The document provides solutions to physics problems for chapter 4 of mathematics 2. It includes solutions for determining derivatives and differentials of various functions with respect to variables like x, y, r, and θ. The highest level of mathematics involved includes taking second order derivatives and solving simultaneous equations. Sample problems include determining derivatives of functions that define relationships between polar and Cartesian coordinates.

1. TUGAS FISIKA UNTUK MATEMATIKA 2
Penyelesaian Soal Soal Chapter 4
Dosen Pengampu : Drs. Pujayanto, MSi
DISUSUN OLEH
MAY NURHAYATI
(K2315048)
PROGRAM STUDI PENDIDIKAN FISIKA
FAKULTAS KEGURUAN DAN ILMU PENDIDIKAN
UNIVERSITAS SEBELAS MARET
SURAKARTA
2016

2. Section 1
4. Untuk ω = x3 – y3 – 2xy +6 tentukan
𝝏 𝟐
𝒘
𝝏𝒙 𝟐 dan
𝝏 𝟐
𝒘
𝝏𝒚 𝟐 pada titik dimana
𝒅𝝎
𝒅𝒙
=
𝒅𝝎
𝒅𝒚
= 𝟎
𝑤 = 𝑥3
− 𝑦3
− 2𝑥𝑦 + 6
𝜕𝑤
𝜕𝑥
= 0
3𝑥2
− 2 = 0
3𝑥2
= 2𝑦
𝑦 =
3𝑥2
2
…….(1)
𝜕𝑤
𝜕𝑦
= 0
−3𝑦2
− 2𝑥 = 0
−3𝑦2
= 2𝑥
Subtitusi persamaan …….(1)
−3 (
3𝑥2
2
)
2
= 2𝑥
−27𝑥4
4
= 2𝑥
−27𝑥3
= 8
𝑥 = −
2
3
𝑦 =
3(−
2
3
)
2
2
𝑦 =
2
3
𝜕2
𝑤
𝜕𝑦2
= −6𝑦
𝜕2
𝑤
𝜕𝑦2
= −6
2
3
= −4
13. Jika 𝒛 = 𝒙 𝟐
+ 𝟐𝒚 𝟐
, dimana 𝒚 = 𝒓𝒔𝒊𝒏 𝜽, 𝒙 = 𝒓𝒄𝒐𝒔𝜽 maka tentukan (
𝝏𝒛
𝝏𝜽
)
𝒙
𝑧 = 𝑥2
+ 𝑟2
𝑠𝑖𝑛2
𝜃
𝑧 = 𝑥2
+
2𝑥2
𝑐𝑜𝑠2 𝜃
𝑠𝑖𝑛2
𝜃
𝑧 = 𝑥2
+ 2𝑥2
𝑡𝑎𝑛2
𝜃
(
𝜕𝑧
𝜕𝜃
)
𝑥
= 2𝑥2
2𝑡𝑎𝑛𝜃𝑠𝑒𝑐2
𝜃

3. 19. Jika 𝒛 = 𝒙 𝟐
+ 𝟐𝒚 𝟐
, dimana 𝒚 = 𝒓𝒔𝒊𝒏 𝜽, 𝒙 = 𝒓𝒄𝒐𝒔𝜽 maka tentukan (
𝝏 𝟐 𝒛
𝝏𝒓𝝏𝒚
)
𝜕2 𝑧
𝜕𝑟𝜕𝑦
=
𝜕
𝜕𝑟
(
𝜕𝑧
𝜕𝑦
) =
𝜕
𝜕𝑟
(
𝜕(( 𝑟 cos 𝜃)2 + 2𝑦2)
𝜕𝑦
)
=
𝜕
𝜕𝑟
(
𝜕 (
𝑦
sin 𝜃
cos 𝜃)
2
+ 2𝑦2
𝜕𝑦
)
=
𝜕
𝜕𝑟
(
𝜕 ((
𝑐𝑜𝑠2 𝜃
𝑠𝑖𝑛2 𝜃
𝑦2) + 2𝑦2)
𝜕𝑦
)
=
𝜕(2𝑦 𝑐𝑜𝑡𝑎𝑛2 𝜃 + 4𝑦)
𝜕𝑟
= 0

4. Section2
5. Carilah 2 variabel Maclaurin dari fungsi √1 + 𝑥𝑦
Jawab :
√1 + 𝑥𝑦 = (1 + 𝑥𝑦)1/2
Deret Maclaurin (1 + 𝑥𝑦)1/2
=1+
1
2
𝑥𝑦 +
1
2
(−
1
2
)
2
( 𝑥2
𝑦2) +
1
2
(−
1
2
)(−
1
2
)
2
( 𝑥3
𝑦3)+.....
= 1+
1
2
𝑥𝑦 −
1
8
( 𝑥2
𝑦2)+
3
48
( 𝑥3
𝑦3)+.....
= 1+
1
2
𝑥𝑦 −
1
8
( 𝑥2
𝑦2)+
1
16
( 𝑥3
𝑦3)+.....

5. Section4
6. The acceleration of gravity can be found from the length l and period T of a pendulum;
the formula is g=4𝜋2
𝑙/𝑇. Find the relative eror in g in the worst case if the relative
eror in l is 5% and the relative eror in T is 2%.
Jawab :
Diketahui :
g=
4𝜋2
𝑙
𝑇
relative eror in l= 5% = 0,05
relative eror in T= 2% = 0,02
Relative eror in g :
ln g = ln 4𝜋2
+ ln l – ln T
𝑑𝑔
𝑔
=
𝑑𝑙
𝑙
− 2
𝑑𝑇
𝑇
|
𝑑𝑔
𝑔
| = |
𝑑𝑙
𝑙
| − 2 |
𝑑𝑇
𝑇
|
= 0,05 + 2(0,02)
= 0,05 + 0,04
= 0,09
= 9%
Jadi nilai kesalah relative g sebesar 9%
10. A force of 500 nt is measured with a possible eror of 1 nt. Its component in a direction 60o
away from its line of action is required, where the angle is subject to an eror of 0,5o. What
is (approximately) the largest possible eror in the component ?
Jawab :
Kesalahan relatif = (
1 𝑛𝑡
500 𝑛𝑡
+
0,5 o
60 o
)x100%
= (
1
500
+
5
600
)x100%
= (
6+25
3000
)x100%
=
31
30
%
= 1,033%
Besar kesalahan relatif=1,033% x 500nt
= 5,15 nt

6. Section 6
6. If 𝒙𝒚 𝟑
− 𝒚𝒙 𝟑
= 𝟔 is the equation of a curve, find the 𝒅 𝟐
𝒚/𝒅𝒙 𝟐
at the point (1,2).
Jawab :
 Turunan pertama :
𝑑𝑦
𝑑𝑥
: 𝑦3
+ x 3𝑦2 𝑑𝑦
𝑑𝑥
− (
𝑑𝑦
𝑑𝑥
. 𝑥3
− 𝑦 3𝑥2
)= 0
𝑦3
+ 3x𝑦2
𝑑𝑦
𝑑𝑥
− 𝑥3
𝑑𝑦
𝑑𝑥
− 3 𝑦𝑥2
= 0
 Substitusi (1,2) :
23
+ 3.1.22 𝑑𝑦
𝑑𝑥
− 13 𝑑𝑦
𝑑𝑥
− 3.12
. 2 = 0
8 + 12
𝑑𝑦
𝑑𝑥
−
𝑑𝑦
𝑑𝑥
− 6 = 0
11
𝑑𝑦
𝑑𝑥
+ 2 = 0
𝑑𝑦
𝑑𝑥
= −
2
11
 Turunan kedua :
𝑑2
𝑦
𝑑𝑥2
=
𝑑
𝑑𝑥
(
𝑑𝑦
𝑑𝑥
)
= 3 𝑦2 𝑑𝑦
𝑑𝑥
+ 6𝑦
𝑑𝑦
𝑑𝑥
𝑥
𝑑𝑦
𝑑𝑥
+ 3 𝑦2 𝑑𝑦
𝑑𝑥
+ 3𝑦2x
𝑑
2
𝑦
𝑑𝑥
2 −3 𝑥2 𝑑𝑦
𝑑𝑥
− 𝑥3 𝑑
2
𝑦
𝑑𝑥
2 − 6𝑥𝑦−3 𝑥2 𝑑𝑦
𝑑𝑥
 Substitusi (1,2) :
3.22
. −
2
11
+ 6.2. (−
2
11
)
2
+ 12.−
2
11
+ 12
𝑑2
𝑦
𝑑𝑥2 − 3.−
2
11
−
𝑑2
𝑦
𝑑𝑥2 − 6.1.2 − 3.1. −
2
11
= 0
−
24
11
+
48
112.
−
24
11
+ 12
𝑑2 𝑦
𝑑𝑥2
+
6
11
−
𝑑2 𝑦
𝑑𝑥2
− 12 +
6
11
= 0
11
𝑑2
𝑦
𝑑𝑥2
=
24
11
−
48
112
+
24
11
−
6
11
−
6
11
+ 12
11
𝑑2
𝑦
𝑑𝑥2
=
36
11
−
48
112
+
132
11
11
𝑑2
𝑦
𝑑𝑥2
=
396 − 48
112
+1452
𝑑2 𝑦
𝑑𝑥2 =
1800
113

8. Section7
4. If w = 𝑒−𝑟2
−𝑠2
, r = uv, s = u+2v, find
𝜕𝑤
𝜕𝑢
and
𝜕𝑤
𝜕𝑣
Jawab :
𝜕𝑤
𝜕𝑢
=
(−2𝑟 𝑑𝑟 − 2𝑠 𝑑𝑠) 𝑤
𝜕𝑢
=
(−2𝑟 ( 𝜕𝑢 𝑣 + 𝑢 𝜕𝑣) − 2𝑠 (𝜕𝑢 + 2 𝜕𝑣)) 𝑤
𝜕𝑢
=
(−2𝑟𝑣 𝜕𝑢 − 2𝑟𝑢 𝜕𝑣 − 2𝑠 𝜕𝑢 − 2 𝑠 𝜕𝑣) 𝑤
𝜕𝑢
=
(−2𝑟𝑣 𝜕𝑢 − 2𝑠 𝜕𝑢) 𝑤
𝜕𝑢
= (−2𝑟𝑣 − 2𝑠) 𝑤
= −2( 𝑟𝑣 + 𝑠) 𝑤
𝜕𝑤
𝜕𝑣
=
(−2𝑟 𝑑𝑟 − 2𝑠 𝑑𝑠) 𝑤
𝜕𝑣
=
(−2𝑟 ( 𝜕𝑢 𝑣 + 𝑢 𝜕𝑣) − 2𝑠 (𝜕𝑢 + 2 𝜕𝑣)) 𝑤
𝜕𝑣
=
(−2𝑟𝑣 𝜕𝑢 − 2𝑟𝑢 𝜕𝑣 − 2𝑠 𝜕𝑢 − 4 𝑠 𝜕𝑣) 𝑤
𝜕𝑣
=
(−2𝑟𝑢 𝜕𝑣 − 4 𝑠 𝜕𝑣) 𝑤
𝜕𝑣
= (−2𝑟𝑢 − 4𝑠) 𝑤
= −2( 𝑟𝑢 + 2𝑠) 𝑤
8. If 𝑥𝑧2
+ 𝑦𝑡2
= 1 and 𝑥2
𝑠 + 𝑦2
𝑡 = 𝑥𝑦 − 4, find 𝜕𝑥/𝜕𝑠, 𝜕𝑥/𝜕𝑡, 𝜕𝑦/𝜕𝑠, 𝜕𝑦/𝜕𝑡 at
(x,y,s,t)=(1, -3, 2, -1). Hint : to simplify the work, subtitutes the numerical value just
after you have taken differential.
Jawab :
xs2 + yt2 = 1
x2s + yt = xy – 4

9. 𝜕𝑥
𝜕𝑠
 dx s2 + x2s ds + dy t2 + y2t dt = 0
s2 dx + t2 dy = - 2yt dt – 2sx ds
 2xs dx – y dx + 2ty dy – x dy = - y2 dt – x2 ds
 Mencari nilai dx
dx =
|
−2𝑦𝑡 𝑑𝑡 𝑡2
−𝑦2 𝑑𝑡− 𝑥2 𝑑𝑠 (2𝑦𝑡−𝑥)
|
| 𝑠2 𝑡2
2𝑥𝑠−𝑦 (2𝑦𝑡−𝑥)
|
dx =
−4𝑦2 𝑡2 dt – 2yx dt – 4sxy ds + 2s𝑥2 ds –(t𝑦2 𝑡2 dt – 𝑡2 𝑥2 ds)
2𝑦𝑡𝑠2−𝑥𝑠2−(2𝑥𝑠𝑡2−𝑦𝑡2)
=
(−4𝑦2 𝑡2 – 2yx+ t𝑦2 𝑡2)dt+(−4sxy + 2s𝑥2+ 𝑡2 𝑥2) ds
2𝑦𝑡𝑠2−𝑥𝑠2−2𝑥𝑠𝑡2+𝑦𝑡2
𝜕𝑥
𝜕𝑠
=
–4sxy + 2s𝑥2+ 𝑡2 𝑥2
2𝑦𝑡𝑠2−𝑥𝑠2−2𝑥𝑠𝑡2+𝑦𝑡2
Substitusi nilai (x,y,s,t) = (1,-3,2,-1)
𝜕𝑥
𝜕𝑠
=
−4 6+2 2+1
2 12−4−2 2+(−3)
𝜕𝑥
𝜕𝑠
= −
19
13
𝜕𝑥
𝜕𝑡
=
(−4𝑦2 𝑡2 – 2yx+ t𝑦2 𝑡2)
2𝑦𝑡𝑠2−𝑥𝑠2−2𝑥𝑠𝑡2+𝑦𝑡2
=
−4 9+2 3+9
2 3 4−4−2 2+(−3)
= −
21
13
 Mencari nilai 𝜕𝑥
𝜕𝑦 =
|
𝑠2
−2𝑦𝑡𝑑𝑡 − 2𝑠𝑥𝑑𝑠
2𝑥𝑠 − 𝑦 −𝑦2
𝑑𝑡 − 𝑥2
𝑑𝑠
|
13
=
(−𝑠2
𝑦2
𝑑𝑡 − 𝑠2
𝑥2
𝑑𝑠) − (−4𝑦𝑡𝑥𝑠𝑑𝑡 − 4𝑥2
𝑠2
𝑑𝑠 + 2𝑦2
𝑑𝑡 + 2𝑠𝑥𝑦𝑑𝑠)
13
𝜕𝑦 =
−𝑠2
𝑦2
𝑑𝑡 − 𝑠2
𝑥2
𝑑𝑠 + 4𝑦𝑡𝑥𝑠𝑑𝑡 + 4𝑥2
𝑠2
𝑑𝑠 + 2𝑦2
𝑡𝑑𝑡 − 2𝑠𝑥𝑦𝑑𝑠
13
=
(−𝑠2
𝑦2
+ 4𝑦𝑡𝑥𝑠 + 2𝑦2
𝑡) 𝑑𝑡 + (−𝑠2
𝑥2
+ 4𝑥2
𝑠2
− 2𝑠𝑥𝑦) 𝑑𝑠
13
 Mencari nilai
∂y
∂s

10. 𝜕𝑦
𝜕𝑠
=
−𝑠2
𝑥2
+ 4𝑥2
𝑠2
− 2𝑠𝑥𝑦
13
=
−4 + (4)(4) − 2(−6)
13
=
−4 + 16 + 12
13
=
24
13
 Mencari nilai
∂y
∂t
𝜕𝑦
𝜕𝑡
=
−𝑠2
𝑦2
+ 4𝑦𝑡𝑥𝑠 − 2𝑦2
𝑡
13
=
−4(9) + (4)(3)(2) − 2(−9)
13
=
−36 + 24 + 18
13
=
6
13

11. Section 8
4. 𝒙 𝟐
− 𝒚 𝟐
+ 𝟐𝒙 − 𝟒𝒚 + 𝟏𝟎 = 𝟎
Jawab :
Syarat 1 :
f’(x) = 0
2x + 2 = 0
2x = -2
X = -1
Syarat 2 :
f’(y) = 0
-2y - 4 = 0
-2y = 4
y = -2
nilai ekstrem pada titik (-1,-2)
fxx = 2 fxx> 0
fyy = -2 fyy > 0
sehingga titik (-1,-2) bukan merupakan titik maksimum/minimum melainkan titik pelana.
9. sebuah aquarium dengn alas dan sisi persegi panjang tanpa tutup tersiri dari 5 sisi.
Tentukan perbandingan rusuknya sehingga diperlukan bahan seminimal mungkin untuk
membuatnya.
Jawab :
Misal rusuk nya : p=a, l=b, t=c
V = a.b.c
c =
𝑉
𝑎𝑏
A = ab + 2 ac +2 bc

12. = ab + 2 a
𝑉
𝑎𝑏
+2 b
𝑉
𝑎𝑏
= ab +
2𝑉
𝑏
+
2𝑉
𝑎
𝑑𝐴
𝑑𝑎
= b −
2𝑉
𝑎2
b =
2𝑉
𝑎2
...............(1)
𝑑𝐴
𝑑𝑏
= a −
2𝑉
𝑏2
a =
2𝑉
𝑏2
...............(2)
 2V = 2V
b𝑎2
= a𝑏2
a = b
 b = b
2𝑉
𝑎2
=
𝑉
𝑎𝑐
c =
𝑎
2
a = 2c
jadi perbandingannya : a = b = 2c
10. Tentukan Garis singgung dari x𝑒 𝑦
+y𝑒 𝑥
=0, substitusikan x=y=0
Jawab :
x𝑒 𝑦
+y𝑒 𝑥
=0
Diturunkan ke x:
𝑒 𝑦
+x𝑒 𝑦 𝑑𝑦
𝑑𝑥
+
𝑑𝑦
𝑑𝑥
𝑒 𝑥
+ y𝑒 𝑥
=0
Substitusi bilai x=y=0 :

13. 𝑒0
+ 0. 𝑒0
𝑑𝑦
𝑑𝑥
+
𝑑𝑦
𝑑𝑥
. 𝑒0
+ 0. 𝑒0
= 0
1 + 0 + 1.
𝑑𝑦
𝑑𝑥
+ 0 = 0
𝑑𝑦
𝑑𝑥
= −1
𝑝𝑒𝑟𝑠𝑎𝑚𝑎𝑎𝑛 𝑔𝑎𝑟𝑖𝑠 𝑠𝑖𝑛𝑔𝑔𝑢𝑛𝑔 ∶
𝑦 − 𝑦1
𝑥 − 𝑥1
=
𝑑𝑦
𝑑𝑥
𝑦 − 0
𝑥 − 0
= −1
y = -x
y + x = 0

14. Section 9
6. A box has three of its faces in the coordinate planes and one vertex on the plane 2x + 3y +
4y = 6. Find the maximum volume for the box!
Answer:
∅ = 2x + 3y + 4z = 6
𝑓 = 8xyz
𝐹( 𝑥, 𝑦, 𝑧) = 𝑓 +⋋ ∅ = 8𝑥𝑦𝑧 +⋋ (2x + 3y + 4y)
→
𝜕𝐹
𝜕𝑥
= 8𝑦𝑧 +⋋ .2 = 0
→
𝜕𝐹
𝜕𝑦
= 8𝑥𝑧 +⋋ .3 = 0
→
𝜕𝐹
𝜕𝑧
= 8𝑥𝑦 +⋋ .4 = 0
3.8𝑥𝑦𝑧 + 3 ⋋ (2x + 3y + 4y) = 0
24𝑥𝑦𝑧 + 13 ⋋ = 0
⋋ = −
24
13
𝑥𝑦𝑧
8𝑦𝑧 −
24
13
𝑥𝑦𝑧. 2 = 0 , 𝑥 =
1
2
8𝑥𝑧 −
24
13
𝑥𝑦𝑧. 3 = 0 , 𝑦 =
13
9
8𝑥𝑦 −
24
13
𝑥𝑦𝑧. 4 = 0 , 𝑧 =
13
12
8𝑥𝑦𝑧 =
1
2
+
13
9
+
13
12
=
109
36
Jadi volume box maximum adalah
109
36
8 . A point moves in the (x, y) plane on the line 2x+3y-4=0. Where will it be when the sum
of the squares of its distances from (1, 0) and (-1, 0) is smallest?
Jawab:
Titik Minimum ( 𝑥 − 1)2 + ( 𝑦)2 + ( 𝑥 + 1)2 + 𝑦2 𝑝𝑎𝑑𝑎 2𝑥 + 3𝑦 − 4 = 0
𝑓 ( 𝑥, 𝑦) = ( 𝑥 − 1)2 + ( 𝑦)2 + ( 𝑥 + 1)2 + l ( 2𝑥 + 3𝑦 − 4)
𝑓 ( 𝑥, 𝑦) = ( 𝑥 − 1)2 + ( 𝑥 + 1)2 + 2( 𝑦)2 + l ( 2𝑥 + 3𝑦 − 4))
𝜕𝑓
𝜕𝑥
= 2( 𝑥 − 1) + 2( 𝑥 + 1) + 2l = 0
4x+ 2 λ = 0
λ = -2x
𝜕𝑓
𝜕𝑦
= 4𝑦 + 3l = 0

15. λ = (-4/3) y
-2x = (-4/3) y
x = (2/3) y
y = (3/2) x
2x+3y-4 = 0
2x+ 3(3/2)x - 4 = 0
2x+(9/2) x - 4 = 0
𝑥 =
8
13
𝑦 =
3
2
8
13
=
12
13
(𝑥, 𝑦) = (
8
13
,
12
13
)
It will be at (8/13, 12/13) when the sum of the squares of its distances from (1, 0) and (-1,
0) is smallest.

16. Section11
6. Reduce the equation 𝑥2
(
𝑑2
𝑦
𝑑𝑥2 ) + 2𝑥 (
𝑑𝑦
𝑑𝑥
) − 5𝑦 = 0
To a differensial equation with constant coefficients in
𝑑2
𝑦
𝑑𝑥2 ,
𝑑𝑦
𝑑𝑧
and y by the change of
variabel 𝑥 = 𝑒 𝑧
.
Jawab :
𝑥2
(
𝑑2
𝑦
𝑑𝑥2
) + 2𝑥 (
𝑑𝑦
𝑑𝑥
) − 5𝑦 = 0
sebenarnya adalah 𝑎2 𝑥2
(
𝑑2 𝑦
𝑑𝑥2
) + 𝑎1 𝑥(
𝑑𝑦
𝑑𝑥
) + 𝑎0 𝑦 = 𝑓( 𝑥)merupakan persamaan
Euler atau Cauchy dengan
𝑎2 = 1 ,
𝑎1 = 2 ,
𝑎0 = −5 ,
𝑥 = 𝑒 𝑧 ,
Lalu kita punya :
𝑥
𝑑𝑦
𝑑𝑥
=
𝑑𝑦
𝑑𝑧
dan 𝑥2 𝑑2 𝑦
𝑑𝑥2
=
𝑑2 𝑦
𝑑 𝑧2
−
𝑑𝑦
𝑑𝑧
Maka :
𝑎2 (
𝑑2
𝑦
𝑑𝑥2
) + ( 𝑎1 − 𝑎2)
𝑑𝑦
𝑑𝑧
+ 𝑎0 𝑦 = 𝑓( 𝑒 𝑧)
1 (
𝑑2
𝑦
𝑑𝑥2
) + (2 − 1)
𝑑𝑦
𝑑𝑧
+ (−5) 𝑦 = 𝑓( 𝑒 𝑧)
(
𝑑2
𝑦
𝑑𝑥2
) +
𝑑𝑦
𝑑𝑧
− 5𝑦 = 𝑓( 𝑒 𝑧)
Section13
11. Given z = 𝑦2
− 2𝑥2
, find a) (
𝜕𝑧
𝜕𝑥
)
𝜃
,b)(
𝜕𝑧
𝜕𝜃
)
𝑥
,c)
𝜕2 𝑧
𝜕𝑥𝜕𝜃
Jawab :
Diketahui : 𝑥 = 𝑟 cos 𝜃
𝑦 = 𝑟 sin 𝜃
z = 𝑦2
− 2𝑥2
= (𝑟sin 𝜃)2
− 2𝑥2
= 𝑟2
𝑠𝑖𝑛2
𝜃 − 2𝑥2

17. a) (
𝜕𝑧
𝜕𝑥
)
𝜃
= (
𝜕( 𝑟2
𝑠𝑖𝑛2
𝜃−2𝑥2
)
𝜕𝑥
)
𝜃
= -4x
b) (
𝜕𝑧
𝜕𝜃
)
𝑥
= (
𝜕( 𝑟2 𝑠𝑖𝑛2 𝜃−2𝑥2)
𝜕𝜃
)
𝑥
=
𝑟2
𝑐𝑜𝑠2 𝜃
𝑠𝑖𝑛2
𝜃 − 2𝑥2
= 𝑥2
𝑡𝑎𝑛2
𝜃 − 2𝑥2
pemisalan ∶ u = tan2
𝜃 = tan𝜃 . tan𝜃
du = sec2
𝜃.tan𝜃 + tan𝜃. sec2
𝜃
= 2 sec2
𝜃.tan𝜃
= 2𝑥2
𝑠𝑒𝑐2
𝜃𝑡𝑎𝑛𝜃
c)
𝜕2 𝑧
𝜕𝑥𝜕 𝜃
=
𝜕
𝜕𝑥
(
𝜕𝑧
𝜕𝜃
)
=
𝜕(2𝑥2
𝑠𝑒𝑐
2
𝜃𝑡𝑎𝑛𝜃)
𝜕𝑥
= 4𝑥 𝑠𝑒𝑐2
𝜃𝑡𝑎𝑛𝜃
9. If z = xy and 2 𝑥3
− 2𝑦3
= 3𝑡2
, find dz/dt
3 𝑥2
− 3𝑦2
= 6𝑡,
Jawab ∶
dz = dxdy
6 𝑥2
𝑑𝑥− 6𝑦2
𝑑𝑦 = 6𝑡 𝑑𝑡,𝑥2 𝑑𝑥 − 𝑦2 𝑑𝑦 = 𝑡 𝑑𝑡
6 𝑥 𝑑𝑥 − 6𝑦 𝑑𝑦 = 6 𝑑𝑡, 𝑥 𝑑𝑥 − 𝑦 𝑑𝑦 = 𝑑𝑡