Задачка

[Гост]

Да се намери дължината на [tex]u = e^{k\theta},0\le\theta\le \pi,k=const.[/tex] върху конуса [tex]\vec{X}=(u\cos\theta,u\sin\theta,u)[/tex].

[Гост]

$$\vec{X}_{\theta } = (–u \sin \theta, u \cos \theta, 0)$$
$$\vec{X}_{u }=(\cos \theta, \sin \theta, 1)$$
$$E = \vec{X}_θ • \vec{X}_θ = u^{2}\sin^{2}θ + u^{2}\cos^{2}θ = u^2$$
$$F = \vec{X}_θ • \vec{X}_u = –u \sin θ \cos θ + u \sin θ \cos θ = 0$$
$$ G = \vec{X}_u • \vec{X}_u = \cos^2θ + \sin^2θ + 1 = 2$$
$$ (du / dθ) = u\cdot k, (dθ / dθ) = 1$$
$$ s = \int\limits_{0}^{\pi} \left[u^2(dθ / dθ)^2 + 0 (dθ / dθ) (du / dθ) + 2(du / dθ)^2\right]^{1/2} dθ = \int\limits_{0}^{\pi} [u^2 + 2u^2 k^2]^{1/2} dθ = \sqrt{1+2k^2}\int\limits_{0}^{\pi} u dθ = \sqrt{1+2k^2}\int\limits_{0}^{\pi} e^{kθ} dθ = \left[\sqrt{1+2k^2} / k\right] e^{kθ} |_{o }^π$$
$$ s = \left[\sqrt{1+2k^2} / k\right] \left(e^{\pi θ}-1\right)$$

[Гост]

*[tex](e^{k\pi}-1)[/tex]

[Гост]

още една: да се намерят геодезичните криви върху правия кръгов конус:
[tex]\vec{X} = (u \sin \alpha \cos \theta, u \sin \alpha \sin \theta, u \cos\alpha), \alpha=const., u>0[/tex]

[Гост]

$$ \vec{X}_u= (\sin \alpha \cos \theta, \sin \alpha \sin \theta, \cos \alpha)$$
$$\vec{X}_θ= (-u\sin \alpha \sin \theta, u\sin \alpha \cos \theta, 0)$$
$$E=\vec{X}_u \cdot \vec{X}_u=\sin^2α \cos^2θ + \sin^2α \sin^2θ + \cos^2α = 1$$
$$F= \vec{X}_u\cdot \vec{X}_θ=–u \sin^2α \cos θ \sin θ + u \sin^2α \sin θ \cos θ = 0$$
$$G=\vec{X}_θ\cdot \vec{X}_θ= u^2 \sin^2α \sin^2θ + u^2 \sin^2α \cos^2 θ = u^2 \sin^2α$$
$$\Gamma^{1}_{11}=0,\Gamma^{1}_{12}=0,\Gamma^{2}_{11}=0,\Gamma^{2}_{22}=0$$
$$\Gamma^{1}_{22}= – (2u^2 \sin^2α \cdot u \sin^2α) / (2u^2 \sin^2α) = – u \sin^2α$$
$$\Gamma^{2}_{12}= (2u \sin^2α) / (2u^2 \sin^2α) = 1/u$$
$$ (d^2θ / ds^2) + (2/u) (du/ds) (dθ/ds) = 0\Rightarrow (d^2θ / ds^2) = – (2/u) (du/ds) (dθ/ds) $$
$$ \varphi = dθ/ds\Rightarrow (1/φ) (dφ/ds) = – (2/u) (du/ds)$$
$$\Rightarrow \ln φ = –2 \ln u + C_1 \Rightarrow φ = dθ/ds = C/u^2 \sin^2α, C = e^{C_1}\sin^2α$$
$$ 1 = ||d\vec{X} / ds||^2 = ||\vec{X}_u (du/ds) + \vec{X}_θ (dθ/ds)||^2$$
$$ 1 = E(du/ds)^2 + 2F (du/ds) (dθ/ds) + G(dθ/ds)^2\Rightarrow 1 = (du/ds)^2 + u^2 \sin^2α(dθ/ds)^2\Rightarrow 1 = (du/ds)^2 + u^2 \sin^2α [C^2 / (u^2 \sin^2 α)^2]$$
$$\Rightarrow (du/ds) = \sqrt{u^2 \sin^2 α-C^2} / (u \sin α)$$
$$(du/dθ) = (du/ds) • (ds/dθ) = (1/C) u \sin α \sqrt{u^2 \sin^2 α-C^2}$$
$$\Rightarrow\int du/(u \sin α \sqrt{u^2 \sin^2 α-C^2}) = \int(dθ/C)$$
$$\Rightarrow u = A \sec [(\sin α)θ + B], A=const., B=const.$$

[Гост]

последна (май): ако [tex]u[/tex] и [tex]v[/tex] са дължината и ширината върху сфера, да се намери ъгълът, под който кривата [tex]v=u[/tex] пресича екватора.

[Гост]

$$\vec{X}=\vec{X}(u,v)=(\cos v\cos u,\cos v\sin u,\sin v)$$
$$\vec{X}_u=(-\sin u \cos v,\cos v \cos u,0)$$
$$\vec{X}_v=(-\sin v\cos u,-\sin v\sin u,\cos v)$$
$$E=\vec{X}_u•\vec{X}_u = \sin^2u \cos^2v + \cos^2v \cos^2u + 0 = \cos^2v$$
$$F=\vec{X}_u•\vec{X}_v= \sin u \sin v \cos u \cos v - \sin v \sin u \cos v \cos u+0=0$$
$$G=\vec{X}_v•\vec{X}_v=\sin^2v \cos^2u + \sin^2v \sin^2u + \cos^2v = 1$$
$$\Rightarrow ds^2 = \cos^2 vdu^2 + dv^2$$
$$u = v = 0: h(u) = u, k(u) = 0; h' = 1, k' = 0$$
$$\Rightarrow \cos θ = (\cos^2v + 0) / (\sqrt{\cos^2v + 1}\cdot\sqrt{\cos^2v})=\cos v / \sqrt{\cos^2v + 1} $$
$$ \cos θ = (\cos 0) / \sqrt{\cos^2 0 + 1} = 1 / \sqrt{2}$$
$$\Rightarrow θ = π/4 $$

[Гост]

пресметнете [tex]\iint_S curl \vec{F} \cdot \vec{n} dS[/tex], ако [tex]\vec{F}=yz\vec{i}+(2x+y-1)\vec{j}+(x^{2}+2z)\vec{k}[/tex], а [tex]S[/tex] е от пресичането на цилиндрите [tex]x^2 + y^2 = a^2[/tex] и [tex]x^2 + z^2 = a^2[/tex] в първи октант.

[Гост]

$$\iint_S (\nabla\times F) \cdot n dS = \oint_C F \cdot dr$$
$$C:=\{AB,BD,DE,EA\}$$
$$\oint_C F ∙ dr = \int\limits_{A}^{B} F ∙ dr + \int\limits_{B}^{D} F ∙ dr + \int\limits_{D}^{E} F ∙ dr + \int\limits_{E}^{A} F ∙ dr$$
$$AB: y = a, x = 0, dy = dx = 0\Rightarrow F=azi + (a – 1) j + 2zk; dr = dxi + dyj + dzk= dzk$$
$$\Rightarrow \int\limits_{A}^{B} F ∙ dr = \int\limits_{z=0}^{a} (azi + (a – 1) j + 2zk) ∙ dzk= \int\limits_{0}^{a} 2zdz = a^2 $$
$$ BD: z = a, x = 0, dz = dx = 0\Rightarrow F = ayi + (y – 1) j + 2ak$$
$$\Rightarrow \int\limits_{B}^{D} F ∙ dr=\int\limits_{0}^{a} (y – 1)dy = [(y^2 / 2) – y]|_0^a = a – (a^2 / 2)$$
$$DE: x^2 = a^2 – z^2, y = 0, dy = 0\Rightarrow\int\limits_{D}^{E} F ∙ dr=\int\limits_{0}^{a} (x^2 + 2z)dz =\int\limits_{0}^{a} (a^2 – z^2 + 2z)dz= [a^2z – (z^3 / 3) + z^2] |_0^a = – (2/3)a^3 – a^2$$
$$EA: x^2 = a^2 – y^2, z = 0, dz = 0\Rightarrow\int\limits_{E}^{A}F ∙ dr=\int\limits_{0}^{a}(2x + y – 1)dy=\int\limits_{0}^{a}(2\sqrt{a^2-y^2} + y – 1)dy$$
$$={y\sqrt{a^2-y^2} + a^2\sin^-1 (y/a) + (y^2 / 2) – y} |_0^a= (a^2 / 2) (π + 1) – a$$
$$\Rightarrow\oint_C F ∙ dr = a^2 + a – (a^2 / 2) – (2/3) a^3 – a^2 + (a^2 / 2)π + (a^2 / 2) – a= (a^2 / 2) π – (2/3) a^3=\iint_S curl F ∙ n dS$$

[Гост]

чрез тензори докажете равенството [tex]\frac{\partial\vec{F}}{\partial v}\cdot\vec{V}\times\frac{\partial\vec{r}}{\partial u}-\frac{\partial\vec{F}}{\partial u}\cdot\vec{V}\times\frac{\partial\vec{r}}{\partial v}=[(\nabla\cdot \vec{F})\vec{V}-(\vec{V}\cdot\nabla)\vec{F}]\cdot\left[\frac{\partial\vec{r}}{\partial u}\times\frac{\partial\vec{r}}{\partial v}\right][/tex], където [tex]\vec{F}=\vec{F}[\vec{r}(u,v,t),t],\vec{r}=\vec{r}(u,v,t),\vec{V}=\frac{\partial\vec{r}}{\partial t}[/tex]

[Гост]

$$(∂F / ∂v) = [(∂r / ∂v) ∙ ∇]F$$
$$(∂F / ∂u) = [(∂r / ∂u) ∙ ∇]F$$
$$(∂F / ∂v) ∙ V × (∂r / ∂u) – (∂F / ∂u) ∙ V × (∂r / ∂v) = (∂F_i / ∂v) [V × (∂r / ∂u)]_i – (∂F_i / ∂u) [V × (∂r / ∂v)]_i$$
$$ = e_{ijk}(∂F_i / ∂v)V_j (∂x_k / ∂u) – e_{ijk}(∂F_i / ∂u)V_j (∂x_k / ∂v)$$
$$ = e_{ijk}[(∂x_ℓ / ∂v)(∂F_i / ∂x_ℓ)]V_j(∂x_k / ∂u) – e_{ijk}[(∂x_ℓ / ∂u)(∂F_i / ∂x_ℓ)]V_j(∂x_k / ∂v)$$
$$ = e_{ijk}V_j (∂F_i / ∂x_ℓ)[(∂x_ℓ / ∂v) (∂x_k / ∂u) – (∂x_ℓ / ∂u)(∂x_k / ∂v)]$$
$$ = e_{ijk}V_j (∂F_i / ∂x_ℓ)(δ_{ℓm}δ_{kn} – δ_{ℓn}δ_{km})(∂x_m / ∂v)(∂x_n / ∂u)$$
$$= e_{ijk}V_j (∂F_i / ∂x_ℓ) e_{ℓkr}e_{mnr} (∂x_m / ∂v)(∂x_n / ∂u)$$
$$= e_{ijk} e_{rℓk}e_{mnr} V_j (∂x_m / ∂v)(∂x_n / ∂u)(∂F_i / ∂x_ℓ)$$
$$ = (δ_{ir}δ_{jℓ} – δ_{iℓ}δ_{jr})e_{mnr} V_j (∂x_m / ∂v)(∂x_n / ∂u)(∂F_i / ∂x_ℓ)$$
$$= e_{mnr}(∂x_m / ∂v)(∂x_n / ∂u)V_ℓ(∂F_r / ∂x_ℓ) – e_{mnj}(∂x_m / ∂v)(∂x_n / ∂u)V_j(∂F_ℓ / ∂x_ℓ)$$
$$= – e_{rnm} V_ℓ (∂F_r / ∂x_ℓ)(∂x_n / ∂u)(∂x_m / ∂v) + e_{jnm} (∂F_ℓ / ∂x_ℓ)V_j(∂x_n / ∂u)(∂x_m / ∂v)$$
$$= – [(V ∙ ∇)F]_r [(∂r / ∂u) × (∂r / ∂v)]_r + [(∇ ∙ F)V]_j [(∂r / ∂u) × (∂r / ∂v)]_j$$
$$ = – [(V ∙ ∇)F] ∙ [(∂r / ∂u) × (∂r / ∂v)] + (∇ ∙ F)V ∙ [(∂r / ∂u) × (∂r / ∂v)] $$
https://www.youtube.com/watch?v=GrCcJOkmMys