Diferencial

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Diferencial označujemo z dx.

$\! y= f(x)$

$\! f(x+dx) \simeq f(x) + dy$

$\! dy = f'(x)dx$

Vsebina

1 Funkcije dveh spremenljivk
2 Diferenciabilnost funkcije
3 Primer
4 Primer 2
5 Primer
6 Drugi odvod
- 6.1 Primer
7 Povezave

Funkcije dveh spremenljivk

$\! z = f(x,y)$

$\! f(x + dx, y + dy) = f(x,y) + df$

     .
     |
.____| dy
  dx

$df = \frac{\partial f}{\partial x} dx +\frac{\partial f}{\partial y} dy$

Diferenciabilnost funkcije

f(x,y) je v točki (x,y) diferenciabilna, če:

obstajata oba parcialna odvoda $\left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right)$
če je razlika $\lim_{dx \rightarrow 0,dy \rightarrow 0} \frac{f(x+dx,y+dx) - f(x,y) - \left( \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy \right) }{ dx^2 + dy^2} = 0$

$f(x+dx,y+dx) - f(x,y) = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \eta (dx^2 + dy^2)$

$\lim_{dx \rightarrow 0,dy \rightarrow 0} \eta = 0$

Izrek: Če je funkcija f(x,y) odvedljiva in sta parcialna odvoda $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}$ zvezna je funkcija diferenciabilna

$z = \frac{x}{y} = f(x,y)$

$dz = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy$

$\frac{1}{y} - \frac{x}{y^2} dy$

$\frac{x+dx}{y+dy} = f(x+dx,y+dy)$

relativna napaka: $\frac{f(x+dx,y+dy)-f(x,y)}{f(x,y)} \simeq \frac{\frac{1}{y} dx - \frac{x}{y^2} dy}{\frac{x}{y}} = \frac{dx}{x} - \frac{dy}{y}$

Primer

Oceni rezultat: $\frac{0.99 \cdot 0.98}{0.97}$

$f(x,y,z) = \frac{xy}{z}$

x = 1, dx = -0,01

y = 1, dy = -0,02

z = 1, dz = -0,03

$\! f(1,1,1) = 1$

$\frac{\partial f}{\partial x} = \frac{y}{z}$

$\frac{\partial f}{\partial y} = \frac{x}{z}$

$\frac{\partial f}{\partial x} = \frac{-xy}{z^2}$

$f(0.99,0.98,0.97) = 1 + \frac{\partial f}{\partial x} dx +\frac{\partial f}{\partial y} dy +\frac{\partial f}{\partial z} dz= 1 + (-0,01 - 0,02 + 0,03) = 1$

Primer 2

Volumen kozarca (stožec)

h = 10cm

R = 5cm (polmer na vrhu)

Formula za prostornino stožca:

$V = \frac{\pi R^2 h}{3} = \frac{\pi \frac14 \cdot 1}{3} \approx \frac14 l$

Ali se splača zamenjati kozarec, če želimo večji volumen s sledečimi spremembami:

dR = 2mm

dh = -2 mm

$\frac{\partial V}{\partial R} = \frac{2 \pi R h}{3} = \frac{2 \pi \frac12 \cdot 1}{3}\approx 1$

$\frac{\partial V}{\partial h} = \frac{\pi R^2}{3} = \frac{\pi \cdot \frac14}{3} \approx\frac14$

$dV = \frac{\partial V}{\partial R} dR + \frac{\partial V}{\partial h} dh = 1 \cdot 0,02 - \frac14 \cdot 0,02 = 0,015$

$\frac{\partial}{\partial x}(f+g) = \frac{\partial f}{\partial x} + \frac{\partial g}{\partial x}$

$\frac{\partial}{\partial x}(f-g) = \frac{\partial f}{\partial x} - \frac{\partial g}{\partial x}$

$\frac{\partial}{\partial x}(f \cdot g) = \frac{\partial f}{\partial x} \cdot g + \frac{\partial g}{\partial x} \cdot f$

$\frac{\partial}{\partial x}\left( \frac{f}{g} \right) = \frac{ \frac{\partial f}{\partial x} \cdot g - \frac{\partial g}{\partial x} \cdot f }{g^2}$

z=f(x,y)

x = x(t), y = y(t)

$z (t) = f (x (t), y (t))$

$z' = \frac{dz}{dt} = \lim_{k \rightarrow 0} \frac{z(t+h) - z(t)}{h} \lim_{k \rightarrow 0} \frac{f(x(t+h),y(t+h)) - f(x(t), y(t))}{h}$

$δ x = x (t + h) - x (t)$

$δ y = y (t + h) - y (t)$

x (t + h) = x (t + δ x)

y (t + h) = y (t + δ y)

$\lim_{h \rightarrow 0} \frac{x(t)+\delta x, y(t) + \delta y) - f(x,y)}{h}$

ocenimo s totalnim diferencialom

$= \lim_{h \rightarrow 0} \frac{\frac{\partial f}{\partial x} \delta x + \frac{\partial f}{\partial y} \delta y + \eta (\delta x^2 + \delta y^2 }{h}$

$= \lim_{h \rightarrow 0} ( \frac{\partial f}{\partial x} \cdot \frac{\delta x}{h} + \frac{\partial f}{\partial y} \cdot \frac{\delta y}{h} + \eta \frac{\delta x^2 + \delta y^2}{h} )$

$\lim_{h \rightarrow 0} \frac{\delta x}{h} = \frac{dx}{dt}$

$\lim_{h \rightarrow 0} \frac{\delta y}{h} = \frac{dy}{dt}$

$= \frac{\partial f}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dt} + \lim \eta \frac{\delta x^2 + \delta y^2}{h}$

$\frac{dz}{dt} = \frac{\partial f}{\partial x} \cdot \frac{d}{dt} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dt}$

$\! z(u,v) =f(x,y) = f(x(u,v),y(u,v))$

x = x(u,v)

y = y(u,v)

$\frac{\partial z}{\partial u} = \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial u} + \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial u}$

$\frac{\partial z}{\partial v} = \frac{\partial f}{\partial x} \cdot \frac{\partial x}{\partial v} + \frac{\partial f}{\partial y} \cdot \frac{\partial y}{\partial v}$

Primer

$z = (1 + t)^{2t^2} = x^y$

$\frac{dz}{dt} = \frac{\partial f}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dt}$

x = 1 + t

y = 2 t 2

$\! f(x,y) = x^y$

$\frac{\partial f}{\partial x} = y \cdot x ^{ y- 1}$

$\frac{\partial f}{\partial y} = x ^ y \log x$

$\frac{dx}{dt}= 1$

$\frac{dx}{dt}= 4t$

$\frac{dz}{dt} = \frac{\partial f}{\partial x} \cdot \frac{dx}{dt} + \frac{\partial f}{\partial y} \cdot \frac{dy}{dt} = y \cdot x ^{ y- 1} + x ^ y \log x \cdot 4t = 2t^2(1 + t)^{2 \cdot t^2-1} + t^{2t^2} \cdot \log (1+t) \cdot 4 t$