Perbaikan UTS

1. Tugas Matematika
Perbaikan UTS
Disusun Oleh :
POLITEKNIK MANUFAKTUR NEGERI BANGKA BELITUNG
TAHUN AJARAN 2014/2015
Industri Air Kantung Sungailiat 33211
Bangka Induk, Propinsi Kepulauan Bangka Belitung
Telp : +62717 93586
Fax : +6271793585 email : polman@polman-babel.ac.id
http://www.polman-babel.ac.id
Nama : Rakam Tiano
Kelas : 1 EB
Prodi : Elektronika
Semester : 2 ( Genap)

2. 1. lim 𝑥→0
8𝑠𝑖𝑛 2 𝑥
𝑥.𝑡𝑎𝑛 11𝑥
−
9𝑥
𝑡𝑎𝑛 10𝑥
= lim 𝑥→0 8
sin 𝑥
𝑥
.
sin 𝑥
𝑥
.
11𝑥
𝑡𝑎𝑛 11𝑥
.
𝑥
11𝑥
−
9𝑥
10𝑥
.
10𝑥
𝑡𝑎𝑛 10𝑥
=
8.1.1.1.
1
11
−
9
10
. 1 =
8
11
−
9
10
=
80−99
110
=
−19
110
2. lim 𝑥→8
𝑥2−11𝑥+24
𝑥−8
= lim 𝑥→8
𝑥−8 𝑥−3
𝑥−8
= 8 − 3 = 5
3. 𝑦 = 𝑥65
+
4
𝑥5
+
2
𝑥3 = 𝑥6 5
+ 4𝑥−5 2
+ 2𝑥3
𝑦′
=
6
5
𝑋
6
5
−1
+ 4
−5
2
𝑋
−5
2
−1
+ 2 −3 𝑋−3−1
=
6
5
𝑥1 5
−
20
2
𝑥−7 2
− 6𝑥−4
=
6
5
𝑥1 5
− 10𝑥−7 2
− 6𝑥−4
4. Dit : y”
𝑦 = cos 5𝑥 + 3 csc 6𝑥 − 4 cot 7𝑥
𝑦′
= −5 sin5𝑥 + 3 −6 csc6𝑥. cot 6𝑥 − 4 −7 𝑐𝑜𝑠𝑒𝑐2
7𝑥
= −5 sin 5𝑥 − 18 csc 6𝑥 . 𝑐𝑜𝑡𝑎𝑛 6𝑥 + 28 𝑐𝑜𝑠𝑒𝑐2
7𝑥
 𝑐𝑜𝑠𝑒𝑐 6𝑥 . 𝑐𝑜𝑡𝑎𝑛 6𝑥
Misalkan :
𝑈 = 𝑐𝑜𝑠𝑒𝑐 6𝑥 → 𝑈′
= −6 𝑐𝑜𝑠𝑒𝑐 6𝑥 . 𝑐𝑜𝑡𝑎𝑛 6𝑥
𝑉 = 𝑐𝑜𝑡𝑎𝑛 6𝑥 → 𝑉′
= −6 𝑐𝑜𝑠𝑒𝑐2
6𝑥
𝑦′
= 𝑢′
𝑣 + 𝑢𝑣′
= −6 𝑐𝑜𝑠𝑒𝑐 6𝑥 . 𝑐𝑜𝑡𝑎𝑛 6𝑥. 𝑐𝑜𝑡𝑎𝑛 6𝑥 + 𝑐𝑜𝑠𝑒𝑐 6𝑥 . −6 𝑐𝑜𝑠𝑒𝑐2
6𝑥
= −6 𝑐𝑜𝑠𝑒𝑐 6𝑥 . 𝑐𝑜𝑡𝑎𝑛2
6𝑥 − 6 𝑐𝑜𝑠𝑒𝑐3
6𝑥
 𝑐𝑜𝑠𝑒𝑐2
7𝑥 = 𝑐𝑜𝑠𝑒𝑐 7𝑥 . 𝑐𝑜𝑠𝑒𝑐 7𝑥
Misalkan :
𝑈 = 𝑐𝑜𝑠𝑒𝑐 7𝑥 → 𝑈′
= −7 𝑐𝑜𝑠𝑒𝑐 7𝑥 . 𝑐𝑜𝑡𝑎𝑛 7𝑥
𝑉 = 𝑐𝑜𝑠𝑒𝑐 7𝑥 → 𝑉′
= −7 𝑐𝑜𝑠𝑒𝑐 7𝑥 . 𝑐𝑜𝑡𝑎𝑛 7𝑥
𝑦 = 𝑢 . 𝑣
𝑦′
= 𝑢′
𝑣 + 𝑢𝑣′
= −7 𝑐𝑜𝑠𝑒𝑐 7𝑥 . 𝑐𝑜𝑡𝑎𝑛 7𝑥 . 𝑐𝑜𝑠𝑒𝑐 7𝑥 + 𝑐𝑜𝑠𝑒𝑐 7𝑥 −7 𝑐𝑜𝑠𝑒𝑐 7𝑥 . 𝑐𝑜𝑡𝑎𝑛 7𝑥
= −7 𝑐𝑜𝑠𝑒𝑐2
7𝑥 . 𝑐𝑜𝑡𝑎𝑛 7𝑥 − 7 𝑐𝑜𝑠𝑒𝑐2
7𝑥 . 𝑐𝑜𝑡𝑎𝑛 7𝑥
𝑦′
= −5 sin5𝑥 − 18 𝑐𝑜𝑠𝑒𝑐 6𝑥 . 𝑐𝑜𝑡𝑎𝑛 6𝑥 + 28 𝑐𝑜𝑠𝑒𝑐2
7𝑥
𝑦" = −5 .5 cos⁡5𝑥 − 18 (−6 𝑐𝑜𝑠𝑒𝑐 6𝑥 . 𝑐𝑜𝑡𝑎𝑛2
6𝑥 − 6 𝑐𝑜𝑠𝑒𝑐3
6𝑥)
+ 28 (−7 𝑐𝑜𝑠𝑒𝑐2
7𝑥 . 𝑐𝑜𝑡𝑎𝑛 7𝑥− ) − 7 𝑐𝑜𝑠𝑒𝑐2
7𝑥. 𝑐𝑜𝑡𝑎𝑛 7𝑥
= −25 cos 5𝑥 + 108 𝑐𝑜𝑠𝑒𝑐 6𝑥 . 𝑐𝑜𝑡𝑎𝑛2
6𝑥 + 108 𝑐𝑜𝑠𝑒𝑐3
6𝑥
− 196 𝑐𝑜𝑠𝑒𝑐2
7𝑥 . 𝑐𝑜𝑡𝑎𝑛 7𝑥 − 196 𝑐𝑜𝑠𝑒𝑐2
7𝑥 . 𝑐𝑜𝑡𝑎𝑛 7𝑥
= −25 cos 5𝑥 + 108 𝑐𝑜𝑠𝑒𝑐 6𝑥 . 𝑐𝑜𝑡𝑎𝑛2
6𝑥 + 108 𝑐𝑜𝑠𝑒𝑐3
6𝑥
− 392 𝑐𝑜𝑠𝑒𝑐2
7𝑥 . 𝑐𝑜𝑡𝑎𝑛 7𝑥

3. 5. 𝑦 = 8𝑥3
. 𝑠𝑖𝑛5𝑥 , 𝑚𝑖𝑠𝑎𝑙𝑘𝑎𝑛: 𝑢 = 8𝑥3
→ 𝑢′
= 24𝑥2
, 𝑣 = 𝑠𝑖𝑛5𝑥 → 𝑣′
=
5 cos 5𝑥 , 𝑦 = 𝑢. 𝑣, 𝑦 = 𝑢′
. 𝑣 + 𝑢. 𝑣′
, = 24𝑥2.
. sin 5𝑥 + 8𝑥3
. 5 cos 5𝑥 =
24𝑥2
. sin5𝑥 + 40𝑥3
cos 5𝑥
6. 𝑦 =
9𝑥
5−𝑥2 , 𝑚𝑖𝑠𝑎𝑙𝑘𝑎𝑛: 𝑢 = 9𝑥 → 𝑢′
= 9 , 𝑣 = 5 − 𝑥2
→ 𝑣′
= −2𝑥 ,
𝑦 =
𝑢
𝑣
, 𝑦′
=
𝑢′ .𝑣+𝑢.𝑣′
𝑣2 =
9. 5−𝑥2 −9𝑥 −2𝑥
5−𝑥2 2 =
45−9𝑥2−9𝑥 −2𝑥
25+𝑥4+2.5 −𝑥2 =
9𝑥2+45
𝑥4−10𝑥2+25
7. 𝑦 = ln 𝑥4 + 3𝑥 + 4 , 𝑚𝑖𝑠𝑎𝑙𝑘𝑎𝑛 ∶ ∗ 𝑢 = 𝑥4
+ 3𝑥 + 4 ,
𝑑𝑢
𝑑𝑥
= 4𝑥3
+ 3 , ∗ 𝑣 =
𝑢
1
2 ,
𝑑𝑣
𝑑𝑢
=
1
2
𝑢−1 2
, ∗ 𝑦 = ln 𝑣 ,
𝑑𝑦
𝑑𝑣
=
1
𝑣
=
1
𝑢1 2 ,
𝑑𝑦
𝑑𝑢
=
𝑑𝑢
𝑑𝑢
.
𝑑𝑣
𝑑𝑢
.
𝑑𝑦
𝑑𝑣
= 4𝑥3
+
3 .
1
2
. 𝑥4
+ 3𝑥 + 4 −1
2 .
1
𝑥4+3𝑥+4
1
2
= 4𝑥3
+ 3 .
1
2
. 𝑥4
+ 3𝑥 + 4 −1
=
1
2
. 4𝑥3
+
3 𝑥4
+ 3𝑥 + 4 −1
8. 4𝑥2
− 5𝑥2
𝑦3
+ 6𝑦3
− sin 2𝑥. 𝑦2
= 0 ,
𝑑
𝑑𝑢
4𝑥2
− 5𝑥2
𝑦3
+ 6𝑦3
− sin 2𝑥. 𝑦2
=
𝑑
𝑢
0 , 20𝑥4 𝑑𝑥
𝑑𝑥
− 10𝑥
𝑑𝑥
𝑑𝑥
𝑦3
+ −5𝑥2
3𝑦2 𝑑𝑦
𝑑𝑥
+ 18𝑦2 𝑑𝑦
𝑑𝑥
− 2 cos 2𝑥
𝑑𝑥
𝑑𝑥
. 𝑦2
+
− sin 2𝑥 . 2𝑦
𝑑𝑦
𝑑𝑥
= 0 , 20𝑥4
− 10𝑥𝑦3
− 5𝑥2
. 3𝑦2 𝑑𝑦
𝑑𝑥
+ 18𝑦2 𝑑𝑦
𝑑𝑥
− 2 cos 2𝑥. 𝑦2
−
sin 2𝑥. 2𝑦
𝑑𝑦
𝑑𝑥
= 0 , − 15𝑥2
𝑦2
+ 18𝑦2
− sin 2𝑥. 2𝑦
𝑑𝑦
𝑑𝑥
= − 20𝑥4
− 10𝑥𝑦3
−
2 cos 2𝑥. 𝑦2
,
𝑑𝑦
𝑑𝑥
=
− 20𝑥4−10𝑥 𝑦3−2 cos 2𝑥.𝑦2
−15𝑥2 𝑦2+18𝑦2−sin 2𝑥.2𝑦
9. 𝑦 = 𝑐𝑜𝑠ℎ 7𝑥 − 4 , 𝑚𝑖𝑠𝑎𝑙𝑘𝑎𝑛 ∶ ∗ 𝑢 = 7𝑥 − 4 ,
𝑑𝑢
𝑑𝑥
= 7 , ∗ 𝑣 = 𝑢
1
2 ,
𝑑𝑣
𝑑𝑢
=
1
2
𝑢−1 2
, ∗ 𝑦 = cos 𝑣,
𝑑𝑦
𝑑𝑣
= sinh 𝑣 = 𝑠𝑖𝑛ℎ𝑈
1
2 , ,,
𝑑𝑦
𝑑𝑥
=
𝑑𝑢
𝑑𝑥
.
𝑑𝑣
𝑑𝑢
.
𝑑𝑦
𝑑𝑣
=
7.
1
2
7𝑥 − 4 1 2
=
7
2
7𝑥 − 4 −1 2
sinh 7𝑥 − 4
1
2
10. Dik : v = 30 liter = 3𝑑𝑚3
Dit : luas kaleng minimum
Jawab :
*V tabung
V=𝜋𝑟2
ℎ
h=
30
𝜋 𝑟2

4. *luas kaleng
L = 2𝜋𝑟ℎ + 2𝜋𝑟2
= 2𝜋𝑟.
30
𝜋 𝑟2 + 2𝜋𝑟2
= 2.
30
𝑟
+ 2𝜋𝑟2
=
60
𝑟
+ 2𝜋𝑟2
Harga Ekstrim
𝐿 = 60 𝑟−1
+ 2𝜋𝑟2
𝐿′
= −60 𝑟−2
+ 4𝜋𝑟
Anggap 𝐿′
= 0
Jadi :
−60 𝑟−2
+ 4𝜋𝑟 = 0
4𝜋𝑟 = 60 𝑟−2
4 .3,14 . 𝑟 = 60 𝑟−2
12,56 𝑟 = 60 𝑟−2
𝑟3
=
60
12,56
= 4,78
𝑟 = 4,78
3
= 1,68
ℎ =
30
𝜋𝑟2
=
30
3,14 . 1,68 2
=
30
3,14 .2,8
=
30
8,792
= 3,4 𝑑𝑚
*Luas kaleng
𝐿 = 2𝜋𝑟ℎ + 2𝜋𝑟2
= 2 .3,14 .1,68 .3,4 + 2 .3,14 . (1,68)2
= 35,87 + 6,28 .2,8
= 35,87 + 17,58
= 53,45 𝑑𝑚2
Jadi, luas kaleng minimum yang diperlukan untuk membuat tabung tersebut adalah 53,45
𝑑𝑚2