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| courses:b0b01ma2:exams [2025/01/14 23:36] – [Řešení] jpelc | courses:b0b01ma2:exams [2025/01/15 00:04] (current) – [Dvojný integrál] jpelc | ||
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| Tedy podezřelé body jsou $a_0 = (x, y, z)$. | Tedy podezřelé body jsou $a_0 = (x, y, z)$. | ||
| + | ==== Konzervativní vektorové pole ==== | ||
| + | Ověřte, že vektorové pole $\vec{F}(x, y, z) = (e^z + z^2 y, \cos(y) + z^2 x, xe^z + 2xyz)$ je konzervativní a | ||
| + | nalezněte jeho potenciál. | ||
| + | |||
| + | \begin{align*} | ||
| + | \frac{\partial f}{\partial x} &= e^z + z^2 y \iff f(x, y, z) = xe^z + xyz^2 + C(y, z) & \\ | ||
| + | \frac{\partial f}{\partial y} &= \cos(y) + z^2x \iff f(x, y, z) = xe^z + xyz^2 + \sin(y) + D(z) & \\ | ||
| + | \frac{\partial f}{\partial z} &= xe^z + 2xyz \iff f(x, y, z) = xe^z + xyz^2 + \sin(y) + k, \, k \in \mathbb{R} | ||
| + | \end{align*} | ||
| + | |||
| + | Potenciál vektorového pole $\vec{F}$ je: | ||
| + | $ f(x, y, z) = xe^z + xyz^2 + \sin(y) + k, \, k \in \mathbb{R}. $ | ||
| + | |||
| + | TODO: zdůvodnění | ||
| + | |||
| + | ==== Dvojný integrál ==== | ||
| + | |||
| + | Vypočtěte | ||
| + | $$ \int_0^1 \int_{2y}^2 e^{x^2} dx\ dy $$ | ||
| + | |||
| + | Prohodíme pořadí integrace na $dy\ dx$. | ||
| + | \begin{align*} | ||
| + | E\text{: } &0 \leq x \leq 2 & \\ | ||
| + | &0 \leq y \leq \frac{x}{2} | ||
| + | \end{align*} | ||
| + | |||
| + | < | ||
| + | \usepackage{tikz} | ||
| + | |||
| + | \begin{document} | ||
| + | |||
| + | \begin{tikzpicture}[> | ||
| + | % x axis | ||
| + | \draw[-> | ||
| + | \foreach \x in {1,2} | ||
| + | \draw[shift={(\x, | ||
| + | | ||
| + | % y axis | ||
| + | \draw[-> | ||
| + | \foreach \y in {1} | ||
| + | \draw[shift={(0, | ||
| + | \node[below left] at (0,0) {\footnotesize $0$}; | ||
| + | |||
| + | % Fill the area manually | ||
| + | \fill[yellow, | ||
| + | |||
| + | % Function plot (piecewise linear) | ||
| + | \draw[thick, | ||
| + | \draw[dashed, | ||
| + | \draw[dashed] (2, 0) -- (2, 1); | ||
| + | \draw[dashed] (2, 1) -- (0, 1); | ||
| + | |||
| + | % Label for the shaded region | ||
| + | \node[font=\scriptsize] at (1.5,0.35) {$E$}; | ||
| + | \end{tikzpicture} | ||
| + | |||
| + | \end{document} | ||
| + | |||
| + | </ | ||
| + | |||
| + | \begin{align*} | ||
| + | \int_0^2 \int_{0}^{\frac{x}{2}} e^{x^2} dy\ dx = \int_{0}^{2} \frac{x}{2} e^{x^2} dx = | ||
| + | \begin{array}{|c|} | ||
| + | u=x^2\\ | ||
| + | du = 2x\ dx\\ | ||
| + | \end{array} | ||
| + | = \frac{1}{4} \int_0^4 e^u\ du = \frac{1}{4} (e^4 - 1)\text{.} | ||
| + | \end{align*} | ||
| + | |||
| + | ==== Gaussova věta ==== | ||
| + | |||
| + | Pomocí Gaussovy věty zjistěte, jaký je tok pole $\vec{F}(x, y, z) = (z^2 -x, -2xy, \frac{3z}{1+x^2})$ hranicí tělesa | ||
| + | omezeného plochami $z = 4 - y^2$, $z=0$, $x=0$ a $x=3$ s vnější orientací. | ||
| + | |||
| + | Ze zadání máme skoro všechny meze útvaru známé, jen si nakreslíme $z$ v závislosti na $y$, abychom zjistili meze pro $y$. | ||
| + | \begin{multicols}{2} | ||
| + | \begin{tikzpicture}[> | ||
| + | %x axis | ||
| + | \draw[-> | ||
| + | \foreach \x in {-2, -1, 1,2} | ||
| + | \draw[shift={(\x, | ||
| + | %y axis | ||
| + | \draw[-> | ||
| + | \foreach \y in {1, 2, 3, 4} | ||
| + | \draw[shift={(0, | ||
| + | \node[below left] at (0,0) {\footnotesize $0$}; | ||
| + | | ||
| + | % Function plot (piecewise linear) | ||
| + | \draw[name path=A, thick, domain=0:2] plot(\x, {4-\x^2} ); | ||
| + | \draw[name path=B, thick, domain=-2: | ||
| + | \draw[name path=C, thick, domain=-2: | ||
| + | \tikzfillbetween[of=A and C, on layer=bg]{yellow}; | ||
| + | \tikzfillbetween[of=B and C, on layer=bg]{yellow}; | ||
| + | \node[font=\scriptsize] at (0.8,1.5) {$E$}; | ||
| + | | ||
| + | \end{tikzpicture} | ||
| + | \columnbreak | ||
| + | \begin{flalign*} | ||
| + | E\text{: } -2 &\leq y \leq 2 & \\ | ||
| + | \phantom{-}0 &\leq z \leq 4-y^2 | ||
| + | \end{flalign*} | ||
| + | \end{multicols} | ||
| + | $$ | ||
| + | \iiint\limits_{(M)} div(\vec{F}) \dif \vec{S} = \int_{0}^{3} \int_{-2}^{2} \int_{0}^{4-y^2} -1 -2x + \frac{3}{1+x^2} | ||
| + | \dif z \dif y \dif x = -\int_{0}^{3} \int_{-2}^{2} 1(4-y^2) + 2x(4-y^2) - \frac{3(4-y^2)}{1+x^2} \dif y \dif x = | ||
| + | $$ | ||
| + | |||
| + | $$ | ||
| + | -\int_{0}^{3} \int_{-2}^{2} 4 - y^2 + 8x -2xy^2 - \frac{12}{1+x^2} + \frac{3y^2}{1+x^2} \dif y \dif x = -\int_{0}^{3} 16 - | ||
| + | \frac{16}{3} + 32x - \frac{32x}{3} - \frac{48}{1+x^2} + \frac{16}{1+x^2} \dif x = | ||
| + | $$ | ||
| + | |||
| + | $$ | ||
| + | -\int_{0}^{3} \frac{32}{3} + \frac{64x}{3} - \frac{32}{1+x^2} \dif x = -(32 + 96 - 32 \arctan(3)) = 32(\arctan(3) - 4)\text{.} | ||
| + | $$ | ||
| ====== Test 10.1.2025 ====== | ====== Test 10.1.2025 ====== | ||
| - Rozviňte funkci $ f(x) = \frac{x}{1+2x^2} $ se středem v $ x_0 = 0 $ do mocninné řady. Určete největší otevřený interval, kde nastává rovnost funkce a řady. | - Rozviňte funkci $ f(x) = \frac{x}{1+2x^2} $ se středem v $ x_0 = 0 $ do mocninné řady. Určete největší otevřený interval, kde nastává rovnost funkce a řady. | ||
