Soal Tentang Matematika

Mama ^:

Nadira 141-1-199

KAIITUIUSZ -02

NPM ^:

2206058053 Uggs Temu 8

3 1) ^let

fix

,y

)

^{= ✗}

2y

⁺

^Fy

dit

^:

Find

each value a)

f- 1211 ) b) f- 13107

Jawab Jawab^:

a)

f- (2/1)=221

⁺

VT

⁼

4-11--5 d)

K= -2 ⁼ ✗²

b) f- 13101

⁼

320

⁺

ro

⁼

is

• K=-4 → -4 ⁼

2) 1¥

y

⁼

-4×2

Jdwab

a)

f- (

^×,

y )

^{= 6-✗}

-24

^{• K = -I -p} y ^{= -✗2}

* _mlsal 2-⁼ FIX>Y ) ^{0 = 0}

2- ⁼ 6-^✗-24 ^{• IT = O -D}

bidang ^✗y ^{→ 2-=0}

bldang

^{✗2- →}9=0

bldangyz

^{-☐ ✗=O • IT = I -D I = ✗2} 0=6 ^- ✗^-zy 2- ⁼ 6-✗ 2- ⁼ 6- 24

Ts

✗ = 6-24 ^{• ✗ --O -D} 2-⁼ 6 ^• 2=0 ^-☐ 9=3

Y=

✗²

• ✗=O ^→

y

^{=3 • 2=0}

-☐✗=6 ^•

9=0-177--6

^•

K=4

^-☐

4

^{= ✗2}

-y

•

9=0

^-☐ ✗⁼ 6

2-

^

4y=X

^'

: kurva ^•

y

⁼ ✗^L

@ ^-

it 4

•6 •

y=✗2

=D

^{- •>} y ^{. •}

^^

. .

9=1-4×2

;

• ^{> y}

% %

^{• •}

/

.

L ✗ _>

☒

✗

0 a • •

b)

^f(×>

4)

⁼ 16 ^-✗^2-

yz

* mlsal 2- = f-(Z,y

)

2- ^{= 16 -✗2-}YZ

, ,^{, o o • ◦}

2-2=16^-✗2-

y2 9=-112

_y=

-1-4×2

✗

2+42+-22=16

bidding

^✗y^-D-2=0 bldang ✗2-^→y=0 bidang _yz^-D ✗⁼ 0

✗

2+92=16

✗2+2-2--16

YZ

^-1-22=16

b)

^K= _2- = ✗^<+

y

-✗=O → y⁼-14 ^{11=0 -D 2-=-1-4 9=0 → 2-=-1-4 •} k= ^-4 ^-D Y ^{= -✗2-}

4

•

y=0

^-☐✗ ⁼ -14 ^7--0-1>✗= -1-4 2-=0 ^-by⁼-1-4 • K^{= -}I ^- D

Y

^{= -✗2- I}

4)

nz ^{n k = O -D} Y ^{= -✗£}

y=^-✗

2-4

• IT ⁼ I ^-☐

Y

^{= I-✗2}

" "

^""

" _"

"

⁼

"

^"

/

' ^{¥ "}

^{" 9k}

. . . ^.

4

4

E-

^{" "}

war

4-

4

✗

<

9=4

^-✗²

•

dz _dy (3×2+4.2)

^-"

3=-1313×2+925

^"

^{? 29}

=

-24

'

33 _(3×2+42)

^" _A

Jawab^:

④ FIX

^,y) :(

2×-414

(=) ^•

8¥ ^/ 2×-414--412×-913.2

⁼

^{812×-9 )} % dlk

:

362=4×2+942

Plane ^{✗ =3 at} ( 31212

)

•

J

^{-2 (2x-}

g) 4=412×-413

^{. -I =}

^-412×-4131

,

dit

^:

_Find

_{the slope} OFthe

tangent

^to

dy

^curve OF intersection OF the 362- ⁼

② ^FIX

^{>g) =}

(4×-42)%2 _4×2+942

= . ..

?

⇐

) ^•

d¥ 14×-9213/2=3,2 ^(4×-92)

^"

^{? 4=6 4×-92}

_, ^Jawab:

•

362=4×2+942

^_☐sybtltusi ^✗=3

•

07

(

4×-421312=3/2

^{(4×-92)"≥ .}

-29

^=-

_{3y 4×-92} 362=41312-1942

Oy

ⁱ

2- ⁼

3%+93-692

④ f(

×

,y )

⁼ ✗^2-

YZ

µ

^(a-f)

'

=a-f^'

z ⁼ I ^-

y2

✗

Y

U⁼ ✗

2-92-1>41--211 4-

•

8¥ (

^✗2✗-

⁴

^Y"

⁾

⁼

^{# ¥-2 (}

^✗

^2×-4

^{=✗ → ✓'=L •}₍

₌₎ ^find

_{-2 =} ^the

₁₊₄₂

^{slope at the point (}

^{31212 )}

:( ty ⁾ /

^{2X-✗ -✗}

1.1×2-441

²

4-

01--12

= ✗

2+4.2 dy

✗

24

_µ

U'

⁼

-24

•

¥ ^{(+2×-12)=1 ;) ! :(}

^✗

^2-92

^Y

^)→v

^{' -- I}

^d± _OU

_{( 3,212)} ⁼

^2-

^{2 =}

^↳

=

( ^{t )} (-29.9-11×2-94) ^y2

^{i. the slope OF} the

tangent

^{to the curve}

= -y^{2. ✗2}

OF

Interaction _of the given survace and the

✗y2 4

Plane at the Point

1312,2 )

^is

11

4

FIX

,g) ⁼ @^✗ cosy (a.f)^'=a.f^'

•

¥gle×

^cosy

⁾

^{= cosy}

.yz(

^ex

⁾

^{= cosy .@× ,}

•

d

-2

(

^ex

cosy )

^{= ex .}

dg±y

^(cosy

⁾

^{= e.×. -}

^{sing ,}

Oy

⑤ ^FIX

^{>g) =}

^EY

^{sin ×}

(a.f)^'=a.f^'

•

dz ₍

e.YSMX

)

=

EY

^.

dt (

^sine

)

^{= EY . cos✗ 11}

Ox dx

•

dz ₍

@YSMX

)

⁼

sina.dz ( EY )

⁼ sin^×. EY

d- _y dy µ

6

f- (

×,y

)

⁼

(3×2+92)

^{_' '3}

•

dz _dx (13×2+92)

^_"

3)

⁼

-1-3 3×2+925%3

^{. 6✗}

= ^- 6x

3

(3×2-142)<113

= ^-2X

3

(3×2+92)

^" ₄