cálculo de várias variáveis - waldecir bianchini
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8/17/2019 Cálculo de Várias Variáveis - Waldecir Bianchini
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Aprendendo Calculo de Varias Variaveis
Waldecir Bianchini
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Ao Daniel
Uma vida minha que segue . . .
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Agradeço, in memoriam, ao Professor Ivo Fernandez Lopez, que nos deixou prematuramente, pela revisão
criteriosa deste livro, bem como muitas observações e sugestões valiosas.
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R3
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y + 2xy = 3x2 , xy + sen x y = ex , 3y + 4y + 5y = cos x
f
f (x) dx
F
f
F
F = f
F = f ⇐⇒
f (x) dx = F (x)
F
F (x) = cos(x)
F (x) =
cos x dx = sen x + c ,
c
m
v
t
a
p = mg
g
F = ma = p = mg =⇒ dv
dt = g =⇒ v(t) = gt + c
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v(0) = 0
v(t) = gt
v(0) = v0
v(t) = gt + v0
v(t)
g
g = 9, 8
dvdt
= g
x(t)
t
dx
dt = v(t) = gt + v0 =⇒ x(t) =
1
2gt2 + v0t + c
x(0) = x0
x(t) = 12
gt2 + v0t + x0
mdv
dt = mg − kv =⇒ dv
dt = g − k
mv
k
g = 9, 8 ms2
m = 20 kg
k = 5 kg
s
dvdt
= 9, 8 − v4
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t × v
v
v = 60
dvdt
= −5, 2
t
t
v = 60
40 m/s
40 m/s
k
m = h
dv
dt = g − hv ⇐⇒ 1
g − hv
dv
dt = 1
t
1
g − hv
dv
dt dt =
dt = t + c1
v = v(t)
dv = v (t)dt
1
g − hv
dv
dt dt =
1
g − hv dv
u = g − hv ⇒ du = −h dv
1
g − hv dv = −1
h
1
u du = −1
h ln |u| + c2 = −1
h ln |g − hv| + c2
−1
h ln |g − hv| + c2 = t + c1 ⇐⇒ ln |g − hv| = −ht − hc1 + c2 = −ht + c3
c3 = −hc1 + c2
|g − hv| = e−ht+c3 =⇒ g − hv = ±e−ht+c3 = ce−ht =⇒ v = v(t) = g
h − ce−ht
c = ±ec3
c
v(0) = 0
v(0) = g
h − c = 0 =⇒ c =
g
h
v(t) = gh
(1 − e−ht)
v(0) = 0
v(0) = 60
v(t) → gh
t → ∞
limt→0
v(t) = g
h
v = g
h
0 = 0
dx
dt = v(t)
x(0) = 0
x(t) = g
ht +
g
h2
e−ht − 1
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v
t
t
v
dy
dx = g(x) h(y)
1
h(y) dy = g(x) dx
1
h(y) dy =
g(x) dx
y = f (x)
dy
dx = g(x) h(y) =⇒ 1
h(f (x)) f (x) = g(x) =⇒ 1
h(f (x)) f (x) dx =
g(x) dx
y = f (x) ⇒ dy = f (x) dx
1
h(y) dy =
g(x) dx
dy
dx = xy
dy
dx =
y cos x
1 + 2y2
dy
dx = 3y + 5
1
y dy =
x dx
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ln |y| = x2
2 + C 1 =⇒ |y| = e
x2
2 ec1 =⇒ y = ±ec1ex2
2 = cex2
2 , 0 = c ∈ R
y = 0
y = ce
x2
2
, c ∈ R
c
y = xy
1 + 2y2
y dy =
cos x dx =⇒ (
1
y + 2y) dy = cos x dx =⇒ ln |y| + y
2
= senx + c
y
x
y = 0
dydx
= y cosx1+2y2
3y + 5
= 0
dy
3y + 5 = dx ⇒ ln |3y + 5| = x + c1 ⇒ y =
−5 ± ec1e3x
3
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y = −5
3
ec1 = c
y = −5 + ce3x
3 ,
c ∈ R
y = 3y + 5
y = −5
3
y(x) + p(x)y(x) = q (x)
y + py = q ,
p = p(x)
q = q (x)
I ⊂ R
1a
p = p(x)
q = q (x)
dv
dt +
k
m v = g .
y + 3y = x
F = f
u
u(y + py) = (uy) = uy + uy ⇔ uy + upy = uy + uy
upy = uy ⇒ u
u = p
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ln |u| =
p dx ⇒ |u| = e
p dx
u = e p dx
y + 2y = cos x
x3y − y = 1
u = e 2dx = e2x
u
ye2x = e2x cos x⇒ ye2x = e2x cos x dx
y = 2
5 cos x +
1
5senx + ce−2x
y = 2
5 cos x + 1
5senx
x3y − y = 1
y + py = q
x = 0
x3y − y = 1 ⇔ y − y
x3 =
1
x3
u = e
x−2
2
u
ex−2
2 y =
x−3e
x−2
2 dx = −ex−2
2 + c ⇒ y = −1 + ce− 1
2x2
x = 0
(0, −1)
x → 0
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xy + 2y = 4x2
y(1) = 2
y + 2
xy = 4x
u = x2
u
(x2y) = 4x3 ⇒ y = x2 + c
x2
y(1) = 2
c = 1
y = x2 + 1
x2
(1, 2)
x = 0
x > 0
y(1) = 1
y = x2
x ∈ R
1a
1a
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y + py = q
y(x0) = y0
p = p(x)
q = q (x)
x0
y = f (x), x ∈ I
p
q
p = 2
x
x = 0
x = 0
f
∂f ∂y
R = {(t, y); |t| < a, |y| < b}
I = {t; |t| < c < a}
y = h(t)
y = f (t, y), y(t0) = y0
|y0| < b |t0| < c
100
80
Q(t)
t
dQ
dt = kQ
k < 0
Q(t)
Q(t) = cekt
Q(0) = 100
Q(t) = 100ekt
k
80 g
Q(7) = 100e7k = 80 ⇒ k = 1
7 ln 0, 8 = −0.031
L
Q(L) = 1
2Q(0) ⇒ 100e−0.031L = 50 ⇒ L =
ln2
0.031 = 21, 74
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Q0
5%
t
Q(t)
t
dQ
dt = kQ =⇒ Q(t) = Q0ekt
Q(20) = 1, 05 Q0 =⇒ Q0e20k = 1, 05Q0 =⇒ k = 1
20 ln 1, 05 = 0, 00243
Q(t) = Q0e0,00243t
t
Q(t) = 2Q0
Q0e0,00243t = 2Q0 =⇒ t = 284, 13
100
10 kg
3 l/min
1/4 kg
Q(t)
t
dQ
dt =
t
=
=
= 1
4
kg
l 3
l
min − Q(t)
100
kg
l 3
l
min
= 3
4 − 3
100Q
dQ
dt +
3
100Q =
3
4 =⇒ Q(t) = 25 + ce−0,03t
Q(0) = 10
c = −15
Q(t) = 25 − 15e−0,03t
t → ∞
25
1/4
100
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m
mdv
dt = mg =⇒ v = gt + c ,
g
v(0) = 0
v = gt
v(30) = 30g
mg
kv
mdv
dt = mg
−kv
⇔ dv
dt +
k
mv = g
kv
u = e
k
mt
v(t) = m
k g + ce−
k
mt
v(0) = 30g
v(t) = m
k g + (30g − mg
k )e−
k
mt
t → +∞ ⇒ v(t) → mg
k
m = 1
v0 m/s
k = 10−3
dv
dt = −10−3v2
1
v2 dv =
−10−3 dt =⇒ v =
1
10−3t + c
v(0) = v0
c = 1v0
v(t) = v0
10−3v0t + 1
x(0) = 0
x(t) = 103 ln (10−3v0t + 1)
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t
v(t) = v0
2
v0
10−3
v0t + 1
= v0
2
=
⇒10−3v0t + 1 = 2 =
⇒t =
103
v0
x
103
v0
= 103 ln (2) = 693, 14
dy
dx = ex+y
dy
dx = 2xy
(t2 − xt2)
dx
dt + x2 + tx2 = 0
xy = 2
√ y − 1
x ln y
dy
dx
= y
(x2 + 1)y + y2 + 1 = 0
y(0) = 1
dy
dx =
ex
y y(0) = 1
dy
dx =
x + y
x
y + 3y = x + e−2x
y + x2y = x2
y
−3y = sen2x
y + 2y = xe−2x
y(1) = 0
xy + 2y = 4x2
y(1) = 2
dy
dx + 2
cos x
senx y = senx
y + y = x2y2
10
3%
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10%
−2√
p
p
1600
m = 1
F
k = 3
F
10
5
m
kv
k
v
v1
m = 0, 01
0, 10
k
k
80
m
12km/h
3km
k = 6m
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m = 1
V km/h
v km/h
k = 2
3
r = 1
3
V
50 km/h
v(t) = 2
3V (1 − e−t)
vellimit = 75 km/h
k
14
p(t)
t
limt
→∞ p(t)
p(t)
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m
x0
F m
k
F v
c
x = x(t)
t
v = v(t)
F m
F v
F m
F v
x(t)
x(t)
x(t)
F v
F v = −cx(t)
2a
F = ma = F m + F v ⇐⇒ mx(t) = −kx(t) − cx(t)
mx(t) + cx(t) + kx(t) = 0
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m = 1
c = 5
k = 6
2a
d2x
dt2 + 5
dx
dt + 6x =
d2x
dt2 + 2
dx
dt + 3
dx
dt + 3.2x
= d
dtdx
dt + 2x
+ 3dx
dt + 2x
dxdt
+ 2x = y
dy
dt + 3y = 0 ⇒ y = ce−3t
dx
dt + 2x = ce−3t ⇒
e2tx
= ce−t
e2tx = ce−t dt =
−ce−t + c1
x(t) = c1e−2t + c2e−3t
c1
c2
x(0) = 5 x(0) = 0
x(t)
x(t) = −2c1e−2t − 3c2e−3t
x(0) = c1 + c2 = 5x(0) = −2c1 − 3c2 = 0
x(t) = 15e−2t − 10e−3t
x(0) = 5
x(0) = 0
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m = 1
c =
4 k = 4
d2x
dt2 + 4
dx
dt + 4x = 0
d2x
dt2 + 4
dx
dt + 4x =
d2x
dt + 2
dx
dt + 2
dx
dt + 4x
= d
dt
dx
dt + 2x
+ 2
dx
dt + 2x
= dy
dt + 2y
y = dx
dt + 2x
dydt
+ 2y = 0 y = ce−2t
dx
dt
+ 2x = ce−2t
⇒(xe2t) = ce−2te2t = c
⇒xe2t = c dt = ct + c1
⇒x = (c1 + ct)e−2t
x(0) = 2
x(0) = −5
x(t) = (2 − t)e−2t
f
f
df
dt
D : f → Df = df
dt = f
D2f = D(Df ) = f
g h
(g + h)(x) = g(x) + h(x) (kg)(x) = kg(x)
D
md2x
dt2 + k
dx
dt + cx = mD2x + kDx + cx = (mD2 + kD + c)x = 0
p(r) = mr2 + kr + c
mr2 + kr + c = 0
⇔r2 + k
m
r + c
m
= 0
r1 r2
p(r) = r2 + k
mr +
c
m = (r − r1)(r − r2)
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md2x
dt2 + k
dx
dt + c
dx
dt = mD2x + kDx + cx = (mD2 + kD + c)x = (D − r1)(D − r2)x = 0
(D
−r2)x = y
(D
−r1)y =
y − r1y = 0 y = cer1t
(D − r2)x = y = cer1t ⇔ x − r2x = cer1t
x(t) = c1er1t + c2er2t
r1 = r2
r1 = r2 = r0
p(r) = (r − r0)(r − r0)
(D − r0)(D − r0)x = 0
(D − r0)x = y
x(t) = c1er1t + c2er2t , r1 = r2
x(t) = (c1 + c2t)ert , r1 = r2 = r
c1 c2
m = 1
k = 2
c = 2
d2x
dt2 + 2
dx
dt + 2x = 0 ⇔ (D2 + 2D + 2)x = 0
r2 + 2r + 2 = 0
r1 =
−1 + i
r2 =
−1−
i
r2 + 2r + 2 = (r − (−1 + i))(r − (−1 − i)) = 0
(D − (−1 + i))(D − (−1 − i))x = 0
(D − (−1 − i))x = y
(D − (−1 + i))y = 0 ⇒
y − (−1 + i)y = 0
u = e (−1+i) dt
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C
f : D ⊂ R → C
f
C
f (t) = 1 + t2 + i(3t − 2)
f (t) = sen(5t) + i cos(5t)
f (t) = e3t(cos(2t) + isen(2t))
f : D ⊂ R → C
f (t) = u(t) + iv(t)
u : D ⊂ R → R v : D ⊂ R → R
f
f (t) = u(t) + iv(t)
f (t) = u(t) + iv(t)
f (t) = u(t) + iv(t)
f (t) dt =
u(t) dt + i
v(t) dt
1 + t2 + i(3t
−2) dt = (1 + t2) dt + i (3t
−2) dt = t +
t3
3
+ i(3
2
t2
−2t)
f (t) = e(a+bi)t = eat(cos bt + i sen bt)
a = 0
b = 1
eit = cos t + i sen t
z = a + bi
d(ezt)
dt = zezt
ezt dt =
1
z ezt + c
•
y − (−1 + i)y = 0 .
u = e
(−1+i) dt = e
(−1+i)t
y = ce(−1+i)t
x + (1 + i)x = ce(−1+i)t ,
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x(t) = c1e(−1+i)t + c2e(−1−i)t
= e−t(c1eit + c2e−it)
= e−t(c1(cos t + i sen t) + c2(cos t + i sen t))
= e−t((c1 + c2)cos t + i (c1 − c2)sen t)
= e−t(c3 cos t + c4 sen t)
c3 = c1 + c2
c4 = i (c1 − c2)
x(0) = 2
x(0) = 0
c3 = c4 = 2
x(t) = 2e−t(cos t + sen t)
c3 cos t+c4 sen t
cos(t − φ)
cos(t − φ) = cos t cos φ + sen t sen φ
c3
c4
c3 cos t + c4 sen t
cos(t − φ)
r
cos(t
−φ) = cos t cos φ + sen t sen φ = rc3 cos t + rc4 sen t
cos φ = rc3
sen φ = rc4
cos2 φ + sen 2φ = 1
r =
1 c23 + c24
r = 1√
8 =
√ 24
φ = π
4
x(t) =√
8e−t cos(t−
π
4)
x(t)
π4
√ 8e−t
−√
8e−t
t
cos(t − π4
) = 1 cos(t − π
4) = −1
x(t)
√ 8e−t
−√ 8e−t
m = 1
c =
0 k = 1
x(0) = 2
x(0) = 2
x + x = 0
x(t) = 2 cos t + 2 sen t =
√ 8cos(t − π
4)
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x(t) + cx(t) + kx(t) = 0
r1
r2 r2 + cr + k = 0
r1 = r2
x(t) = c1er1t + c2er2t
r1 = r2 = r
x(t) = (c1 + c2t)ert
r1 = a + bi
r2 = a − bi
x(t) = eat(c1 cos bt + c2 sen bt)
g(t)
mx(t) + cx(t) + kx(t) = g(t)
m = 1
c = 5
k = 6
g(t) = 4t
r2
+ 5r + 6 = 0
r1 = −2
r2 = −3
(D + 2)(D + 3)x = 4t
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(D + 3)x = y
(D + 2)y = 4t
1a
y + 2y = 4t
u = e2t
(ye2t) = 4te2t ⇒ ye2t = 4 te2t dt ⇒ y = (2t − 1) + c1e−2t
x + 3x = y
x + 3x =
(2t − 1) + c1e−2t
u = e3t
x(t) = 2
3t − 5
9 + c1e−2t + c2e−3t
c1
c2 c1 = c2 = 0
x(t) = 2
3t − 5
9
x(t) = x p(t) + xh(t)
x p(t)
xh(t)
mx + cx + kx = 0
c1
c2
x(0) = 1 x(0) = 0
x(t) = 2
3 − 2c1e−2t − 3c2e−3t
x(0) = 2
3 − 2c1 + c2
−59 + c1 + c2 = 12
3 − 2c1 − 3c2 = 0
c1 = 4
c2 = −22
9
x(t) = 2
3t − 5
9 + 4e−2t − 22
9 e−3t
mx(t) + cx(t) + kx(t) = 0
ad2x
dt2
+ bdx
dt
+ cx = g(t)
g(t) = P n(t)eαt
cos(βt)
sen(βt) P n(t) = a0 + a1t + a2t2 + · · · + antn
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x p
g(t)
ax p+bx p+cx p
g(t)
x p
g(t)
x + x − 2x = 4t2
r2 + r − 2 = 0
r1 = 1
r2 = −2
xh = c1et + c2e−2t
x p(t)
x p(t)
x p + x p − 2x p
4t2
x p
2
−2x p
x p(t) = at2 + bt + c ⇒ x p(t) = 2at + b ⇒ x p(t) = 2a
2a + 2at + b − 2(at2 + bt + c) = 4t2 ⇔ 2a + b − 2c + (2a − 2b)t − 2at2 = 4t2
2a + b
−2c = 0
2a − 2b = 0−2a = 4
a = b = −2
c = −3
x p(t) = −2(t2 + t) − 3
x(t) = −2(t2 + t) − 3 + c1et + c2e−2t
x + x = 4t2
r2 + r = 0
r1 = 0
r2 = −1
xh = c1 + c2e−t
x p
x p + x p 4t2
x p
2
x p = at2 + bt + c
1
x p
3
2
x p = at3 + bt2 + ct
x p
x p
6at + 2b + 3at2 + 2bt + c = 4t2
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2b + c = 06a + 2b = 0
3a = 4
a = 43 b = −4 c = 8
x p(t) = 8t − 4t2 + 4
3t3 + c1 + c2e−t
x + x − 2x = e3t
x p
x p + x p − 2x p
e3t
x p
x p = ae3t
9ae3t + 3ae3t − 2ae3t = e3t
a = 1
10
x(t) = 1
10e3t + c1et + c2e−2t
x + x − 2x = e−2t
x p = ae−2t
4ae−2t − 2ae−2t − 2ae−2t = e−2t ⇒ 0 = e−2t
a
e−2t
x p = ae−2t
u(t)e−2t
(u − 4u + 4u)e−2t + (u − 2u)e−2t − 2ue−2t = e−2t
u − 3u = 1
u
1
1
e−2t
x p = ate−2t
a = −1
3
t2
t
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x + x − 2x = 2 sen t
x p
x p + x p − 2x p = 2 sen t
x p = a sen t + b cos t
x p = a cos t − b sen t
x p = −a sen t − b cos t
a cos t − b sen t − a sen t − b cos t − 2(a sen t + b cos t) = 2 sen t
a = −3
5 b = −1
5
x − 4x = te2t
r2−4 = 0
±2
xh = c1e2t + c2e−2t
x p
x p = (at + b)e2t
be2t
x p
x p
t
x p = (at2
+ bt)e2t
x p
a = 18
b = − 1
16
x(t) = (1
8t2 − 1
16t)e2t + c1e2t + c2e−2t
g(t)
ax + bx + cx = g1 + g2
x p1 ax + bx + cx = g1
x p2 ax + bx + cx = g2 x p1 + x p2
ax + bx + cx = g1 + g2
x + 4x = tet + t sen2t
x(0) = 0, x(0) = 1
5
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r2+4 = 0
±2i
xh = c1 cos 2t + c2 sen 2t
x + 4x = tet
x + 4x = t sen2t
x p1 = (at + b)et
x p1 a = 1
5 b = − 2
25
sen2t
x p2 = (at + b)sen2t + (ct + d)cos2t
b sen2t
d cos2t
x p2
t
x p2 = (at2 + bt)sen2t + (ct2 + dt)cos2t
x p2
a = 0 b = 1
16 c = − 1
8 d = 0
x(t) = (t
5 − 2
25)et +
t
16 sen 2t − t2
8 cos2t + c1 cos 2t + c2 sen 2t
x(t) = (t
5 − 2
25)et +
t
16 sen 2t − t2
8 cos2t +
2
25 cos 2t +
1
25 sen 2t
g(t) xp
P n(t) ts(A0 + A1t + · · · + Antn)
P n(t)eαt ts(A0 + A1t + · · · + Antn)eαt
P n(t)eαt
cos(βt)sen(βt)
tseαt[(A0 + A1t + · · · + Antn)cos βt+
(B0 + B1t + · · · + Bntn)sen βt]
(s = 0, 1, 2)
x p
y − y − 2y = 0
y − 7y = 0
y + 4y = 0
y + 2y + 3y = 0
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e−2t
2e2t − 5e−t
(4 − 3t)e2t
y − 8y + 7y = 14
2y − 4y + 2y = 0, y(0) = 1 y(0) = 0
y + 6y + 9y = t + 2, y(0) = 0, y(0) = 0
y − 7y + 12y = −e4x
y − 4y + 4y = 2e2x, y(0) = y (0) = 1
y + y − 6y = xe2x
y + y = cos x
y − y = 2xsenx, y(0) = 0, y(0) = 1
y − 2y + 10y = ex + sen3x
y − 3y = x + cos x
x + 16x = 2 sen 4t , x(0) = −1
2 , x(0) = 0
10 kg
0, 7m
1m/s
f (t) = 50sent
1 kg
0, 2 m
14 m/s
28cos7t + 56sen7t
t
g = 9, 8 m/s2)
2 Kg
k = 10 N/m
8 N/(m/s)
2sen2t + 16 cos2t
1 m
2 m/s
x(t)
x(t) = xh(t) + x p(t)
limt→∞ |x(t) − x p(t)|
x(t) − x p(t)
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x + 25x = 20sen5t x(0) = 1, x(0) = 0
1 Kg
k = 9 N/m
c = 6 N
m/s
F (t) = 6te−3t
t
t = 0
1 m/s
x(t)
limt→∞ x(t)
g = 10
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2a
d2x
dt2 = 0 ⇒ x = x0 + vx t
vx
x0
2
a
m = 1
a = d2y
dt2 = −g ⇒ y = y0 + vy t − g t2
2
vy
y0
x = x0 + vxt
y = y0 + vyt − gt2
2
(x, y)
γ
P = (x, y)
t
x = x(t) y = y(t)
γ
t
t
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y = f (x) = ax2 + bx + c
(x, y)
x = t y = at2 + bt + c
y = f (x)
x = t y = f (t)
−→AB
A
B
A
B
−→AB
−→OP
−→EF
−−→DC −−→GH
−→AB
−→OP
−→EF
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−→OP
−→AB
−→EF
P = (a, b)
v =
−→OP
v =
−→OP
P
v = (a, b)
a
b
v
u
v
w
P
Q
R
. . .
0 = (0, 0)
v = (a, b)
−v = (−a, −b)
v
−v
v
v = (a, b)
|v| =√
a2 + b2
P
u = (u1, u2)
k ∈ R
k
u
ku = (ku1, ku2)
k > 0
ku
u
k < 0
ku
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|k| u
|ku| =
(ku1)2 + (ku2)2 = |k|
u21 + u2
2 = |k| · |u|
u = (u
1, u
2)
v = (v
1, v
2)
u + v
u + v = (u1 + v1, u2 + v2)
u
u
v
v
u
v
u
v
u + v = (u1 + v1, u2 + v2) = (u1 + v1, 0) + (0, u2 + v2)
u = (u1, u2)
v = (v1, v2)
u − v
u
−v = u + (
−v) = (u1
−v1, u2
−v2)
u − v
u = (u1, u2)
v = (v1, v2)
u
v
u · v = u1v1 + u2v2
u
v
w
h
k
u + v = v + u
(u + v) + w = u + (v + w)
u + 0 = 0
u + (−u) = 0
0u = 0
h(ku) = (hk)u
k(u + v) = ku + kv
(h + k)u = hu + ku
u · v = v · u
u · (v + w) = u · v + u · w
k(u · v) = (ku) · v = u · (kv)
v · v = |v|2
0 · v = 0
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u
v
u
u = 1
|v|v = v
|v|
v
k = 1
|v|
|u| = |kv| = |k| |v| = 1
|v| |v| = 1
u
i
j
u = (a, b)
i = (1, 0)
j = (0, 1)
u = (a, b) = (a, 0) + (0, b) = a(1, 0) + b(0, 1) = ai + bj
a
u
b
u {i, j}
R2
u
v
u
v
α
u
v
0
≤α
≤π
u
v
u
v
α
u · v = |u| |v| cos α
u
v
v − u
|v − u|2 = |u|2 + |v|2 − 2|u| |v| cos α
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|v − u|2 = (v − u) · (v − u)
= (v − u) · v − (v − u) · u
= v · v − u · v − v · u + u · u= |v|2 − 2u · v + |u|2
|v|2 − 2u · v + |u|2 = |u|2 + |v|2 − 2|u| |v| cos α
u · v = |u| |v| cos α
α
cos α = u · v
|u| |v|
0
u
v
u · v = 0
u v
I ⊂ R
r t ∈ I
r(t) = (x(t), y(t)) ∈ R2
r : t
∈I
⊂R
→(x(t), y(t))
∈R2
t
P = (x(r), y(t))
γ
γ
r(t)
γ
γ
t
r(t)
r(t)
r(t)
−→OP
x = x(t) , y = y(t)
γ
t
γ
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x0 = 0
t = x
vx
y = y0 + vyvx
x − g
2vx2 x2
x = 3 cos t y = 2 sen t , 0 ≤ t ≤ 2π
x
3 = cos t
y
2 = sen t
x3
2 +
y2
2 = 1
t
0
2π
P = (3 cos t, 2sen t)
(3, 0)
x = 3 cos t y = 2 sen t , 0 ≤ t ≤ 3π
P = (3cos t, 2sen t)
(3, 0)
(−3, 0)
γ
x = 3 + t2
y = 1 + 2t4 , t ∈ R
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t2 = x
−3
⇒y = 1 + 2(x
−3)2
γ
A = (3, 1)
x = 3 + t2
x ≥ 3
t
γ
x(t) = 8 cos(1
4t) − 7cos t
y(t) = 8 sen( 14
t) − 7sen t
x(t) = 8 sen (34
t)y(t) = 7 sen t
s
P 0 = (x0, y0)
v = (a, b)
a
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P = (x, y)
−→OP = −−→OP 0 + t −→v
t
(x, y) = (x0, y0) + t(a, b)
(x, y)
s(t) = (x0 + at, y0 + bt)
s :
x = x0 + aty = y0 + bt
s
(x0, y0)
(a, b)
a
θ
(a cos θ, a sen θ)
θ
x = a cos θy = a sen θ
dθ
dt = 1 rd/s
θ(t) = t
t
x = a cos ty = a sen t
dθ
dt = 2 rd/s
θ(t) = 2t
x = a cos2ty = a sen2t
a
P
P
P
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θ
θ = 0
P
−→OP =
−→OC +
−→CP
θ
OT
|OT | = arcoPT = aθ
−→OC
−→CP
−→OC =
−→OT +
−→T C = (aθ, 0) + (0, a) = (aθ,a)
−→CP = (−a sen θ, −a cos θ)
−→OP = a(θ − sen θ, 1 − cos θ)
x(θ) = a(θ − sen θ)y(θ) = a(1 − cos θ)
r(θ) = (x(θ), y(θ)) = (a(θ − sen θ), a(1 − cos θ))
θ ∈ R
θ
3π
θ = 3πt
r(t) = (x(t), y(t)) = (a(3πt − sen3πt), a(1 − cos3πt)), t ∈ R
t
I ⊂ R
R2
r(t) = (x(t), y(t)), , t ∈ I ,
x = x(t)
y = y(t)
I
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r(t) = (x(t), y(t))
limt→t0
x(t) = L1 lim
t→t0y(t) = L2 ,
limt→t0 r(t) =
limt→t0 x(t), limt→t0 y(t)
= (L1, L2)
r(t) =
sen t
t , t2
limt→0 r(t)
limt→0
r(t) = limt→0
sen t
t , t2
=
limt→0
sen t
t , lim
t→0t2
= (1, 0)
r(t)
r(t) = ( sen t
t , t2)
r(t) t0
limt→t0
r(t) = r(t0)
r(t) = (x(t), y(t))
x(t) y(t)
r(t)
x(t)
y(t)
t = 0
t = 0
r(t) = (x(t), y(t))
x(t) = sen t
t , t = 0
1 , t = 0
y(t) = t2
t = 0
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r(t) = dr
dt = lim
∆t→0
r(t + ∆t) − r(t)
∆t
r(t) = (x(t), y(t))
x(t)
y(t)
r(t) = lim∆t→0
r(t + ∆t) − r(t)
∆t
= lim∆t→0
1
∆t [(x(t + ∆t), y(t + ∆t)) − (x(t), y(t))]
=
lim∆t→0
x(t + ∆t) − x(t)
∆t , lim
∆t→0
y(t + ∆t) − y(t)
∆t
= (x(t), y(t))
r(t) = (2 + 3t2, 1 + sen 2t3)
r(t) = ((2 + 3t2), (1 + sen 2t3)) = (6t, 6t2 cos2t3)
−→P Q = r(t + ∆t) − r(t)
−→P S = r(t+∆t)−r(t)
∆t
∆t > 0
∆t → 0 −→P S
r(t)
t
r(t) = (x(t), y(t))
r(t)
v(t) = r (t) = lim∆t→0
r(t + ∆t) − r(t)
∆t
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t
a(t) = v (t) = r (t)
P 0 = (x0, y0)
v = (a, b)
x = x(t) y = y(t)
r(t) =
x(t), y(t))
r(t) = v(t) = (a, b)
a
1 rd/s
r(t) = (a cos t, a sen t)
v(t) = r(t) = (−a sen t, a cos t)
a(t) = v (t) =
(−a sen t, −a cos t)
x2
9 +
y2
4 = 1
T = (3√ 2
2 ,
√ 2)
r(t) = (x(t), y(t))
x = x(t) = 3 cos(t) y = y(t) = 2 sin(t)
0 ≤ t ≤ 2π
t
T
3√
2
2 = 3 cos(t)
√ 2 = 2 sin(t)
t =
π
4
t = π
4
v =−
3 (2)
2 ,
(2)
T
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s(t) =:
x(t) = 3
√ 2
2 − 3
√ 2
2 t
y(t) =√
2 +√
2t
r(t) = (x(t), y(t))
a ≤ t ≤ b
ba
r(t) dt = limn→∞
ni=1
r(ci)∆t
a = t0
≤t1
≤. . . ti
−1
≤ti ≤ · · · ≤
tn = b
ti−1
≤ci ≤
ti
r(t)
ba
r(t) dt =
ba
(x(t) dt,
ba
y(t) dt
r
[a, b]
R
r
R(t) = r(t)
R(t) =
ta
r(s) ds
ba
r(t) dt = R(b) − R(a)
r(t) dt = R(t) + C
r
C
r(t) = (6 cos(2t), 6 sin(2t))
r(t) dt =
6 cos(2t) dt,
6 sin(2t) dt
= (3 sin(2t), −3cos(2t)) + C
π/40
r(t) dt = (3 sin(2t), −3 cos(2t))|π/40 = (3 sin(√
2), −3cos(√
2))−(0, −3) = 3(sin(√
2), −1−cos(√
2))
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C
r(t) = (x(t), y(t))
t
[a, b]
C
C
C
n
n
C
n
C
y = f (x)
L
f
a
≤x
≤b
L =
ba
1 + (f (x))2 dx
x = x(t)
y = y(t)
r(t) = (x(t), y(t)) a ≤ t ≤ b
L =
ba
(x(t))2 + (y(t))2 dt
r(t) = (cos(t), sen(t))
0
≤t
≤2π
L =
2π0
(− sen(t))2 + (cos(t))2 dt =
2π0
dt = 2π
r(t) = (x(t), y(t))
a ≤ t ≤ b
v(t) = r(t)
t = a
t
s(t) =
ta
(x(t))2 + (y(t))2 dt =
ta
|v(t)| dt
ds(t)
dt =
(x(t))2 + (y(t))2 = |v(t)|
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P 0
t0
r(t) = (t2
, 4
t2 )
t ∈ (0, +∞)
t0 = 2
r(t) = (−2 + t2, 3 + 4t2)
t ∈ [1, 5]
t0 = 2
r(t) = (4 sen t, 3cos t)
0 ≤ t ≤ π
t = π
3
r(t) = (3et, 4e2t)
t ∈ (−∞, +∞)
t0 = 0
r(t) =
t2, sen
t2
π
t ∈ [0, 2π]
t0 = π
r(t) = (4 + cos t, 3 + sen t)
t ∈ [0, π]
t0 = π
4
y =
f(x)
x2 + y2 − 2 x − 2 y − 4 = 0
(R, 0)
x2
a2 + y2
b2 = 1
4 x2 + 9 y2 − 8 x − 36 y + 4 = 0
xy
y = tx
x = 6
S
y2
= −6x+24
Q
t
P
QP S
QS
α(t)
α
α
xy
y = tx
x = 8
S
x2 + y2 = 4
Q
t
P
QP S
α(t)
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α
α
σ1(t) =
2 + t, −2 +
t2
2
σ2(t) =
−8 + 7t, −1 +
7
2t
, t ≥ 0
t = 2
A = (1, 0)
σ(t) = (x(t), y(t))
v(t) = (y(t) + et, x(t) + e−t)
σ(t)
P = (1, 2)
y
5 m/s
r
P
A = (r, 0)
θ
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R3
R2
P
a
b
(a, b)
(a,b,c)
R3
P = (x,y,z )
u
O = (0, 0, 0)
P
u = (x,y,z )
u × v
u
v
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R3
u = (u1, u2, u3)
v = (v1, v2, v3)
u
v
u × v = (u2b3 − u3b2, a3b1 − a1b3, a1b2 − a2b1)
i j k
u1 u2 u3
v1 v2 v3
=
u2 u3
v2 v3
i −u1 u3
v1 v3
j +
u1 u2
v1 v2
k
u × v
u
v
(u × v) · u = 0 (u × v) · v = 0
n = u × v
u
v
v
n
P 0 = (x0, y0)
u = (u1, u2)
x = x0 + u1t
y = y0 + u2t
L
P 0 = (x0, y0, z 0)
u = (u1, u2, u3)
x = x0 + u1ty = y0 + u2tz = z 0 + u3t
t ∈ R
L
P = P 0 + tu
t
x − x0
u1
= y − y0
u2
= z − z 0
u3
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P 0 = (x0, y0, z 0)
Q = (x1, y1, z 1)
R = (x2, y2, z 2)
u = Q
−P 0
v = R
−P 0
n = (a,b,c) = u × v
u
v
P = (x,y,z )
P
P 0
−−→P 0P = P − P 0
n
n · (P − P 0) = 0
(a,b,c) · (x − x0, y − y0, z − z 0) = 0
a(x−
x0) + b(y
−y0) + c(z
−z 0) = 0
P 0 = (x0, y0, z 0)
n = (a,b,c)
A = (1, 0, 2)
B = (0, 2, 1)
C = (0, 0, 3)
u
v
u = B − A = (−1, 2, −1) v = C − A = (−1, 0, 1)
n = u × v =
i j k
−1 2 −1−1 0 1
=
2 −10 1
i −−1 −1−1 1
j +
−1 2−1 0
k = 2i + 2 j + 2k = (2, 2, 2)
A
B
C
2(x − 1) + 2(y − 0) + 2(z − 2) = 0 ⇔ x + y + z = 3
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R3
r
x = 1 + 2t , y = 2 + t , z = −1 − 2t
2x + y + 4z = 1
r
P = (1 + 2t, 2 + t, −1 − 2t)
t
2(1 + 2t) + (2 + t) + 4(−1 − 2t) = 1
t = −1
3 P = (
1
3, 5
3, −1
3)
γ R3
t
t
x = x(t) y = y(t) z = z (t)
γ
r
t ∈ I ⊂ R
(x(t), y(t), z (t)) R
3
r : t ∈ I ⊂ R → (x(t), y(t), z (t)) ∈ R3
I
⊂R
R3
r
γ
r : R −→ R2
r : R −→ R3
z (t)
x(t) = 2 cos(t), y(t) = 2sen (t) z (t) = 3
x2 + y2 = 4
xy
z
z = 3
(0, 0, 3)
x(t) = 2 cos(t), y(t) = 2 sen(t) z (t) = t
2
x2 + y2 = 4
P = (x,y,z )
2
xy
2
z =
t
2
t
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x2 + y2 = 4
z = 3
x2 + y2 = 4
xy
xy
P = (x,y,z )
xy
x2 + y2 = 4
P = (x,y, 0)
z
x2 + y2 = 4
z = y2
yz
zy
P = (x,y,z )
zy
z = y2
P = (0, y , z )
x
z = y2
E
x2 + y2
4 = 1
x + 2y + z = 6
√ 2
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R3
x2 + y2 = 4
z = y2
E
T =
√ 2
2
,√
2, 7
√ 2
2
E
E
E
P = (x(t), y(t), z (t))
E
x(t)
y(t)
P
x(t) = cos(t) y(t) = 2 sen (t)
0
≤t
≤2π
z = z (t)
x + 2y + z = 6√ 2
z (t) = 6√
2 − cos(t) − 4sen(t)
E
r(t) = (cos(t), 2sen(t), 6
√ 2 − cos(t) −
4sen(t))
v(t) = r(t) = (− sin(t), 2cos(t), sen(t) − 4 cos(t))
T
t
r(t) = T
cos(t) =
√ 2
2
2sen(t) = √ 26√
2 − cos(t) − 4sen(t) = 7
√ 2
2
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t =
π
4
v(π
4) =
−
√ 2
2 ,
√ 2, −3
√ 2
2
E
E
T
x =
√ 2
2 −
√ 2
2 t y =
√ 2 +
√ 2 t z =
7√
2
2 − 3
√ 2
2 t
x
y
z
Ax2
+ By2
+ Cz 2
+ Dxy + Eyz + F xz + Gx + Hy + Iz + J = 0
A , B , C , D, E , F , G, H , I, J
D
E
F
D = E = F = 0
Ax2 + By2 + Cz 2 + Gx + Hy + Iz + J = 0
Ax2+By2+Cz 2+J = 0
x = 0
y = 0
z = 0
a
b
c
x2
a2 +
y2
b2 +
z 2
c2 = 1
z = 0
xy
x2
a2 + y2
b2 = 1
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R3
x = 0
yz
y2
b2 + z2
c2 = 1
y = 0
xz
x2
a2 + z2
c2 = 1
z = k
x2
a2 +
y2
b2 = 1 − k2
c2 ⇔ x2
a2(1 − k2
c2)
+ y2
b2(1 − k2
c2)
= 1
−c ≤ k ≤ c
c2 − k2 > 0 ⇐⇒ −c <
k < c
k ≤ −c k ≥ c
xz
yz
x = k
y2
b2 + z
2
c2 = 1 − k
2
a2 ≥ 0 ⇔ −a ≤ k ≤ a
y = k
x2
4 +
z 2
c2 = 1 − k2
b2 ≥ 0 ⇔ −b ≤ k ≤ b
x2
4 + y2
9 + z 2 = 1
x2
a2 +
y2
b2 − z 2
c2 = 1
z = 0
x2
a2 + y2
c2 = 1
x = 0
y = 0
y2
a2 − z 2
c2 = 1 :
yz
x2
a2 − z 2
c2 = 1 :
xz
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z = k
k ∈ R
x2
a2 +
y2
b2 = 1 +
k2
c2 ⇔ x2
a2(1 + k2
c2)
+ y2
b2(1 + k2
c2)
= 1
x2
4 + y2
9 − z 2 = 1
z
z
−x2
a2 − y2
b2 + z 2
c2 = 1
x = 0
y = 0
−y2
b2 + z2
c2 = 1
−x2
a2 + z2
c2 = 1
z
z = 0
−x2
a2 − y2
b2 = 1
z = k
x2
a2 +
y2
b2 =
k2
c2 − 1
k2
c2
−1
≥0
⇔k
≤ −c
k
≥c
z = x2
a2 +
y2
b2
z = 0
x2
a2 +
y2
b2 = 0 ⇒ x = y = 0
O = (0, 0, 0)
xy
z = k > 0
x2
a2 +
y2
b2 = k ⇔ x2
a2k +
y2
b2k = 1
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R3
−4x2 − y2 + z 2 = 1
z = k
k > 0
x = 0 y = 0 z = y2b2
z = x2a2
z = x2 + y2
4
z = −x2
a2 +
y2
b2
z = 0
−x2
a2 +
y2
b2 = 0 ⇒ y2
b2 =
x2
a2 ⇒ y = ± b
ax
x = 0
z = y2
b2
y = 0
z = −x2
a2
z = k
xy
−
x2
a2
+ y2
b2
= k
⇔ −
x2
a2
k
+ y2
b2
k
= 1
k > 0
z = k
y
k < 0
x
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z = −x2
4 +
y2
9
z2
c2 =
x2
a2 +
y2
b2
z = 0
x2
a2 +
y2
b2 = 0 ⇒ x = y = 0
O = (0, 0)
y = 0
z 2
c2
= x2
a2
⇒z =
±c
a
x
x = 0
z 2
c2 =
y2
b2 ⇒ z = ±c
by
z = k
z 2 = x2 + y2
A
B
C
Ax2 + By2 + Cz 2 + Gx + Hy + Iz + J = 0
(x − x0) + (y − y0) + (z − z 0) = r2
O = (x0, y0, z 0)
r
O = (0, 0, 0)
r
(x0, y0, z 0)
x = x − x0 , y = y − y0 , z = z − z 0
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R3
x2 + y2 + z 2 = 4
(x − 2)2
+ (y − 3)2
+ (z − 3)2
= 4
L
2x + y + 3z = 1 x − 4y − 2z = 1
A = (
−1, 0, 2)
B = (2, 1, 1)
C = (0,
−1,
−2)
Π
P (1, 2, −1)
Π
γ
z = x2 + y2
x2 + y2 = 4
γ
T = (
√ 2,
√ 2, 4)
S : 4x2
+ y2
= 1
π : 4x + 4y + z = 16
C = S
π
C
P =√
24
,√ 22
, 16 − 3√
2
γ
z =
4 − x2 − y2
x2 + (y − 1)2 = 1
γ
T =√ 2
2 , 1 +
√ 22 ,
2 − √ 2
r(t) = (t cos(t), t sen(t), t)
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t =
π
6
(xo, yo, z o)
v = (a, b, c)
σ1(t) = (1 + t, 2 + t2, 3) σ2(t) = (1 − t2, 1 + t2, 3), t ≥ 0
t = 2
(1, 2, 3)
σ(t) = (x(t), y(t), z (t))
σ(t) +
2σ(t) − 3σ(t) = (cos t, 4, 4e−3t)
(0, 1, 3)
(0, 5, 4)
t
A
t
σ1(t) = (300t, 1670t +
10t2, 500 + 60t)
B
σ2(t) = (100 + 100t, −80 + 880t, 280t)
A
t = 0
B = (0, 0, 1)
σ(t) = (x(t), y(t), z (t))
σ(t) = (y(t), t − x(t), 0)
t ∈ [0, 2π]
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R3
t1
t2 ∈ [0, 2π]
v(t) = |σ(t)| = 1
[t1, t2]
σ(t) = (x(t), y(t), z (t))
σ(t) = 2σ(t) − σ(t) t ∈ R
P = (1, 0, 1)
v0 = (2, 1, 2)
σ(t)
(x,y,z ) = (1, 2, 3) +
t(1, 0, 3)
P = (1, 2, −1)
x + y + 2z − 3 = 0
P = (1, 2, −1)
x + y + 2z − 3 = 0
A = (2, 0, 1) B = (−2, −1, 1)
C = (1, −2, 2)
A = (1, −2, 3)
x = 2t
y = 1+t
z = −1 − 2t
r
P = (3, 1, 2)
π : x − 2y + z = 1
r
π
r
xy
z = ln y
y2 + (z − 2)2 = 4
y = sen (x)
(x − 2)2 + (y − 1)2 + (z − 3)2 = 4
y2 = 4x2 + z 2
(z − 2)2 = x2 + z 2
−x2 + y2 − z 2 = 1
x = 2y2 + z 2
x = −z 2 + y2
x2
4 − y2 + z2
9 = 1
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y = f (x)
D ⊂ R2
f
(x, y) ∈D
f (x, y)
D
f
{f (x, y); (x, y) ∈ D} ⊂ R
D
f
R2
f
z = f (x, y)
f
(x, y)
x
y
z
z = f (x, y) =
16 − x2 − y2
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f
(x, y)
16 − x2 − y2 ≥ 0
D = {(x, y); x2 + y2 ≤ 16}
(0, 0)
f
{z ; z =
16 − x2 − y2
(x, y) ∈ D} = [0, 4]
z = f (x, y) =
√ x − 1
x − y
f
D = {(x, y); x ≥ 1
x = y}
z = f (x, y) = ln (y − x2)
ln (y − x2)
(x, y)
y − x2 > 0
y > x2
y = x2
R2
y < x2
y > x2
y > x2
T = (0, 2)
y > x2
y−x2
> 0
ln (y − x2
)
R
y > x2
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z = f (x, y)
D ⊂ R2
{(x,y,z ) ∈ R3; z = f (x, y), ∀(x, y) ⊂ D}
z = f (x, y)
z = k
x = k
y = k
k
xy
(x, y)
z
z = f (x, y)
z = k
xy
f (x, y) = k
f
k
xy
z = f (x, y)
z = 3, z = 5, z = 7, z = 9
z = 11
z = 13
z = 2, 3, 4, 5
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z = −x2 + y2
z =
−4x
2x2 + 2y2 + 1
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z = −4x
2x2+2y2+1
z = −4x
2x2+2y2+1
y = f (x)
limx→a
f (x) = L
x
a
f (x)
L
ε > 0
δ > 0
x
0 < |x − a| < δ ⇒ |f (x) − L| < ε
0 < |x − a| < δ x a δ
a
lim(x,y)→(a,b)
f (x, y) = L
f (x, y)
L
|f (x, y) − L|
(x, y)
(a, b)
(x, y)
(a, b)
(x, y)
(a, b)
δ
(x, y)
(a, b)
δ
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z = f (x, y)
D ⊂ R2
Dr(a, b) = {(x, y); (x − a)2 + (y − b)2 < r2}
r > 0
lim(x,y)→(a,b)
f (x) = L
ε > 0
δ > 0
|f (x, y) − L| < ε ∀ (x, y) ∈ Dδ(a, b)
z = sen
√ x2+y2
√
x2+y2
f
Dδ(a, b)
(L − ε, L + ε)
δ
(x, y)
(a, b)
(x, y)
(a, b)
(a, b)
(x, y) (a, b)
z = f (x, y) =
sen
x2 + y2
x2 + y2
f
R2
(0, 0)
sen
x2 + y2
x2 + y2
= k
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r =
x2 + y2
r
sen
x2 + y2
x2 + y2
= sen(r)
r → 1
(x, y) (0, 0)
1
z =
sen (√
x2+y2)√ x2+y2
z = x2−y2x2+y2
lim(x,y)→(0,0)
x2 − y2
x2 + y2
(0, 0)
lim(x,0)→(0,0)
x2 − 0
x2 + 0 = lim
x→0
x2
x2 = 1
lim(0,y)→(0,0)
0 − y2
0 + y2 = lim
y→0
−y2
y2 = −1
lim(x,y)→(0,0)
x2 y
x4 + y2
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x
y
z = x2y
x4+y2
y = kx
lim(x,kx)→(0,0)
x2kx
x4 + (kx)2 = lim
x→0
kx
x2 + k = 0
k
y = x2
lim(x,x2)→(0,0)
x2x2
x4 + x4 = lim
x→0
x4
2x4 =
1
2
(x, y)
12
lim(x,y)→(0,0)
2x2 y
3x2 + 3y2
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y = kx
y = x2
f
f (x, y) = 2x2 y
3x2+3y2
y = kx
limx→0
2x2kx
3x2 + 3(kx)2 = lim
x→0
2kx
3(1 + k2) = 0
k
y = x2
limx→0
2x2x2
3x2 + 3x4 = lim
x→0
2x2
3(1 + x2) = 0
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y = kx
y = x2
2x2y
3(x2 + y2)
≤ 2|y|3
= 2
y2
3 ≤ 2
x2 + y2
3
(x, y) → 0
f (x, y) → 0
|f (x, y)|
23|y|
23
x2 + y2
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z = f (x, y)
P 0 = (x0, y0)
lim(x,y)→(x0,y0)
f (x, y) = f (x0, y0)
g(x, y) =
sen
x2 + y2
x2 + y2
, (x, y) = (0, 0)
1 , (x, y) = (0, 0)
R2
f (x, y) =
x2 + y2 , (x, y) ∈ {(x, y); x2 + y2 ≤ 4}
4 , (x, y) ∈ {(x, y); x2 + y2 > 4}
R2
f (x, y) = x2 + y2 , (x, y) ∈ {(x, y); x2 + y2 ≤ 4}0 , (x, y) ∈ {(x, y); x
2
+ y2
> 4}
(x, y)
T (x, y) =
9x2 + 4y2
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(2, 1)
(2, 1)
x
y = 1
z = 80
x
y
T (2, 1) = 9.22 + 4.1 = 40
(2, 1)
9x2 + 4y2 = 40
(2, 1)
z = 80
y = 1
9x2 + 4 = 80 ⇒ x = 2, 9
(2, 1)
(2.9, 1)
80 − 40
2, 9
−2
= 40
0, 9 = 44, 4
graus
metro
x
(2, 1)
x
y
1
x
x
lim∆x→0
T (2 + ∆x, 1) − T (2, 1)
∆x = lim
∆x→0
9(2 + ∆x)2 + 4 − 40
∆x
= lim∆x→0
36∆x + 9∆x2
∆x
= lim∆x→0 36 + 9∆x
= 36 graus
metro
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x
y
x
lim∆x
→0
T (2, 1 + ∆y) − T (2, 1)
∆y = lim
∆x
→0
40 + 4(1 + ∆y)2 − 40
∆y
= lim∆x→0
8∆y + 4(∆y)2
∆y= lim
∆x→08 + 4∆y
= 8 graus
metro
y
x
y
y
x
f
x
(x, y)
f x(x, y) = ∂f
∂x(x, y) = lim
∆x→0
f (x + ∆x, y) − f (x, y)
∆x
f
y
(x, y)
f y(x, y) = ∂f
∂y(x, y) = lim
∆y→0
f (x, y + ∆y) − f (x, y)
∆y
∂z∂x
∂z∂y
z = f (x, y) = x2y3 sen xy
f
x
y
f
x
∂z
∂x = (x2)y3 sen xy + x2y3(sen xy) = 2xy3 sen xy + x2y3 sen xy.(xy) = 2xy3 sen xy + x2y4 cos xy
f
y
x
∂z
∂y = 3x2y2 sen xy + x3y3 cos xy
∂z
∂x
∂z
∂y
z =
x2 + 3xy2
z = ex3y + (x5 + 10xy) ln xy2
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∂z
∂x =
2x + 3y2
2
x2 + 3xy2
∂z
∂y =
6xy
2
x2 + 3xy2
∂z
∂x = 3x2yex3y + (5x4 + 10y) ln xy2 + (x5 + 10xy)
1
x∂z
∂y = x3ex
3y + 10x ln xy2 + (x5 + 10xy)2
y
f
x
(x0, y0)
∂f
∂x(x0, y0)
x
y
x
z = f (x, y0)
x
C 1
f
y = y0
C 1 T 1
P
f
y = y0
∂f
∂x(x0, y0)
α
T 1
y = y0 x
z = f (x, y0)
C 1
P
y = y0
f
y
(x0, y0)
∂f
∂y(x0, y0)
x = x0
f
C 2
z = f (x0, y)
C 2
β
x = x0
f
y
∂f
∂y(x0, y0)
β
z = f (x0, y)
C 2
P
x = x0
∂f
∂x
∂f
∂y
z = f (x, y)
z = f (x, y)
f xx = ∂ 2f
∂x2 = ∂
∂x
∂f
∂x
f yy = ∂ 2f
∂y2 = ∂
∂y
∂f
∂y
f yx = ∂ 2f
∂x∂y =
∂
∂x∂f
∂y
f xy = ∂ 2f
∂y∂x =
∂
∂y
∂f
∂x
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∂f
∂x(x0, y0)
∂f
∂y
(x0, y0)
3a
4a
z = f (x, y) =
x3 + 2y3 + 3x2y2
∂z
∂x = 3x2 + 6xy2 ∂ 2z
∂x2 = 6x + 6y2
∂z ∂y
= 6y2 + 6x2y ∂ 2
z ∂y2 = 12y + 6x2
∂ 2z
∂x∂y =
∂ 2z
∂y∂x = 12xy
z = x3 + 2y3 + 3x2y2
∂ 2z∂x∂y
= 12xy
∂z∂x
= 3x2 + 6xy2
z = xy + sen xy
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∂ 2z∂x2 = 6x + 6y2
∂z∂y
= 6x2 + 6x2y
∂ 2z∂y2
= 12y + 6x2
∂z
∂x = y + y cos xy ⇒ ∂ 2z
∂y∂x = 1 + cos xy − yx sen xy
∂z
∂y = x + x cos xy ⇒ ∂ 2z
∂y∂x = 1 + cos xy − xy sen xy
∂ 2f
∂y∂x
∂ 2f
∂y∂x
z = f (x, y)
∂ 2z
∂y∂x =
∂ 2z
∂x∂y
P
z 2 = x2 + y2
S
z = f (x, y)
P 0 = (x0, y0, z 0)
z 0 = f (x0, y0)
z = f (x, y)
rx ry
γ x γ y
S
x = x0 y = y0
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x
y
S
γ x
γ y
γ x :=
x = xy = y0
z = f (x, y0)
γ y :=
x = x0
y = yz = f (x0, y)
γ x(x0) :=
x = 1y = 0
z = ∂f ∂x
(x0, y0)
γ y(y0) :=
x = 0y = 1
z = ∂f ∂y
(x0, y0)
ux = (1, 0,
∂f
∂x (x0, y0))
uy = (0, 1,
∂f
∂y (x0, y0))
rx
ry
n
P 0
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n = ux × uy =
−→i
−→ j
−→k
1 0 ∂f
∂x(x0, y0)
0 1 ∂f
∂y(x0, y0)
=
−→k − ∂f
∂x(x0, y0)
−→i − ∂ f
∂y(x0, y0)
−→ j
=
−∂f
∂x(x0, y0), −∂f
∂y(x0, y0), 1
P = (x,y,z )
n
P
−P 0 = (x
−x0, y
−y0, z
−z 0)
n · (P − P 0) =
−∂f
∂x(x0, y0), −∂f
∂y(x0, y0), 1
· (x − x0, y − y0, z − z 0) = 0
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z = f (x0, y0) + ∂ f
∂x(x0, y0)(x − x0) +
∂ f
∂y(x0, y0)(y − y0)
z = f (x, y) = 1 + 4x2 + y2
P = (1
2,√
3, 5)
Q
f
P = (1
2,√
3, 5)
xy
P = (1
2,√
3, 5)
z = f (1
2 , √ 3) + ∂ f
∂x (1
2 , √ 3)(x − 1
2 ) + ∂ f
∂y (1
2 , √ 3)(y − √ 3)
f (x, y) = 1 + 4x2 + y2
∂f
∂x = 8x
∂f
∂x = 2y
∂f
∂x(
1
2,√
3) = 4
∂f
∂y(
1
2,√
3) = 2√
3
z = 5 + 4(x − 1
2) + 2
√ 3(y −
√ 3)
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P
n = (−∂f
∂x(
1
2,√
3), −∂f
∂y(
1
2,√
3) = (−4, −2√
3, 1)
P
x(t) = 1
2 − 4t, y(t) =
√ 3 − 2
√ 3 t
z (t) = 5 + t
Q
xy
z = 0
t = −5
Q = (41
2 , 11
√ 3, 0)
z = f (x, y)
f
f
f (x, y) =
x y x2
+ y2
, (x, y)
= (0, 0)
0 , (x, y) = (0, 0)
x
y
x y
∂f
∂x(0, 0) = 0
∂f
∂x(0, 0) = 0
y = x
x = t
y = t
z = |t|√
2
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z = xy√
x2+y2
f : R → R
x0
f (x0) = limx
→x0
f (x) − f (x0)
x−
x0
P 0 = (x0, f (x0))
f
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(x0, f (x0))
f (x)
x
x0 P 0
f (x) = f (x0) + f (x0)(x − x0) + ε
x
x0
limx→x0
f (x) − f (x0) = limx→x0
(f (x0)(x − x0) + ε) ⇐⇒ limx→x0
f (x) − f (x0)
x − x0= f (x0) + lim
x→x0
ε
x − x0
f
x0
f (x0) = limx→x0
f (x) − f (x0)
x − x0
,
limx→x0
ε
x − x0
= 0
f : R
→R
x
0 f (x) = f (x
0) + m(x
−x0) + ε
m, ε ∈ R
limx→x0
ε
x − x0
= 0
f
x0
m = f (x0)
f
x0 x
x0
t(x) = f (x0) + f (x0)(x − x0)
f (x) − t(x) = ε → 0
x → x0
z = f (x, y)
(x0, y0)
f
(x0, y0, f (x0, y0))
x − x0 = ∆x
y − y0 = ∆y
z = f (x, y)
P 0 = (x0, y0)
m
n
f (x0 + ∆x, y0 + ∆y) = f (x0, y0) + m∆x + n∆y + ε ,
lim(∆x,∆y)→(0,0)
ε ∆x2 + ∆y2
= 0
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m =
∂f
∂x(x0, y0)
n =
∂f
∂y(x0, y0)
ε
P 0
∆x
∆y
ε = ε(x0 + ∆x, y0 + ∆y) = f (x0 + ∆x, y0 + ∆y)
−f (x0, y0)
−m∆x
−n∆y
lim
(∆x,∆y)→(0,0)
ε(x0 + ∆x, y0 + ∆y)
||(∆x, ∆y)|| = 0
lim∆x→0
ε(x0 + ∆x, y0)
||(∆x, 0)|| = lim∆x→0
f (x0 + ∆x, y0) − f (x0, y0) − m∆x
|∆x| = 0
lim∆x→0
f (x0 + ∆x, y0) − f (x0, y0)
|∆x| − m ∆x
|∆x|
= 0
∆x > 0 ⇒ |∆x| = ∆x
∆x < 0 ⇒ |∆x| = −∆x
0 = lim∆x→0
f (x0 + ∆x, y0) − f (x0, y0)|∆x| − m ∆x|∆x|
= lim
∆x→0
f (x0 + ∆x, y0) − f (x0, y0)
∆x − m
m = lim
∆x→0
f (x0 + ∆x, y0) − f (x0, y0)
∆x =
∂f
∂x(x0, y0)
n =
∂f
∂y(x0, y0)
P
P
ε
(x, y)
(x0, y0)
(x0, y0, f (x0, y0))
f (x, y) =
x y x2 + y2
, (x, y) = (0, 0)
0 , (x, y) = (0, 0)
f
lim(∆x,∆y)→(0,0)
ε(∆x, ∆y)
||(∆x, ∆y)|| = lim(∆x,∆y)→(0,0)
∆x ∆y
∆x2 + ∆y2
y = x
lim(∆x,∆y)→(0,0)
∆x ∆y
∆x2 + ∆y2 = lim∆x→0
∆x2
2∆x2 = 1
2
x = 2y
lim(∆x,∆y)→(0,0)
∆x ∆y
∆x2 + ∆y2 = lim∆y→0
2∆y2
4∆y2 + ∆y2 = 2
5
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z = f (x, y) = x2 + y2
R2
(x, y)
ε(x + ∆x, y + ∆y) = f (x + ∆x, y + ∆y) − f (x, y) − ∂ f
∂x(x, y)∆x − ∂f
∂y(x, y)∆y
= (x + ∆x)2 + (y + ∆y)2 − x2 − y2 − 2x∆x − 2y∆y= ∆x2 + ∆y2
lim(∆x,∆y)→(0,0)
ε(x + ∆x, y + ∆y)
||(∆x, ∆y)|| = lim(∆x,∆y)→(0,0)
∆x2 + ∆y2 ∆x2 + ∆y2
= lim(∆x,∆y)→(0,0)
∆x2 + ∆y2 = 0
f
(x, y) ∈ R
2
=⇒
z = f (x, y)
P 0 = (x0, y0)
f
P 0
lim(∆x,∆y)→(0,0)
f (x0 + ∆x, y0 + ∆y) = f (x0, y0)
f
P 0
m n
lim(∆x,∆y)→(0,0) f (x0 + ∆x, y0 + ∆y) == lim
(∆x,∆y)→(0,0)(f (x0, y0) + m∆x + n∆y + ε(x0 + ∆x, y0 + ∆y))
= f (x0, y0) + lim(∆x,∆y)→(0,0)
(m∆x + n∆y) + lim(∆x,∆y)→(0,0)
ε(x0 + ∆x, y0 + ∆y)
= f (x0, y0) + lim(∆x,∆y)→(0,0)
ε(x0 + ∆x, y0 + ∆y)
= f (x0, y0) + lim(∆x,∆y)→(0,0)
ε(x0 + ∆x, y0 + ∆y)
||(∆x, ∆y)|| ||(∆x, ∆y)||
= f (x0, y0) + lim(∆x,∆y)
→(0,0)
ε(x0 + ∆x, y0 + ∆y)
||(∆x, ∆y)
|| lim(∆x,∆y)
→(0,0)
||(∆x, ∆y)||= f (x0, y0) + 0 · 0
= f (x0, y0)
f (x, y) = x2 y
x4 + y2 , (x, y) = (0, 0)
0 , (x, y) = (0, 0)
f
(0, 0)
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lim(x,y)→(0,0)
x2 y
x4 + y2
f
(0, 0)
f
(0, 0)
f
(x, y) → (0, 0)
f
x
y
∂f
∂x
∂f
∂y
P 0 = (x0, y0)
P 0
f
P 0
lim(∆x,∆y)→(0,0)
ε(x0 + ∆x, y0 + ∆y)
||(∆x, ∆y)|| = 0
f (x0+∆x, y0+∆y)−f (x0, y0) = f (x0+∆x, y0+∆y)−f (x0, y0+∆y)+f (x0, y0+∆y)−f (x0, y0)
g(x) = f (x, y0 + ∆y)
c
∈(x0, x0 + ∆x)
g(x0 + ∆x) − g(x0) = g (c)∆x ⇐⇒ f (x0 + ∆x, y0 + ∆y) − f (x0, y0 + ∆y) = ∂f
∂x(c, y0 + ∆x)∆x
h(x) = f (x0, y)
d ∈ (y0, y0 + ∆y)
h(y0 + ∆y) − h(y0) = h(d)∆y ⇐⇒ f (x0, y0 + ∆y) − f (x0, y0) = ∂f
∂y(x0, d)∆y
ε(x0 + ∆x, y0 + ∆y) =
∂f
∂x (c, y0 + ∆y)∆x +
∂f
∂y (x0, d)∆y − ∂ f
∂x (x0, y0)∆x − ∂f
∂y (x0, y0)∆y
ε(x0 + ∆x, y0 + ∆y)
||(∆x, ∆y)|| =
∂f
∂x(c, y0 + ∆y) − ∂f
∂x(x0, y0)
∆x
||(∆x, ∆y)|| +∂f
∂y(x0, d) − ∂f
∂y(x0, y0)
∆y
||(∆x, ∆y)||
(∆x, ∆y) → (0, 0)
c → x0
d → y0
∂f ∂x
∂f ∂y
P 0 ∆x
||(∆x, ∆y)|| ≤ 1
∆y
||(∆x, ∆y)|| ≤ 1
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lim(∆x,∆y)→(0,0)
ε(x0 + ∆x, y0 + ∆y)
||(∆x, ∆y)|| = 0
(2, 1)
x = x(t)
y = y(t)
z = T (x, y)
t
t
(x(t), y(t))
T (x(t), y(t))
t
y = F (x)
x = x(t)
y = F (x(t))
dF
dt = dF
dx
dx
dt
y = F (x)
x = x(u, v) ⇒ y = F (x(u, v))
z = f (x, y)
x = x(t)
y = y(t) ⇒ z = f (x(t), y(t))
z = f (x, y)
x = x(u, v)
y = y(u, v) ⇒ z = f (x(u, v), y(u, v))
y = F (x) = x5 + 3x2 + 10
x = x(u, v) = u2 cos v
z = f (x, y) = x5y2 + 3x2y + 10
x = x(t) = 2et
y = y(t) = 3 ln t
z = f (x, y) = x5y2 + 3x2y + 10
x = x(u, v) = u cos v
y = y(u, v) = veu
y = F (x) = (u2 cos v)5 + 3(u2 cos v)2 + 10 = u10 cos5 v + 3u4 cos2 v + 10
z = f (x, y) = x5y2 + 3x2y + 10 = (2et)5(3ln t)2 +3(2et)23 ln t + 10 = 288e5t ln2
t + 36e2t ln t + 10
z = f (x, y) = x5y2 + 3x2y + 10 = u5v2 cos5 v e2u + 3u2v cos2 v eu + 10
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f (x(t), y(t))
t
x(t)
y(t)
(x(t), y(t))
f
z = f (x(u, v), y(u, v))
(u0, v0)
(u, v)
(u0, v0)
x(u, v)
y(u, v)
x(u, v), y(u, v))
x(u0, v0), y(u0, v0)
f
f (x(u, v), y(u, v))
f (x(u0, v0), y(u0, v0))
(u0, v0)
y = F (x)
x = x(t)
y = F (x(t))
dF
dt =
dF
dx
dx
dt
x = x(u, v)
(u, v) y = F (x)
x(u, v)
y = F (x(u, v))
(u, v)
∂y∂u
= dydx
∂x∂u
∂y
∂v =
dy
dx
∂x
∂v
u
v
y = F (x(u, v))
u
x = x(t)
y = y(t)
t
z = f (x, y) (x(t), y(t)) z = f (x(t), y(t))
t
df
dt =
∂f
∂x
dx
dt +
∂f
∂y
dy
dt
dz
dt =
∂z
∂x
dx
dt +
∂z
∂y
dy
dt
∆x = x(t + ∆t) − x(t) , ∆y = y(t + ∆t) − y(t) ∆z = f (x(t + ∆t), y(t + ∆t)) − f (x(t), y(t))
dz
dt = lim
∆t→0
∆z
∆t ,
dx
dt = lim
∆t→0
∆x
∆t
dy
dt = lim
∆t→0
∆y
∆t
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(∆x, ∆y) = (0, 0)
0 = 0
f
(x(t), y(t))
∆z = ∂z
∂x∆x +
∂z
∂y∆y + ε
(x(t), y(t))
lim(∆x,∆y)→(0,0)
ε ∆x2 + ∆y2
= 0
∆t
∆z
∆t =
∂z
∂x
∆x
∆t +
∂z
∂y
∆y
∆t +
ε
∆t
lim∆→0
∆x2 + ∆y2
|∆t| = lim∆t→0
∆x
∆t
2
+
∆y
∆t
2
=
lim∆t→0
∆x
∆t
2
+
lim∆t→0
∆y
∆t
2
=
dx
dt
2
+
dy
dt
2
(∆x, ∆y) → (0, 0)
∆t → 0
x = x(t)
y = y(t)
lim∆t→0
ε
∆t = lim
∆t→0 ε
∆x2 + ∆y2 ∆x2 + ∆y2
∆t = lim
(∆x,∆y)→(0,0) ε
∆x2 + ∆y2 lim∆t→0
∆x2 + ∆y2
|∆t|
= 0 ·
dx
dt
2
+
dy
dt
2
= 0
∆t → 0
dz
dt =
∂z
∂x
dx
dt +
∂z
∂y
dy
dt
x = x(u, v) y = y(u, v)
(u, v)
z = f (x, y)
(x(u, v), y(u, v))
z = f (x(u, v), y(u, v))
(u, v)
∂f
∂u(x(u, v), y(u, v)) =
∂f
∂x(x(u, v), y(u, v))
∂x
∂u(u, v)
+ ∂f
∂y(x(u, v), y(u, v))
∂y
∂u(u, v)
∂f
∂v(x(u, v), y(u, v)) =
∂f
∂x(x(u, v), y(u, v))
∂x
∂v(u, v)
+ ∂f
∂y(x(u, v), y(u, v))
∂y
∂v(u, v)
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∂f
∂u =
∂f
∂x
∂x
∂u +
∂f
∂y
∂y
∂u
∂f
∂v =
∂f
∂x
∂x
∂v +
∂f
∂y
∂y
∂v
u
v
f (x(u, v), y(u, v))
u
z = f (x, y)
∂f
∂x
(1, 2) = 5
∂f
∂y
(1, 2) =
−2
x = x(u, v) =
u
π + sen v
y = y(u, v) =
v2
π2 − cos u
∂f
∂u
∂f
∂v
u = π
v = π
x = u
π + sen v ⇒ ∂x
∂u =
1
π
∂x
∂v = cos v
y = v2
π2 − cos u ⇒ ∂y
∂u = sen u
∂y
∂v =
2v
π2
u = π
v = π
∂f
∂u = 5.
1
π − 2 . 0 =
5
π
∂f
∂v = 5.(−1) − 2.
2
π = −5 − 4
π
(x, y)
T (x, y) =
x2 + y2
8
t
x(t) = 1 + 2t
y(t) = t2
3
dT
dt =
∂T
∂x
dx
dt +
∂T
∂y
dy
dt = 2x · 2 +
y
4 · 2t
3
(7, 3)
dT
dt (x(3), y(3)) = 2 · 7 · 2 +
3
4 · 2 · 3
3 = 29, 5
0C
s
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(2, 1)
T (x, y) = 9x2 + 4y2
y = x
2
(2, 1)
z = 80
y = x
2
y = x
2
(2.8, 1.4)
(2, 1)
(2.8, 1.4)
80 − 40 (2.81 − 2)2 + (1.41 − 1)2
= 44, 1 graus
metro
y = x
2
y = x
2
lim(x,y)→(2,1)
T (x, y) − T (2, 1) (x − 2)2 + (y − 1)2
(x, y)
y = x
2
t
u = 1√ 5
(2, 1)
y = x
2
x(t) = 2 + 2√ 5
t
y(t) = 1 + 1√ 5
t
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lim(x,y)→(2,1)
T (x, y) − T (2, 1) (x − 2)2 + (y − 1)2
= limt→0
T (2 + 2√ 5
t, 1 + 1√ 5
t) − T (2, 1)
t
= limt→0
16√
5t + 8t2
t= lim
t→016
√ 5 + 8t
= 16√
5 graus
metro
z = f (x, y)
P = (x0, y0)
u = (u1, u2)
r
x(t) = x0 + tu1
y(t) = y0 + tu2
f
r
f (x(t), y(t))
C
V
r
S
f
f
(x(t), y(t))
t
f (x(t), y(t)) = F (t)
F (0) = f (x(0), y(0)) = f (x0, y0)
(t, F (t))
V
(x0, y0)
(x, y)
r
(x0, y0)
|t|
t
(x(t), y(t))
f
P 0 = (x0, y0)
u = (u1, u2)
Duf (P 0) = dF
dt (0) = lim
t→0
F (t) − F (0)
t = lim
t→0
f (x0 + tu1, y0 + tu2) − f (x0, y0)
t
f (x0 + tu
1, y
0 + tu
2)−
f (x0, y
0)
t
C
C
P = (x0, y0, f (x0, y0))
u = (1, 0)
Duf (x0, y0) = limt→0
f (x0 + t, y0) − f (x0, y0)
t =
∂f
∂x(x0, y0)
u = (0, 1)
Duf (x0, y0) = limt→0
f (x0, y0 + t) − f (x0, y0)
t =
∂f
∂y(x0, y0)
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v
u =
v
||v||
f
v
Dvf (x, y) = Duf (x, y)
z = f (x, y)
P 0 = (x0, y0)
z = f (x, y)
P 0 = (x0, y0)
x = x(t) = x0 + tu1
y = y(t) = y0 + tu2 z = f (x0 + tu1, y0 + tu2) = F (t)
F (t)
t = 0
Duf (x0, y0) = dF
dt (0) =
∂f
∂x(x0, y0)
dx
dt(0) +
∂f
∂y(x0, y0)
dy
dt(0)
= ∂f
∂x
(x0, y0), ∂f
∂y
(x0, y0) ·dx
dt
(0), dy
dt
(0)=
∂f
∂x(x0, y0),
∂f
∂y(x0, y0)
· (u1, u2)
=
∂f
∂x(x0, y0),
∂f
∂y(x0, y0)
· u
P 0 = (x0, y0)
∇f (P 0) =
∂f
∂x(P 0),
∂f
∂y(P 0)
∇
f
f
z = f (x, y)
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∇f (P 0) = 0
α ∇f (P 0)
u
f
P 0 = (x0, y0)
Duf (P 0) = ∇f (P 0) · u
= |∇f (P 0)| |u| cos α
= |∇f (P 0)| cos α
|∇f (P 0)|
cos α = 1
α = 0
u ∇f (P 0)
Duf (P 0) = |∇f (P 0)|
cos α = −1
α = π
u −∇f (P 0)
Duf (P 0) = −|∇f (P 0)|
α = π2
f
f
z = f (x, y)
(x, y)
∇f (P 0) = 0
Duf (P 0) = 0
P 0
f (x, y) = xy
P 0 = (2, 1)
v = (4, 3)
|v| =
√ 42 + 32 = 5
u = (4
5, 35
)
v
Duf (P 0) = (1, 2)·
(4
5, 3
5) =
4
5 +
6
5 = 2
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Duf (2, 1) = ∇f (2, 1) · u
z = f (x, y)
P 0 = (x0, y0)
∇f (P 0) = 0 ∇f (P 0)
P 0
∇f (P 0) = 0
f (x, y) = c
P 0
P 0
x = x(t)y = y(t)
x0 = x(0)
y0 = y(0)
dxdt
(0), dydt
(0)
z = f (x(t), y(t)) = c
t
∂f
∂x(x(t), y(t))
dx
dt(t) +
∂ f
∂y(x(t), y(t))
dy
dt(t) =
d
dtc = 0
t = 0
∂f
∂x(x0, y0)
dx
dt(0) +
∂ f
∂y(x0, y0)
dy
dt(0) = 0
∂f
∂x(x0, y0),
∂f
∂y(x0, y0)
·
dx
dt(0),
dy
dt(0)
= 0
∇f (P 0) ·dx
dt(0),
dy
dt(0) = 0
f
P 0
f (x, y) = c
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z = f (x, y) = xy
f
P = (2, 1
2)
∇f (P )
f
P
Q = (1, 2)
R = (−2, 1)
f
z = xy
z = f (x, y)
f
D = (1.1, 1)
E = (−1.1, 1)
I = (−1.1, −1) J = (1.1, −1)
f (x, y)
∇f (D)
D
D
r
r
D
0.8
f (D) = 0.8
f (x, y) = 1
f = 0.8
f = 1
0.1
r
f = 0.8
f = 1
1 − 0.80.1
= 2
|∇f (D)|
2
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z = f (x, y) = 36 − 2x2 − 4y2
A = (2, 1, 24)
?
v = (−3, 4)
∩
z = 0 ⇒ 36 − 2x2 − 4y2 = 0
∩ x = 0 ⇒ z = 36 − 2x2
∩
y = 0 ⇒ z = 36 − 4y2
∇f (2, 1) =
∂f
∂x(2, 1),
∂f
∂y(2, 1)
= (−8, −8)
(2, 1)
v
v
v
u = v|v| =
15
(−3, 4)
Duf (2, 1) = ∇f (2, 1) · u = −(8, 8) · 1
5(−3, 4) =
1
5(24 − 32) = −8
5
(2, 1)
−85
v = (−3, 4)
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u1 = (8, −8)
u2 = (−8, 8)
A
n ∈ N
D ⊂ Rn
f
n
(x1, x2, . . . , xn) ∈ D
f (x1, x2, . . . , xn)
D
f
{f (x1, x2, . . . , xn); (x1, x2, . . . , xn) ∈ D} ⊂ R
f : D ⊂ R
n
−→ R
(x1, x2, . . . , xn) −→ w = f (x1, x2, . . . , xn)
f : D ⊂ R3 −→ R
(x1, x2, x3) −→ w = f (x1, x2, x3)
w = f (x,y,z ) = x2 +√
y − z
f
y − z ≥ 0
f
D = {(x,y,z ) ∈ R3; y > z }
D y = z x
f
[0, ∞)
R4
{(x , y, z, w); w = f (x,y,z )}
w = f (x,y,z )
f (x,y,z ) = k
k
(x,y,z )
T (x,y,z ) = x2 + y2 + z 2
x2 + y2 − z 2 = k
k = −2, 0, 2
w = f (x,y,z ) = x2 + y2 − z 2
k > 0 k = 0
k < 0
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x2 + y2 − z 2 = k
k = −2, 0, 2
w = f (x,y,z )
z
∂f ∂z
f
z
x
y
w = sen xyz
∂w
∂x = yz cos xyz
∂w
∂y = xz cos xyz
∂w
∂z = xy cos xyz
z = f (x,y,z )
x = x(t)
y = y(t)
z = z (t)
df dt = ∂f ∂x dxdt + ∂f ∂y dydt + ∂ f ∂z dz dt
w = f (x,y,z )
P 0 =(x0, y0, z 0)
Duf (P 0) = ∇f (P 0) · u = |∇f (P 0)| cos α
u = (u1, u2, u3)
α ∇f (P 0)
u
−1 ≤ cos α ≤ 1 |∇f (P 0)|
u
f
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−|∇f (P 0)|
∇f (P 0)
P 0
(x,y,z )
T (x,y,z ) =
x2 + y2 − z 2
Q = (3, 1, 2)
v = (1, 1, 1)
u = v
|v| = 1√ 3
(1, 1, 1)
∇f = (2x, 2y, −2z )
Duf (Q) = ∇f (Q) · u = (6, 2, −4) · 1√ 3
(1, 1, 1) = 4√
3=
4√
3
3
4√ 3
3
0C m
(−6, −2, 4)
−|∇f (Q)| = −√
62 + 22 + 42 = −√
560C
m
w = (a,b,c)
∇f (Q) · u = (6, 1, −4) · (a,b,c) = 0 =⇒ 6a + b − 4c = 0
b = 4c − 6a a = 0 c = 1
b = 4
w = (0, 4, 1)
(3, 1, 2)
P 0 = (x0, y0, z 0)
P 0
P 0
∇f (P 0) =
∂f
∂x(P 0),
∂f
∂y(P 0),
∂f
∂z (P 0)
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∇f (P 0) · (P − P 0) = 0 ⇔ ∂f
∂x(P 0)(x − x0) +
∂f
∂y(P 0)(y − y0) +
∂ f
∂z (P 0))(z − z 0) = 0
S
F (x,y,z ) = k
P 0 = (x0, y0, z 0)
P 0
S
P 0
F
∇F (P 0)
x(t) = x0 + ∂F
∂x(P 0) t , y(t) = y0 +
∂F
∂y (P 0) t , z (t) = z 0 +
∂F
∂z (P 0) t
x − x0
∂F
∂x(P 0)
= y − y0∂F
∂y (P 0)
= z − z 0∂F
∂z (P 0)
z = f (x, y)
w = F (x,y,z ) = z − f (x, y) k = 0
F (x,y,z ) = z − f (x, y) = 0
F
P 0
∇F (P 0) = (∂f
∂x(P 0),
∂f
∂y(P 0),
∂f
∂z (P 0)) = (−∂f
∂x((x0, y0), −∂f
∂y((x0, y0), 1)
z = f (x0, y0) + ∂ f
∂x(x0, y0)(x − x0) +
∂ f
∂y(x0, y0)(y − y0)
z = 1 +
x2 + y2
P 0 = (1, 1, 1 +√
2)
z = 1 +
x2 + y2
(z − 1)2 =
x2 + y2
F (x,y,z ) = x2 + y2 − (z − 1)2
x2 + y2 − (z − 1)2 = 0
∇F = (2x, 2y, −2(z − 1)) ⇒ ∇F (1, 1, 1 +
√ 2) = (2, 2, −2
√ 2)
(x − 1) + (y − 1) −√
2(z − (1 +√
2)) = 0
x = 1 + 2t , y = 1 + 2t
z = 1 + √ 2 − 2√ 2t
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f (x, y) =
9 − x2 − y2
f (x, y) =
9 − x2 − y2
x − y
f (x, y) = 2 + 1
2
x2 + 4y2
f
f
x = 0
y = 0
z = 3
f
(3, 2)
D = {(x, y) ∈ R2; x2 + y2 ≤ 1}
f (x, y) =
x2 + y2 (x, y) ∈ D
0 (x, y) ∈ R2 \ D
f
f
x = 0
y = 0
z = 0
f
z = 1/2
z = 1
f
R2
f
f (x, y) = 4 − x2 − y2
f
f x = 0 y = 0 z = 0
f
z = −2, 0, 2, 4
f
z = x2 − y2
z = −x3 + xy2
z = 2(−x3y + xy3)
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z =
sen xy
xy
z = e−x2
+ e−y2
z = cos x sen y
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f (x, y) =
xy2
3x2
−y4
(x, y) = (0, 0)
0 (x, y) = (0, 0)
f
(x, y) → (0, 0)
y = mx
m ∈ R
(0, 0)
z = f (x, y) = 1 + 4x2 + y2
f
z = 1, 2, 3, 4
γ
f
z = 5
f
P = (1
2,√
3, 5)
f
P = (1
2,√
3, 5)
f (x, y) = 100 − 4x2 − 5y2
A = (3, 2, 44)
v = (1, −2)
∇ f (3, 2)
z = f (x, y) = 36 − 2x2
− 4y2
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A = (2, 1, 24)
?
v = (−3, 4)
D
xy
(x, y)
f (x, y) = 300 − 2x2 − 3y2
P = (4, 9)
u = (3, 4)
z = f (x, y) = 16 − y2 − x
z = 5
z = 10
z = 14
z = 16
Q = (0, 0, 16)
C
y = x
T = (3, 3, 4)
y = f (x) = x2
F : R2 → R
F (x, y) = 0
f
z = f (x, y) = x2 + y2
F : R3 → R
F (x,y,z ) = 0
f
w = f (x,y,z ) = x2 + y2 − z 2
S
P = (3, 1, 2)
S
4x + 12y − 8z + 10 = 0
S P
y = 0.
S
z = 3x2 + y2 − 2
S
P = (1, 1, 2)
S
z = 0
S
P
T (x, y) = 16
1 + x2 + 2y2
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P = (1, 1)
P
v = (−3, 4)
r(t) = (√
1 + t, 1 + 2t)
S
x2 + 2y2 + 3y2 = 1
S
S
x − 2y + 6z = 1
z = f (x, y) =
4x2 + 10y2
x = 5et−1
y = 2 ln t
dz
dt
t = 1
f (x, y) =
(x − y)2
x2 + y2 (x, y) = (0, 0)
0 (x, y) = (0, 0)
f (x, y)
R2
df
dt(g(t))|t=0
g(t) = (1 − t2, 2 + t)
f (x, y) =
xy(x2 − y2)
x2 + y2 , (x, y) = (0, 0)
0 (x, y) = (0, 0)
(0, 0)
f (x, y) =
(x2 + y2)sen
1
x2 + y2
, (x, y) = (0, 0)
0 (x, y) = (0, 0)
∂f
∂x
∂f
∂y
∂f
∂x
∂f
∂y (0, 0)
f
(0, 0)
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x = x(t) = 0.8t
y = y(t) = 0, 2 t
t
E (x, y) =12
x + 3y
x
y
t = 5
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z = f (x, y)
P 0 = (x0, y0)
f
Dr(P 0) = {(x, y) ∈ R2; (x − x0)2 + (y − y0)2 < r2}
f (x, y) ≤ f (x0, y0)
(x, y) ∈ Dr(P 0)
f (P 0)
f
f
Dr(P 0)
f (x, y) ≥
f (x0, y0)
(x, y) ∈ Dr(P 0)
f (P 0)
f
f (x, y) ≤ f (x0, y0)
(x, y)
f
P 0 f
f
f (x, y) ≥ f (x0, y0)
(x, y)
f
P 0
f
f
P 0
= (x0, y
0)
f
∂f
∂x(P 0) =
∂f
∂y(P 0) = 0
z = f (x, y)
f
P 0
P 0
f
(x0, y0, f (x0, y0))
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P 0
P 0
g(x) = f (x, y0)
dg
dx(x0) =
∂f
∂x(P 0) = 0
h(y) = f (x0, y)
dh
dx(y0) =
∂f
∂x(P 0) = 0
P 0
f (x, y) = x2 + y2 − 2x − 4y + 7
z = x2 + y2 − 2x − 4y + 7
∂f
∂x = 2x − 2
∂f
∂y = 2y − 4
x = 1
y =
2
P 0 = (1, 2)
f (x, y) = x2 + y2−2x−4y +7 = (x−1)2 + (y −2)2 + 2
(x − 1)2 ≥ 0
(y − 2)2 ≥ 0
f (x, y) ≥ 2
(x, y)
P 0 = (1, 2)
f
P 0 = (1, 2, 2)
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f (x, y0)
f (x0, y)
z = f (x, y) = y2
−x2
∂f
∂x = −2x
∂f
∂y = 2y
x = 0
y = 0
P 0 = (0, 0)
(0, 0)
f (0, y) = y2
f (x, 0) = −x2
f
(a, b)
(a, b)
(x, y)
f
f (x, y) > f (a, b)
f (x, y) < f (a, b)
(a,b,f (a, b))
z = f (x, y)
z = y2 − x2
z = f (x, y) = 1−x2+4xy−y2
z = f (x, 0)
z = f (0, y)
x0 = 0
y0 = 0
(x0, y0)
f
(0, 0)
f
x = y
z = 0
P 0 = (x0, y0)
P 0
z = g(x, y) = 3x4 + y2 − 4x2y
xy
y = kx
γ
k ∈ R
xy
y
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z = 1 − x2 + 4xy − y2
z = 3x4 + y2 − 4x2y
z = 3x4 + k2x2 − 4kx3
z = 3x4 + y2 − 4x2y
z = −0.75x4
φ
g
y = 1
z = 3x4
+1−4x2
x = 0.81
x = −0.81
(0.81, 1)
(−0.81, 1)
φ
xy
y = 1.5x2
φ(t) = (t, 1.5t2, g(t, 1.5t2)) =
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(t, 1.5t2, −0.75t4)
xz
z = −0.75x4
g
xy
φ
xy
Dr(0, 0)
Dr(0, 0)
Dr(0, 0)
φ
g
f
(a, b)
f
A = f xx(a, b) , B = f xy(a, b) , C = f yy D = B2 − AC
D < 0
A > 0
(a, b)
f
D < 0
A < 0
(a, b)
f
D > 0
(a, b)
Dr(a, b)
P
Dr(a, b)
C
f
P
2r
u = (u1, u2)
Duf = f xu1 + f yu2
D2uf = Du(Duf ) =
∂ (Duf )
∂x u1 +
∂ (Duf )
∂y u2 = ((f xxu1 + f yxu2)u1 + (f xyu1 + f yyu2)u2
= f xxu21 + 2f xyu1u2 + f yyu2
2
u1
D2uf = f xx
u1 +
f xyu2
f xx
2 − u22(f 2xy − f xxf yy)
f xx
f xx
f 2xy − f xxf yy A > 0
D < 0
Dr(a, b)
f xx > 0
D(x, y) < 0
(x, y)
Dr(a, b)
D2
uf > 0
(x, y)
Dr(a, b)
C
f
(a, b)
u = (u1, u2)
C
2r
u = (u1, u2)
Dr(a, b)
f
(a,b,f (a, b))
f (a, b) ≤ f (x, y)
(x, y)
Dr(a, b)
(a, b)
f
D = 0
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f (x, y) = 2x2 − xy + y2 − 7y
f x
= 4x−
y f
y =
−x + 2y
−7
4x − y = 0 − x + 2y − 7 = 0
y = 4x
x = 1
y = 4
P = (1, 4)
f xx = 4 , f yy = 2 f xy = −1
A = 4 B = −1 C = 2 D = 1 − 4.2 = −7 < 0
(1, 4)
f (x, y) = x3 + 3xy2 − 15x − 12y
f x = 3x2 + 3y2 − 15 f y = 6xy − 12
3x2 + 3y2 − 15 = 0
6xy − 12 = 0
P 1 = (1, 2) , P 2 = (2, 1) , P 3 = (−1 − 2) , P 4 = (−2, −1)
f xx = 6x , f yy = 6x ,
f xy = 6y
B2 − AC AP 1 122 − 6.6 > 0
P 2 62 − 12.12 < 0 12 > 0
P 3 (−12)2 − (−6).(−6) > 0
P 4 (−6)2 − (−12).(−12) < 0 −12 < 0
f
f (x, y) = x4 + y4
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z = x3 + 3xy2 − 15x − 12y
z = x4 + y4
f x = 4x3 = 0 , f y = 4y3 = 0
P = (0, 0)
f xx = 12x2 , f xy = 0 f yy = 12y2
P = (0, 0) B2 − AC = 0
f (0, 0) = 0 < f (x, y)
(x, y) =
(0, 0)
P = (0, 0)
z = f (x, y) =
3
4y2 +
1
24y3 −
1
32y4 − x2
f x = −2x = 0
f y =
3
2y +
1
8 y2
− 1
8y3
= 0
P 1 = (0, 0) , P 2 = (0, −3) , P 3 = (0, 4)
f xx = −2 , f xy = 0 f yy =
3
2 +
1
4y − 3
8y2
B2 − AC A
P 1 0 − (−2).32 > 0
P 2 0 − (−2).(−218 ) < 0 −2 < 0
P 3 −(−2).(−72
) < 0 −2 < 0
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z = 3
4y2 + 1
24y3 − 1
32y4 − x2
x + y + z = 1
x > 0 y > 0
z > 0
V = xyz
P = (x,y,z )
x + y + z = 1
z = 1 − x − y
V
V = V (x, y) = xy(1 − x − y) = xy − x2y − xy2
V x = y − 2xy − y2 = 0 V y = x − x2 − 2xy = 0
x = 1
3 y =
1
3
V xx = −2y , V xy = 1 − 2x − 2y , V yy = −2x
B2 − AC =
1
3
2
−
2
3
2
< 0 A = −2
3 < 0
13
, 13
z = 1 − x − y
x = 1
3 y = 1
3
z = 13
x = 13
y = 13
z = 13
V = xyz
x + y + z = 1
z = 1 − x − y
V = xyz
x
y
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w = f (x,y,z )
(x,y,z )
g(x,y,z ) = 0
z = f (x, y)
(x, y)
g(x, y) = 0
f
2
C
g(x, y) = 0
g(x, y) = 0
y = φ(x)
x = ψ(y)
z = f (x, φ(x))
z = f (ψ(y), y)
P = (x0, y0)
z = f (x, y)
C
g(x, y) = 0
∇f (P )
∇ g(P )
f
C
f
f = 50
P
P C f = 50
f
P
∇f (P )
C
∇g(P )
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C
P
r(t) = (x(t), y(t))
t0
r(t0) = P
f
C
h(t) = f (x(t), y(t))
f
P = (x0, y0)
h
t0
h(t0) = 0 f
0 = h(t0) = ∂f
∂x(x0, y0)
dx
dt(t0) +
∂f
∂y(x0, y0)
dy
dt(t0) =
∂f
∂x(x0, y0),
∂f
∂y(x0, y0)
·
dx
dt(t0),
dy
dt(t0)
= ∇f (x0, y0) · r(t0)
∇f (P )
r(t0)
C
P
C
g(x, y) = 0
z = g(x, y)
g
P
∇g(P )
r(t0) ∇f (P )
∇g(P )
λ ∈ R
∇f (P ) = λ∇g(P )
C
∇g(P ) = 0
f
z = f (x, y)
z = g(x, y)
f
g(x, y) = 0
P
∇g(P ) = 0
∇f (P ) = λ ∇g(P )
λ
λ
z = f (x, y)
g(x, y) = 0
x
y
λ
∇f (x, y) = λ ∇g(x, y) g(x, y) = 0 (∇g = 0)
∂f
∂x = λ
∂ g
∂x∂f
∂y = λ
∂g
∂y
g(x, y) = 0
f
(x, y)
f
f
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g(x, y) = 0
z = f (x, y) = x2 + y2
−2x
−2y + 3
C : 4x2 − 8x + y2 − 2y + 1 = 0
z = 1 + (x − 1)2 + (y − 1)2
(x
−1)2 +
(y − 1)2
4
= 1
2x − 2 = λ (8x − 8)
2y − 2 = λ (2y − 2)4x2 − 8x + y2 − 2y + 1 = 0
⇐⇒ x − 1 = 4λ (x − 1)
y − 1 = λ (y − 1)4x2 − 8x + y2 − 2y + 1 = 0
y = 1
λ = 1
x = 1
y = 3
y = −1
P 1 = (1, 3) P 2 = (1, −1)
y = 1
x = 2
x = 0
P 3 = (2, 1)
P 4 = (0, 1)
f (P 1) = f (P 2) = 5 f (P 3) = f (P 4) = 2
P 1 = (1, 3)
P 2 = (1, −1)
P 3 = (2, 1)
P 4 = (0, 1)
z = f (x, y) = x2 + (y − 2)2
C : x2 − y2 = 1
z = x2 + (y − 2)2
C : x
2
− y
2
= 1
g(x, y) = x2 − y2 − 1 = 0
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2x = 2λ x
2(y
−2) =
−2λ y
x2 − y2 − 1 = 0 ⇐⇒
x = λ x
y
−2 =
−λ y
x2 − y2 − 1 = 0
x2 − y2 = 1
x = 0
λ = 1
y = 1
x =√
2 x = −√
2
P 1 = (√
2, 1) P 2 = (
−
√ 2, 1)
f (P 1) = f (P 2) = 3
P 1
P 2
f (√
2 + h, 1 + k) ≥ f (√
2, 1) = 3
f (−√
2 + h, 1 + k) ≥ f (√
2, 1) = 3
h k (±√ 2 + h, 1 + k)
C : x2 − y2 = 1
f (√
2 + h, 1 + k) = (√
2 + h)2 + (1 + k − 2)2 = 3 + 2√
2h + h2 − 2k + k2
(√
2 + h, 1 + k)
C : x2 − y2 = 1
(√
2 + h)2 − (1 + k)2 = 1 ⇒ −2k = k2 − 2√
2 h − h2
f (√
2 + h, 1 + k) = 3 + 2√
2 h + h2 + k2 + k2 − 2√
2 h − h2 = 3 + 2k2 ≥ 3
k
w = f (x,y,z )
x, y
z
g(x,y,z ) = 0
x
y
z
λ
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x
y
z
λ
∇f (x,y,z ) = λ∇g(x,y,z ) g(x,y,z ) = 0 (∇g = 0)
∂f
∂x = λ
∂ g
∂x∂f
∂y = λ
∂g
∂y
∂f
∂z = λ
∂g
∂z
g(x,y,z ) = 0
f (x,y,z )
f
f
f
g = 0
a
a
(x, 0, 0) (0, y, 0)
(0, 0, z )
V (x,y,z ) = 8xyz
P = (x,y,z )
a
x2 + y2 + z 2 = a2
x > 0
y > 0
z >
0
8yz = λ 2x8xz = λ 2y8xy = λ 2z
x2 + y2 + z 2 = a2
⇐⇒
4xyz = λ x2
4xyz = λ y2
4xyz = λ z 2
x2 + y2 + z 2 = a2
λ x2 = λ y2 = λ z 2
λ > 0
x
y
z
x = y = z
x =
a
√ 3
a
2a√ 3
8√
3a3
9
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w = f (x,y,z )
g(x,y,z ) = 0
h(x,y,z ) = 0
f
C
g = 0
h = 0
f
P = (x0, y0, z 0)
∇f
C
P
∇g
∇h
g = 0 h = 0
C
∇f (P )
∇g(P ) ∇h(P )
λ
µ
∇f (P ) = λ∇g(P ) + µ∇h(P )
x
y
z
λ
µ
x
y
z
λ
µ
∇f (x,y,z ) = λ∇g(x,y,z ) + µ∇h(x,y,z ) g(x,y,z ) = 0 , h(x,y,z ) = 0
∇g = 0, ∇h = 0
∂f
∂x = λ
∂ g
∂x + µ
∂h
∂x∂f
∂y = λ
∂g
∂y + µ
∂h
∂y
∂f
∂z
= λ ∂g
∂z
+ µ ∂h
∂z g(x,y,z ) = 0
h(x,y,z ) = 0
f
(x,y,z )
f
f
γ
x + 2y + z = 1
x2
4 + z 2 = 1
γ
y = 0
γ
y = 0
|y|
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g(x,y,z ) =
y2
f (x,y,z ) = y2
x + 2y + z = 1
x2
4 + z 2 = 1
x
y
z
λ
µ
0 = λ + 12
xµ
2y = 2λ0 = λ + µ2z x + 2y + z = 1x2
4 + z 2 = 1
⇐⇒
λ + 12
µx = 0
λ = yλ + 2µz = 0x + 2y + z = 1x2
4 + z 2 = 1
=⇒
y + 1
2µx = 0y + 2µz = 0x + 2y + z = 1x2
4 + z 2 = 1
µ(x − 4z ) = 0
µ = 0
x = 4z
µ = 0
y = 0
x + z = 1x2
4
+ z 2 = 1
{x = 0, z = 1}
x =
8
5, z = −3
5
P 1 = (0, 0, 1) P 2 =
8
5, 0, −3
5
x = 4z
z =±
1
√ 5
x =±
4
√ 5
x + 2y + z = 1
P 3 =
4√
5
5 ,
1 − √ 5
2 ,
√ 5
5
P 4 =
−4
√ 5
5 ,
1 +√
5
2 , −
√ 5
5
P 1
P 2
y = 0
P 4
y
γ
x + 2y + z = 1
x2
4 + z 2 = 1
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y = f (x)
y = f (x)
[a, b]
(x0, y0)
A ∈ R
2
(x0, y0)
r
Dr(x0, y0) = {(x, y) ∈ R2; (x − x0)2 + (y − y0)2 < r2}
A
A ∈ R2
(x0, y0)
F ∈ R2
(x0, y0)
r
Dr(x0, y0) = {(x, y) ∈ R2; (x − x0)2 + (y − y0)2 < r2}
F
F R
2 \ F
F ⊂ R2
F
R
2 \ F
M ⊂ R2
Dr(0, 0)
M ⊂ Dr(0, 0)
z = f (x, y)
D ⊂ R2
D
D
D
D
D
D
T (x, y) = x2 + y2 − 2x − 2y + 3
D
D = {(x, y) ∈ R2; 0 ≤ x ≤ 2 0 ≤ y ≤ 3}
D
E = {(x, y) ∈ R2; 4x2 + y2 − 8x − 2y + 1 = 0}
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z = T (x, y) = x2 + y2 − 2x − 2y + 3 = (x − 1)2 + (y − 1)2 + 1
R
T
R
∂T
∂x = 2x − 2 = 0
∂T
∂y = 2y − 2 = 0
x = 1
y = 1
P 1 = (1, 1)
R
T
l1
l2
l3
l4
R
T
T
R
l1
z = T (x, y) = T (x, 0) = x2 − 2x + 3 ⇒ dz
dx = 2x − 2 = 0 ⇒ x = 1
l2
z = T (2, y) = y
2
− 2y + 3 ⇒ dz
dy = 2y − 2 = 0 ⇒ y = 1
l3
z = T (x, 3) = x2 − 2x + 6 ⇒ dz
dx = 2x − 2 = 0 ⇒ x = 1
l4
z = T (0, y) = y2 − 2y + 3 ⇒ dz
dy = 2y − 2 = 0 ⇒ y = 1
P 2 = (1, 0) , P 3 = (2, 1) , P 4 = (1, 3) P 5 = (0, 1)
T R
P 6 = (0, 0) , P 7 = (2, 0) , P 8 = (2, 3) P 9 = (0, 3)
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T (P 1) = 1 , T (P 2) = 2 , T (P 3) = 2 , T (P 4) = 5 , T (P 5) = 2 , T (P 6) = T (P 7) = 3
T (P 8) = T (P 9) = 6
P 1 P 8 P 9
T
E
4x2 + y2 − 8x − 2y + 1 = 0
T
P 1 = (1, 1)
(x − 1)2 + (y − 1)2
4 = 1
P 1
T 4(x − 1)2 + (y − 1)2 = 4
2x − 2 = λ (8x − 8)2y − 2 = λ (2y − 2)4x2 + y2 − 8x − 2y + 1 = 0
⇐⇒
x − 1 = 4λ (x − 1)y − 1 = λ (y − 1)4x2 + y2 − 8x − 2y + 1 = 0
x−1 = 0
λ =
1
4
y = 1
x = 0
x = 2
P 2 = (0, 1) P 3 = (2, 1)
x − 1 = 0 ⇒ x = 1
y = 3
y = −1
P 4 = (1, 3) P 5 = (1, −1)
T
T (P 1) = 1 T (P 2) = T (P 3) = 2 T (P 4) = T (P 5) = 5
E
P 1
P 4 P 5
z = x5 + y4 − 5x − 32y − 3
z = 3x − x3 − 3xy2
z =
x5
5 − x3
3 + y2 − 2xy + x2
z = x4 − y4 − 2x2 + 2y2
z = 2y3 − 3x4 − 6x2y + 1
16
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4(x − 1)2 + (y − 1)2 = 4
x + 2y + 3z = 6
24 cm3
R$ 8, 00
cm2
R$ 3, 00
cm2
R$ 2, 00
R
4x2 + 36y2 + 9z 2 = 36
3x + 2y + z − 14 = 0
T (x,y,z ) = xy +z 2
y − x = 0 x2 + y2 + z 2 = 4
D
x − y = −2, x − y = 2, x + y = −2
x + y = 2
T (x, y) = 2x2 + y2 − y
(x, y) ∈ R2
f (x, y) = − 1
12x3 + x − 1
4y2 + 1
2
f
f
(x + 2)2 + y2 = 4
f
D = {(x, y); (x + 2)2 + y2 ≤ 4}
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f (x, y) = 2y2 − 8y +
1
3x3 +
5
2x2 − 6x + 8
f
f
(x + 2)2 + y2 = 4
f
D = {(x, y) ∈ R2; (x − 1)2 + (y − 2)2 ≤ 9}
16x2 + 4y2 + 9y2 = 144
3x2 + 4xy + 6y2 = 140
ax + by + cz = 1
a
b
c
a
b
c
(1, 2, 3)
(x, y)
T (x, y) = 64(3x2 − 2xy + 3y2 + 2y + 5)
f (x, y) = 4y2e−(x2+y2)
P 1 = (−2, 2)
P 2 = (2, 2)
P 3 = (2, −2)
P 4 = (−2, −2)
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F
d
P
P
(P, F ) =
(P, d)
d
F
(0, p)
p > 0
y =
− p
P (x, y) (x − 0)2 + (y − p)2 =
(x − x)2 + (y + p)2
x2 + (y − p)2 = (y + p)2
x2 + y2 − 2 py + p2 = y2 + 2 py + p2
x2 = 4 py
y = 1
4 px2 = ax2
a = 1
4 p > 0
p < 0
x
y
y2 = 4 px
( p, 0)
x = − p
p > 0
p < 0
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y = ax2
y = −ax2
x = ay2
x = −ay2
F 1
F 2
P
P
a > 0
(P, F 1) +
(P, F 2) = 2a
x
P (x, y) (x + c)2 + y2 +
(x − c)2 + y2 = 2a
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(x + c)2 + y2 = 2a −
(x − c)2 + y2
x2 + 2cx + c2 + y2 = 4a2 − 4a (x − c)2 + y2 + x2 − 2cx + c2 + y2
a
(x − c)2 + y2 = a2 + cx
a2(x2 + 2cx + c2 + y2) = a4 + 2a2cx + c2x2
(a2 − c2)x2 + a2y2 = a2(a2 − c2)
c < a
a2
−c2 > 0
b2
a2
−c2 = b2
b2x2 + a2y2 = a2b2
a2b2
x2
a2 +
y2
b2 = 1
b2 = a2
−c2 < a2
b < a
y = 0
(a, 0)
(−a, 0)
x = 0
(0, b)
(0, −b)
(a, 0)
(−a, 0)
(0, b) (0, −b)
y
F 1(0, −c)
F 2(0, c)
x
y
x2
a2 + y2
b2 = 1
x2
b2 + y2
a2 = 1
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F 1 F 2
P
P
a > 0
| (P, F 1) − (P, F 2)| = 2a
F 1 = (−c, 0)
C 2 = (c.0)
x
x2
a2 − y2
b2 = 1
c2
= a2
+ b2
x
V 1 = (−a, 0)
V 2 = (a, 0)
y
x = 0
y2 = −b2
x2
a2 = 1 +
y2
b2 ≥ 1 =⇒ x2 ≥ a2 ⇔ |x| ≥ a
x ≥ a
x ≤ −a
x
y2 = b2
a2
(x2
−a2) =
⇒y =
b
a
√ x2
−a2
y =
−b
a
√ x2
−a2
x → +∞
b
ax − b
a
√ x2 − a2 → 0
y = ± b
a
OAV 2
c2 = a2 + b2
y
F 1 = (0, −c)
F 2 = (0, c)
x
y
−x2
b2 +
y2
a2 = 1
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x2
a2 − y2
b2 = 1
−x2
b2 + y2
a2 = 1
x2 + y2 = r2
C = (m, n)
P = (x, y)
C
(x − m)2 + (y − n)2 = r2
P
xy
(x, y)
xy
(x)2 + (y)2 = r2
(x, y)
P
xy
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x = m + x
y = n + y ⇐⇒
x = x − m
y = y − n
(m, n)
(m, n)
(m, n)
x
(y − n)2 = 4 p(x − m)
(y − n)2 = −4 p(x − m)
(m, n)
y
(x − m)2 = 4 p(y − n)
(x − m)2 = −4 p(y − n)
(m, n)
(x − m)2
a2 +
(y − n)2
b2 = 1
(m, n)
(x − m)2
a2 − (y − n)2
b2 = 1
x
y
Ax2 + By2 + Cx + Dy + E = 0
x
xy
Ax2 + By2 + Cxy + Dx + Ey + F = 0
4x2 + 9y2 − 32x − 54y + 109 = 0
2x2 + 12x − y + 17 = 0
4x2 − 9y2 − 16x − 18y − 29 = 0
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x
y
4x2 + 9y2 − 32x − 54y + 109 = 4(x2 − 8x + 16) + 9(y2 − 6y + 9) − 64 − 81 + 109
= 4(x − 4)2 + 9(y − 3)2 − 36
4(x − 4)2 + 9(y − 3)2 − 36 = 0 ⇐⇒ (x − 4)2
9 +
(y − 3)2
4 = 1
(4, 3)
(x−4)24
+ (y−3)29
= 1 y + 1 = 2(x + 3)2
x
y
2x2 + 12x − y + 17 = 2(x2 + 6x + 9) − y − 18 + 17 = 2(x + 3)2 − y − 1
2x2 + 12x − y + 17 = 0 ⇐⇒ y + 1 = 2(x + 3)2
(−3, −1)
x
y
4x2 − 9y2 − 16x − 18y − 29 = 4(x2 − 4x + 4) − 9(y2 + 2y + 1) − 16 + 9 − 29
= 4(x − 2)2 − 9(y + 1)2 − 36
4x2 − 9y2 − 16x − 18y − 29 = 0 ⇐⇒ (x − 2)2
9 − (y + 1)2
4 = 1
(2, −1)
x
y = −1 ± 23
(x − 2)
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(x−2)29
− (y+1)2
4 = 1
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y = ln
−1c+ex
y = cex2
t+xtx
+ ln |xt| = c
x = 0
y = (ln |x| + c)2 + 1
y = 1
(ln y)2 = ln x2 + c
y = 1−x1+x
y =
√ 2ex − 1
y = cx + x ln |x|
y = x3− 1
9 + e−2x + ce−3x
y = ce−
x3
3 + 1
y = ce3x − −313
sen2x − −213
cos2x
y = x2
2 e−2x − 1
2e−2x
y = x2 + x−2
y = c−3cos x+cos3 x3sen2x
y + p(x)y = q (x)yn
y−n
u = y1
−n
⇒y−ny
= u
1−n
k = 1
10 ln (103
100)
t = ln 2
k
− ln 2
ln 0, 9 = 6, 6
F = 30
t = ln 21/3
F = kv1 t = m
k ln 2
k = ln(5/2)
10
t = 3
4.104
k
v(t) = 23
V (1 − e−t) vellimit = 75 km/h
y = c1e−x + c2e2x
y = c1e
√ 7x + c2e−
√ 7x
y = c1 cos 2x + c2sen2x
y =
c1e−x cos(√
2x) + c2e−xsen(√
2x)
x + 2x = 0
x − x − 2x = 0
x − 4x + 4x = 0
y = c1e7x+c2ex+2
y = et−tet
y =
− 427
− 59
t
e−3t+ t9
+ 427
y = c1e3x+c2e4x−xe4x
y = (1 − x)e2x + x2e2x
y = c1e2x + c2e−3x + xe2x
x10
− 125
y = c1 cos x + c2senx + x2
senx
y = ex − xsenx − cos x
y = ex(c1 cos 3x + c2sen3x) + 1
9ex + 1
37(sen3x + 6 cos 3x)
y =
c1 + c2e3x − 110 cos x −
310senx −
16x2 −
19x
x(t) = −1
5e−2t + 1
5e−7t
g = 9, 8
x(t) = 1
50 (−11e−7t + 20e−2t − 9cos t + 13sent)
x(t) =
187
+ 2t
sen7t − 4t cos7t
x(t) = e−2t(cos t + 2 sen t) + sen2t
x(t) = 2
5 sen 5t + (1 − 2t)cos5t
x(t) = e−3t(t3 − 2t2 + t)
v(t) = 10 + e−t + et
x = 4+4t
y = 1
−t
x = 2+4t
y = 19+ 16t
x = 2
√ 3 +2t
y = 32 − 3√ 3
2 t
x = 3 + 3t
y = 4 + 8t
x = π2 + 2πt
y = −2t
x = 4 +
√ 22 −
√ 22
t y = 3 +
√ 22
+√ 22
t
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x = t
y = 1 + t
x = 2t
y = 1 + 2t
y = 1 + x
x = 1 +√
6cos t
y = 1 +√
6sen t
x = R cos wt
y = R sen wt
x = a cos t
y = b sen t
x = 1 + 3 cos t
y = 2 + 2 sen t
x = −3+√ 9+24t2
t2
y = 6t
x2y2 + 216x − 864 = 0
x = 2√
1+t2 y = 8t
64x2 + x2y2 = 256
x > 0
2 + t = −8 + 7t
−2 + t2/2 = −1 + 7/2t
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(6, 6)
σ(2) = (1, 2)
σ(t) = (( t
2 + 1)et − t
2e−t, 1
2(t + 1)et + 1
2(t − 1)e−t)
y = 2 + ln x
x(t) = e5t
y(t) = 5t + 2
x(θ) = r(cos θ + θ sen θ)
y(θ) = r(sen θ − θ cos θ)
x = −5/9 + 10t
y = −8/9 + 7t
z = 1 − 9t
−5(x + 1) + 11y − 4(z − 2) = 0
x = 1 − 5t
y = 2 + 4t
z = −1 − 4t
x = 2 cos t
y = 2 sen t
z = 4
x =
√ 22 − √
2 t x =
√ 22
+√
2 t z = 4
x = 1
2 cos t
y = sen t
z = 16 − 2cos t − 4sen t
x =
√ 24 −
√ 24
t
x = cos t
y = 1 + sen t
z =
√ 2 − 2sen t
x =
√ 22 −
√ 22
t y = 1 +
√ 22
+√ 22
t z =
2 − √ 2 −
√ 2
2√
2−√ 2t
v(π
6) =
32 − π
12, 12
+ π12
, 1
x = x0 + kat y = y0 + kbt z = z 0 + kct k ∈ R
t
σ(2) = (1, 4, 0)
σ(2) = (−4, 4, 0)
√ 17 − 1
4 ln (−4 +
√ 17)
x(t) = 1
40(3e−3t + 5et + 4 sen t − 8cos t)
y(t) = −2
3e−3t + 3et − 4
3 z (t) = 1
12(e−3t + 51et − 16
t
(300, 1680, 560)
σ(0) = (300, 1670, 60)
x = t − sen t
y = 1 − cos t
z = 1
t1 = π
3 t2 = 5π
3
4√
3
x = (1 + t)et
y = tet
z = (1 + t)et
x + y + 2z = 0
−2x + z + 3 = 0
x − 4y − 7z + 5 = 0
2x + 10y + 7z − 3 = 0
x = 3 + t
1
17
(17, 15, 10)
(1, 5, 0)
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dom(f ) = D3(0, 0) = {(x, y); x2 + y2 ≤ 9}
img(f ) = [0, 3]
dom(f ) = {(x, y); x2 + y2 ≤9
x = y}
img(f ) = [0, 3]
dom(f ) = R2
img(f ) = [2.5, ∞)
f ∩
x = 0 ⇒ z = 2 + |y| f
∩
y = 0 ⇒ z = 2 + |x|
f ∩ z = 0 f ∩ z = 3 ⇒ x2 + 4y2 = 4
x2 + 4y2 = 25
dom(f ) = R
2
img(f ) = [0, 1]
x2 + y2 = 1/2 x2 + y2 = 1
f
R2 \ S 1(0, 0) = {(x, y); x2 + y2 = 1}
f
S 1(0, 0)
(a, b) ∈ S 1(0, 0)
R2 \ S 1(0, 0)
lim(x,y)→(a,b) = 0
1
D1(0, 0)
dom(f ) = R2 img(f ) = (−∞, 4]
z = x2 − y2 ←→
←→
z = −x3 + xy2 ←→
←→
z = 2(−x3y + xy3) ←→ ←→
←→
←→
←→
←→
←→
←→
f
(0, 0)
f
(0, 0)
x = y2
x = cos t
y = 2sen t
t ∈ [0, 2π]
z = 4x + 2
√ 3y − 3
x = 1
2 − 4t
y =
√ 3 − 2
√ 3t
z = 5 + t
∇f (3, 2) = (
−24,
−20)
D
∇f (3, 2) = 4
√ 61
−64
√ 5
(20,
−24)
5x2 + y2 = 56
∇f (2, 1) = (−8, −8)
8
√ 2
−8/3
(8, −8)
(16, 54)
−52, 8
2x2 + 3y2 = 275
y2 + x = 11
y2 + x = 6
y2 + x = 2
y2 + x = 0
∇f (0, 0) = (−1, 0) −7/
√ 2
F (x, y) = y − x2
F (x,y,z ) = z − x2 − y2
x2 + y2 − z 2 = 6
(x − 1)+3(y − 3) − 2(z − 2) = 0
(x +1)+3(y + 3) − 2(z + 2) = 0
x = 3 + 6t
y = 1 + 2t
z = 2 − 4t
y = 0
(0, 0, 4)
6x + 2y − z − 6 = 0
2√ 41
x2 + 2y2 = 3
−∇T (1, 1) = (2, 4)
−||∇T (1, 1)|| = −2√
5
−2
dT dt
(x(3), y(3)) = 0, 085
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(x − 1√
15) − 2(y + 1√
15) + 6(z − 2√
15) = 0
(x + 1√
15) − 2(y − 1√
15) + 6(z + 2√
15) = 0
10.1
f
(0, 0)
f
(0, 0)
625
E (x(5), y(5)) = 5 dE dt = 1
(1, 2)
(−1, 2)
(−1, 0)
(1, 0)
(0, 1), (0, −1)
(1, 1)
(0, 0)
(−1, −1)
(1, 0)
(−1, 0)
(0, 1)
(0, −1)
(0, 0)
(1, 1)
(1, −1)
(−1, 1)
(−1, −1)
(−1, −1)
(1, −1)
(0, 0)
x = 2
y = 1
z = 2/3
4/3
x = 3
y = 2
z = 4
√ 2R
16
√ 3
(3 2 1)
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