Školska 2015/16, julski ispitni rok, I kolokvijum grupa 2

Koristimo metodu parcijalne integracije…

\( U=\arcsin x,\quad V=\int \sqrt{1-x^{2}}\,dx = \frac{1}{2}(\arcsin x + x\sqrt{1-x^{2}}) \)

\( dU=\frac{1}{\sqrt{1-x^{2}}} dx,\quad dV=\sqrt{1-x^{2}} dx \)

\( I = \arcsin x \cdot \frac{1}{2}(\arcsin x + x\sqrt{1-x^{2}}) – \frac{1}{2} \int (\arcsin x + x\sqrt{1-x^{2}}) \frac{dx}{\sqrt{1-x^{2}}} \)

\( = \frac{1}{2}(\arcsin x)^2 + \frac{1}{2}(\arcsin x \cdot x \sqrt{1-x^{2}}) – \frac{1}{2} \int \frac{\arcsin x}{\sqrt{1-x^{2}}} dx – \frac{1}{2} \int \frac{x \sqrt{1-x^{2}}}{\sqrt{1-x^{2}}} dx \)

Uvodimo smenu:

\( t = \arcsin x, \quad \sqrt{1-x^{2}} = \cos t, \quad x = \sin t, \quad dx = \cos t\, dt \)

Sledi:

\( \frac{\arcsin x}{\sqrt{1-x^{2}}} dx = t\, dt \)

\( I = \frac{1}{2}(\arcsin x)^2 + \frac{1}{2}(\arcsin x \cdot x \sqrt{1-x^{2}}) – \frac{1}{2} \int t\, dt – \frac{1}{2} \int x\, dx \)

\( = \frac{1}{2}(\arcsin x)^2 + \frac{1}{2}(\arcsin x \cdot x \sqrt{1-x^{2}}) – \frac{1}{2} \cdot \frac{t^2}{2} – \frac{1}{2} \cdot \frac{x^2}{2} + C \)

\( = \frac{1}{2}(\arcsin x)^2 + \frac{1}{2}(\arcsin x \cdot x \sqrt{1-x^{2}}) – \frac{1}{4} (\arcsin x)^2 – \frac{1}{4} x^2 + C \)

\(\displaystyle I = \frac{1}{4}(\arcsin x)^2 + \frac{1}{2}(\arcsin x \cdot x \sqrt{1-x^2}) – \frac{x^2}{4} + C \)

Zapremina: \( V = \pi \int_{-1}^{1} \left( \left(\frac{6}{x^2+1}\right)^2 – (x^2+2)^2 \right) dx \)

Dakle:

\( V = \pi \int_{-1}^{1} \left( \frac{36}{(x^2+1)^2} – (x^4 + 4x^2 + 4) \right) dx \)

Prvi integral označimo sa I:

\( I = \pi \int_{-1}^{1} \frac{36}{(x^2+1)^2} dx = 36\pi \int_{-1}^{1} \frac{1}{(x^2+1)^2} dx \)

Koristimo da je

\( \frac{d}{dx} \arctan x = \frac{1}{x^2+1}, \quad \frac{d}{dx} \left( \frac{x}{x^2+1} \right) = \frac{1 – x^2}{(x^2+1)^2} \)

Pokušamo da izrazimo:

\( \frac{36}{(x^2+1)^2} = A \cdot \frac{1 – x^2}{(x^2+1)^2} + B \cdot \frac{1}{x^2+1} \)

Pomnožimo sa \( (x^2+1)^2 \):

\( 36 = A(1 – x^2) + B(x^2 + 1) = (A + B) + (-A + B) x^2 \)

Rešenje: \( A = B = 18 \)

Dakle:

\( 36\pi \int_{-1}^{1} \frac{1}{(x^2+1)^2} dx = 18\pi \int_{-1}^{1} \frac{1 – x^2}{(x^2+1)^2} dx + 18\pi \int_{-1}^{1} \frac{1}{x^2+1} dx \)

Primitivne funkcije:

\( \int \frac{1 – x^2}{(x^2+1)^2} dx = \frac{x}{x^2+1}, \quad \int \frac{1}{x^2+1} dx = \arctan x \)

\( 36\pi \int_{-1}^{1} \frac{1}{(x^2+1)^2} dx = 18\pi \left[ \frac{x}{x^2+1} \right]_{-1}^1 + 18\pi \left[ \arctan x \right]_{-1}^1 \)

Izračunamo vrednosti: \[ \left. \frac{x}{x^2+1} \right|_{-1}^1 = \frac{1}{1+1} – \frac{-1}{1+1} = \frac{1}{2} + \frac{1}{2} = 1 \] \[ \left. \arctan x \right|_{-1}^1 = \arctan 1 – \arctan(-1) = \frac{\pi}{4} – \left(-\frac{\pi}{4}\right) = \frac{\pi}{2} \]

Dakle, \[ I = 18\pi \cdot 1 + 18\pi \cdot \frac{\pi}{2} = 18\pi + 9\pi^2 \]

Drugi integral je:

\[ \pi \int_{-1}^{1} (x^4 + 4x^2 + 4) dx = \pi \left[ \int_{-1}^{1} x^4 dx + 4 \int_{-1}^{1} x^2 dx + 4 \int_{-1}^{1} dx \right] \]

Računamo pojedinačno (funkcije su parne pa možemo duplirati integral od 0 do 1):

\[ \int_{-1}^{1} x^4 dx = 2 \int_0^1 x^4 dx = 2 \cdot \frac{x^5}{5}\Big|_0^1 = \frac{2}{5} \]

\[ \int_{-1}^{1} x^2 dx = 2 \int_0^1 x^2 dx = 2 \cdot \frac{x^3}{3}\Big|_0^1 = \frac{2}{3} \]

\[ \int_{-1}^{1} dx = 2 \]

Zbir: \[ \pi \left( \frac{2}{5} + 4 \cdot \frac{2}{3} + 4 \cdot 2 \right) = \pi \left( \frac{2}{5} + \frac{8}{3} + 8 \right) \]

Izračunamo zbir : \[ \frac{2}{5} + \frac{8}{3} + 8 = \frac{2}{5} + \frac{8}{3} + \frac{40}{5} = \frac{2}{5} + \frac{8}{3} + \frac{40}{5} \] \[ = \frac{2}{5} + \frac{40}{5} + \frac{8}{3} = \frac{42}{5} + \frac{8}{3} = \frac{126}{15} + \frac{40}{15} = \frac{166}{15} \]

Dakle: \[ \pi \cdot \frac{166}{15} = \frac{166\pi}{15} \]

Zapremina je razlika ova dva integrala: \[ V = I – \frac{166\pi}{15} = \left(18\pi + 9\pi^2 \right) – \frac{166\pi}{15} = 9\pi^2 + 18\pi – \frac{166\pi}{15} \]

Izračunamo zbir konstanti uz \(\pi\): \[ 18\pi – \frac{166\pi}{15} = \pi \left( 18 – \frac{166}{15} \right) = \pi \left( \frac{270}{15} – \frac{166}{15} \right) = \pi \frac{104}{15} \]

Dakle, konačan rezultat: \[ \boxed{ V = 9\pi^2 + \frac{104\pi}{15} } \]

Zapremina tela je:
\( V = 9\pi^2 + \frac{104\pi}{15} \)

Formula: \( P=2\pi\int y(t)\sqrt{(x'(t))^2+(y'(t))^2} \, dt \)

Izvodi su:

\( x'(t) = \frac{d}{dt} (2\cos^3 t) = -6 \cos^2 t \sin t \)

\( y'(t) = \frac{d}{dt} (2\sin^3 t) = 6 \sin^2 t \cos t \)

Dakle:

\( P = 2\pi \int 2\sin^3 t \sqrt{(-6 \cos^2 t \sin t)^2 + (6 \sin^2 t \cos t)^2} \, dt = 4\pi \int \sin^3 t \sqrt{36 \sin^2 t \cos^4 t + 36 \sin^4 t \cos^2 t} \, dt \)

\( = 4\pi \int \sin^3 t \sqrt{36 \sin^2 t \cos^2 t (\cos^2 t + \sin^2 t)} \, dt = 4\pi \int \sin^3 t \sqrt{36 \sin^2 t \cos^2 t} \, dt \)

\( = 24\pi \int \sin^3 t |\sin t \cos t| \, dt \)

U prvom kvadrantu, \( t \in [0, \pi/2] \), pa je \( \sin t \geq 0 \), \( \cos t \geq 0 \), dakle:

\( P = 4 \cdot 24 \pi \int_0^{\pi/2} \sin^3 t \sin t \cos t \, dt = 96 \pi \int_0^{\pi/2} (\sin^2 t)^2 \cos t \, dt \)

Uvedemo smenu: \( z = \sin t \), \( dz = \cos t \, dt \), za \( t=0 \Rightarrow z=0 \), za \( t=\pi/2 \Rightarrow z=1 \)

\( \displaystyle P = \frac{96\pi}{5} \)

\( A\left(\dfrac{8}{\sqrt{10}}, \dfrac{4}{\sqrt{10}}, z_0 > 0 \right) \)

Funkcija elipsoida:

\( F(x,y,z) = \dfrac{x^{2}}{8} + \dfrac{y^{2}}{16} + 4z^{2} – 1 \)

Jednačina tangentne ravni u tački \( A(x_0, y_0, z_0) \):

\( F_x(x_0, y_0, z_0)(x – x_0) + F_y(x_0, y_0, z_0)(y – y_0) + F_z(x_0, y_0, z_0)(z – z_0) = 0 \)

Izvodi:

\( F_x = \dfrac{2x}{8} = \dfrac{x}{4}, \quad F_y = \dfrac{2y}{16} = \dfrac{y}{8}, \quad F_z = 8z \)

Prvo nalazimo \( z_0 \) iz tačke A:

\( \dfrac{\left(\frac{8}{\sqrt{10}}\right)^2}{8} + \dfrac{\left(\frac{4}{\sqrt{10}}\right)^2}{16} + 4 z_0^2 = 1 \)

\( \dfrac{64/10}{8} + \dfrac{16/10}{16} + 4 z_0^2 = 1 \)

\( \dfrac{8}{10} + \dfrac{1}{10} + 4 z_0^2 = 1 \)

\( \dfrac{9}{10} + 4 z_0^2 = 1 \quad \Rightarrow \quad 4 z_0^2 = \dfrac{1}{10} \quad \Rightarrow \quad z_0 = \dfrac{1}{2 \sqrt{10}} \)

Izračunavamo koordinate normalnog vektora:

\( F_x(A) = \dfrac{\frac{8}{\sqrt{10}}}{4} = \dfrac{2}{\sqrt{10}}, \quad F_y(A) = \dfrac{\frac{4}{\sqrt{10}}}{8} = \dfrac{1}{2 \sqrt{10}}, \quad F_z(A) = 8 \cdot \dfrac{1}{2 \sqrt{10}} = \dfrac{4}{\sqrt{10}} \)

Jednačina tangentne ravni:

\( \dfrac{2}{\sqrt{10}} \left( x – \dfrac{8}{\sqrt{10}} \right) + \dfrac{1}{2 \sqrt{10}} \left( y – \dfrac{4}{\sqrt{10}} \right) + \dfrac{4}{\sqrt{10}} \left( z – \dfrac{1}{2 \sqrt{10}} \right) = 0 \)

Daljim sređivanjem dobijamo:

\(\displaystyle \alpha: 2x + \tfrac{1}{2} y + 4z – \tfrac{20}{\sqrt{10}} = 0\)

Vektor normale ima koordinate:

\(\vec{n} = \bigl( F_x(A), F_y(A), F_z(A) \bigr) = \left( \dfrac{2}{\sqrt{10}}, \dfrac{1}{2 \sqrt{10}}, \dfrac{4}{\sqrt{10}} \right)\)

Parametarska jednačina vektora normale:

\[ \frac{x – \frac{8}{\sqrt{10}}}{\frac{2}{\sqrt{10}}} = \frac{y – \frac{4}{\sqrt{10}}}{\frac{1}{2 \sqrt{10}}} = \frac{z – \frac{1}{2 \sqrt{10}}}{\frac{4}{\sqrt{10}}} \]

Tangentna ravan: \( 2x + \frac{1}{2} y + 4z – \frac{20}{\sqrt{10}} = 0 \)

Normalni vektor: \( \vec{n} = \left( \frac{2}{\sqrt{10}}, \frac{1}{2 \sqrt{10}}, \frac{4}{\sqrt{10}} \right) \)

\( \displaystyle \int\frac{2x-x^{2}-3}{x^{3}+3x}\,dx \)

\( \displaystyle \int\frac{2x-x^{2}-3}{x^{3}+3x}\,dx = \int\left(-\frac{1}{x} + \frac{2}{x^{2}+3}\right) dx = -\int \frac{dx}{x} + 2 \int \frac{dx}{x^{2}+3} \)

Primenjujemo formulu za standardni integral:

\( \displaystyle \int \frac{dx}{x^{2}+a^{2}} = \frac{1}{a} \arctan \frac{x}{a} \)

Ovde je \( a = \sqrt{3} \), pa dobijamo:

\(\displaystyle \int \frac{2x – x^{2} – 3}{x^{3} + 3x} dx = -\ln|x| + 2 \cdot \frac{1}{\sqrt{3}} \arctan \frac{x}{\sqrt{3}} + C = -\ln|x| + \frac{2 \sqrt{3}}{3} \arctan \frac{x}{\sqrt{3}} + C \)

b) Izračunati površinu ravnog lika ograničenog funkcijama \( y=0, \quad x = \sqrt{y}, \quad x + y = 2. \)

Prvo izražavamo sve u obliku \( y = f(x) \):

Parabola: \( x = \sqrt{y} \quad \Leftrightarrow \quad y = x^{2} \)

Prava: \( x + y = 2 \quad \Leftrightarrow \quad y = 2 – x \)

Pronalazimo presečne tačke parabole i prave:

\( x^{2} = 2 – x \quad \Rightarrow \quad x^{2} + x – 2 = 0 \)

Rešenja: \( x = 1 \) (uzimamo \( x > 0 \)), tada \( y = x^{2} = 1 \).

Takođe, parabola seče \( y = 0 \) u tački \( (0,0) \), a prava seče \( y = 0 \) u tački \( (2,0) \).

Formula za površinu ravnog lika:

\( P = \int_{0}^{1} x^{2} dx + \int_{1}^{2} (2 – x) dx \)

Računamo integrale:

\( \displaystyle \int_{0}^{1} x^{2} dx = \left. \frac{x^{3}}{3} \right|_{0}^{1} = \frac{1}{3} \)

\( \displaystyle \int_{1}^{2} (2 – x) dx = \left. \left( 2x – \frac{x^{2}}{2} \right) \right|_{1}^{2} = (4 – 2) – \left( 2 – \frac{1}{2} \right) = 2 – \frac{3}{2} = \frac{1}{2} \)

Dakle, ukupna površina je:

\(\displaystyle P = \frac{1}{3} + \frac{1}{2} = \frac{5}{6}\)

c) Naći totalni diferencijal funkcije \( z = \frac{y}{x^{4}} \ln(y^{2} + 1) + 3xy – 6 \) u tački \( M(1, 0) \).

Totalni diferencijal prvog reda:

\( dz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy \)

Izračunavamo parcijalne izvode:

\[ \frac{\partial z}{\partial x} = \frac{\partial}{\partial x} \left( \frac{y}{x^{4}} \ln(y^{2} + 1) + 3xy \right) = -\frac{4y \ln(y^{2} + 1)}{x^{5}} + 3y \]

\[ \frac{\partial z}{\partial y} = \frac{\partial}{\partial y} \left( \frac{y}{x^{4}} \ln(y^{2} + 1) + 3xy \right) = \frac{1}{x^{4}} \left( \ln(y^{2} + 1) + \frac{2 y^{2}}{y^{2} + 1} \right) + 3x \]

Vrednosti parcijalnih izvoda u tački \( M(1,0) \):

\( \frac{\partial z}{\partial x}(1,0) = -\frac{4 \cdot 0 \cdot \ln(0^{2} + 1)}{1^{5}} + 3 \cdot 0 = 0 \)

\( \frac{\partial z}{\partial y}(1,0) = \frac{1}{1^{4}} \left( \ln(0^{2} + 1) + \frac{2 \cdot 0^{2}}{0^{2} + 1} \right) + 3 \cdot 1 = 0 + 3 = 3 \)

Dakle, totalni diferencijal prvog reda je:

\(\displaystyle dz = 0 \cdot dx + 3 \, dy = 3 \, dy \)

a) \(-\ln|x| + \frac{2 \sqrt{3}}{3} \arctan\frac{x}{\sqrt{3}} + C\)

b) \( P = \frac{5}{6} \)

c) \( dz = 3 \, dy \)