Differential Calculus-II - New Age International

2.1 JACOBIAN
UNIT II
Differential Calculus-II
The Jacobians* themselves are of great importance in solving for the reverse (inverse functions)
derivatives, transformation of variables from one coordinate system to another coordinate system
(cartesian to polar etc.). They are also useful in area and volume elements for surface and volume
integrals.
2.1.1 Definition
If u = u (x, y) and v = v (x, y) where x and y are independent, then the determinant
∂u
∂u
∂x
∂y
∂v
∂v
∂x
∂y
is known as the Jacobian of u, v with respect to x, y and is denoted by
∂auv
, f or J (u, v)
∂bxy
, g
Similarly, the Jacobian of three functions u = u (x, y, z), v = v (x, y, z), w = w (x, y, z) is defined
as
J (u, v, w) =
a f
b g =
∂ uvw , ,
∂ xyz , ,
∂u
∂x
∂v
∂x
∂w
∂x
* Carl Gustav Jacob Jacobi (1804–1851), German mathematician.
95
∂u
∂y
∂v
∂y
∂w
∂y
∂u
∂z
∂v
∂z
∂w
∂z

96 A TEXTBOOK OF ENGINEERING MATHEMATICS—I
2.1.2 Properties of Jacobians
1. If u =u (x, y) and v = v (x, y), then
a f
b g
∂ uv ,
∂ xy ,
b g
a f
× ∂ xy ,
∂ uv ,
= 1 or JJ′ = 1 (U.P.T.U., 2005)
a f
b g
b g
a f
where, J =
∂ uv ,
∂
and J′ =
∂ xy ,
∂
xy ,
uv ,
Proof: Since u = u (x, y) ...(i)
v = v (x, y) ...(ii)
Differentiating partially equations (i) and (ii) w.r.t. u and v, we get
∂u
= 1
∂u
=
∂u
∂x
∂u
∂y
· + ×
∂x
∂u
∂y
∂u
...(iii)
∂u
= 0
∂v
=
∂u
∂x
∂u
∂y
· + ×
∂x
∂v
∂y
∂v
...(iv)
∂v
= 0
∂u
=
∂v
∂x
·
∂x
∂u
+
∂v
∂y
∂y
×
∂u
...(v)
∂v
= 1
∂v
=
∂v
∂x
∂v
∂y
· + ×
∂x
∂v
∂y
∂v
...(vi)
Now,
∂auv
, f ∂bxy
, g ×
∂bxy
, g ∂auv
, f =
∂u
∂x
∂v
∂x
∂u
∂y
∂v
∂y
×
∂x
∂u
∂y
∂u
∂x
∂v
∂y
∂v
=
∂u
∂x
∂v
∂u
∂y
∂v
×
∂x
∂u
∂x
∂y
∂u
∂y
∂x
∂y
∂v
∂v
(By interchanging rows and columns in II determinant)
=
∂u
∂x
∂u
∂y
+
∂x
∂ u ∂y
∂u
∂u
∂x
∂u
∂y
+
∂x
∂ v ∂y
∂v
∂v
∂x
∂v
∂y
+ .
∂x
∂ u ∂y
∂u
∂v
∂x
∂v
∂y
. +
∂x
∂ v ∂y
∂v
Putting equations (iii), (iv), (v) and (vi) in above, we get
a f
b g
∂ uv ,
∂ xy ,
b g
a f =
∂ xy ,
×
∂ uv ,
1 0
0 1
= 1
or J J ′ = 1. Hence proved.
(multiplying row-wise)
2. Chain rule: If u, v, are function of r, s and r, s are themselves functions of x, y i.e.,
u = u (r, s), v = v (r, s) and r = r (x, y), s = s (x, y)
a f
b g =
a f
a f
then
∂
∂
uv ,
x, y
∂
∂
⋅ ∂ uv ,
ars , f
r, s
∂ bx, yg
Proof: Here u = u (r, s), v = v (r, s)
and r = r (x, y), s = s (x, y)

DIFFERENTIAL CALCULUS-II 97
Differentiating u, v partially w.r.t. x and y
∂u
∂x
=
∂u
∂r
∂u
∂s
· + ·
∂r
∂x
∂s
∂x
∂u
∂y
=
∂u
∂r
∂u
∂s
· + ·
∂r
∂y
∂s
∂y
∂v
∂x
=
∂v
∂r
∂v
∂s
· + ·
∂r
∂x
∂s
∂x
∂v
∂y
=
∂v
∂r
∂v
∂s
· + ·
∂r
∂y
∂s
∂y
Now,
a f
a f
∂ uv ,
∂ rs ,
a f
b g =
∂ rs ,
·
∂ xy ,
=
=
∂u
∂r
∂v
∂r
∂u
∂r
∂v
∂r
∂u
∂s
∂v
∂s
∂u
∂s
∂v
∂s
.
.
∂r
∂x
∂s
∂x
∂r
∂x
∂r
∂y
∂r
∂y
∂s
∂y
∂s
∂x
∂s
∂y
...(i)
...(ii)
...(iii)
...(iv)
By interchanging the rows and columns in second determinant
∂u
∂r
∂u
∂s
+
∂r
∂ x ∂s
∂x
∂v
∂r
∂v
∂s
+
∂r
∂ x ∂s
∂x
Using equations (i), (ii), (iii) and (iv) in above, we get
a f
a f
∂ uv ,
∂ rs ,
a f
b g =
∂ rs ,
.
∂ xy ,
a f
b g =
auv , f
bxy , g
∂ uv ,
∂ xy ,
∂u
∂x
∂v
∂x
a f
a f
∂u
∂y
∂v
∂y
Or
= ∂ uv ,
∂ xy ,
a f
b g
∂
⋅
∂
∂ uv , rs ,
.
rs , ∂ xy ,
∂u
∂r
∂u
∂s
+
∂r
∂ y ∂s
∂y
∂v
∂r
∂v
∂s
+
∂r
∂ y ∂s
∂y
a f
b g .
Hence proved.
Example 1. Find ∂
, when u = 3x + 5y, v = 4x – 3y.
∂
Sol. We have u = 3x + 5y
v = 4x – 3y
∴
∂u
∂x
= 3, ∂u
∂v
∂v
= 5, = 4 and
∂y
∂x
∂y
Thus,
a f
b g =
∂ uv ,
∂ xy ,
∂u
∂x
∂v
∂x
∂u
∂y
∂v
∂y
= 3 5
4 −3
= – 3
= – 9 – 20 = – 29.
|multiplying row-wise

98 A TEXTBOOK OF ENGINEERING MATHEMATICS—I
a f
Example 2. Calculate the Jacobian ∂ uvw , ,
of the following:
∂bxyz
, , g
u = x + 2y + z, v = x + 2y + 3z, w = 2x + 3y + 5z. (U.P.T.U., 2007)
Sol. We have u = x + 2y + z
v = x + 2y + 3z
w =2x + 3y + 5z
∴
∂u
∂u
∂u
∂v
∂v
∂v
= 1, = 2, = 1, = 1, = 2, = 3,
∂x
∂y
∂z
∂x
∂y
∂z
∂w
∂x
∂w
∂w
= 2,
∂y
= 3 and = 5.
∂z
Now,
∂auvw
, , f
∂bxyz
, , g =
∂u
∂u
∂u
∂x
∂v
∂x
∂w
∂y
∂v
∂y
∂w
∂z
∂v
∂z
∂w
=
1
1
2
2
2
3
1
3
5
∂x
∂y
∂z
= 1 (10 – 9) – 2 (5 – 6) + 1 (3 – 4) = 2.
Example 3. Calculate ∂bxyz
, , g 2yz 3zx 4xy
if u = , v = , w =
∂auvw
, , f x y z .
Sol. Given u = 2yz 3zx 4xy
, v = , w =
x y z
∴
∂u
−2yz
= 2 ,
∂x
x
∂u
2z ∂u
2y ∂v
3z ∂v
3zx
= , = , = , = − 2 ,
∂y
x ∂z
x ∂x
y ∂y
y
∂
∂ =
v 3 x
,
z y
∂w
∂x
=
4y ∂w
4x ∂w
4xy
, = and = – 2
z ∂y
z ∂z
z
Now,
∂auvw
, , f
∂bxyz
, , g =
∂u
∂x
∂v
∂x
∂w
∂u
∂y
∂v
∂y
∂w
∂u
∂z
∂v
∂z
∂w
=
2yz – 2
x
3z y
4y 2z x
−3zx
2
y
4x
2y
x
3x
y
–4xy
∂x
∂y
∂z
z z 2
z
⇒
∂ uvw , ,
∂ xyz , ,
But, we have
∴
∂
∂
=– 2yz
2
x
a f
b g
∂bxyz
, , g
∂auvw
, , f
bxyz , , g
auvw , , f
L
M
NM
2
2
12x yz 12x
−
2 2
yz yz
= 0 + 48 + 48 = 96.
= 1
96 .
a f
b g
∂ uvw , ,
×
∂ xyz , ,
O
P
QP
– 2z
x
L
M
NM
= 1 (Property 1)
−12xyz
12xy
−
2
yz yz
O
P
QP
+ 2y
x
L
M
NM
12xz 12xyz
+
2 yz zy
O
P
QP

DIFFERENTIAL CALCULUS-II 99
Example 4. If u = xyz, v = x2 + y2 + z2 , w = x + y + z find J (x, y, z).
Sol. Here we calculate J(u, v, w) as follows:
(U.P.T.U., 2002)
∂u
∂u
∂u
J(u, v, w) =
∂x
∂v
∂x
∂w
∂y
∂v
∂y
∂w
∂z
∂v
∂z
∂w
=
yz
2x 1
zx
2y 1
xy
2z
1
∂x
∂y
∂z
= yz (2y – 2z) – zx (2x – 2z) + xy (2x – 2y)
= 2 [y2z – yz2 – zx2 + z2x + xy (x – y)]
= 2 [– z (x2 – y2 ) + z2 (x – y) + xy (x – y)]
= 2 (x – y)[– zx – zy + z2 + xy]
= 2 (x – y)[z (z – x) – y (z – x)]
= 2 (x – y) (z – y) (z – x)
= – 2 (x – y) (y – z) (z – x)
2 2
As x – y = ( x– y)( x+ y)
But J(x, y, z)· J(u, v, w) = 1
∴ J (x, y, z) =
1
−
2 x−y y− z z−x b gb ga f .
Example 5. If x =
v =
vw , y = wu , z = uv and u = r sin θ cos φ,
r sin θ sin φ, w = r cos θ, calculate ∂bxyz
, , g
∂br,
θφ , g
.
Sol. Here x, y, z are functions of u, v, w and u, v, w are functions of r, θ, φ so we apply IInd
property.
∂bxyz
, , g
∂ r,
θφ ,
= ∂bxyz
, , g ∂auvw
, , f ·
∂ uvw , , ∂ r,
θφ ,
...(i)
Consider
b g
b g
a f =
∂ xyz , ,
∂ uvw , ,
=
= 1
8
a f
∂x
∂u
∂y
∂u
∂z
∂u
1
2
1
2
0
L
NM
w
u
v
u
∂x
∂v
∂y
∂v
∂z
∂v
w
v
1
2
1
2
∂x
∂w
∂y
∂w
∂z
∂w
0
v u
u w
w
v
u
v
+
b g
1
2
1
2
0
v w
w u
v
w
u
w
u
v
O
QP
= 1
8
1+ 1 = 2
8
= 1
4

100 A TEXTBOOK OF ENGINEERING MATHEMATICS—I
b g
a f
⇒
∂ xyz , ,
=
∂ uvw , ,
1
4
...(ii)
∂u
∂u
∂u
Next
∂auvw
, , f
∂br,
θφ , g =
∂r
∂v
∂r
∂θ
∂v
∂θ
∂φ
∂v
∂φ
=
sinθcosφ sinθsin φ
rcosθcosφ rcosθsin φ
−rsinθsinφ
rsinθcosφ
∂w
∂w
∂w
cosθ −r
sin θ 0
∂r
∂θ ∂φ
= sinθ cos φ (r2 sin2 θ cos φ) – r cos θ cos φ (–r2 sin θ cos θ cos φ)
+ r2 sin θ sin φ (sin2 θ sin φ + cos2 θ sin φ)
= r2 sin θ cos2 φ (sin2 θ + cos2 θ) + r2 sin θ sin2 ⇒
φ
∂auvw
, , f
= r
∂br,
θφ , g
2 sin θ cos2 φ + r2 sin θ sin2 φ = r2 sin θ ...(iii)
Using (ii) and (iii) in equation (i), we get
b g
b g
∂ xyz , ,
∂ r,
θφ ,
= 1
4 × r2 sin θ = r2 sin θ
.
4
Example 6. If u = x (1 – r 2 ) −1/2 , v = y (1 – r 2 ) −1/2 , w = z (1 – r 2 ) −1/2
a f
b g
where r = 2 2 2
x + y + z , then find ∂ uvw , ,
.
∂ xyz , ,
and
Sol. Since r 2 = x 2 + y 2 + z 2
∴
∂r
∂x
= x ∂r
y ∂r
z
, = , =
r ∂y
r ∂z
r
Differentiating partially u = x (1 – r2 ) –1/2 w.r.t. x, we get
∂u
∂x
= 1 − r
= 1
⇒
∂u
∂x
=
1 − r + x
3
2
1 − r 2 e j
Differentiating partially u w.r.t. y, we get
∂u
F −
= x
∂y
1
−3
2
1 − r 2
2
⇒
∂u
∂y
=
∂u
∂z
=
e
−1
2 2 j + x − F I
HG KJ 1
−3
2
(– 2r) 1 − 2 e r j .
2
∂
∂
e
−1
−3
2
− r 2 j + rx 2
1 − r 2 e j . x 1
2
HG I K J
2 2
xy
2 e1−rj xz
2 e1−rj 3
2
3
2
e j · (– 2r) ∂
r =
r
∂y
=
1 − r
r
x
xr
2 e1−rj 3
2
+
2
x
2 e1−rj · y
r
3
2

DIFFERENTIAL CALCULUS-II 101
Similarly,
Thus,
∂v
∂x
∂w
∂x
∂ uvw , ,
∂ xyz , ,
=
=
a f
= b g
=
yx
, 3
2
1 − r 2 e j
∂
2 2
v 1 − r + y
= 3 ,
∂y
2
1 − r 2 e j
∂v
∂z
=
yz
3
2
1 − r 2 e j
zx
3 ,
2
1 − r 2 e j
∂w
∂y
=
zy
2
1 − r
, 3
2
∂
2 2
w 1 − r + z
= 3 .
∂z
2
1 − r 2 e j
= 1
2 2
1 − r + x
e j
xy
3
3
3
2
2
2
e j e j e j
1−r21−r21−r2 yx
2 2
1 − r + y yz
3
2
1 2 e − r j
3
2
1 2 e −rj 3
2
1 2 e −rj
zx zy
2 2
1 − r + z
3
2
1− 2 e r j
3
2
1−r2
e j
3
2
1 − r 2 e j
1
9
2 2
1 − r + x
yx
xy
2 2
1 − r + y
xz
yz
2
1 − r 2 zx zy 1 − r + z
e j
2 2
e
−9
2
− r 2 [(1 – r j 2 + x2 ) {(1 – r2 + y2 )(1 – r2 + z2 )– y2z2 }
= 1 − r
xz
– xy {xy (1 – r 2 + z 2 ) – xyz 2 } + xz{xy 2 z – zx(1 – r 2 + y 2 )}]
−9
2 2 [(1 – r e j 2 + x2 ) (1 – r2 + y2 ) (1 – r2 + z2 )
−9
2 2 e j [(1 – r2 ) 3 + (1 – r2 ) 2 (x2 + y2 + z2 )]
−9
2 2
= 1 − r
e j [(1 – r2 ) 3 + (1 – r2 ) 2 r2 ]
−9
2
− r 2 e j . (1 – r2 ) 2 [1 – r2 + r2 −5
] = 2
1 − r 2 e j .
= 1 − r
= 1
– (1 – r 2 ) (y 2 z 2 + x 2 y 2 + x 2 z 2 ) – x 2 y 2 z 2 ]
Example 7. Verify the chain rule for Jacobians if x = u, y = u tan v, z = w. (U.P.T.U., 2008)
Sol. We have x = u ⇒ ∂x
∂x
∂x
= 1, =
∂u
∂v
∂w
= 0
y = u tan v ⇒ ∂y
∂y
= tan v,
∂u
∂v
= u sec2 v, ∂y
= 0
∂w
z = w ⇒ ∂z
∂z
=
∂u
∂v
= 0 , ∂z
= 1
∂w
b g
b g
J = ∂ x, y, z
∂ u, v, w
=
1 0 0
tan v
2
usecv 0
0 0 1
= u sec 2 v ...(i)

102 A TEXTBOOK OF ENGINEERING MATHEMATICS—I
∂w
∂z
Solving for u, v, w in terms of x, y, z, we have
u = x
y –1 v = tan
u
=
− y
tan
x
1
w = z
∴ ∂
∂
=
∂
∂
= ∂
∂
=
∂
∂
= −
+
∂
∂
=
+
∂
∂
=
∂
∂
= ∂
u
x
u
1, y
u
z
0, v
x
y
, 2 2
x y
v
y
x
, 2 2
x y
v
z
w
0, x
w
∂y
= 0 and
=
1
b g
J′ = ∂
=
∂
−
+ +
=
+
=
F I +
HG KJ =
u, v, w
bx, y, zg
1
y
2 2
x y
0
0
x
2 2
x y
0
0
0
1
x
2 2
x y
1
2
y
x 1 2
x
1
2
usec v
Hence from (i) and (ii), we get
2.1.3 Jacobian of Implicit Functions
J.J′ = u sec2 1
v. = 1 2 .
usec v
If the variables u, v and x, y be connected by the equations
f (u, v, x, y) 1 = 0 ...(i)
f (u, v, x, y) 2 = 0 ...(ii)
i.e., u, v are implicit functions of x, y.
Differentiating partially (i) and (ii) w.r.t. x and y, we get
∂f1
∂f1
∂u
∂f1
∂v
+ · + ·
∂x
∂u
∂x
∂v
∂x
= 0 ...(iii)
∂f1
∂y
+ ∂f1
∂u
∂f1
∂u
· + ·
∂u
∂y
∂v
∂y
= 0 ...(iv)
∂f2
∂f2
∂u
∂f
2 ∂v
+ · + · = 0
∂x
∂u
∂x
∂v
∂x
...(v)
∂f
2 ∂f
2 ∂u
∂f
2 ∂v
+ · + · = 0
∂y
∂u
∂y
∂v
∂y
...(vi)
Now,
b g
a f
∂ f1, f2
∂ uv ,
a f
b g =
∂ uv ,
×
∂ xy ,
=
∂f
∂u
∂f
∂u
∂f
∂v
1 1
∂f
∂v
2 2
∂u
∂x
∂v
∂x
∂f
∂u
∂f
+
∂u
∂ x ∂v
∂v
∂x
∂f
∂u
∂f
+
∂u
∂ x ∂v
∂v
∂x
∂u
∂y
∂v
∂y
∂f
∂u
∂f
∂v
+
∂u
∂ y ∂v
∂y
1 1 1 1
∂f
∂u
∂f
∂v
+
∂u
∂ y ∂v
∂y
2 2 2 2
...(ii)

DIFFERENTIAL CALCULUS-II 103
Using (iii), (iv), (v) and (vi) in above, we get
Thus,
⇒
b g
a f
∂ f1, f2
∂ uv ,
a f
b g
∂ uv ,
×
∂ xy ,
a f
b g
∂ uv ,
∂ xy ,
=
− ∂
∂
− ∂
f
x
f
∂x
− ∂f
∂y
1 1
− ∂f
∂y
2 2
b g
b g
bf1, f2g
∂bxy
, g
bf1, f2g
∂auv
, f
∂ f1, f2
= (– 1)2
∂ xy ,
= (– 1)2
Similarly for three variables u, v, w
a f
b g
∂ uvw , ,
∂ xyz , ,
= (– 1)3
∂
∂
b 1 2 3g
∂ f , f , f
∂bxyz
, , g
∂ f , f , f
∂ uvw , ,
b 1 2 3g
a f
= (– 1) 2
and so on.
Example 8. If u3 + v3 = x + y, u2 + v2 = x3 + y3 , show that
and
a f
2 2
∂ f
∂ x
∂ f
∂ x
∂ f
∂ y
1 1
∂ f
∂ y
2 2
∂ uv ,
∂bxy
, g =
y − x
.
2uvau−vf
(U.P.T.U., 2006)
Sol. Let f1 ≡ u3 + v3 f2 − x − y = 0
≡ u2 + v2 − x3 − y3 = 0
Now,
b g
b g =
∂ f1, f2
∂ xy ,
b 1 2g
∂ f , f
∂ uv ,
a f =
∂f
∂x
∂f
∂x
∂f
∂y
1 1
∂f
∂y
2 2
∂f1
∂f1
∂u
∂v
∂f
2 ∂f
2
∂u
∂v
b 1 2g
=
= 3 3
–1 –1
2 2
–3 x –3y
2 2
u v
2u 2v
= 3 (y 2 – x 2 )
= 6 uv (u – v)
Thus,
∂auv
, f
∂bxy
, g =
∂ f , f
∂bxy
, g =
∂bf1,
f2g
∂auv
, f
3
2 2 ey − x j 2 2
( y − x )
=
. Hence Proved.
6uvau– vf
2uv
( u − v)
Example 9. If u, v, w are the roots of the equation in k,
x
a+ k
+
y
b+ k
+
z
c+ k
= 1, prove
that ∂bxyz
, , g au−vfav−wfaw−uf = –
∂ uvw , , a−b b−c c−a a f
a fa fa f .

104 A TEXTBOOK OF ENGINEERING MATHEMATICS—I
x
Sol. We have
a+ k
+
y
b+ k
+
z
c+ k
= 1
or x (b + k) (c + k) + y (a + k) (c + k) + z (a + k) (c + k) = (a + k) (b + k) (c +k)
or k3 + k2 (a + b + c – x – y – z) + k{bc + ca + ab – (b + c) x – (c + a)
y – (a + b)z} + (abc – bcx – cay – abz) = 0
Since its roots are given to be u, v, w, so we have
u + v + w = – (a + b + c – x – y – z)
uv + vw + wu = bc + ca + ab – (b + c) x – (c + a)y – (a + b)z
uvw =– (abc – bcx – cay – abz)
Let f ≡ u + v + w + a + b + c – x – y – z = 0
1
f ≡ uv + vw + wu – bc – ca – ab + (b + c) x + (c + a) y + (a + b)z = 0
2
f ≡ uvw + abc – bcx – cay – abz = 0
3
and
Now,
Thus,
∴
b 1 2 3g
∂ f , f , f
∂ xyz , ,
b g =
b 1 2 3g
∂ f , f , f
∂ uvw , ,
=
a f =
a f
b g
∂ uvw , ,
∂ xyz , ,
b g
a f
∂ xyz , ,
∂ uvw , ,
∂f1
∂f1
∂f1
∂x
∂y
∂z
∂f2
∂f2
∂f2
∂x
∂y
∂z
∂f3
∂f3
∂f3
∂x
∂y
∂z
=
1 0 0
b+ c a−b a−c bc c a −bba−c ∂f
∂u
∂f
∂u
∂f
∂u
a f a f
∂f
∂v
∂f
∂v
∂f
∂v
∂f
∂w
∂f
∂w
∂f
∂w
1 1 1
2 2 2
3 3 3
1 0 0
= v+ w u−v u−w vw w u −vvu−w =(u – v)(u – w)(v – w)
= (–1)3
∂ f , f , f
∂ xyz , ,
∂ f , f , f
∂ uvw , ,
=
a f a f
b 1 2 3g
–1 –1 –1
b+ c c+ a a+ b
– bc – ca – ab
a f a f a f
= (a – b)(a – c)(b – c)
1 1 1
v + w w + u u+ v
vw wu uv
a f a f a f
b g aa−bfaa−cfab−cf = – b 1 2 3g
au−vfau−wfav−wf a f
a − fa − fa − f
aa−bfab−cfaa−cf =– u v v w u w
(C 2 → C 2 – C 1 , C 3 → C 3 – C 1 )
. Hence proved. As JJ′ = 1.
Example 10. If u = 2axy, v = a (x 2 – y 2 ) where x = r cosθ, y = r sin θ, then prove that
a f
a f
∂ uv ,
∂ r,
θ =– 4a2 r 3 .

DIFFERENTIAL CALCULUS-II 105
or
and
Sol. We have u =2axy, v = a (x 2 – y 2 )
Now,
Hence
a f
b g =
∂x
∂v
∂y
∂v
=
∂x
∂y
2 2 ay
2ax ax
−2ay
= – 4a2 (x2 + y2 )
auv , f
bxy , g = − 4a2r2 As +
2
=
bxy , g
ar, θ f =
∂x
∂r
∂y
∂x
∂θ
∂y
=
cos θ
sin θ
– r sin θ
r cos θ
= r
∂r
∂θ
auv , f ∂auv
, f ∂bxy
, g
= · ar, θ f ∂bxy
, g ∂ar,
θ f
= (− 4a2r2 ).r = − 4a2r3 . Hence proved.
∂ uv ,
∂ xy ,
∂
∂
∂
∂
∂
∂
∂u
∂u
2 2
x y r
Example 11. If u3 + v3 + w3 = x + y + z, u2 + v2 + w2 = x3 + y3 + z3 , u + v + w = x2 + y2 + z2 ,
then show that
∂auvw
, , f =
∂ xyz , ,
bx−ygby−zgaz−xf ⋅
u−v v−w w−u and
b g
a fa fa f
Sol. Let f 1 ≡ u 3 + v 3 + w 3 – x – y – z = 0
f 2 ≡ u 2 + v 2 + w 2 – x 3 – y 3 – z 3 = 0
f 3 ≡ u + v + w – x 2 – y 2 – z 2 = 0
Now,
⇒
b 1 2 3g
∂ f , f , f
∂ xyz , ,
b g
b 1 2 3g
∂ f , f , f
∂ xyz , ,
b g
∂ f1, f2, f3
∂ uvw , ,
=
=
b g = a f
∂f
∂x
∂f
∂x
∂f
∂x
∂f
∂y
∂f
∂y
∂f
∂y
∂f
∂z
∂f
∂z
∂f
∂z
1 1 1
2 2 2
3 3 3
–1 0 0
2
–3x
3
2 2
x −y 3
2 2
x −z
–2x
2 x−y 2 x−z =
e j e j
b g a f
–1 –1 –1
2
–3 x
2
–3 y
2
–3z
–2 x –2 y –2z
= – 6[(x 2 – y 2 ) (x – z) – (x 2 – z 2 ) (x – y)]
= – 6 (x – y) (x – z) [(x + y) – (x + z)]
= 6 (x – y) (y – z) (z – x)
∂f
∂u
∂f
∂u
∂f
∂u
∂f
∂v
∂f
∂v
∂f
∂v
∂f
∂w
∂f
∂w
∂f
∂w
1 1 1
2 2 2
3 3 3
=
3u 3v 3w
2u 2v 2w
1 1 1
C →C −C , C →C −C
2 2 2
2 2 1 3 3 1

106 A TEXTBOOK OF ENGINEERING MATHEMATICS—I
=
b g b g
a f a f C →C −C , C →C −C
2 2 2 2 2
3u 3 v − u 3 w − u
2u 2 v− u 2 w − u
1 0 0
Expand it with respect to third row, we get
= 6[(v2 – u2 ) (w – u) – (w2 – u2 ) (v – u)]
⇒
∂bf1,
f2, f3g
∂ uvw , ,
= 6 (v – u) (w – u) [(v + u) – (w + u)]
Hence
= – 6 (u – v) (v – w) (w – u)
a f
∂auvw
, , f = (–1)
∂bxyz
, , g
3
∂bf1,
f2, f3g
∂bxyz
, , g = +
∂bf1,
f2, f3g
∂auvw
, , f
6bx−ygby−zgaz−xf
6au−vfav−wfaw−uf
bx−ygby−zgaz−xf =
· Hence proved.
au−vfav−wfaw−uf Example 12. u, v, w are the roots of the equation
(x – a) 3 + (x – b) 3 + (x – c) 3 = 0, find ∂
b g
b g
uvw , ,
.
∂ abc , ,
2 2 1 3 3 1
Sol. We have (x – a) 3 + (x – b) 3 + (x – c) 3 = 0
x3 – a3 – 3xa (x – a) + x3 – b3 – 3xb (x – b) + x3 – c3 – 3xc (x – c) = 0
or 3x3 – 3x2 (a + b + c) + 3x (a2 + b2 + c2 ) – (a3 + b3 + c3 ) = 0
Since u, v, w are the roots of this equation, we have
u + v + w = a + b + c
uv + vw + wu = a2 + b2 + c2 3 3 3
a + b + c
uvw =
3
Let f1 ≡ u + v + w – a – b – c = 0
f2 ≡ uv + vw + wu – a2 – b2 – c2 As α+ β+ γ
αβ + βγ + γα
αβγ
= −ba
= ca
= − da
= 0
and
Now
b 1 2 3g
∂ f , f , f
∂ abc , ,
b g
b 1 2 3g
∂ f , f , f
∂ u, v, w
b g
=
f 3 = uvw –
−1 −1 −1
−2a −2b −2c
−a −b −c
3 3 3
a + b + c
3
=
a f a f F
e j e j
HG ,
−1
0 0
−2a 2 a−b 2 a−c −a a −b a −c
2 2 2 2 2 2 2 2
c → c −c
c → c −c
2 2 1
3 3 1
= – 2{(a – b) (a 2 – c 2 ) – (a – c) (a 2 – b 2 )} = – 2(a – b) (b – c) (c – a)
=
F
a f a f HG ,
1 1 1 1 0 0
v+ w u+ w v+ u = v+ w u−v u−w vw wu uv vw w u −vvu−w I
KJ
c → c −c
c → c −c
2 2 1
3 3 1
I
KJ

DIFFERENTIAL CALCULUS-II 107
Thus
a f
a f
∂ uvw , ,
∂ abc , ,
= (u – v) v(u – w) – (u – w) w(u – v)
= – (u – v) (v – w) (w – u)
a f
= −
2.1.4 Functional Dependence
1 3
b 1 2 3g
∂ f , f , f
∂aabc
, , f
∂ f , f , f
∂ uvw , ,
b 1 2 3g
a f
a fa fa f
a fa fa f
−2a−b b−c c−a = −
− u−v v−w w−u a fa fa f
a fa fa f .
a−b b−c c−a = −2
u−v v−w w−u Let u = f (x, y), v = f (x, y) be two functions. Suppose u and v are connected by the relation
1 2
f (u, v) = 0, where f is differentiable. Then u and v are called functionally dependent on one
another (i.e., one function say u is a function of the second function v) if the ∂u
∂u
∂v
, ,
∂x
∂y
∂x
and
∂v
∂y
are not all zero simultaneously.
Necessary and sufficient condition for functional dependence (Jacobian for functional
dependence functions):
Let u and v are functionally dependent then
f (u, v) = 0 ...(i)
or
Differentiate partially equation (i) w.r.t. x and y, we get
∂f
∂u
∂f
· +
∂u
∂x
∂v
∂v
∂x
= 0 ...(ii)
∂f
∂u
∂f
∂v
· + ·
∂u
∂y
∂v
∂y
= 0 ...(iii)
There must be a non-trivial solution for ∂f
∂u
Thus,
∂u
∂x
∂u
∂y
∂u
∂x
∂v
∂x
∂v
∂x
∂v
∂y
∂u
∂y
∂v
∂y
a f
∂f
≠ 0,
∂v
≠ 0 to this system exists.
= 0 For non-trivial solution xAx = 0
= 0 Changing all rows in columns
∂ uv ,
or
= 0
∂bxy
, g
Hence, two functions u and v are “functionally dependent” if their Jacobian is equal to
zero.

108 A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Note: The functions u and v are said to be “functionally independent” if their Jacobian is not equal
to zero i.e., J(u, v) ≠ 0
Similarly for three functionally dependent functions say u, v and w.
J(u, v, w) = ∂auvw
, , f = 0.
∂ xyz , ,
b g
Example 13. Show that the functions u = x + y – z, v = x – y + z, w = x 2 + y 2 + z 2 – 2yz are
not independent of one another. Also find the relation between them.
Sol. Here u = x + y – z, v = x – y + z and w = x 2 + y 2 + z 2 – 2yz
Now,
a f
b g =
∂ uvw , ,
∂ xyz , ,
=
∂u
∂x
∂v
∂x
∂w
∂x
∂u
∂y
∂v
∂y
∂w
∂y
∂u
∂z
∂v
∂z
∂w
∂z
1 1 0
1 –1 0
2x 2y−2z 0
=
1 1 –1
1 –1 1
2x 2y−2z 2z−2y (C 3 → C 3 + C 2 )
= 0. Hence u, v, w are not independent.
Again u + v = x + y – z + x – y + z = 2x
u – v = x + y – z – x + y – z = 2 (y – z)
∴ (u + v) 2 + (u – v) 2 = 4x2 + 4 (y – z) 2
= 4(x2 + y2 + z2 – 2yz) = 4w
⇒ (u + v) 2 + (u – v) 2 = 4w
or 2(u2 + v2 ) = 4w or u2 + v2 = 2w.
Example 14. Find Jacobian of u = sin –1 x + sin –1 y and v = x
relation between u and v.
2
1 − y + y 1 − x . Also find
Sol. We have u = sin –1 x + sin –1 y, v = 2
x 1 y
Now,
= −
xy
a f
b g =
∂ uv ,
∂ xy ,
1−x 1−
y
∂u
∂x
∂v
∂x
+ 1− 1 +
∂u
∂y
∂v
∂y
xy
2 2 2 2
=
1−x 1−
y
1 −y− − + y 1 − x
1
1
1 −x 1 −y
2
2 2
xy
−
xy
1−x1−y 2
2 2
2
+ 1 −x
= 0. Hence u and v are dependent.
{ }
{ − + − }
Next, u = sin –1 x + sin –1 y ⇒ u = sin –1 2 2
x 1− y + y 1−x
⇒ sin u = x
2
1 − y + y
2
1 − x = v
or v = sin u.
−1 −1 −1
2 2
As sin A + sin B = sin A 1 B B 1 A
2

DIFFERENTIAL CALCULUS-II 109
or
Example 15. Show that ax2 + 2hxy + by2 and Ax2 + 2Hxy + By2 are independent unless
a
A =
h b
=
H B .
Sol. Let u = ax2 + 2hxy + by2 v = Ax 2 + 2Hxy + By 2
a f
b g
If u and v are not independent, then ∂ uv ,
∂ xy ,
∂u
∂x
∂v
∂x
∂u
∂y
∂v
∂y
=
= 0
2ax+ 2hy 2hx + 2by
2Ax+ 2Hy 2Hx + 2By
⇒ (ax + hy) (Hx + By) – (hx + by) (Ax + Hy) = 0
⇒ (aH – hA) x2 + (aB – bA) xy + (hB – bH) y2 = 0
But variable x and y are independent so the coefficients of x2 and y2 must separately vanish
and therefore, we have
aH – hA = 0 and hB – bH = 0 i.e., a h
=
A H
i.e,
a
A =
h b
= · Hence proved.
H B
Example 16. If u = x2 e –y cos hz, v = x2 e –y sin hz and w = 3x4 e –2y then prove that u, v, w are
functionally dependent. Hence establish the relation between them.
Sol. We have u = x2 e –y cos hz, v = x2 e –y sin hz, w = 3x4 e –2y
a f
b g =
∂ uvw , ,
∂ xyz , ,
∂u ∂x ∂u ∂y ∂u ∂z
∂v ∂x ∂v ∂y ∂v ∂z
∂w ∂x ∂w ∂y ∂w ∂z
=
= 0
and h
H
= b
B
−y 2 −y 2 −y
2xe
cos hz −x
e cos hz x e sin hz
−y 2xe
sin hz
2 −y −x
e sin hz
2 −y
x e coshz
3 −2y 12xe 4 −2y
−6xe
0
= 2x e –y cos hz{0 + 6x6 e –3y cos hz} + x2 e –y cos hz{0 – 12x5 e –3y cos hz}
+ x2 e –y sin hz{–12x5 e –3y sin hz + 12x5 e –3y sin hz}
= 12x7 e –4y cos h2 z – 12x7 e –4y cos h2 z = 0
Thus u, v and w are functionally dependent.
Next, 3u2 – 3v2 = 3(x4 e –2y cos h2 z – x4 e –2y sin h2 z) = 3x4 e –2y (cos h2 z – sin h2 z)
= 3x4 e –2y
⇒ 3u 2 – 3v 2 = w.
1. If x = r cos θ, y = r sin θ find ∂ xy ,
.
∂ r,
θ
2. If y = 1 xx 2 3
x1
EXERCISE 2.1
b g
a f
L
NM
Ans. 1
r
x3x1 , y = , y = 2 3
x2
xx 1 2
show that the Jacobian of y , y , y with respect to x ,
1 2 3 1
x3
x , x is 4. (U.P.T.U., 2004(CO), 2002)
2 3
O
QP

110 A TEXTBOOK OF ENGINEERING MATHEMATICS—I
3. If x = r sin θ cos φ, y = r sin θ sin φ, z = r cos θ, show that
b g
b g
∂ xyz , ,
∂ r,
θφ ,
= r 2 sin θ. (U.P.T.U., 2000)
b g
4. If u = x + y + z, uv = y + z, uvw = z, evaluate ∂ xyz , ,
. (U.P.T.U., 2003) Ans. uv ∂auvw
, , f 2
2 2 e j
, find
b g
5. If u = y
2
, v =
2 x
x + y
2x
∂ u, v
.
∂ bx, yg
y
Ans. −
NM 2x
6. If x = a cos h ξ cos η, y = a sin h ξ sin η, show that
b g
∂ xy , 1
=
∂bξη
, g 2 a2 (cos h 2ξ – cos 2η).
7. If u3 + v + w = x + y2 + z2 , u + v3 + w = x2 + y + z2 , u + v + w3 = x2 + y2 + z, then evaluate
a f
b g
∂ uvw , ,
.
∂ xyz , ,
L
M
NM
Ans.
1− 4xy − 4yz − 4zx + 6xyz
2 2 2 2 2 2
27u v w + 2 − 3 u + v + w
L
O
QP
b g O
P
e jP
QP
auvw , , f .
∂bxyz
, , g
L −2( x−y) ( y−z) ( z−x) Ans.
N
M
O
a − − − Q
P
u vfav wfaw uf
8. If u, v, w are the roots of the equation (λ – x) 3 + (λ – y) 3 + (λ – z) 3 = 0 in λ, find ∂
(U.P.T.U., 2001)
9. If u = x + x + x + x , uv = x + x + x , uvw = x + x and uvwt = x , show that
1 2 3 4 2 3 4 3 4 4
∂bx1,
x2, x3, x4g
∂ uvwt , , ,
a f
b g
a f = u 3 v 2 w.
b g
a f
10. Calculate J = ∂ uv ,
∂ xy ,
and J′ = . Verify that JJ′ = 1 given
∂ xy ,
∂ uv ,
(i) u = x + y
2
x
, v = y
2
.
x
L
N
M
Ans. = 2y
J
x J
x
, ′=
2y
2u −2u
(ii) x = eu cos v, y = eu sin v. Ans. J = e , J′ = e .
a f
a f
11. Show that ∂ uv ,
∂ r,
θ = 6r3 sin 2θ given u = x2 – 2y2 , v = 2x2 – y2 and x = r cos θ, y = r sin θ.
12. If X = u2v, Y = uv2 and u = x2 – y2 , v = xy, find ∂aXY
, f
∂bxy
, g . L
2 2 2 2 2 2
2
Ans. 6x
yex + yjex – y
NM
j O
QP
a f
13. Find ∂ uvw , ,
,
∂ xyz , ,
b g if u = x2 , v = sin y, w = e –3z . Ans. –6 3 − z
a f
b g
O
Q
P
e xcos y
14. Find ∂ uvw , ,
, if u = 3x + 2y – z, v = x – y + z, w = x + 2y – z. Ans. –2
∂ xyz , ,
b gb ga f
15. Find J (u, v, w) if u = xyz, v = xy + yz + zx, w = x + y + z. Ans. x− y y−z z−x

DIFFERENTIAL CALCULUS-II 111
16. Prove that u, v, w are dependent and find relation between them if u = xe y sin z, v = xe y
cos z, w = x 2 e 2y . Ans. dependent,
b g
2 2
u + v = w
2
3x
2 y+ z x−y 2
17. u = , v =
2by+
z
2 , w = . Ans. dependent, uvw = 1
g 3bx−yg
x
18. If X = x + y + z + u, Y = x + y – z – u, Z = xy – zu and U = x2 + y2 – z2 – u2 , then show
that J = ∂a
XYZU , , , f
= 0 and hence find a relation between X, Y, Z and U.
∂ xyzu , , ,
19. If u =
b g
x
, v =
y− z
y
, w =
z− x
Ans. XY = U +2 Z
z
, then prove that u, v, w are not independent and also
x−y find the relation between them. Ans. uv + vw + wu + 1= 0
20. If u = x + 2y + z, v = x – 2y + 3z, w = 2xy – xz + 4yz – 2z2 , show that they are not
2 2
independent. Find the relation between u, v and w. Ans. 4w
= u − v
x+ y
21. If u =
1 − xy
uv ,
. Are u and v functionally related? If
∂bxy
, g
yes find the relationship.
22. If u = x + y + z, uv = y + z, uvw = z, show that
[Ans. yes, u = tan v]
∂bxyz
, , g
∂ uvw , ,
=
2
uv.
and v = tan –1 x + tan –1 y, find ∂
a f
23. If x 2 + y 2 + u 2 – v 2 = 0 and uv + xy = 0 prove that ∂
a f
auv , f
∂bxy
, g
=
2 2
2 2 .
x − y
u + v
24. Find Jacobian of u, v, w w.r.t. x, y, z when u = yz
x v
zx
y w
xy
, = , = . [Ans. 4]
z
2.2 APPROXIMATION OF ERRORS
Let u = f (x, y) then the total differential of u, denoted by du, is given by
du =
∂f
∂f
dx +
∂x
∂y
dy ...(i)
If δx and δy are increments in x and y respectively then the total increment δu in u
is given by δu = f (x + δx, y + δy) – f (x, y)
or f (x + δx, y + δy) = f (x, y) + δu ...(ii)
But δu ≈ du, δx ≈ dx and δy ≈ dy
∴ From (i) δu ≈
∂f
∂f
δx + δy ...(iii)
∂x
∂y

112 A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Using (iii) in (ii), we get the approximate formula
f (x + δx, y + δy) ≈ f (x, y) + ∂f
∂x
δx +
∂f
∂y
δy ...(iv)
Thus, the approximate value of the function can be obtained by equation (iv). Hence
δx or dx = Absolute error
δx dx
or
x x
= Proportional or Relative error
and 100 × dx
dx
or 100 ×
x x
= Percentage error in x.
Example 1. If f (x, y) = x y
2
Sol. We have f (x, y) = x y
2
∴
∂f
∂x
1
10 , compute the value of f when x = 1.99 and y = 3.01.
(U.P.T.U., 2007)
1
10
1
10
= 2xy
, ∂f
1 2
= x y
∂y
10
−
x+ δx=2+
–0.01 199 .
Let x = 2, δx = – 0.01 As
y+ δy=
1+ 201 . = 301 .
y = 1, δy = 2.01
Now, f (x + δx, y + δy) = f (x, y) + ∂f
∂x
δx +
9
10
∂f
∂y
δy
1
⇒ f{2 + (– 0.01), 1 + 2.01} = f (2, 1) + 2× 2af 1 10 × (– 0.01) + 1
10 (2)2 . 1
1
9
af 10
−
a f
a f =
× (2.01)
⇒ f (1.99, 3.01) ≈ 22 × 110
+ (– 0.04) + 0.804
≈ 4 – 0.04 + 0.804 = 4.764.
Example 2. The diameter and height of a right circular cylinder are measured to be 5 and
8 cm. respectively. If each of these dimensions may be in error by ± 0.1 cm, find the relative
percentage error in volume of the cylinder.
Sol. Let diameter of cylinder = x cm.
height of cylinder = y cm.
then V =
πx y
2
(radius =
4
x
2 )
∴
∂V
∂x
=
πxy ∂V
πx2
, =
2 ∂y
4
⇒ dV = ∂V
∂x
∂V
dx +
∂y
dy

DIFFERENTIAL CALCULUS-II 113
⇒ dV = 1
1
π xy. dx +
2 4 πx2 . dy
or
dV
V =
1
πxy.
dx
2
2
πx
y
+
1 2
πxdy
4
2 = 2.
πx
y
4 4
dx dy
+
x y
or 100 × dV
= 2 F dxI
100 ×
V HG KJ + 100 ×
x
dy
y
Given x = 5 cm., y = 8 cm. and error dx = dy = ± 0.1. So 100 × dV
V = ± 100 2 01 F . 01 .
× + HG 5 8
= ± 5.25.
Thus, the percentage error in volume = ± 5.25.
Example 3. A balloon is in the form of right circular cylinder of radius 1.5 m and length
4 m and is surmounted by hemispherical ends. If the radius is increased by 0.01 m and the length
by 0.05 m, find the percentage change in the volume of the balloon. [U.P.T.U., 2005 (Comp.), 2002]
Sol. Let radius = r = 1.5 m, δr = 0.01 m
height = h = 4 m, δh = 0.05 m
1.5
volume (V) =πr 1.5 m
2 h + 2
3 πr3 + 2
3 πr3 = πr2h + 4
3 πr3
or
∴
∂V
∂r
=2πrh + 4πr2 , ∂V
= πr2
∂h
dV = ∂V
∂V
dr +
∂r
∂h
dh = (2πrh + 4πr2 ) dr + πr2dh dV
2
2πrh a + 2rf
πr
= dr +
V 2F
4rI
2F
4rI
πr
h+
πr
h+
3 H 3 K dh
= 3 2 2
=
= 1
9
H K
ah rf
dr + a3 + 4f
× +
r h r
3
15 . 12+ 6
3
3h+ 4r
a f
dh =
3
r 3h+ 4r
a f
a f [2(4 + 3)(0.01) + 1.5 (0.05)] δ
0215 .
[0.14 + 0.075] =
9
⇒ 100 × dV
V
0215 .
= 100 × = 2.389%
9
Thus, change in the volume = 2.389%.
4 m
1.5
Fig. 2.1
[2(h + 2r) dr +rdh]
r = dr
δh=
dh
I K J

114 A TEXTBOOK OF ENGINEERING MATHEMATICS—I
Example 4. Calculate the percentage increase in the pressure p corresponding to a reduction
of 1
2 % in the volume V, if the p and V are related by pV1.4 = C, where C is a constant.
Sol. We have pV1.4 Taking log of equation (i)
= C ...(i)
log p + 1.4 log V =
Differentiating
log C
1
p
14
· dp + .
V
or 100 × dp
p
· dV = 0 ⇒ dp
p
F
HG
= −1.4 100 ×
= −1.4 × −1
2
= 0.7
Hence increase in the pressure p = 0.7%.
= – 1.4 dV
V
F I
HG KJ dVI
KJ V
dV
V = = 1 1
%
2 2× 100
Example 5. In estimating the cost of a pile of bricks measured as 6′ × 50′ × 4′, the tape is
stretched 1% beyond the standard length. If the count is 12 bricks to one ft3 , and bricks cost Rs.
100 per 1000, find the approximate error in the cost. (U.P.T.U., 2004)
Sol. Let length (l) = x ...(i)
breadth (b) = y
height (h) = z
∴ V = lbh = xyz
or log V = log x + log y + log z
On differentiating
1
V
or 100 × dV
V
⇒ 100 × dV
V
dV = 1
x
dx + 1
y
= 100 × dx
x
= 1 + 1 + 1 = 3
y + 1
z dz
+ 100 × dy
y
+ 100 × dz
z
a f = 36 cube fit
dV =
3V
3 6× 50× 4
=
100 100
∴ Number of bricks in dV = 36 × 12 = 432
Hence, the error in cost = 432 × .100
= Rs. 43.20.
1000
Example 6. The angles of a triangle are calculated from the sides a, b, c of small changes δa,
δb, δc are made in the sides, show that approximately δA = a
[δa – δb cos C – δc. cos B]
2∆
where ∆ is the area of the triangle and A, B, C are the angles opposite to a, b, c respectively. Verify
that δA + δB + δC = 0. (U.P.T.U., 2001)

DIFFERENTIAL CALCULUS-II 115
Sol. From trigonometry, we have
2 2 2
cos A =
b + c −a
2bc
⇒ b2 + c2 – a2 = 2bc cos A
⇒ a2 = b2 + c2 – 2bc cos A
On differentiating, we get
a · da = b · db + c · dc – db · c cos A – b · dc cos A + bc sin A · dA
⇒ bc sin A · dA = a · da – b · db – c · dc + db · c cos A + b · dc cos A
⇒ 2∆ · dA = a · da – (b – c cos A) · db – (c – b cos A) dc
⇒ 2 ∆ · dA = a da – (a cos C + c cos A – c cos A) db – (a cos B + b cos A
– b cos A) dc
1
As ∆= bc sin A
2
a = bcos C+ ccos B
or 2 ∆ · dA = a da – db.a cos C – dc.a cos B
⇒ δA =
a
[δa – δb. cos C – δc. cos B]
2∆
Hence proved.
δa As δA
≈ dA
≈ da, δb
≈
δc
≈ dc
db
Similarly, δB =
b
[δb – δc. cos A – δa. cos C]
2∆
and δC =
c
[δc – δa. cos B – δb. cos A]
2∆
Adding δA, δB and δC, we get
δA + δB + δC =
1
[(a – b cos C – c cos B) δa + (b – a cos C – c cos A) δb
2∆
+ (c – a cos B – b cos A) δc]
=
1
[(a – a) δa + (b – b) δb + (c – c) δ c]
2∆
⇒ δA + δB + δC = 0. Verified.
Example 7. Show that the relative error in c due to a given error in θ is minimum when
θ = 45° if c = k tan θ.
Sol. We have
On differentiating, we get
c = k tanθ ...(i)
dc = k sec2 θ dθ ...(ii)
From (i) and (ii), we get
2 dc sec θd θ 2dθ
= =
c tan θ sin 2θ
Thus, dc
will be minimum if sin2θ is maximum
c
Since sin θ lies
i.e., sin 2θ = 1 = sin 90
between – 1 and 1
⇒ 2θ = 90 ⇒ θ = 45°. Hence proved.