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Interaktivni zadaci – GeoGebra + MathJax
Zadatak 1: Izračunati integral
Neka je dato:
\[
I=\int \frac{dx}{\big(\sin x+\cos x+2\big)^2}.
\]
Smernice. Upotrebiti trigonometrijsku smenu
\(t=\mathrm{tg}\,\frac{x}{2}\) uz relacije
\(\displaystyle \sin x=\frac{2t}{1+t^2},\ \cos x=\frac{1-t^2}{1+t^2},\ dx=\frac{2\,dt}{1+t^2}\).
Postupak (detaljno)
\[
\begin{aligned}
I&=\int \frac{dx}{(\sin x+\cos x+2)^2},\qquad t=\mathrm{tg}\frac{x}{2},\\[2pt]
\sin x&=\frac{2t}{1+t^2},\ \ \cos x=\frac{1-t^2}{1+t^2},\ \ dx=\frac{2\,dt}{1+t^2}.\\[4pt]
\sin x+\cos x+2&=\frac{2t+1-t^2}{1+t^2}+2=\frac{t^2+2t+3}{1+t^2}.\\[4pt]
\Rightarrow\quad
I=2\int \frac{1+t^2}{(t^2+2t+3)^2}\,dt.
\end{aligned}
\]
Sređivanje kvadrata: \(t^2+2t+3=(t+1)^2+2\), \(p=t+1\Rightarrow t=p-1\).
\[
I=2\int\frac{p^2-2p+2}{(p^2+2)^2}\,dp
=2\!\left(\int\frac{dp}{p^2+2}-2\int\frac{p\,dp}{(p^2+2)^2}\right).
\]
\[
\int\frac{p\,dp}{(p^2+2)^2}=-\frac{1}{2(p^2+2)},\quad
\int \frac{dp}{p^2+2}=\frac{1}{\sqrt2}\arctan\!\frac{p}{\sqrt2}.
\]
\[
\boxed{\,I=\sqrt2\,\arctan\!\frac{p}{\sqrt2}+\frac{2}{p^2+2}+C\,},\quad p=\mathrm{tg}\frac{x}{2}+1.
\]
\begin{aligned}
I&=\int \frac{dx}{(\sin x+\cos x+2)^2},\qquad t=\mathrm{tg}\frac{x}{2},\\[2pt]
\sin x&=\frac{2t}{1+t^2},\ \ \cos x=\frac{1-t^2}{1+t^2},\ \ dx=\frac{2\,dt}{1+t^2}.\\[4pt]
\sin x+\cos x+2&=\frac{2t+1-t^2}{1+t^2}+2=\frac{t^2+2t+3}{1+t^2}.\\[4pt]
\Rightarrow\quad
I=2\int \frac{1+t^2}{(t^2+2t+3)^2}\,dt.
\end{aligned}
\]
Sređivanje kvadrata: \(t^2+2t+3=(t+1)^2+2\), \(p=t+1\Rightarrow t=p-1\).
\[
I=2\int\frac{p^2-2p+2}{(p^2+2)^2}\,dp
=2\!\left(\int\frac{dp}{p^2+2}-2\int\frac{p\,dp}{(p^2+2)^2}\right).
\]
\[
\int\frac{p\,dp}{(p^2+2)^2}=-\frac{1}{2(p^2+2)},\quad
\int \frac{dp}{p^2+2}=\frac{1}{\sqrt2}\arctan\!\frac{p}{\sqrt2}.
\]
\[
\boxed{\,I=\sqrt2\,\arctan\!\frac{p}{\sqrt2}+\frac{2}{p^2+2}+C\,},\quad p=\mathrm{tg}\frac{x}{2}+1.
\]
Zadatak 2: Tangenta na elipsu i površina trougla
Elipsa: \(\displaystyle \frac{x^2}{9}+\frac{y^2}{4}=1\),
tačka \(A\!\left(\tfrac{3}{2}, \sqrt{3}\right)\).
Naći tangentu u \(A\), preseke sa osama i površinu trougla koji zatvara sa osama.
Postupak (detaljno)
\[
\frac{x^2}{9}+\frac{y^2}{4}=1\ \Rightarrow\ \frac{2x}{9}+\frac{2y\,y’}{4}=0
\ \Rightarrow\ y’=-\frac{4x}{9y}.
\]
U \(A(3/2,\sqrt3)\): \(\displaystyle k=y’_A=-\frac{2}{3\sqrt3}\).
Tangenta:
\[
y-\sqrt3=k\,(x-\tfrac32)\ \Longleftrightarrow\ 2x+3\sqrt3\,y=12.
\]
Preseci: \(B=(0,\tfrac{4}{\sqrt3})\), \(C=(6,0)\). Površina: \(P=\tfrac{12}{\sqrt3}\).
\frac{x^2}{9}+\frac{y^2}{4}=1\ \Rightarrow\ \frac{2x}{9}+\frac{2y\,y’}{4}=0
\ \Rightarrow\ y’=-\frac{4x}{9y}.
\]
U \(A(3/2,\sqrt3)\): \(\displaystyle k=y’_A=-\frac{2}{3\sqrt3}\).
Tangenta:
\[
y-\sqrt3=k\,(x-\tfrac32)\ \Longleftrightarrow\ 2x+3\sqrt3\,y=12.
\]
Preseci: \(B=(0,\tfrac{4}{\sqrt3})\), \(C=(6,0)\). Površina: \(P=\tfrac{12}{\sqrt3}\).
Zadatak 3: Površina obrtnog tela
Obrtati oko \(x\)-ose deo u \([0,1]\) između \(y=x\) i \(y=x^3\).
Postupak (detaljno)
Površina pri obrtanju krive \(y=f(x)\) oko \(x\)-ose na \([a,b]\):
\[
S=2\pi\int_a^b f(x)\sqrt{1+\big(f'(x)\big)^2}\,dx.
\]
Za \(y=x\) na \([0,1]\): \(f'(x)=1\Rightarrow \sqrt{1+(f’)^2}=\sqrt2\).
\[
S_1=2\pi\!\int_0^1 x\sqrt2\,dx=2\pi\sqrt2\cdot\frac12=\boxed{\pi\sqrt2}.
\]
Za \(y=x^3\): \(f'(x)=3x^2\Rightarrow \sqrt{1+(f’)^2}=\sqrt{1+9x^4}\).
\[
S_2=2\pi\!\int_0^1 x^3\sqrt{1+9x^4}\,dx.
\]
Neka \(t=1+9x^4\Rightarrow dt=36x^3dx\).
\[
S_2=\frac{2\pi}{36}\int_{t=1}^{10}\!\sqrt{t}\,dt
=\frac{\pi}{27}\Big[t^{3/2}\Big]_{1}^{10}
=\boxed{\frac{\pi}{27}\big(10\sqrt{10}-1\big)}.
\]
\[
\boxed{S=S_1+S_2=\pi\sqrt2+\frac{\pi}{27}\big(10\sqrt{10}-1\big)}.
\]
\[
S=2\pi\int_a^b f(x)\sqrt{1+\big(f'(x)\big)^2}\,dx.
\]
Za \(y=x\) na \([0,1]\): \(f'(x)=1\Rightarrow \sqrt{1+(f’)^2}=\sqrt2\).
\[
S_1=2\pi\!\int_0^1 x\sqrt2\,dx=2\pi\sqrt2\cdot\frac12=\boxed{\pi\sqrt2}.
\]
Za \(y=x^3\): \(f'(x)=3x^2\Rightarrow \sqrt{1+(f’)^2}=\sqrt{1+9x^4}\).
\[
S_2=2\pi\!\int_0^1 x^3\sqrt{1+9x^4}\,dx.
\]
Neka \(t=1+9x^4\Rightarrow dt=36x^3dx\).
\[
S_2=\frac{2\pi}{36}\int_{t=1}^{10}\!\sqrt{t}\,dt
=\frac{\pi}{27}\Big[t^{3/2}\Big]_{1}^{10}
=\boxed{\frac{\pi}{27}\big(10\sqrt{10}-1\big)}.
\]
\[
\boxed{S=S_1+S_2=\pi\sqrt2+\frac{\pi}{27}\big(10\sqrt{10}-1\big)}.
\]
Zadatak 4: Oblast definisanosti funkcije
\(\displaystyle f(x,y)=\ln\sqrt{\,9y-x^2y-y^3\,}\).
Naći domenu funkcije i vizuelizovati regione.
Postupak (detaljno)
Za \(\ln\sqrt{\cdot}\) zahtev je \(\;9y-x^2y-y^3>0\).
\[
9y-x^2y-y^3=y(9-x^2-y^2)>0.
\]
Dakle, proizvod je pozitivan kada su faktori istog znaka:
\[
\begin{cases}
y>0\ \text{i}\ 9-x^2-y^2>0\ \ \Longleftrightarrow\ \ y>0,\ x^2+y^2<9,\\[2pt]
y<0\ \text{i}\ 9-x^2-y^2<0\ \ \Longleftrightarrow\ \ y<0,\ x^2+y^2>9.
\end{cases}
\]
Zato je domen:
\[
\boxed{D=\{(x,y):\ y>0,\ x^2+y^2<9\ \ \text{ili}\ \ y<0,\ x^2+y^2>9\}.}
\]
\[
9y-x^2y-y^3=y(9-x^2-y^2)>0.
\]
Dakle, proizvod je pozitivan kada su faktori istog znaka:
\[
\begin{cases}
y>0\ \text{i}\ 9-x^2-y^2>0\ \ \Longleftrightarrow\ \ y>0,\ x^2+y^2<9,\\[2pt]
y<0\ \text{i}\ 9-x^2-y^2<0\ \ \Longleftrightarrow\ \ y<0,\ x^2+y^2>9.
\end{cases}
\]
Zato je domen:
\[
\boxed{D=\{(x,y):\ y>0,\ x^2+y^2<9\ \ \text{ili}\ \ y<0,\ x^2+y^2>9\}.}
\]
Zadatak 5 – Mešovita parcijalna integracija
(a) Za \(f(x,y)=x\,\arctan(y/x)\) naći \(\partial^2 f/\partial y\,\partial x\) u \((1,\sqrt3)\).
(b) Dodatno: \(\displaystyle \int_{1/e}^{e} x\ln x\,dx\) и дужина \(x=2\cos t,\ y=2\sin t,\ t\in[\pi/2,\pi]\).
(b) Dodatno: \(\displaystyle \int_{1/e}^{e} x\ln x\,dx\) и дужина \(x=2\cos t,\ y=2\sin t,\ t\in[\pi/2,\pi]\).
≈1.75
Postupak (detaljno)
\[
f(x,y)=x\,\arctan\!\frac{y}{x},\quad u=\frac{y}{x}.
\]
\[
f_y=\frac{x^2}{x^2+y^2},\qquad
f_{yx}=\frac{2x\,y^2}{(x^2+y^2)^2},\quad
f_{yx}(1,\sqrt3)=\frac{3}{8}.
\]
\(\displaystyle\int_{1/e}^{e}x\ln x\,dx=\frac{e^2}{4}+\frac{3}{4e^2}\).
\quad
Dužina luka \(x=2\cos t,\ y=2\sin t,\ t\in[\pi/2,\pi]\): \(\ L=\pi\).
f(x,y)=x\,\arctan\!\frac{y}{x},\quad u=\frac{y}{x}.
\]
\[
f_y=\frac{x^2}{x^2+y^2},\qquad
f_{yx}=\frac{2x\,y^2}{(x^2+y^2)^2},\quad
f_{yx}(1,\sqrt3)=\frac{3}{8}.
\]
\(\displaystyle\int_{1/e}^{e}x\ln x\,dx=\frac{e^2}{4}+\frac{3}{4e^2}\).
\quad
Dužina luka \(x=2\cos t,\ y=2\sin t,\ t\in[\pi/2,\pi]\): \(\ L=\pi\).