\
 &= r^2 (\hat{x} \times \hat{x}').
 \end{align*}
+$$
 
 
 From this, we can compute $\vec{a} \times \vec{c}$:
 
 
-
+$$
 \begin{align*}
 \vec{a} \times \vec{c} &= (-\frac{GM}{r^2})\hat{x} \times r^2(\hat{x} \times \hat{x}')\\
 &= -GM \hat{x} \times (\hat{x} \times \hat{x}') \\
 &= GM (\hat{x} \times \hat{x}')\times \hat{x}.
 \end{align*}
+$$
 
 
 The last line by anti-commutativity.
@@ -948,22 +954,24 @@ The last line by anti-commutativity.
 But, the triple cross product can be simplified through the identify $(\vec{u}\times\vec{v})\times\vec{w} = (\vec{u}\cdot\vec{w})\vec{v} - (\vec{v}\cdot\vec{w})\vec{u}$. So, the above becomes:
 
 
-
+$$
 \begin{align*}
 \vec{a} \times \vec{c} &=  GM ((\hat{x}\cdot\hat{x})\hat{x}' - (\hat{x} \cdot \hat{x}')\hat{x})\\
 &= GM (1 \hat{x}' - 0 \hat{x}).
 \end{align*}
+$$
 
 
 Now, since $\vec{c}$ is constant, we have:
 
 
-
+$$
 \begin{align*}
 (\vec{v} \times \vec{c})' &= (\vec{a} \times \vec{c})\\
 &= GM \hat{x}'\\
 &= (GM\hat{x})'.
 \end{align*}
+$$
 
 
 The two sides have the same derivative, hence differ by a constant:
@@ -979,7 +987,7 @@ As $\vec{x}$ and $\vec{v}\times\vec{c}$ lie in the same plane - orthogonal to $\
 Now
 
 
-
+$$
 \begin{align*}
 c^2 &= \|\vec{c}\|^2 \\
 &= \vec{c} \cdot \vec{c}\\
@@ -989,6 +997,7 @@ c^2 &= \|\vec{c}\|^2 \\
 &= GMr + r \hat{x} \cdot \vec{d}\\
 &= GMr + rd \cos(\theta).
 \end{align*}
+$$
 
 
 Solving, this gives the first law. That is, the radial distance is in the form of an ellipse:
@@ -1397,12 +1406,15 @@ plotly();
 In [Arc length](../integrals/arc_length.html) there is a discussion of how to find the arc length of a parameterized curve in $2$ dimensions. The general case is discussed by [Destafano](https://randomproofs.files.wordpress.com/2010/11/arc_length.pdf) who shows:
 
 
-> *Arc-length*: if a curve $C$ is parameterized by a smooth function $\vec{r}(t)$ over an interval $I$, then the arc length of $C$ is:
->
-> $$
-> \int_I \| \vec{r}'(t)  \| dt.
-> $$
+::: {.callout-note icon=false}
+## Arc-length
 
+If a curve $C$ is parameterized by a smooth function $\vec{r}(t)$ over an interval $I$, then the arc length of $C$ is:
+
+$$
+\int_I \| \vec{r}'(t)  \| dt.
+$$
+:::
 
 
 If we associate $\vec{r}'(t)$ with the velocity, then this is the integral of the speed (the magnitude of the velocity).
@@ -1519,12 +1531,13 @@ $$
 As before, but further, we have if $\kappa$ is the curvature and $\tau$ the torsion, these relationships expressing the derivatives with respect to $s$ in terms of the components in the frame:
 
 
-
+$$
 \begin{align*}
 \hat{T}'(s) &=                    &\kappa \hat{N}(s)  &\\
 \hat{N}'(s) &= -\kappa \hat{T}(s) &                   &+ \tau \hat{B}(s)\\
 \hat{B}'(s) &=                    &-\tau \hat{N}(s)   &
 \end{align*}
+$$
 
 
 These are the [Frenet-Serret](https://en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas) formulas.
@@ -1637,12 +1650,13 @@ end
 Levi and Tabachnikov prove in their Proposition 2.4:
 
 
-
+$$
 \begin{align*}
 \kappa(u) &= \frac{d\alpha(u)}{du} + \frac{\sin(\alpha(u))}{a},\\
 |\frac{dv}{du}| &= |\cos(\alpha)|, \quad \text{and}\\
 k &= \frac{\tan(\alpha)}{a}.
 \end{align*}
+$$
 
 
 The first equation relates the steering angle with the curvature. If the steering angle is not changed ($d\alpha/du=0$) then the curvature is constant and the motion is circular. It will be greater for larger angles (up to $\pi/2$). As the curvature is the reciprocal of the radius, this means the radius of the circular trajectory will be smaller. For the same constant steering angle, the curvature will be smaller for longer wheelbases, meaning the circular trajectory will have a larger radius. For cars, which have similar dynamics, this means longer wheelbase cars will take more room to make a U-turn.
@@ -1657,13 +1671,14 @@ The last equation, relates the curvature of the back wheel track to the steering
 To derive the first one, we have previously noted that when a curve is parameterized by arc length, the curvature is more directly computed: it is the magnitude of the derivative of the tangent vector. The tangent vector is of unit length, when parametrized by arc length. This implies its derivative will be orthogonal. If $\vec{r}(t)$ is a parameterization by arc length, then the curvature formula simplifies as:
 
 
-
+$$
 \begin{align*}
 \kappa(s) &= \frac{\| \vec{r}'(s) \times \vec{r}''(s) \|}{\|\vec{r}'(s)\|^3} \\
 &= \frac{\| \vec{r}'(s) \times \vec{r}''(s) \|}{1}                           \\
 &= \| \vec{r}'(s) \| \| \vec{r}''(s) \| \sin(\theta)                         \\
 &= 1 \| \vec{r}''(s) \| 1 = \| \vec{r}''(s) \|.
 \end{align*}
+$$
 
 
 So in the above, the curvature is $\kappa = \| \vec{F}''(u) \|$ and $k = \|\vec{B}''(v)\|$.
@@ -1691,7 +1706,7 @@ $$
 It must be that the tangent line of $\vec{B}$ is parallel to $\vec{U} \cos(\alpha) + \vec{V} \sin(\alpha)$. To utilize this, we differentiate $\vec{B}$ using the facts that $\vec{U}' = -\kappa \vec{V}$ and $\vec{V}' = \kappa \vec{U}$. These coming from $\vec{U} = \vec{F}'$ and so it's derivative in $u$ has magnitude yielding the curvature, $\kappa$, and direction orthogonal to $\vec{U}$.
 
 
-
+$$
 \begin{align*}
 \vec{B}'(u) &= \vec{F}'(u)
 -a \vec{U}' \cos(\alpha) -a \vec{U} (-\sin(\alpha)) \alpha'
@@ -1703,12 +1718,13 @@ a (\kappa) \vec{U} \sin(\alpha) - a \vec{V} \cos(\alpha) \alpha' \\
 + a(\alpha' - \kappa) \sin(\alpha) \vec{U}
 - a(\alpha' - \kappa) \cos(\alpha)\vec{V}.
 \end{align*}
+$$
 
 
 Extend the $2$-dimensional vectors to $3$ dimensions, by adding a zero $z$ component, then:
 
 
-
+$$
 \begin{align*}
 \vec{0} &= (\vec{U}
 + a(\alpha' - \kappa) \sin(\alpha) \vec{U}
@@ -1721,6 +1737,7 @@ a(\alpha' - \kappa) \cos(\alpha)\vec{V} \times \vec{U} \cos(\alpha) \\
 a(\alpha'-\kappa) \cos^2(\alpha)) \vec{U} \times \vec{V} \\
 &= (\sin(\alpha) + a (\alpha' - \kappa)) \vec{U} \times \vec{V}.
 \end{align*}
+$$
 
 
 The terms $\vec{U} \times\vec{U}$ and $\vec{V}\times\vec{V}$ being $\vec{0}$, due to properties of the cross product. This says the scalar part must be $0$, or
@@ -1733,7 +1750,7 @@ $$
 As for the second equation, from the expression for $\vec{B}'(u)$, after setting $a(\alpha'-\kappa) = -\sin(\alpha)$:
 
 
-
+$$
 \begin{align*}
 \|\vec{B}'(u)\|^2
 &= \| (1 -\sin(\alpha)\sin(\alpha)) \vec{U} +\sin(\alpha)\cos(\alpha) \vec{V} \|^2\\
@@ -1742,6 +1759,7 @@ As for the second equation, from the expression for $\vec{B}'(u)$, after setting
 &= \cos^2(\alpha)(\cos^2(\alpha) + \sin^2(\alpha))\\
 &= \cos^2(\alpha).
 \end{align*}
+$$
 
 
 From this $\|\vec{B}(u)\| = |\cos(\alpha)\|$. But $1 = \|d\vec{B}/dv\| = \|d\vec{B}/du \| \cdot |du/dv|$ and $|dv/du|=|\cos(\alpha)|$ follows.
@@ -1778,11 +1796,12 @@ Consider a parameterization of a curve by arc-length, $\vec\gamma(s) = \langle u
 Consider two nearby points $t$ and $t+\epsilon$ and the intersection of $l_t$ and $l_{t+\epsilon}$. That is, we need points $a$ and $b$ with: $l_t(a) = l_{t+\epsilon}(b)$. Setting the components equal, this is:
 
 
-
+$$
 \begin{align*}
 u(t) - av'(t) &= u(t+\epsilon) - bv'(t+\epsilon) \\
 v(t) + au'(t) &= v(t+\epsilon) + bu'(t+\epsilon).
 \end{align*}
+$$
 
 
 This is a linear equation in two unknowns ($a$ and $b$) which can be solved. Here is the value for `a`:
@@ -1801,24 +1820,26 @@ out[a]
 Letting $\epsilon \rightarrow 0$ we get an expression for $a$ that will describe the evolute at time $t$ in terms of the function $\gamma$. Looking at the expression above, we can see that dividing the *numerator* by $\epsilon$ and taking a limit will yield $u'(t)^2 + v'(t)^2$. If the *denominator* has a limit after dividing by $\epsilon$, then we can find the description sought. Pursuing this leads to:
 
 
-
+$$
 \begin{align*}
 \frac{u'(t) v'(t+\epsilon) - v'(t) u'(t+\epsilon)}{\epsilon}
 &= \frac{u'(t) v'(t+\epsilon) -u'(t)v'(t) + u'(t)v'(t)- v'(t) u'(t+\epsilon)}{\epsilon} \\
 &= \frac{u'(t)(v'(t+\epsilon) -v'(t))}{\epsilon} + \frac{(u'(t)- u'(t+\epsilon))v'(t)}{\epsilon},
 \end{align*}
+$$
 
 
 which in the limit will give $u'(t)v''(t) - u''(t) v'(t)$. All told, in the limit as $\epsilon \rightarrow 0$ we get
 
 
-
+$$
 \begin{align*}
 a &= \frac{u'(t)^2 + v'(t)^2}{u'(t)v''(t) - v'(t) u''(t)} \\
 &= 1/(\|\vec\gamma'\|\kappa) \\
 &= 1/(\|\hat{T}\|\kappa) \\
 &= 1/\kappa,
 \end{align*}
+$$
 
 
 with $\kappa$ being the curvature of the planar curve. That is, the evolute of $\vec\gamma$ is described by:
@@ -1844,13 +1865,14 @@ plot_parametric!(0..2pi, t -> (rₑ₃(t) + Normal(rₑ₃, t)/curvature(rₑ₃
 We computed the above illustration using $3$ dimensions (hence the use of `[1:2]...`) as the curvature formula is easier to express. Recall, the curvature also appears in the [Frenet-Serret](https://en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas) formulas: $d\hat{T}/ds = \kappa \hat{N}$ and $d\hat{N}/ds = -\kappa \hat{T}+ \tau \hat{B}$. In a planar curve, as under consideration, the binormal is $\vec{0}$. This allows the computation of $\vec\beta(s)'$:
 
 
-
+$$
 \begin{align*}
 \vec{\beta}' &= \frac{d(\vec\gamma + (1/ \kappa) \hat{N})}{ds}\\
 &= \hat{T} + (-\frac{\kappa '}{\kappa ^2}\hat{N} + \frac{1}{\kappa} \hat{N}')\\
 &= \hat{T} - \frac{\kappa '}{\kappa ^2}\hat{N} + \frac{1}{\kappa} (-\kappa \hat{T})\\
 &= - \frac{\kappa '}{\kappa ^2}\hat{N}.
 \end{align*}
+$$
 
 
 We see $\vec\beta'$ is zero (the curve is non-regular) when $\kappa'(s) = 0$. The curvature changes from increasing to decreasing, or vice versa at each of the $4$ crossings of the major and minor axes - there are $4$ non-regular points, and we see $4$ cusps in the evolute.
@@ -1915,11 +1937,12 @@ $$
 If $\vec\gamma(s)$ is parameterized by arc length, then this simplifies quite a bit, as the unit tangent is just $\vec\gamma'(s)$ and the remaining arc length just $(s-a)$:
 
 
-
+$$
 \begin{align*}
 \vec\beta_a(s) &= \vec\gamma(s) - \vec\gamma'(s) (s-a) \\
 &=\vec\gamma(s) - \hat{T}_{\vec\gamma}(s)(s-a).\quad (a \text{ is the arc-length parameter})
 \end{align*}
+$$
 
 
 With this characterization, we see several properties:
@@ -1940,11 +1963,12 @@ $$
 In the following we show that:
 
 
-
+$$
 \begin{align*}
 \kappa_{\vec\beta_a}(s) &= 1/(s-a),\\
 \hat{N}_{\vec\beta_a}(s) &= \hat{T}_{\vec\beta_a}'(s)/\|\hat{T}_{\vec\beta_a}'(s)\| = -\hat{T}_{\vec\gamma}(s).
 \end{align*}
+$$
 
 
 The first shows in a different way that when $s=a$ the curve is not regular, as the curvature fails to exists. In the above figure, when the involute touches $\vec\gamma$, there will be a cusp.
@@ -1953,7 +1977,7 @@ The first shows in a different way that when $s=a$ the curve is not regular, as
 With these two identifications and using $\vec\gamma'(s) = \hat{T}_{\vec\gamma(s)}$, we have the evolute simplifies to
 
 
-
+$$
 \begin{align*}
 \vec\beta_a(s) + \frac{1}{\kappa_{\vec\beta_a}(s)}\hat{N}_{\vec\beta_a}(s)
 &=
@@ -1962,6 +1986,7 @@ With these two identifications and using $\vec\gamma'(s) = \hat{T}_{\vec\gamma(s
 \vec\gamma(s) + \hat{T}_{\vec\gamma}(s)(s-a) + \frac{1}{1/(s-a)} (-\hat{T}_{\vec\gamma}(s)) \\
 &= \vec\gamma(s).
 \end{align*}
+$$
 
 
 That is the evolute of an involute of $\vec\gamma(s)$ is $\vec\gamma(s)$.
@@ -1970,12 +1995,13 @@ That is the evolute of an involute of $\vec\gamma(s)$ is $\vec\gamma(s)$.
 We have:
 
 
-
+$$
 \begin{align*}
 \beta_a(s) &= \vec\gamma - \vec\gamma'(s)(s-a)\\
 \beta_a'(s) &= -\kappa_{\vec\gamma}(s)(s-a)\hat{N}_{\vec\gamma}(s)\\
 \beta_a''(s) &= (-\kappa_{\vec\gamma}(s)(s-a))' \hat{N}_{\vec\gamma}(s) + (-\kappa_{\vec\gamma}(s)(s-a))(-\kappa_{\vec\gamma}\hat{T}_{\vec\gamma}(s)),
 \end{align*}
+$$
 
 
 the last line by the Frenet-Serret  formula for *planar* curves which show $\hat{T}'(s) = \kappa(s) \hat{N}$ and $\hat{N}'(s) = -\kappa(s)\hat{T}(s)$.
@@ -1984,11 +2010,12 @@ the last line by the Frenet-Serret  formula for *planar* curves which show $\hat
 To compute the curvature of $\vec\beta_a$, we need to compute both:
 
 
-
+$$
 \begin{align*}
 \| \vec\beta' \|^3 &= |\kappa^3 (s-a)^3|\\
 \| \vec\beta' \times \vec\beta'' \| &= |\kappa(s)^3 (s-a)^2|,
 \end{align*}
+$$
 
 
 the last line using both $\hat{N}\times\hat{N} = \vec{0}$ and $\|\hat{N}\times\hat{T}\| = 1$. The curvature then is $\kappa_{\vec\beta_a}(s) = 1/(s-a)$.
@@ -2672,13 +2699,14 @@ radioq(choices, answ)
 The evolute comes from the formula $\vec\gamma(T) - (1/\kappa(t)) \hat{N}(t)$. For hand computation, this formula can be explicitly given by two components $\langle X(t), Y(t) \rangle$ through:
 
 
-
+$$
 \begin{align*}
 r(t) &= x'(t)^2 + y'(t)^2\\
 k(t) &= x'(t)y''(t) - x''(t) y'(t)\\
 X(t) &= x(t) - y'(t) r(t)/k(t)\\
 Y(t) &= y(t) + x'(t) r(t)/k(t)
 \end{align*}
+$$
 
 
 Let $\vec\gamma(t) = \langle t, t^2 \rangle = \langle x(t), y(t)\rangle$ be a parameterization of a parabola.