huong dan giai bai tap_hinh hoc vi phan

8/8/2019 Huong Dan Giai Bai Tap_hinh Hoc Vi Phan

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HNG DN GII BI TP 1

HNG DN GII BI TP

BI TP CHNG 1Bi tp 1.1. Ta c

lim(x,y)0

|f(a + x, b + y) f(a, b)(x, y)|(x, y)

= lim(x,y)0

| sin(a + x) sin a cos a.x|x2 + y2

= lim(x,y)0

|2cos 2a+x2 sin x2 cos a.x|

x

2

+ y

2

= limx0

| cos 2a+x2 x cos a.x|x2 + y2

.

Ta li c

0 | cos2a+x

2 x cos a.x|x2 + y2

| cos2a+x

2 x cos a.x||x|

Ta c nh gi

lim

x

0

| cos 2a+x2 x cos a.x|

|x|= lim

x0cos

2a + x

2 cos a = 0

limx0

| cos 2a+x2 x cos a.x|x2 + y2

= 0

Df(a, b) = 0.

Bi tp 1.2. chng minh f kh vi ti x = 0 ta cn ch ra tn ti mt nhx tuyn tnh i t Rn vo R tha gi thit.

Tht vy, xt nh x tuyn tnh O : Rn R. Do hm f tha:

|f(0)| 02 = 0 f(0) = 0.

nn ta c|f(0 + h) f(0)O(h)|

h =|f(x)|h

h2h = h

nnlimh0

|f(0 + h) f(0)O(h)|h = limh0 h = 0.


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2 Hng dn gii bi tp chng 1

Vy f kh vi ti x = 0 v Df(0) = 0.

Bi tp 1.3.

(a) D1f(x, y) = limx0

f(x +x, y) f(x, y)x

D1f(0, 0) = limx0

f(0 + x, 0) f(0, 0)x = limx0

x.0 0x = 0.

Tng t:

+ D2f(x, y) = limy0

f(x, y +y) f(x, y)y

D2f(0, 0) = limy0

f(0, 0 + y) f(0, 0)y = 0.

(b) Gi s f kh vi ti (0, 0)

Df(0, 0) = (0, 0).

Ta c:

lim(x,y)(0,0)

|f(0 + x, 0 + y) f(0, 0) (Df(0, 0)(x,y))|(x)2 + (y)2 = 0

lim(x,y)(0,0)

|f(x,y)|(x)2 + (y)2 = 0

lim(x,y)(0,0)

x|y|

(x)2 + (y)2 = 0. (1)

Chn x = y > 0.Suy ra:

lim(x,y)(0,0)

x|y|(x)2 + (y)2 = limx0

(x)22(x)2 =

1

2= 0 (>< (1)).

Vy f khng kh vi ti (0, 0).

Bi tp 1.4.

(a) f(x,y,z) =

fx

fy

fz

=

y.xy1 (lnx).xy 0

(b) t f1 = xy, f2 = 0.

= f(x,y,z) =

f1x

f1y

f1z

f2x

f2y

f2z

=

y.xy1 (lnx).xy 0

0 0 0

(c) f

(x,y,z) =

f

x

f

y

f

z

=

sin y. cos(x. sin y) x. cos y cos(x. sin y)


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HNG DN GII BI TP 3

(d) t f1 = sin(xy), f2 = sin(x sin y), f3 = xy.

= f(x, y) =

f1x

f1y

f2x

f2y

f3x

f3y

=

y. cos(x.y) x. cos(x.y)sin y. cos(x. sin y) x. cos y cos(x. sin y)y.xy1 (lnx).xy

Bi tp 1.5. Ta c

f(0) = limx0

f(x) f(0)x 0 = limx0

x2

+ x2 sin(1/x)

= 0

Vi x = 0 ta cf(x) =

1

2+ 2x sin

1

x cos 1

x

nn f khng lin tc ti x = 0.

By gi ta chng minh trong mi ln cn ca 0, hm f khng th c nh x

ngc. Tht vy chn 2 dy:

xk =1

2kv yk =

1

(4k + 1)

2

k N.

Ta c

f(xk) = 12

< 0, f(yk) =1

2+

4

(4k + 1)> 0.

Suy ra f khng n iu trong mt ln cn no ca 0, nn khng th tn ti

hm ngc f1.

Ni cch khc, iu kin lin tc khng th b c trong nh l hm ngc.

Bi tp 1.6.

(a) Ta c cng thc xc nh hm h l:

h(t) =

t.x.g

x

x

nu t > 0

t.x.gxx

nu t < 0

0 nu t = 0

hay

h(t) = t.x.g

xx

nu x = 0

0 nu x = 0


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Xt cc trng hp sau

+ x = 0 : Do x.g

x

x

l hng s nn suy ra:

h(t) = x.g xx

, t = 0.

Khi t = 0 ta c:

limt0

|h(t) h(0)||t| = x.g

x

x

.

Hay h kh vi trn R.

+ x = 0: Khi x = 0 nn h = 0 trn R. Suy ra h kh vi trn R.Nh vy trong mi trng hp ta c hm h kh vi trn R.

(b) Ta c:

D1f(0, 0) = limh0

|f(h, 0) f(0, 0)||h| = limh0

f(h, 0)

h

= limh0

h.gh, 0

hh =

limh0

h.g

h, 0

h

h

vi h > 0

limh0

h.g

h, 0h

h

vi h < 0

Suy ra D1f(0, 0) = 0.

Tng t ta cng tnh c:

D1f(0, 0) = 0 = limk0

|f(0, k) f(0, 0)|

|k

|

= 0

By gi gi s f kh vi ti im (0, 0), ta c:

Df(0, 0) = (0, 0).

m

lim(h,k)(0,0)

|f(h, k) f(0, 0) (h, k)|

(h, k)

= lim

(h,k)(0,0)

|f(h, k)|h2 + k2

= lim(h,k)(0,0)

g

h, k

h2 + k2

= 0


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HNG DN GII BI TP 5

Nu tn ti (x0, y0) S1 sao cho g(x0, y0) = 0 th ta c th gi s x0 > 0.Khi vi h > 0, k = h

y0x0

ta c:

g

h,h. y0x0

h2 + h2.y20x20

= g (x0, y0)

x20 + y20

.x0

= g

1, y0x0

x20 + y20

.x0

= g(x0, y0) = 0!!

Vy f khng th kh vi ti im (0, 0).

Bi tp 1.7. Nu vi mi (x, y) R2, ta c f(x, y) = 0 th f l hm hng nnf khng th n nh.

By gi gi s tn ti (x0, y0) R2 sao cho: f(x0, y0) = 0. Ta c th gi s

f

x(x0, y0) = 0.

Khi tn ti mt tp m A cha (x0, y0) sao cho

D1f(x, y) = 0, (x, y) A.

Xt hm s g : R2 R2, g(x, y) = (f(x, y), y), (x, y) R2.Ta c:

g

(x, y) =

f

x

f

y

0 1

nn

det g

(x, y) =f

x

(x, y)

= 0,

(x, y)

A.

Suy ra tn ti hm ngc g1 : g(A) A, g1(f(x, y), y) = (x, y).Ta c:

g(x, y) = g(x

, y

) y = y

f(x, y) = f(x

, y)

nn nu f n nh trn R2 th g n nh kh vi trn R2. Suy ra tn ti g1 n

nh kh vi trn R2 (mu thun vi gi thit ca hm g).

Vy f khng th n nh.


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Bi tp 1.8. Vi mi vx = (v, x) Rnx, x Rn, ta c

(g f)x(vx) = [D(g f)(x)(v)](gf)(x)

= [Dg(f(x))Df(x)(v)](gf)(x)= gf(x)[Df(x)(v)]f(x)

= g[f(vx)] = (g f)(vx)

Suy ra (g f) = g f.

Bi tp 1.9. Ta c |L(x) L(y)| = |L(x y)| L|x y|, t suy ra nhx L lin tc. Chng minh DL = L.


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HNG DN GII BI TP 7

BI TP CHNG 2

Bi tp 2.1. Hnh 2.0.1

Hnh 2.0.1:

Bi tp 2.2. (t) = sin t, cos t

Bi tp 2.3. t f(t) = 2(t). Theo gi thit th

(t0) = min f(t)

=f(t0) = 0=2.(t0).

(t0) = 0 (1)

Do khng i qua gc ta nn (t) = 0, t. Do t (1) ta c (t0) trcgiao vi

(t0).

Bi tp 2.4. Nu (t) = 0 (t) = c, t. Vy vt ca (t) l mt im. Nu (t) = c = 0 (t) = ct + a, t. Vy vt ca (t) l mt ng

thng hoc mt phn ca ng thng.

Bi tp 2.5. Theo gi thit ta c:

(t).v = 0

t0

(t).v.dt =t0

0.dt

v.t

0

(t).dt = 0

v. ((t) (0)) = 0 v.(t) v.(0) = 0 (1)


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Do (0) trc giao vi v nn v.(0) = 0

(1) = v.(t) = 0

Vy (t) trc giao vi v , t I.Bi tp 2.6. Vi : I R3, (t) = 0, t I, ta c

|(t)| = a 2(t) = a2

=2.(t).(t) = 0=(t)(t), t I.

Bi tp 2.7.(a) Ta c

x2 + y2 = a2t2 cos2 t + a2t2 sin2 t

= a2t2(cos2 t + sin2 t)

= a2t2 = 2z.

Vy vt ca ng tham s nm trn mt mt nn.

(b)

C(t) = (sin 2t, 1 cos2t, 2cos t)= (2 sin t cos t, 2sin2 t, 2cos t)

Ta c

x2 + y2 = (2sin t cos t)2 + (2 sin2 t)2

= 4 sin2 t(cos2 t + sin2 t)

= 4 sin2 t

= 4(1 cos2 t)= 4 4cos2 t= 4 z2

Suy ra x2 + y2 + z2 = 4.

Vy vt ca ng tham s C(t) nm mt mt cu c tm O(0, 0, 0) v bn

knh R = 2. Chng ta cng chng minh c vt ca C(t) nm trn mt tr


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HNG DN GII BI TP 9

Hnh 2.0.2:

(Hnh 2.0.2).

Bi tp 2.8. Tip tuyn ng tham s (t) = (3t, 3t2, 2t3) nhn t = (t) =

(3, 6t, 6t2) lm vector ch phng

ng thng (d) :

y = 0z = x c VTCP u = (1, 0, 1).

gc ((), d) = gc(t, u)Ta c :

cos(t, u) =t.u

|t| . |u| =3t.6t

2.

9 + 36t2 + 36t4=

3t.6t2

(3 + 6t2).

2=

2

2.

Bi tp 2.9.

(a)

Theo Hnh 2.0.3, ta c

cos =IH

IM= IK IH = 1 cos

MH = | cos |OK = l(KM) = IK. =

x = OB = OK MH = sin C() = ( sin , 1 cos )


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Hnh 2.0.3:

Suy ra

|C()| =

(1 )2 + sin2 =

1 2cos + cos2 + sin2

= 2(1 cos ) |C()| = 0

1 cos = 0 cos = 0

= k2 , k Z

Do C(k2) = (k2, 0)

Vy nhng im (k2, 0) l nhng im k d ca C().

(b) di mt nhp ca ng Cycloit.

l =

20

2(1 cos )d =

20

4sin2

2d =

20

2.

sin 2 d

= 4

20

sin

2d = 4.2. cos

2|0

= 8.

Bi tp 2.10.(a) Ta c c(t) = (t, t2), c

(t) = (1, 2t), |c(t)| = 1 + 4t2


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HNG DN GII BI TP 11

Vy l =BA

|c(t)|dt =BA

1 + 4t2dt

t u =

1

4+ t2 du = t1

4 + t2

dt

dv = dt v = t

12

l = t

1

4+ t2

B

A

B

A

t1

4+ t2

dt

= t

1

4+ t2

B

A

BA

1

4+ t2dt +

BA

1

4.

11

4+ t2

dt

t x = t +1

4 + t2 1

xdx =dt

1

4+ t2

Vy l = (t

1

4+ t2 +

1

4ln(t +

1

4+ t2))|BA

(b) c : t (t, ln t)c

(t) = (1,1

t) |c(t)| =

1 +

1

t2

l =

B

A |

c

(t)

|dt =

B

A1 +

1

t2

dt

t u =

1 +1

t2 du = 2

t3.1

1 +1

t2du = dt v = t l = t

1

t2+ 1

B

A

+ 2BA

t

t3

1

t2+ 1

dt

= t

1t2

+ 1B

A

+ 2BA

1

t2

1

t2+ 1

dt

t x =1

t dx = 1

t2dt

BA

1

t2

1

t2+ 1

dt = B

A

dx1 + x2

t y = x +

1 + x2 dyy

=dx

1 + x2

Vy l =

t

1 +1

t2 2 ln1

t +

1 +1

t2

B

A(c) c : t (t, cosh t

a)


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c

(t) = (1, sinh ta

)

l =B

A|c(t)|dt =

B

A

1 + sinh2t

adt

=BA

cosh2 tadt =

BA

cosh ta dt=

BA

cosht

adt

= asinh ta

B

A(d)

c : t (a sin t, a(1 cos t)) a > 0= C(t) = (a cos t, a sin) a > 0

= l =B

A

|C(t)|dt =B

A

a2(cos2 t + sin2 t)dt

=

BA

a.dt = a(B A)

(e)

c : t a(lntant

2+ cos t), a sin t a > 0

= C(t) =

a(1

sin t sin t), a cos t

= |c(t)| = a2

cos4

sin2 t+ cos2 t = a| cos t|.

cos2

sin2 t+ 1

=a| cos t|| sin t| = a. cot t

=B

A

= |c(t)|dt =B

A

a cot t.dt = a.

BA

cos tsin t

dt

= a.

BA

d(sin t)

sin t= a lnsin t|BA

= a lnsin B

sin A

Bi tp 2.11.

(a) c(t) giao vi mt phng y = 0 khi 1 cos t = 0 t = k2, k ZChn k = 0, 1 ta c t1 = 0, t2 = 2.


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HNG DN GII BI TP 13

Ta c c

(t) =

a(1 cos t), a sin t, 2a sin t/2 |c(t)| =

a2(1 cos t)2 + a2 sin2 t + 4a2 sin2 t/2

= a

2 2cos t + 2(1 cos t)= 2a

1 cos t

l =20

|c(t)|dt = 2a20

1 cos tdt

= 2

2a

20

| sin t/2|dt = 8

2a.

(b) c : t (cos3 t, sin3 t, cos2t). D thy ng tham s cho c chu k 2.Ta c c

(t) = (3cos2 t sin t, 3sin2 t cos t, 2sin2t)

l =20

|c(t)|dt =20

9cos4 t sin2 t + 9 sin4 t cos2 t + 4 sin2 2tdt

=

20

25 cos2 t sin2 tdt =

5

4

20

sin2 2tdt

=5

2

20

sin2tdt

2

sin2tdt +

3

2

sin2tdt 2

3

2

sin2tdt

=

5

4

cos2t|

2

0 + cos 2t|2

cos2t|2 + cos 2t|232

= 10.

Bi tp 2.12. Ta c

x3

= 3a2

y2xz = a2

= y =

x3

3a2

z =a2

2x

Suy ra ng cong (c) c tham s ha l

c(t) =

t,

t3

3a2,

a2

2t

Khi y =

a

3 =t3

3a2 =

a

3 = t = a.Khi y = 9a = t

3

3a2= 9a = t = 3a.


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Vy di phn ng cong cn tm bng

l(c) =

3a

a

|c(t)|dt =3a

a

1 +

t4

a4+

a4

4a4dt =

3a

a

(2t4 + at)2

4a4t4dt

=

3aa

2a4 + a4

4a2t2dt =

3aa

t2

a+

a2

2t2

dt

=

t3

3a a

2

2t

3aa

= 9a.


15/50

HNG DN GII BI TP 15

Bi tp 2.13.

(a) Ly P nm trn ng xixoit. Ta c P 2at2

1 + t2 ,

2at3

1 + t3

Phng trnh ng thng OP : tx y = 0.Giao im B ca (OP) vi ng thng x = 2a c ta B(2a, 2at).

Giao im C ca (OP) vi ng trn (x a)2 + y2 = a2 c ta

C

2a

1 + t2,

2at

1 + t2

.

Ta c:

OP =

4a2t4 + 4a2t6

(1 + t2)2BC =

4a2t4 + 4a2t6

(1 + t2)2.

Vy OP = BC.

Phng trnh tham s ca ng trn (C) :

x = a cos t + a

y = a sin tt [0, 2]

C (C) : C(t) = (a cos t0 + a, sin t0).

Phng trnh (OC) :

x = t(a cos t0 + a)

y = t sin t0, t R

P (OC) = P(t1(cos t0 + a), t1 sin t)B = OC AV = B

2a,

2a sin t0cos t0 + 1

.

Suy raCB =

a a cos t0, a sin t

1 + cos t0(1 cos t0)

P (OC) = P(t1(cos t0 + 1), t1 sin t0).

Suy ra CB2 = 2a2(1 cos t0)2.1

1 + cos t0 , OP2 = 2t2(1 + cos t0).

Do CB2 = OP2 nn

2a2(1 cos t0)2. 11 + cos t0

= 2t2(1 + cos t0)

= t21 =a2(1 cos t0)2

(1 + cos t0)2

= t1 = a(1 cos t0)1 + cos t

0

= P =

a(1 cos t0), a sin t0.1 cos t01 + cos t0


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t t2 =1 + cos t

sin t0. Khi , ta c:

2at

2

21 + t22

, 2at

3

21 + t22

=

a(1 cos t0), a sin t0.1 cos t01 + cos t0 P

(b) Ta c

(t) = 4at

(1 + t2)2,

6at2 + 2at4

(1 + t2)2

= (t) = (0, 0).

Vy O(0, 0) l im k d duy nht ca ng xixoit.

(c) Chn M

2at2

1 + t20,

2at3

1 + t20

(t)

d(M, ) =

2at201 + t20

2a12 + 02 =

2a

1 + t20

Suy ra

limt0

d(M, ) = limt0

2a

1 + t20= 0

Ta c

limt0

(t) = limt0

4at

(1 + t2)2,

6at2 + 2at4

(1 + t2)2

= (0, 2a).

Vy, khi t th c(t) dn v ng thng x = 2a v (t) (2a, 0).

Bi tp 2.14.

(a) Ta c x(t) = sin t l hm s cp kh vi trn (0, ) v y(t) = cos t+ln(tant

2)

xc nh trn (0, ) v kh vi trn (0, ).

Do (t) kh vi trn (0, ) .


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HNG DN GII BI TP 17

Ta cng c (t) =

cos t, sin t + 1sin t

(t) = (0, 0) cos t = 0

sin t + 1sin t

= 0

cos t = 0

1 + sin2 t

sin t= 0

cos t = 0

cos2 t = 0

cos t = 0 t =

2+ k (k Z) ()

V t (0, ) nn ta c ti t = 2

th (t) = 0

Do (t) khng chnh quy ti t =

2.

(b) Ly M

sin t0, cos t0 + ln(tant02

)

(t)

Ta c

t(t0) =

|| =1

1 + 1sin2 t0

cos t0, sin t0 + 1

sin t0

Tip tuyn i qua M nhn t(t0) lm vector ch phng c phng trnh l

(d) :

x = sin t0 + t cos t0

y = cos t0 + ln(tant0

2) + (

sin t0 +

1

sin t0)t

, t R

Gi N = d Oy. Khi xN = 0 = yN = ln(tan t02

)

Suy ra N

0, ln(tant02

)

MN =

sin2 t0 + (ln tan

t02

cos t0 lntan t02

)2 =

sin2 t0 + cos2 t0 = 1.

Bi tp 2.15.

(a) Ta c

(t) =

3a 6at3

(1 + t3)2, 6at 33at

4

(a + t3)2

.

Ti t = 0, ta c (0) = (0, 0).


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(0) = (3a, 0) tip xc vi Ox.

(b) Ta c: limt

3at

1 + t3= lim

t

3a

t21

t3

+ 1= 0

limt

3at2

1 + t3= lim

t

3a

t1

t3+ 1

= 0 limt

(t) = (0, 0)

Tng t ta cng c

limt

3a 6at3(1 + t3)2

= limt

3a

t3 6a

1

t3

2

+ t3

2

2 = 0

limt

6at 6at4(1 + t3)

2 = limt

6at3

3a1

t2+ t2

2 = 0

limt

(t) = (0, 0).

(c) Tham s ha ca ng vi nh hng ngc li l

(t) = 3at

1 t3 ,3at2

1 t3

Khong cch t (t) n ng thng

d =

3at(1 + t3)

+3at2

(1 + t3)+ a

2

=3at + 3at2 + at3 + a

(1 + t3) 2

=

|a|(1 + t)

3

1 + t3

2=

|a| (1 + t)

2

1 t + t2

2

Do

limt1

d = limt1

|a|

(1 + t)2

1 t + t2

2

= 0

Ta c vector ch phng ca tip tuyn ti (t) l vector cng phng vi

vector

(t) l vector u = (3a 6at3, 6at 3at4)


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HNG DN GII BI TP 19

Khi limt1

u = limt1

(3a 6at3, 6at 3at4) = (9a, 9a). Vector (9a, 9a)cng phng vi vector (1, 1) cng l vector ch phng ca ng thng

(l) : x + y + a = 0.

Vy khi t 1. ng cong v tip tuyn ca n tin ti ng thng

x + y + a = 0

Bi tp 2.16.

(a) (t) = (aebt cos t, aebt sin t), t R, a > 0, b < 0Ta c

0 < aebt cos t < aebt

0, (t

)

= aebt cos t 0 khi t Tng t ta c

aebt sin t 0 khi t Vy (t) O(0, 0) khi t

(b) (t) = (abebt cos t

aebt sin t,abebt sin t + aebt cos t)

Ta c

limt

(abebt cos t aebt sin t) = limt

(abebt cos t) limt

(aebt sin t) = 0

Tng t

limt

abebt sin t + aebt cos t) = 0

Vy (t) (0, 0) khi t .Mt khc ta c

|(t)

|= aebt.

b2 + 1

= limt0

tt0

aebt.

b2 + 1dt = a

b2 + 1. limt0

ebt

t0

tt0

=a

b2 + 1

blimt0

(ebt ebt0)

= a

b2 + 1

b.ebt0

Vy limt

tt0

|(t)|dt l hu hn.


20/50


Bi tp 2.17. p dng nh l gi tr trung bnh cho cc hm x,y,z.

Bi tp 2.18.

(a) Ta c

(q p)v = ((b) (a))v

= (t)vba

=

ba

(t)v + (t)v dt

Do v l hng nn v = 0.

Suy ra (q p)v = ba

(t).v dt (1)

p dng bt ng thc Bunhiacopki ta c

|(t)v | |(t)|.|v | = |(t)|b

a

(t).v dt |b

a

(t).v dt| b

a

|(t).v |dt b

a

|(t)|dt (2)

T (1), (2), suy ra

(q p)v =b

a

(t).v dt b

a

|(t)|dt.(b) t v = (q p)|p q| , theo Cu (a) ta c

b

a

|(t)|dt (q p)v = (q p).q pp

q

=(q p)2p q = |p q|

=b

a

|(t)|dt |(b) (a)|.

Bi tp 2.19. Gi s = (x, y) vi || = 1. Khi ta c: x

(s) = cos (s)y

(s) = sin (s) x

(s) =

(s)sin (s)y

(s) =

(s)cos (s)


21/50

HNG DN GII BI TP 21

Theo gi thit:

k =

(s)cos2 (s) +

(s)sin2 (s) =

(s)

(s) = const(s) = cs + a

x

(s) = cos(cs + a)

y

(s) = sin(cs + a)

Vy, =1

ccos(cs + a) + c1,

1

csin(cs + a) + c2

.

Do c vt nm trn ng trn C(I, r) vi I(c1, c2), r =1

|c|=

1

|k|.

Bi tp 2.20.

(a) c

= (2t, 1, 3t2), c

= (2, 0, 6t), c

= (0, 0, 6).

Suy ra c c = (6t, 6t2, 2),

c2 = 4t2 + 1 + 9t4, |c| = 4t2 + 1 + 9t4

(c c)2 = 36t2 + 36t4 + 4, (c, c, c) = 12

t =

2t

4t2 + 1 + 9t4,

14t2 + 1 + 9t4

,3t2

4t2 + 1 + 9t4

b = 3t

9t4 + 9t2 + 1,

3t29t4 + 9t2 + 1

,1

9t4 + 9t2 + 1

n =1

9t4 + 9t2 + 1

4t2 + 1 + 9t4

9t4 1, t(2 + 9t2), 3t(1 + 2t2)k =

2

9t4 + 9t2 + 1

(4t2 + 1 + 9t4)3/2

=3

9t4 + 9t2 + 1(b) c(t) = a cosh t, a sinh t,at)

c

= (a sinh t, a cosh t, a)

c

= (a cosh t, a sinh t, 0)

c

= (a sinh t, a cosh t, 0)

c c = (a2 sinh t, a2 cosh t, a2)

c2 = 2a2 cosh2 t

|c| = a2 cosh t(c

c

)2 = 2a4 cosh2 t

|c c| = a22 cosh t(c

, c

, c

) = a3


22/50


t =

sinh t2cosh t

,1

2,

12 cosh t

b = sinh t

2 cosh t,

12

,1

2cosh t

n =

1cosh t

, 0, sinh tcosh t

k =

1

2cosh2 t

=1

2a cosh2 t(c) c(t) = (et, e(t),

2t)

c

= (et, e(t), 2)c

= (et, e(t), 0)

c

= (et, e(t), 0)c

c = (2e(t), 2et, 2)|c| = e

2t + 1

et

|c c| =

2(e2t + 1)

et

(c

, c

, c

) = 22t =

e2t

e2t + 1,

1e2t + 1

,

2et

e2t + 1

b = 1

e2t + 1,

e2t

e2t + 1,

2ete2t + 1

n =

2et

e2t + 1,

2et

e2t + 1,

1 e2te2t + 1

k =

2e2t

(e2t + 1)2

=2e2t

(e2t + 1)2

(d) c(t) = (cos3

t, sin3

t, cos2t)c

= (3cos2 t sin t, 3sin2 t cos t, 2sin2t)c

= (6sin2 t cos t 3cos3 t, 6cos2 y sin t 3sin3 t, 4cos2t)c

= (21 sin t cos2 t 6sin3 t, 21cos y sin2 t + 6 cos3 t, 8sin2t)c

c = (12 sin2 t cos3 t, 12sin3 t cos2 t, 9sin2 t cos2 t)|c| = 5

sin2 t cos2 t

|c c| = 15 sin2 t cos2 t

(c

, c

, c

) = 36 sin

3

t cos

3

tt =

3cos2 t sin t

sin2 t cos2 t,

3sin2 t cos tsin2 t cos2 t

,2sin2tsin2 t cos2 t


23/50

HNG DN GII BI TP 23

b =

4cos t

5,4sin t

5,35

n =

sin2 t cos tsin2 t cos2 t

,sin t cos2 tsin2 t cos2 t

, 0

k = 325

sin2 t cos2 t

=4

25sin t cos t(e) c(t) = (2t, ln t, t2)

c

= (2,1

t, 2t), c

= (0,1t2

, 0), c

= (0,2

t3, 0).

c c = (4

t, 4, 2

t2), |c| = 1 + 2t

2

t.

|c

c

| =2(1 + 2t2)

t2 , (c

, c

, c

) = 8

t3 .

t =

2t

1 + 2t2,

1

1 + 2t2,

2t2

1 + 2t2

b =

2t

1 + 2t2,

2t21 + 2t2

,1

1 + 2t2

n =

1 2t21 + 2t2

,2t

1 + 2t2,

2t

1 + 2t2

k =2t

(1 + 2t2)2, =

2t(1 + 2t2)2

.

Bi tp 2.21.

(a) Ta c

(s) =

a

csin

s

c,

a

ccos

s

c,

b

c

= |(s)| =

a2

c2

sin2

s

c+ cos2

s

c

+

b2

c2

=

a2 + b2

c2= 1.

Vy tham s s l di cung.

(b) Ta c

(s) =

a

c2cos

s

c, a

c2sin

s

c, 0

Hm cong k(s) = |(s)| =a2

c4 cos2

s

c +a2

c4 sins

c =a

c2

Tm hm xon


24/50


Ta c

n =(s)

|(s)| =c2

a

a

c2cos

s

c, a

c2sin

s

c, 0

t = (s) = (s) =a

csin s

c, a

ccos s

c, b

c

b = t n =

b

csin

s

c,

b

ccos

s

c,

a

c

= b = b

c2cos

s

c, b

c2cos

s

c, 0

.

Suy ra = bc2

(c) Mt phng mt tip ca (s) qua im

(s0) =

a cos

s0c

, a sins0c

, bs0c

v nhn

b(s0) =

b

csin

s0c

,b

ccos

s0c

,a

c

lm vector php tuyn nn n c phng trnh:

bc

sins0c

x a cos s0

c

+

b

ccos

s0c

y a sin s0

c

+

a

c

z bs0

c

= 0

bc

sins0c

.x +b

ccos

s0c

.y +a

c.z +

abs0c2

= 0.

(d) Ta c n(t) =

a coss

c a cos s

c.t, sin

s

c a sin s

c.t,

bs

c

t N = n(t)

Oz, suy ra N(0, 0,bs/c).

= cos(n(t), Oz) =0 a cos

s

c+ 0a sin s

c+ 1.0

|a| = 0.

gc gia n(s) v Oz bng /2.(e) Ta c

cos(t(s), Oz) =

0 ac

sins

c

+ 0

a cos

s

c

+ 1.

b

c

|1.1|

=b/c

= const.


25/50

HNG DN GII BI TP 25

Bi tp 2.22. Ta c:

c(t) =

a(1 cos t), a sin t, 2a sin t

2

=

a(2 2cos2 t2), 2a sin t2 cos t2 , 2a sin t2

c(t) =

2a sint

2cos

t

2, a(2cos2

t

2 1), a cos t

2

Khi ta c:

c(t) c(t) =

4a2 sin6t

2+ 4a2 sin4

t

2cos2

t

2+ 4a4 sin4

t

2

= 8a2 sin4 t

2= 2

2a sin2

t

2.

|c(t)|3 =

8a2 sin4t

2

3= 16

2a3 sin3

t

2.

Suy ra cong ca ng tham s trn l:

k(t) =c(t) c(t)

|c

(t)|3

=2

2a sin2t

2

162a3 sin3t

2

=1

8a sint

2

.

Khi bn knh cong ca ng cho bng

r(t) =1

k(t)= 8a sin

t

2

bn knh cong t cc tr a phng th

r(t) = 0 cos t2

= 0 t = + k2

Suy ra cc im lm cho bn knh cong t cc tr a phng ng vi

t = + k2, k Z.Bi tp 2.23. Gi s tham s l di cung v gi e l vector c nh. Theo

gi thit ta c:

e.t = 0, = e.t = 0, = k.(e.n) = 0


26/50


Do gi thit song chnh quy nn k = 0. T suy ra e.n = 0 (1), ly ohm hai v biu thc (1), ta c

e.n

= 0 =

e(

k.t + .b) = 0.

Do e.t = 0 nn e..b = 0.

Mt khc e.n = e.t = 0 nn e.b = 0. Suy ra = 0.Vy ng cong cho l mt ng cong phng.

Bi tp 2.24.

(a) Gi s c l tham s ha vi tham s di cung v gi l im c nh.

Theo gi thuyt ta c: (c(s)

a) = c

(s). T y suy ra:

c

(s) = c

(s) +

c

(s).

iu ny c ngha l: k =c

n

|c|2 = 0.Do vt l mt ng thng hoc mt phn ng thng.

(b) Do b = 0 nn = 0.

Bi tp 2.25.

(a) 120x 597y + 108z + 1752 = 0.(b) 61/16(x 25/8) 3(y 2) 5/4(z 9/4) = 0.


27/50

HNG DN GII BI TP 27

Bi tp 2.26.

(a) Ta c

c(t) = (a sin t, a cos t, b)c(t) = (a cos t, a sin t, 0)

= t = c(t)

|c(t)| =1

a2 + b2(a sin t, a cos t, b)

b =c c|c c| =

a

a

1 + b2(b sin t, b cos t, a)

=

c

c = (ab sin t,

ab cos t, a2)

n = b t = a2 b2

a2 + b2

1 + b2(cos t, sin t, 0).

Php tuyn n ca c(t) nhn n lm vector ch phng qua im c(t0) c

phng trnh tham s l

x = a cos t0 + cos t0t

y = a sin t0 + sin t0t

z = bt0

, t R

Tng t ta c phng trnh tham s ca cc ng thng sau

Tip tuyn (t) :

x = a cos t0 a sin t0ty = a sin t0 + a cos t0t

z = bt0 + bt

Trng php tuyn (v) :

x = a cos t0 + b sin t0t

y = a sin t0 b cos t0tz = bt0 + at

Mt phng mt tip qua c(t) nhn b(t0) lm vector php tuyn c phng

trnh l

b sin t0(x a cos t0) b cos t0(y a sin t0) + a(z bt0) = 0b sin t0x b cos t0y + az abt0 = 0.

Mt phng trc t nhn n(t0) lm vector ch phng v qua c(t) c phng

trnh l

cos t0x + sin t0y a2 = 0.


28/50


(b) Gi l gc to bi tip tuyn (t) v trc Oz

cos = |t.e3

||t|.|e3 | = |b

|1 = |b|Gi s M = n Oz th ta M l nghim ca h

a cos t0 + cos t0t = 0

a sin t0 + sin t0t = 0

bt0 = c

=

cos t0t = a cos t0

sin t0t = a sin t0()

H phng trnh () lun c nghim nn cc php tuyn ca c(t) lun cttrc Oz.

Bi tp 2.27. Xt nh x :< a, b >< 0, 1 >, t (t) = t ab a . D thy

l mt nh x vi phi. Khi ng cong = c 1 l ng cong tham stng ng cn tm.

Bi tp 2.28.

(a) Do f(t), g(t) l cc hm trn nn n kh vi v x(t) = t l hm s cp nnkh vi.

T ta c: c

(t) =

1, f

(t), g

(t) = 0 nn c l ng tham s chnh qui.

(b) Vi f(t) = sin t + t2 v g(t) = et(1 t3) th c(t) = t, sin t + t2, et(1 t3).T :

c

(t) = (1, cos t + 2t, et(1 t3) 3t2et)Suy ra vector tip xc cn tm l:

c

(t) =

1, cost + 2t, et(1 3t2 t3).Bi tp 2.29. chng minh iu kin cn, ly o hm ca ng thc

|(s)|2 = const 3 ln. chng minh iu kin , chng ta ly o hm

(s) = (s) + Rn RTb

chng ta c

(s) = t + R (kt b) + R

n (T R

)

b R

n

=

R + (T R)

b


29/50

HNG DN GII BI TP 29

Theo gi thuyt ta c R2 +

R2

T2 = const, ly o hm hai v ta c

2RR + 2 (T R) (T R)

= 0

=2R

R + (T R

)

= 0.

Do k, = 0 nn = 0. T suy ra iu phi chng minh.

Bi tp 2.30.

(a) t R =1

k, T =

1

. chng minh

c a = 1k

.n ( 1k

).1

.b = Rn RTb

ta chng minh c + Rn + RTb l hm hng.Tht vy, ta c:

(c + Rn + RTb) = c + Rn + Rn + (RT).b + (RT).b

= t + Rn + R(kt + b) + (RT)b + R 1

.(n)

= t + Rn 1k

t + Rb + (RT)b Rn

= Rb + (R

T)

b= (R + RT)b (1)

p dng Bi tp 2.29, do vt ca c(I) nm trn mt cu nn

R2 + (RT)2 = const. (2.0.1)

Ly o hm hai v ng thc 2.0.1, ta c

2RR

+ 2(R

T)(R

T)

= 0

R

R + T(RT)

= 0 (do = 0)

R + 1

(RT) = 0

R + RT = 0 (2)

T (1) & (2) suy ra:

(c + Rn + RTb) = 0

c + Rn + RTb = a (const)c a = Rn RTb.


30/50


(b) Chng minh tng t Bi tp 2.29.

Bi tp 2.31. Gi s c1 c2, tc l tn ti mt vi phi : (0, /2)

(0, 1),

sao cho c2(t) = (c1)(t), t (0, /2). Khi , ta c

cos t = (t)

sin t = 1 2(t)

cos t =

2(t)

sin t = 1 cos t

cos t = 2(t) (1)

sin t + cos(t) = 1. (2)

Do phng trnh (2) khng c nghim vi mi t (0, /2) nn c1 v c2 lkhng tng.

Chng minh tng t cho cc trng hp cn li.

Bi tp 2.32. Gi a l vector ch phng ca ng thng v (t) l gc gia

hai vector a v (t).

Gi s = const, tc l t.a = const, suy ra n.a = 0. Do

a = t cos + b. sin .

Ly o hm hai v ta c

k.n. cos + n. sin = 0

=k

= tan = const

Ngc li, nu k

= const = tan th chng minh c

a = t cos + b. sin .

Bi tp 2.33. Gi s s = 0, xt biu din ca ca trong ln cn ca s = 0.

Khi P phi c dng z = cy hay y = 0. Mt phng (P) : y = 0 khng tha

mn iu kin th 2 ca bi. By gi xt trong ln cn rt nh ca s sao cho

y(s) > 0 v z(s) cng du vi s. Theo iu kin th hai c = z/y va dng li

va m, do P l mt phng c phng trnh z = 0.


31/50

HNG DN GII BI TP 31

Bi tp 2.34.

(a) Xem Bi tp 2.32

(b) Theo chng minh ca Cu (a), vector a vung gc vi n nn php tuyn

song song vi mt phng nhn vector a lm php vector.

(c) Nu l gc gia a v tip tuyn th vector trng php tuyn b to vi

vector a mt gc /2 .

Bi tp 2.35. S dng c s a phng.

Bi tp 2.36. Ma trn ca mt php bin i ng c c nh thc bng 1v cng thc i bin ca tch phn bi.

Bi tp 2.37. Gi s l tham s ha vi tham s di cung v a R3 lim c nh.

Theo bi ta c

(t) a.(t) = 0

(t) a2

= 0

(t) a2 = r2Suy ra vt ca () l ng trn hoc mt phn ng trn tm a bn knh r.

Bi tp 2.38.

(a) ng cong phng.

(b) ng xon c.

(c) ng thng hoc mt phn ca ng thng.Bi tp 2.39. Cc tnh cht ny tng ng vi iu kin ng xon c

tng qut.

Bi tp 2.40. Tng t Bi tp 2.24.

Bi tp 2.41.

(a) ng tractrix c phng trnh tham s l

c(t) =

sin t, cos t + ln tan t

2


32/50


Ta c

x(t) = sin t = x(t) = cos t; x(t) = sin t

y(t) = cos t + ln tan

t

2

= y(t) = cos2 t

sin t; y

(t) = cos t cos tsin2(t)

.

Gi (t) = (X, Y) l ng tc b ca th c phng trnh tham s l

X = x x2 + y

2

xy xy .y

Y = y +x

2 + y2

x

y

x

y

.x

X =1

sin t

Y = ln

tan

t

2

(b) ng Hypebol c phng trnh tham s ha l:

c(t) :

x = a cosh t

y = b sinh t

x

(t) = a sinh t; x

(t) = a cosh t

y

(t) = b cosh t; y

(t) = b sinh t

Do c(t) c ng tc b c xc nh nh sau:

X = x x2 + y

2

xy xy .y

Y = y +x

2 + y2

x

y

x

y

.x

X = a cosh t.(1 + cosh 2t)

Y = b sinh t.

1 a

2

b2cosh2t

Nu l ng tc b ca c(t) th c(t) l ng thn khai ca (t).

(c) ng cycloid c(t) =

R(t sin t), R(1 cos t)x

(t) = R(1 cos t); x(t) = R sin ty

(t) = R sin t; y

(t) = R cos t)

l ng tc b ca c th c phng trnh tham s l

X = R(t + sin t)Y = R(1 cos t)Bi tp 2.42.


33/50

HNG DN GII BI TP 33

(a) Ta c

k =xy xy

(x2 + y2)3

2

=cosh t

1 + sinh2t3

2

=1

cosh2t.

(b) Theo cng thc xc nh tham s ha ca tc b

X = x x

2 + y2

xy

xyy

Y = y +x2 + y2

xy xyx

X =t

sinh t cosh t

Y =2 cosh t

Bi tp 2.43. Ellipsex2

a2+

y2

b2= 1 c tham s ha l

x(t) = a cos t

y(t) = b sin tt (0, 2).

Cc nh n ng vi cc gi tr t = /2, , 3/2. T , chng ta xc nh

c cc cong tng ng l b/a2, a/b2.

Bi tp 2.44. Vi ng cong c(t) =

(t), t(t), ta c

c

(t) =

(t), (t) + t

(t)

c

(t) = (t), 2(t) + t(t)ng cong c l mt cung thng khi v ch khi k = 0. iu ny tng ng

vi ng thc sau:

(t)(2

(t) + t

(t)) (t)((t) + t(t)) = 022(t) + t.(t).(t) (t).(t) t.(t).(t) = 0

2

2(t)

(t).(t) = 0

Nu tn ti t0 I sao cho (t0) = 0 th c khng chnh qui ti t0. Do o chng


34/50


ta c:

1

(t)

=

(t)

2(t)

=

(t)2(t) + 2(t)2(t)4(t)

=(t)(t) + 2(t)

3(t)= 0

1(t)

= at + b (t) = 1at + b

Vy vi (t) =1

at + bth c l mt cung thng.

Bi tp 2.45. Sinh vin t gii.

Bi tp 2.46. Tm biu thc ta ca i vi mc tiu {(t) : t(t),n(t),b(t)}.Bi tp 2.47. Gi s ng cong cho c tham s ha t nhin. Khi

x = cos v y = sin . S dng cng thc tnh cong i s thit lp

phng trnh vi phn thng theo .

Bi tp 2.48. S dng biu thc quan h gia h ta cc v h ta

Decarter:

x = ()cos y = ()sin

Thay cc ng thc trn vo cng thc tnh di v cong i s.

Bi tp 2.49. S dng bt ng thc ng cho l2 4A. Suy ra khng tnti ng cong.

Bi tp 2.50. i xng min cho qua AB v s dng kt qu bi ton ng

cho suy ra l chnh l na ng trn ng knh AB.

Bi tp 2.51. Tnh ton trc tip.

Bi tp 2.52. Do l mt ng cong n ng nn ta cl0

k(s)ds = (l) = (0) = 2.

Do k(s) < c, nn ta c

2 =

l0

k(s)ds l0

cds = cl


35/50

HNG DN GII BI TP 35

T suy ra l 2/c.

Bi tp 2.53. Theo nh l Jordan v ng cong th bao mt tp hp K.

Nu K l mt tp khng li th tn ti hai im p, q

K sao cho on thng

pq cha mt s im khng nm trong K. Khi on thng pq ct ti mt

im r = p, q. Chng minh rng qp l mt tip tuyn ca ti cc im p,q,r,t dn n mt iu mu thun.

Bi tp 2.54. Nu ng cong l lm th ly hai im p, q nm hai pha

ca phn lm . Di chuyn hai im p, q v pha lm sao cho pq tr thnh tip

tuyn ti p v q. Chng minh ng cong thu c gim chiu di tng din tch.

Bi tp 2.55.(a) Gi L l ng thng qua im q (I). Nu vt ca nm mt pha

ca L th ng thng L l mt tip tuyn. Nu vt ca n nm hai pha so vi

L th s ct L ti im th hai.

(b) Chng minh tng t nh chng minh hm Cauchy-Crofton.


36/50


BI TP CHNG 3

Bi tp 3.1. Xt f : R3 R, (x,y,z) x2 + y2. Khi , ta c

fx

(x,y,z) = 0

f

y(x,y,z) = 0

f

z(x,y,z) = 0

2x = 0

2y = 0

0 = 0

x = 0

y = 0.

Vy f c gi tr ti hn l duy nht l 0. Suy ra C l mt mt chnh qui.

C th chn h bn

f1 : (0, 2) R R3, (u, v) (cos u, sin u, v)f2 : (, ) R R3, (u, v) (sin u, cos u, v)

Khi h

(0, 2)R, f1

,

(, )R, f2

l mt h bn ph C. Lu

h ny khng duy nht.

Bi tp 3.2. Tp {(x,y,z) R3 : z = 0, x2 + y2 1} khng phi l mt chnhqui. Tp

{(x,y,z)

R

3 : z = 0, x2 + y2 < 1}

l mt chnh qui.

Bi tp 3.3. D thy (0, y , z) l cc im ti hn ca f v f1(0) l mt phng

Oyz nn n l mt chnh qui.

Bi tp 3.4. R rng X tha mn cc iu kin: kh vi, ng phi (do x > y).

Ta s chng minh X l n nh

Gi s X(u1, v1) = X(u2, v2). T biu thc ca X ta thy {u1, v1} v {u2, v2}l trng nhau v cng l nghim ca phng trnh bc hai X2(u+v)X+uv = 0.Nhng u1 > v1 v u2 > v2 nn ta c u1 = u2 v v1 = v2.

Vy X l n nh. Do X l mt tham s ha ca mt mt chnh qui.


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HNG DN GII BI TP 37

Bi tp 3.5.

(a) Ta c

f

x= 2(x + y + z 1) = 0

f

y= 2(x + y + z 1) = 0

f

z= 2(x + y + z 1) = 0

x + y + z 1 = 0

Vy cc im ti hn ca f l mt phng (P)x + y + z 1 = 0.

(b) Nu c = 0 th c l mt gi tr chnh qui ca f. Suy ra f(x,y,z) = c lmt mt chnh quy. Nu c = 0 th f(x,y,z) = c x + y + z 1 = 0.Do khi c = 0 th nhng im M(x,y,z) tha f(x,y,z) = c cng l 1 mt

chnh quy.

Tm li vi mi c R th f1(c) l mt mt chnh qui.(c) Ta c

fx

= yz2 = 0

fy = xz2 = 0fz

= 2xyz = 0

z = 0

x = y = 0

Vy tp hp tt c cc im ti hn nm trn mt phng z = 0 v ng

thng x = y = 0. T , suy ra f c gi ti hn duy nht l 0.

Nu c = 0 th tp hp cc im M(x,y,z) tha f(x,y,z) = c l mt mtchnh qui.

Nu c = 0 th tp cc imM(x,y,z) tha f(x,y,z) = 0 khng phi l mt

mt chnh qui (do ti O(0, 0, 0) khng trn).


Bi tp 3.7. Xt nh x X : V R2 R3, (u, v) (u,v, 0). Khi (V, X)l mt bn ca S. Do , tp cho l mt mt chnh qui.

Bi tp 3.8. Xt nh x f(x,y,z) = x2 y2 z, chng minh (0, 0, 0) l mtgi tr chnh qui. T suy ra S l mt mt chnh qui.

Xt X : R2 R3, (u, v)

u,v,u2 v2. Khi x,y,z l cc hm khvi. Mt khc, ta c Xu = (1, 0, 2u), Xv = (0, 1,2v) c lp tuyn tnh do


38/50


1 00 1 = 1 = 0. Do X l mt tham s ha.

(a) Ta thy

limuv

x (u, v) = , limu+v+

x (u, v) = +lim

uv

y (u, v) = , limu+v+

y (u, v) = +

limuv0

z (u, v) = +, limu0v+

z (u, v) =

(b) Tham s ny ph phn S {R3 : z > 0}.

Bi tp 3.9. X(u, v) = (a sinh u cos v, a sinh u sin v, c cosh u).

Bi tp 3.10. S khng phi l mt chnh qui do n c 1 ng thng k d.

Bi tp 3.11. Chng minh trc tip. Cc ng cong X(const, v) l giao ca

S vi mt phng z = c cos u.

Bi tp 3.12. X(u, v) = (0, 0, bu) +

(a,bu, 0) (0, 0, bu)v = av, buv, bu(1 v). (S) l mt mt chnh qui.

Bi tp 3.13.

(a) Tnh ton trc tip.

(b) Dng php chiu t cc bc v cc nam ln mt phng R2.

Bi tp 3.14.

(a) Chng minh tng t nh mt chnh qui.

(b) Tng t nh Cu (a)

(c) Khng chnh qui ti O.

Bi tp 3.15. D thy A2 = idS2 nn A = A1.

Bi tp 3.16. f(x, y) = x2 + y2. Khi f bin mt phng R2 thnh parabolid.

Bi tp 3.17. Xt f : (E) (S), (x,y,z) (x/a,y/b,z/c).

Bi tp 3.18. Chng minh theo nh ngha.

Bi tp 3.19. S dng tnh cht kh vi ca php i tham s.


39/50

HNG DN GII BI TP 39

Bi tp 3.20. Kim tra trc tip 3 iu kin phn x, i xng, bc cu.

Bi tp 3.21. Chng minh trc tip.

Bi tp 3.22. Nu p = (x,y,z) th F(p) nm trn giao ca H vi ng thngt (tx,ty,z), t > 0. Do

F(p) = 1 + z2

x2 + y2x,

1 + z2x2 + y2

y, z

Chn U l ton b tr i trc Oz. Khi hm F : U R3 xc nh nh trnl mt hm kh vi.

Bi tp 3.23. ng cong C ch c cc im k d nm trn trc quay.

Bi tp 3.24. Chng minh trc tip bng nh ngha.


Bi tp 3.26. S dng nh ngha ca hm kh vi trn R3 v trn mt chnh

qui chng minh.

Nu f l thu hp ca mt nh x kh vi th f kh vi (v d). chng minh

iu ngc li, chng ta xt X : U R3 l mt tham s ha ca S ti p.Chng ta c th m rng X thnh nh x F : U R R3. Ly W l mtln cn ca p trong R3 sao cho F1 l mt vi phi trn n. nh ngha hm

g : W R c xc nh bi g = fX F1 trn W, vi l php chiut nhin t U R ln U. Khi , g l mt nh x kh vi v g|WS = f.


Bi tp 3.28. nh x F kh vi trn S2 \ {N}, do n l hp ca cc nh xkh vi. chng minh F kh vi ti N, xt php chiu ni t cc nam v t

Q = S F 1S . Chng minh rng N 1S () = 4/. T ta c

Q() =n

a0 + a1 + + ann .

Do , Q kh vi ti 0. Suy ra F = 1S F S kh vi ti S.

Bi tp 3.29. Vi mi v TpS, ta c v = (t) = (x(t), y(t), z(t)) vi(t) = (x(t), y(t), z(t)) l mt ng cong nm trn S v (0) = p.


40/50


Do (t) nm trn S nn f((t)) = 0. Tc l ta c:

f(x(t), y(t), z(t)) = 0,t I=

fx

(p)x(0) + fy

(p)y(0) + fz

(p)z(0) = 0

=n(fx(p), fy(p), fz(p))v=(T pS) : fx(p)(x x0) + fy(p)(y y0) + fz(p)(z z0) = 0.

Bi tp 3.30. Phng trnh mt phng tip xc ti (a,b, 0) ca mt phng

chnh qui cho bi phng trnh f(x,y,z) = x2 + y2 z2 1 = c dng

TpS :f

x(p)(x a) + fy(p)(y b) + fz(p)(z 0) = 0

2a(x a) + 2b(y b) = 02ax + 2bx 2a2 2b2 = 0.

Php vector ca (T pS) l n = (2a, 2b, 0) vung gc vi e3. Suy ra mt phng

(TpS) vung gc vi trc Oz.

Bi tp 3.31.

C th xem S l F1(0) vi F(x,y,z) = z f(x, y) hoc gii theo cch sau.

(a) Tham s ha ca S l X(u, v) = (u,v,f (u, v)). Khi chng ta c

Xu = (1, 0, fu), Xv = (0, 1, fv)

Suy ra n = (fu,fv, 1).Phng trnh mt phng tip xc ca S ti p = (x0, y0, f(x0, y0)) vi c dng

fu(p)(x x0) + fv(p)(y y0) + (z f(x0, y0)) = 0

z = fx(x0, y0)(x

x0) + fy(x0, y0)(y

y0) + f(x0, y0))

(b) Ta c F = Dfq(x, y) =f

x(q)x +

f

y(q)y. Suy ra

Gr(F) =

x, y,f

x(q)x +

f

y(q)y

: (x, y) R2

T suy ra iu phi chng minh.

Bi tp 3.32. p dng Bi tp 3.31.

Bi tp 3.33. Ta c X(u, v) = (u) + (v)

ng ta th I: X(t, c), t I.


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HNG DN GII BI TP 41

t p = X(u, c). Khi ta c

Xu = (u), Xv =

(v)

=N(p) =

(u)

(c)

Suy ra cc mt phng tip xc ca (S) dc theo ng ta th I song

song vi cc ng thng c vector ch phng l (c).

Trng hp th hai chng minh tng t.

Bi tp 3.34. t p1 = X(u0, v1); p2 = X(u0, v2)

Ta s chng minh TP1S = TP2S. Tht vy

Xu =

(u) + v

(u), Xv =

(u)

= Xu Xv = ((u) + v(u)) (u)= (u) (u) + v(u) (u)= v(u) (u).

Suy ra

Tp1S : v1.((u0)

(u0)) ((u) + v

(u)

(u0)

v1

(u0)) = 0

Tnh ton tng t, chng ta c

Tp2S : v2

(u0) (u0)

((u) + v(u) (u0) v2(u0)) = 0.

Mt khc, ta c

v1.((u0) (u0)) [(u0) + v2(u0) (u0) v1(u0)] =

= v1.((u0)

(u0)).(v2 v1).

(u0)= v1.(v2 v1) [((u0) (u0)).(u0)]= v1.(v2 v1) ((u0), (u0), (u0)) = 0

Suy ra p2 l mt im ca Tp1S. T ta c iu phi chng minh.

Bi tp 3.35. Ly v TPS, (t) =

x(t), y(t), z(t), (0) = p, (0) = v.

Ta c: f (t) = ((t) p0)2.

Do

Dfp(v) =d

dt[f (t)]t=0

= 2(0)((0) p0) = 2v(p p0), v TPS.


42/50



Bi tp 3.37. Tnh ton trn Maple

with(LinearAlgebra);

X := [v*cos(u), v*sin(u), a*u];

XU := convert(diff(X, u), Vector);

XV := convert(diff(X, v), Vector);

N := simplify(&x(XU, XV));

X := convert([x, y, z]-X, Vector);assuming([simplify(X.N)],

[x > 0, y > 0, z > 0, 0 < u and u < 2*Pi, a > 0]);

Mt phng tip xc cn tm c phng trnh

a sin(u0) x + a cos(u0) y + vau vz = 0

Bi tp 3.38. Ta c

Xs = (s) + r(n(s)cos v + b(s)sin v)

Xv = r(n(s)sin v + b(s)cos v)= N(s, v) = Xs Xv

= ((s) + r(n(s)cos v + b(s)sin v) r(n(s)sin v + b(s)cos v)

= [T + r(

k.T+) cos v + (

.N)sin v][r(

n sin v + b cos v)]

= T rNsin v + T B cos v + r(kT cos vNsin v) + r(k.TB)cos v+ r(.B cos vN. sin v) + r(.BB)cos v r(NN)sin v + r(N. sin vB. cos v)

S dng cng thc Frenet ta suy ra iu phi chng minh.

Bi tp 3.39. Ta c

Xu = (f

(u)cos v, f

(u)sin v, g

(u));Xv = (f(u)sin v, f(u)cos v, 0);


43/50

HNG DN GII BI TP 43

Suy ra phng trnh php tuyn ca S l

f

(u0)cos v0(x x0) + f(u0)sin v0(y y0) + g(u0)(z z0) = 0

f(u0)sin v0(x x0) + f(u0)cos v0(y y0) = 0

f

(u0)cos v0(x x0) + f(u0)sin v0(y y0) + g(u0)(z z0) = 0(f(u0)sin v0)x + (f(u0)cos v0)y = 0

D thy h phng trnh sau c nghim

f

(u0)cos v0(x x0) + f(u0)sin v0(y y0) + g(u0)(z z0) = 0

(f(u0)sin v0)x + (f(u0)cos v0)y = 0x = 0, y = 0

Suy ra php tuyn ca S lun ct trc Oz.

Bi tp 3.40. Nhng im p = (x0, y0, z0) thuc giao tuyn ca hai mt S1 v

S2 c tnh cht ax0 = by0.

Php vector ca S1 ti im p l n1 = (2x0 a, 2y0, 2z0); cn php vector

ca S2 ti im p l n2 = (2x0, 2y0 b, 2z0). T suy ra n1 n2.Bi tp 3.41.

(a) Ly (t) l ng cong nm trn mt S sao cho (0) = p v (0) = v.

Khi , ta c

Dfp(w) =w, (t) (0)|(t) (0)| = 0

Suy ra p l mt im ti hn ca f khi v ch khi Dfp(w) = 0. Ta suy rac iu phi chng minh.

(b) Tng t cu (a).

Bi tp 3.42.

(a) S dng tnh cht lin tc ca hm f, chng minh trong mi khong

(, c), (c, b), (b, a) cha mt nghim ca f.(b) iu kin cn v hai mt f(t1) = 1 v f(t2) = 1 trc giao vi nhau:

fx(t1)fx(t2) + fy(t1)fy(t2) + fz(t1)fz(t2) = 0


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S dng nh ngha ca hm f v cc ti suy ra iu phi chng minh.

Bi tp 3.43. Chng minh rng d(X(u, v), I) = const t gi thuyt

(X I).Xu = (X I).Xv = 0T suy ra im c nh I chnh l tm ca mt cu.

Bi tp 3.44. Mi ln cn a phng ca mt mt chnh qui l nghch nh

ca gi tr chnh qui ca mt hm kh vi. Do ta c th gi s S1 = f1(0)

v S2 = g1(0), vi 0 l gi tr chnh qui ca cc hm f v g. Trong ln cn ca

p, S1 S2 l nghich nh ca hm F : R3 R2, q (f(q), g(q)). Do S1 v S2

c giao ngang nhau nn hai php vector (fx, fy, fz) v (gx, gy, gz) c lp tuyntnh. Do (0, 0) l gi tr chnh qui ca hm F v S1 S2 l mt ng congchnh qui.

Bi tp 3.45. Chng minh bng nh ngha.

Bi tp 3.46. S dng X(u, v) = X

u(u, v), v(u, v)

v cng thc o hm

ca hm hp.

Bi tp 3.47. Chng minh rng cc ng cong trn S ct mt phng (P) ti1 im hoc nm trn mt phng (P). T suy ra (P) l mt mt phng tip

xc ca S.

Bi tp 3.48. Phng trnh mt phng tip xc ti im (x0, y0, z0) c dng

xx0a2

+yy0b2

+zz0c2

= 1

ng thng (d) vung gc vi mt phng tip xc ca c phng trnh

xa2

x0=

yb2

y0=

zc2

z0

T biu thc trn, chng ta c

x2a2

xx0+

y2b2

yy0+

z2c2

zz0=

x2a2 + y2b2 + z2c2

xx0 + yy0 + zz0

Tng t ta c

xx0x20/a

2=

yy0y20/a

2=

zz0z20 /a

2=

xx0 + yy0 + zz01

.


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HNG DN GII BI TP 45

T suy ra iu phi chng minh.

Bi tp 3.49. Tng t nh chng minh ca hm nhiu bin.

Bi tp 3.50. Gi r l ng thng c nh v p l mt im nm trn S.Mt phng P1 cha im p v ng thng r, cha tt c cc php tuyn ti

cc im S P1. Xt mt phng P2 qua im p v trc giao vi r. Do phptuyn i qua p ct r nn P2 c lp vi TpS. T suy ra P2 S = C l mtng cong phng chnh qui. Hn na P1 P2 trc giao vi ATpS P2; Do P1 P2 trc giao vi C. T suy ra cc php tuyn ca C i qua mt im cnh q = rP2. S dng tnh cht lin thng ca Ssuy ra iu phi chng minh.


Bi tp 3.52. Gi v l vector tip xc ca C1 v C2 ti p, chng minh rng

(C1) v (C2) c chung vector ch phng l(v).

Bi tp 3.53. Chn p l gc mc tiu, Xu, Xv l trc honh v trc tung,

N = Xu Xv l trc cao.

Bi tp 3.54.

(a) Cho q R, gi (U, ) l mt h ta a phng ca S ti p = 1(q).Nu q l mt gi tr chnh qui ca hm 1 th q c gi l gi tr chnhqui ca hm .

(b) Suy ra trc tip t nh ngha gi tr chnh qui v ng cong chnh qui.

Bi tp 3.55. Lnh tnh ton vi Maple

[>restart;

with(linalg);

X := [a*sin(u)*cos(v), b*sin(u)*sin(v), c*cos(u)];

XU := diff(X, u); 1; XV := diff(X, v);

dk := a > 0, b > 0, c > 0, 0 < u and u < 2*Pi,

0 < v and v < Pi;

E = simplify((convert(XU, Vector).convert(XU, Vector)))assuming dk;

F = simplify((convert(XU, Vector).convert(XV, Vector)))assuming dk;

G = simplify((convert(XV, Vector).convert(XV, Vector)))assuming dk;


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(a)

E = a2 (cos(u))2 (cos(v))2 + b2 (cos (u))2 b2(cos(u))

2(cos(v))

2+ c2 c2 (cos(u))2

F = cos(u)cos(v)sin(u)sin(v) a2 b2

G = (sin(u))2 (cos (v))2 a2 b2 (cos(v))2 a2

(b) E = 4 u2 + a2

F = 0

G = a2u2

(c) E = a2 (cosh (v))2 + 4 u2 b2 + b2 (cosh(v))2

F = cosh (v) u sinh(v)

a2 + b2

G = u2

b2

(cosh (v))

2

a2

+ a2

(cosh (v))

2(d) E = a2 (cosh(v))2 + 4 u2 b2 + b2 (cosh (v))2

F = cosh (v) u sinh(v)

a2 + b2

G = u2

b2 (cosh (v))2 a2 + a2 (cosh (v))2

Bi tp 3.56. Cc h s ca dng c bn th nht

E = 16/(u2 + v2 + 4)2

F = 0

G = 16/(u2 + v2 + 4)2

Bi tp 3.57. Ly hai ng cong 1 = X(ui, v), i = 1, 2, ng cong

= X(u, v0). Xc nh giao im v ta c di ca on chn bng

2|u2u1|.

Bi tp 3.58. p dng cng thc tnh din tch ca mt tham s chnh qui

theo cc h s ca dng c bn th nht.

Bi tp 3.59. l cc mt trn xoay c ng sinh l cc ng tham s

ha di cung.

Bi tp 3.60. I = d2 + 2d2.

Bi tp 3.61.

(a) X(u, v) = (3u sin v, 3u cos v, 3u2), u R, v (0, 2)(b) N =

6u2 sin(v),6u2 cos(v),u


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HNG DN GII BI TP 47

(c) Cc h s dng c bn th nht v th hai

E = 36 u2 + 1, F = 0, G = u2

e = 6/36u2

+ 1, f = 0, g = 6u2

/36u2

+ 1.

(d) cong Gauss v cong trung bnh

K =36

(36u2 + 1)2, H =

108u2

(36u2 + 1)(3/2).

Bi tp 3.62. Cc im paraboloic nm trn ng trn X(/2, v) v X(3/2, v),

cc im eliptic nm trn phn X(u, v), u (0, /2) (3/2, 2), phn cn licha cc im hyperboloic.

Bi tp 3.63.

(a) K =1

(1 + 2u2)2v H =

1 + u2(1 + 2u2)(3/2)

.

(b) H(0, 0) = 1 v K(0, 0) = 1. Suy ra k1 = k2.

Bi tp 3.64.(a) Tm tham s ha ca mt trn xoay v p dng cng thc tnh din tch

theo cc h s ca dng c bn th nht.

(b) A = 42Ra.

Bi tp 3.65. S dng cng thc tnh din tch.



Bi tp 3.68. Chng minh hm N lin tc.

Bi tp 3.69. Ly v d l Mobius.

Bi tp 3.70. nh hng trn S1 c cm sinh t 1.

Bi tp 3.71. Tng t bi 3.70.

Bi tp 3.72. S dng bi 3.71.


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Bi tp 3.73. Gi S l mt mt chnh qui tip xc vi mt mt phng dc

theo ng cong . N l php vector ca mt phng dc theo ng cong .

Suy ra N = const. Ta c:

= a.Xu + b.Xv

= u(t).Xu + v(t).Xv

= DN((t)) = (aNu + bNv).(t)= u(t)Nu + v

(t).Nv

= N((t)) = N(t)

= DN((t)) = N(t) = 0,(t) S.

Suy ra (t) ng vi gi tr ring = 0.

T suy ra (t) l im paraboloic hoc l im phng.

Bi tp 3.74. Xem chng minh cng thc Euler.

Bi tp 3.75. Khng ng, v d mt cu.


Bi tp 3.77. S dng nh ngha cong php dng.

Bi tp 3.78. S dng cng thc tnh cong chnh v phng chnh theo cc

h s c bn, cong Gauss v cong trung bnh.

Bi tp 3.79.

(a) Na mt cu di khng k ng xch o.

(b) Mt cu tr i cc Bc v cc Nam.

(c) Mt cu tr i cc Bc v cc Nam.


Bi tp 3.81. Chng minh vector trng php tuyn ca C l hng.

Bi tp 3.82. Xem ngha ca ch Dupin.

Bi tp 3.83. Xt module

|1N2

2N1

|v s dng kt qu

|sin

|=

|N1

N2

|suy ra iu phi chng minh.


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HNG DN GII BI TP 49


Bi tp 3.85. Tm phng chnh ca ti cc im nm trn ng trn trung

tm ca xuyn.


Bi tp 3.87. Tham s ha l X(u, v) = (u,v,auv) v dng cng thc tnh

cc cong theo h s ca dng c bn.

Bi tp 3.88. Tham s ha l X(u, v) = (u,v,uv) v s dng cng thc xc

nh cc ng tim cn v ng chnh khc.

Bi tp 3.89. ng tim cn u = const v v = const.ng chnh ln(v +

v2 + c2) u = const.

Bi tp 3.90. u v = const.

Bi tp 3.91. Tnh ton trc tip


Bi tp 3.93. Chng minh theo nh ngha.

Bi tp 3.94. Cc ng sinh v ng trn trc giao vi trc quay.

Bi tp 3.95. Ly mt mt cu cha mt mt (S) pha trong, gim bn knh

ca mt cu mt cch lin tc, xt cc lt ct chun tc ti cc giao im ca

mt cu v mt (S).


Bi tp 3.97. Khng c im rn nu a = b = c.Bi tp 3.98. Chng minh trc tip.

Bi tp 3.99. S dng nh ngha ca mt k, ng tht v tham s phn b.

Bi tp 3.100. S dng nh ngha ca ng tht.

Bi tp 3.101. S dng nh ngha ng cong chnh.



50/50



Bi tp 3.104. Mt phng l mt mt cc tiu nhng khng b chn. Do

n khng compact.


Bi tp 3.106. S dng gi thuyt cong trung bnh ca S, S bng khng

chng minh cong trung bnh ca mt cho l mt mt cc tiu.

huong dan giai bai tap_hinh hoc vi phan

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