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Poslovilno predavanje
Matematicne teme z didaktiko
Marko Razpet, Pedagoska fakulteta
Ljubljana, 20. november 2014
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Nase skupne ure
Matematicne tehnologije — 2011/12Funkcije vec spremenljivk — 2011/12Diferencialne enacbe — 2012/13Zgodovina matematike — 2013/14Matematicne teme z didaktiko — 2014/15
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Matematicne tehnologije — 2011/12Na pojem dolocenega integrala naravno pridemo, ce se na primervprasamo, kako izracunati ploscino lika, omejenega z grafom tefunkcije, osjo x in premicama x = a in x = b. Doloceni integrallahko povezemo s primitivno funkcijo F funkcije f (F ′ = f ):∫ b
af (x)dx = F (x)|ba = F (b)− F (a).
Poglejmo naslednji primer. Iz svoje vpisne stevilke student vzamezadnje tri stevke in jih oznaci s p, q in r po sledecem vrstnem redu:
p = 4, q = 8, r = 5.Potem iz teh treh stevk zapise
P =∫ 1
0x4(1− x8)5 dx +
∫ 1
0x8(1− x5)4 dx +
∫ 1
0x5(1− x4)8 dx .
Uporabi program derive in dobi rezultat:
P = 2621444393935 .
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Ploscina lika pod parabolo
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Primer izracuna integrala
Ploscina lika pod parabolo y = f (x) = x2 na intervalu [0, a].Interval [0, a] razdelimo na n enakih delov:
0 < an <
2an <
3an < . . . <
nan = a, ∆xk = a
n ,
Ik = [(k − 1)a/n, ka/n].
mk = inf{f (x), x ∈ Ik} = ((k − 1)a/n)2,
Mk = sup{f (x), x ∈ Ik} = (ka/n)2.
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Spodnja integralska vsota
Spodnja integralska vsota:
Sn = (f (0) + f (a/n) + f (2a/n) + . . .+ f ((n − 1)a/n)) · (a/n) =
= ((a/n)2 + (2a/n)2 + . . .+ ((n − 1)a/n)2) · (a/n) =
= (a/n)3(12 + 22 + . . .+ (n − 1)2) =
= a3
n3 ·(n − 1)n(2n − 1)
6 = a3
6 ·(
1− 1n
)(2− 1
n
).
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Zgornja integralska vsota
Zgornja integralska vsota:
Sn = (f (a/n) + f (2a/n) + . . .+ f (na/n)) · (a/n) =
= ((a/n)2 + (2a/n)2 + . . .+ (na/n)2) · (a/n) =
= (a/n)3(12 + 22 + . . .+ n2) =
= a3
n3 ·n(n + 1)(2n + 1)
6 = a3
6 ·(
1 + 1n
)(2 + 1
n
).
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Limitni prehod
Sn − Sn = a3
n ,
limn→∞
(Sn − Sn) = 0,
limn→∞
Sn = limn→∞
a3
6 ·(
1− 1n
)(2− 1
n
)= a3
3 ,
limn→∞
Sn = limn→∞
a3
6 ·(
1 + 1n
)(2 + 1
n
)= a3
3 .
Nazadnje imamo: ∫ a
0x2 dx = a3
3 .
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Newton–Leibnizeva formula — zacetek izpeljave
Naj bo F primitivna funkcija funkcije f na intervalu [a, b], karpomeni F ′ = f . Vzamemo poljubno delitev intervala [a, b]:
a = x0 < x1 < x2 < . . . < xn−1 < xn = b.
Oznacimo∆xk = xk − xk−1, % = max
1≤k≤n∆xk .
Zapisemo:
F (b)− F (a) = F (xn)− F (x0) =n∑
k=1(F (xk)− F (xk−1)).
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Newton–Leibnizeva formula — nadaljevanje in konecizpeljave
Uporabimo Lagrangev izrek:
F (b)− F (a) =n∑
k=1F ′(ξk)∆xk =
n∑k=1
f (ξk)∆xk .
V limiti, ko %→ 0 dobimo
F (b)− F (a) =∫ b
af (x) dx .
To je osnovna formula integralskega racuna ali Newton–Leibnizevaformula. Obicajno jo zapisemo v obliki∫ b
af (x) dx = F (x)|ba = F (b)− F (a).
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Funkcije vec spremenljivk — 2011/12
Telo G je omejeno s ploskvama:
z = 1a (x2 + y2) in z = c − 1
b (x2 + y2),
pri cemer so a, b in c pozitivne konstante.1 Skicirajte telo G.2 Dolocite pravokotno projekcijo D telesa G na ravnino z = 0.
Skica!3 Izrazite prostornino V (G) telesa G z dvojnim integralom.4 V dobljeni dvojni integral vpeljite polarne koordinate.5 Preverite, da je
V (G) = πabc2
2(a + b) .
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Telo v prostoru
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Telo v prostoru – anaglifna slika
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Telo v prostoru – osni presek
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Telo v prostoru kot vrtenina – prostornina po srednjesolsko
y = x2
a , y = c − x2
b .
Presecisce:
x2
a = c − x2
b , x2 = abca + b , x0 =
√abc
a + b , y0 = bca + b .
V1 = π
∫ y0
0x2 dy = πa
∫ y0
0y dy = πay2
02 ,
V2 = π
∫ c
y0x2 dy = πb
∫ c
y0(c − y) dy = πb(c − y0)2
2 ,
V = V1 + V2 = π
2 (ay20 + b(c − y0)2) = πabc2
2(a + b) .
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Natecaj za najboljso novoletno jelko
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Diferencialne enacbe — 2012/13
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VeriznicaDiferencialna enacba:
y ′′ = 1a
√1 + y ′2.
Resitev:y = a cosh x − x0
a + y0.
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Chaos
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Zgodovina matematike — 2013/14
Izracunajte na egipcanski nacin1 949× 567.
Resitev
Manjsi faktor 567 zlahkazapisemo kot
567 = 1 + 2 + 4 + 16 + 32 + 512.
Nato zapisemo preglednico:
1949× 1 1 949√
2 3 898√
4 7 796√
8 15 59216 31 184
√
32 62 368√
64 124 736128 249 472256 498 944512 997 888
√
1949× 567 1 105 083
Odgovor:1 949× 567 = 1 105 083.
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Otvoritev razstave plakatov — junij 2014 — A
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Otvoritev razstave plakatov — junij 2014 — B
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Otvoritev razstave plakatov — junij 2014 — C
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Otvoritev razstave plakatov — junij 2014 — D
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Matematicne teme z didaktiko — 2014/15
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Vsota geometrijske vrste
Na geometrijski nacin smo izracunali:
1 + q + q2 + q3 + . . . = 11− q , |q| < 1.
Rezultat smo uporabili pri izracunu ploscin na poseben nacinpobarvanih kvadratov.
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Enakorazmerno temperirana lestvica
α = 12√2.
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Kaj zna GeoGebra 3D
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Kaj zna GeoGebra 3D — anaglifne slike
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Hvala za vaso prisotnost in pozornost!
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