CHAPTER 4 (SECTION 4)
8.
About how much (in percent) does an error of 1% in a and b affect a 2b3 ? Jawab : Dengan
dan
a2b3 = 2 ln a + 3 ln b =2
+ 3
= 2 (0,01) + 3 (0,01) = 0,02 + 0,03 = 0,05 = 5%
9.
Show that the approximate relatif error
() ) of a product f = gh gh is the sum of the
approximate relative errors of the factors.
(
+ )
Jawab : f = gh ln f ln f = ln g ln g..h [ln f [ln f = ln g ln g + ln h]
∫ = ∫
+
∫
+
10. A force of 500 nt is measured with a possible error of 1 nt. Its component in a direction 60˚ away from its line of action is required, where the angle is subject to an error of 0,5˚. What is (approximately) the largest possible error in the component?
11. Assuming that your calculator isn’t handy, show how to make a quick estimate (to two decimal places) of () () . Hint : Consider f (x, y) = . Jawab :
(4,98)2 mendekati (5)2 Maka : f () = 2
f ‘() = , Dengan = 5 dan
= (4,98 – 5) = - 0,02
= (3,03 – 3) = 0,03
Sehingga : (4,98)2 = f ( + )
f ( ) + f ‘()
= 2 +
= (5)2 + 2(5)(- 0,02) = 25 – 0,2 = 24,8
(3,03)2 mendekati (3)2 Maka : f () = 2
f ‘() = , Dengan = 3 dan
Sehingga : (3,03)2 = f ( + )
f ( ) + f ‘()
= 2 +
= (3)2 + 2(3)(0,03) = 9 – 0,18 = 9,18 Sehingga () () = () = Pendekatan nilai adalah . . .
mendekati √ f () = √
f ‘( ) =
, Dengan = 16 dan √
dimana (- 0,38) (- 0,4)
Sehingga : (15,62)2 = f ( + )
f ( ) + f ‘()
= (15,62 – 16) = - 0,38
= √ +
√
= √ +
(-0,4) √
= 4 – 0,05 = 3,95
12. As in problem 11, estimate
() () .
Jawab :
(2,05)2 mendekati (2)2 Maka : f () = 2
f ‘() = , Dengan = 2 dan
= (2,05 – 2) = 0,05
= (1,98 – 2) = - 0,02
Sehingga : (2,05)2 = f ( + )
f ( ) + f ‘()
= 2 +
= (2)2 + 2(2)(0,05) = 4 – 0,2 = 4,2
(1,98)2 mendekati (2)2 Maka : f () = 2
f ‘() = , Dengan = 2 dan
Sehingga : (1,98)2 = f ( + )
f ( ) + f ‘()
= 2 +
= (2)2 + 2(2)(-0,02) = 4 – 0,02 = 3,92 Sehingga
() () = √ = √
Pendekatan nilai
√ adalah . . .
√ mendekati √
f () = √ Sehingga :
f ‘( ) = , Dengan = 8 dan √
= (8,12 – 8) = 0,12
√ = f ( + ) f ( ) + f ‘()
= =
√ + √
√ +
(0,12) √
= 2 + 0,01 = 2,01
13. Without using a calculator, estimate the change in length of a space diagonal of a box whose dimensions are changed from 200 x 200 x 100 to 201 x 202 x 99 Jawab : Dimensions of a box = 200 x 200 x 100 or 2 x 2 x 1 Then, dimensions of a box change to 201 x 202 x 99 or 2,01 x 2,02 x 0,99 Length of a space diagonal of a box can write : dr = Maka : Panjang diagonal ruang dengan dimensi = 2 x 2 x 1 dr = = √ = √ =3
Setelah dimensi dari balok dirubah menjadi 2,01 x 2,02 x 0,99. Maka :
p = 2,01; sehingga pendekatan nilai pendekatan (2,01)2 adalah … (2,01)2 mendekati (2)2
Maka : f () = 2
f ‘() = , Dengan = 2 dan
= (2,01 – 2) = 0,01
Sehingga : (2,05)2 = f ( + ) f
( ) + f ‘()
= 2 +
= (2)2 + 2(2)(0,01) = 4 + 0,04 = 4,04
l = 2,02; sehingga pendekatan nilai pendekatan (2,02)2 adalah … (2,02)2 mendekati (2)2 Maka : f () = 2
f ‘() = , Dengan = 2 dan
= (2,02 – 2) = 0,02
Sehingga : (2,02)2 = f ( + ) f
( ) + f ‘()
= 2 +
= (2)2 + 2(2)(0,02) = 4 + 0,08 = 4,08
t = 0,99; sehingga pendekatan nilai pendekatan (0,99)2 adalah … (0,99)2 mendekati (1)2 Maka : f () = 2
f ‘() = , Dengan = 1 dan
Sehingga : (0,99)2 = f ( + )
f ( ) + f ‘()
= 2 +
= (1)2 + 2(1)(-0,01) = 1- 0,02 = 0,98
= (0,99 – 1) = - 0,01
Perubahan panjang diagonal ruang pada balok : dr =
= √ = √ Pendekatan nilai √ adalah . . .
√ mendekati √ f () =√
f ‘( ) =
, Dengan = 9 dan √
= (9,1 – 9) = 0,1
Sehingga : √ = f ( + ) f
( ) + f ‘( )
= √ + (0,1) √ = √ + (0,1) √
= 3,0167
14. Estimate the change in
f( ) = ∫ dt if change from 0,7 to 0,71
Jawab :
Ketika diganti dari 0,7 ke 0,71 ; = . Then df = ( ) d
= 0,7 ; d = (0,71 – 0,7) = 0,01 df =
() (0,01) ()
(0,01) () df = () df =
df =
15. For an ideal gas law of N molecules, the number of molecules with speeds ≤ is given by
dt () = √ ∫ where is a constant and is the total number of molecules. If , estimate the number of molecules with speeds between = and .
Jawab :
; = Ketika diganti dari sampai
= , d =
=
2 () dn = ( ) dn =
seingga :
() =
dn √
=
√
=
√
= = 0,83 . = 8,3 .
. Then dn = ( ) d
CHAPTER 4 (SECTION 5)
1.
Diberikan fungsi z = dengan
= dan = , tentukan
Jawab :
=
=
+
( )
( ) +
( ) ( )
= ( ) + (-1) (- ) = ( ) + ( ) = ( ) + ( )
2.
Diberikan fungsi Tentukan
√ , = ( ) , = [ ( ) .
.
Jawab : Misal :
( ) , sehingga = dan =
= ( ) () Ingat bahwa :
= √
= √ =1 Maka : w = 1
= 0 3.
Diberikan fungsi Jawab :
. , dan , Tentukan
= +
( ) ( ) ( ) ( ) = +
) ( ) ( ) ( ) ( = +
= ( ) + () ( ) = () + () (–) = () + () = ( )
4.
Diberikan fungsi
( ) dan , tentukan
.
Jawab :
= +
( ) ( ) ( ) = + = + () ( ) ( ) () = + ( ) ( ) ( ()) = ( )
5.
Diberikan fungsi
() () Tunjukkan bahwa aturan rantai ditunjukkan
oleh persamaan :
= +
Jawab :
= +
6.
Diberikan ( ) ,
() , tentukan .
Jawab :
= + () () () = + = () (1)+ () (1)[ () . 10 = () + () . 10 [ () = () [1+ 10 ()]
7.
Diberikan
( ), , tentukan .
Jawab :
= + () () ( ) = + () = ( ) + (-1) ( ) (
) ( )
= ( ) - ( ) ( ) = ( ) – ( ) ( )– ( ) ( ) = ( ) ( 1- )
. + . = -
CHAPTER 4 (SECTION 6)
1.
= ( )
+ = - (
Jawab : Turunan pertama
( ) :
+ (
( ) ( ) + . = C .
= +
) =
=
+ = 0
)
= -
=
=
=
=
() = Dimana =
= Turunan kedua (
.
2.
) :
dimana = = Sehingga :
=
= + = 0 ( + ) = (0) ( ) + =0
= Turunan pertama (
)
- = 0 ( - ) = 0
+
–
() =0
+ ( ) – = 0 + – = 0 =
=
Pada (0,0) :
3.
- = 0
()()
() = ()()
( - ) = 0
= 1 Turunan kedua (
=
( ) ( ) = 0
) :
- = 0 +
dimana = =
( ) = 0 +
Sehingga :
=
=
Pada (2,4) :
=
– () = – ()
+ = 0 ( + ) = 0 ( ) - ( ) + ( ) = 0
+
+ +
+ = 0
+ (y )+ + + = 0
+ - + + = 0 + - + = 0
=
Pada (0,0) :
() ()() () () () = ()()
= 0
– () = – () () – = – ( – ) = ( – ) ( – ) = ( – )
