4. Odrediti totalni diferencijal funkcije
\( z(x, y) = \cos\left( \frac{x^2 + y^2}{x^3 + y^3} \right) + exy \cdot \sin(xy) \)
u tački \( A(1, 1) \).
\( z(x,y)=\cos\!\left(\tfrac{x^2+y^2}{x^3+y^3}\right)+e^{xy}\sin(xy) \), A(1,1)
1) \(u=\dfrac{x^2+y^2}{x^3+y^3},\quad z_1=\cos u.\)
\[
du=\frac{(2x\,dx+2y\,dy)(x^3+y^3)-(x^2+y^2)(3x^2\,dx+3y^2\,dy)}{(x^3+y^3)^2}
\]
\[
du=\frac{2x(x^3+y^3)-3x^2(x^2+y^2)}{(x^3+y^3)^2}\,dx
+\frac{2y(x^3+y^3)-3y^2(x^2+y^2)}{(x^3+y^3)^2}\,dy.
\]
Pošto \(z_1=\cos u\), dobija se \(dz_1=-\sin u\,du\).
2) \(v=xy,\quad z_2=e^{v}\sin v.\)
\[
\frac{d}{dv}(e^v\sin v)=e^v(\sin v+\cos v),\quad d(xy)=y\,dx+x\,dy.
\]
\[
dz_2=e^{xy}(\sin(xy)+\cos(xy))(y\,dx+x\,dy).
\]
- \(x^2+y^2=2,\;x^3+y^3=2,\;u(1,1)=1,\;xy=1\)
- Za \(du\) koeficijenti se pojednostavljuju do \(-\tfrac12\) za oba \(dx\) i \(dy\).
\[
dz_1(1,1)= -\sin(1)\Big(-\tfrac12\,dx-\tfrac12\,dy\Big)
=\tfrac{\sin(1)}{2}\,dx+\tfrac{\sin(1)}{2}\,dy.
\]
\[
dz_2(1,1)=e(\sin 1+\cos 1)(dx+dy).
\]
5.a) Izračunati površinu ravnog lika ograničenog sa
\( y = -x^2 + 6x – 8 \), \( y = x – 2 \) (skica)
\[
\begin{aligned}
&\begin{gathered}
P=? \quad y=-x^2+6x-8 \quad y=x-2 \\
x=0 \Rightarrow y=-8 \\
y=0 \Rightarrow x_{1/2}=\frac{-6 \pm \sqrt{36 – 32}}{-2} \\
x_1 = 2 \quad x_2 = 4
\end{gathered} \\
&\begin{gathered}
y’ = 0 \Rightarrow -2x + 6 = 0 \\
x = 3
\end{gathered} \\
&\text{presek:} \\
&\begin{aligned}
x – 2 &= -x^2 + 6x – 8 \\
x^2 – 5x + 6 &= 0 \quad \Rightarrow \quad x_{1/2} = \frac{5 \pm \sqrt{25 – 24}}{2} \\
x_1 = 3 \quad x_2 = 2
\end{aligned} \\
&P = \int_2^3 \left(-x^2 + 6x – 8 – x + 2\right) \, dx = \int_2^3 \left(-x^2 + 5x – 6\right) \, dx = \\
&\begin{aligned}
= -\left.\frac{x^3}{3}\right|_2^3 + \left.\frac{5x^2}{2}\right|_2^3 – \left.6x\right|_2^3
&= -\left(9 – \frac{8}{3}\right) + \frac{5}{2}(9 – 4) – 6(3 – 2) \\
&= -\frac{19}{3} + \frac{25}{2} – 6 = \frac{-38 + 75 – 36}{6} = \frac{1}{6}
\end{aligned}
\end{aligned}
\]
5.b) Odrediti \( \int \frac{x}{2} \sin\left(\frac{x}{2}\right) dx \)
\[
\begin{aligned}
& \int \frac{x}{2} \sin \left(\frac{x}{2}\right) d x = \left\{
\begin{array}{ll}
u=\sin \left(\frac{x}{2}\right) & dv=\frac{x}{2}dx \\
du=\cos \left(\frac{x}{2}\right) \cdot \frac{1}{2} dx & v=\frac{x^2}{6}
\end{array}
\right\} \\
=\ & \frac{x^2 \sin \left(\frac{x}{2}\right)}{6} – \frac{1}{12} \int x^2 \cos \left(\frac{x}{2}\right) dx =
\begin{cases}
u = \frac{x}{2} & dv = \sin \left(\frac{x}{2}\right) dx \\
du = \frac{dx}{2} & v = \int \sin \left(\frac{x}{2}\right) dx
\left\lvert
\begin{array}{l}
t = \frac{x}{2} \\
dt = \frac{dx}{2}
\end{array}
\right.
\\
& v = 2 \int \sin t dt = -2 \cos t = -2 \cos\left( \frac{x}{2} \right)
\end{cases} \\
=\ & -2 \cos\left(\frac{x}{2}\right) \cdot \frac{x}{2} + \int \cos\left(\frac{x}{2}\right) dx
\left|
\begin{array}{l}
z = \frac{x}{2} \\
dz = \frac{dx}{2} \\
dx = 2dz
\end{array}
\right| \\
=\ & -x \cos\left(\frac{x}{2}\right) + 2 \int \cos z dz \\
=\ & -x \cos\left(\frac{x}{2}\right) + 2 \sin z = -x \cos\left(\frac{x}{2}\right) + 2 \sin\left(\frac{x}{2}\right) + c
\end{aligned}
\]
5.c) Izračunati \( \int_4^5 \frac{4 dx}{x^2 – 4x + 3} \)
\[
\begin{aligned}
& \int_4^5 \frac{4\,dx}{x^2 – 4x + 3}
= 4 \int_4^5 \frac{dx}{(x – 2)^2 – 1}
\left|
\begin{array}{l}
t = x – 2 \\
dt = dx \\
g: 3 \quad d: 2
\end{array}
\right| = 4 \int_2^3 \frac{dt}{t^2 – 1} \\
=\ & -4 \int_2^3 \frac{dt}{1 – t^2}
= -\left. 4 \cdot \frac{1}{2} \ln \left| \frac{1 + t}{1 – t} \right| \right|_2^3
= -2 \ln \left| \frac{-2}{-3} \right|
= 2 \ln \frac{3}{2}
\end{aligned}
\]