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Title: Partial derivatives
Description: All concepts related to Partial derivatives
Description: All concepts related to Partial derivatives
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4
PARTIAL DERIVATIVES AND THEIR
A PPLICATIONS
aaaaa
4
...
Here in this
chapter, we extend the concept of functions of two or more variables
...
…(1)
wherein x, y are independent variables and z is the dependent variable
...
In actual life we frequently come across such functions for instance the area A of a rectangle
of length x(as one independent variable) and breadth y (as the 2nd independent variable)
given by A = xy
...
The function z = f (x, y) of two independent variables represents a surface in three
dimensional space referred to a set of co-ordinate axes x, y, z is completely analogous to a
real function of a single independent variable x, say, y = f (x) represented geometrically by a
curve in the xy-plane
...
2 PARTIAL DERIVATIVES
Consider the function z = f(x, y) of two independent variables x and y and extend the concept
of ordinary derivative of the function of one variable to the function z = f(x, y) by keeping y
constant while taking derivative with respect to x and keeping x constant while taking
derivative with respect to y
...
∂f
∂f
Further, in general
and ∂y are functions of both x and y and, therefore can be
∂x
differentiated for higher order derivatives with respect to x or y
...
∂z
∂z
Example 1: If z = eax + by f(ax – by), prove that b ∂x + a ∂y = 2 abz
[VTU, 2004]
Solution:
∴
Here
∂ z ( ax + by)
· a f (ax – by)+ e( ax + by) · f'(ax − by)· a
=e
∂x
Similarly,
∂z
= e(ax + by)· b · f (ax – by)+ e(ax +by) f'(ax − by)(−b)
∂y
{
}
∂z
∂z
(ax + by)
· a f (ax − by) + e(ax +by) f'(ax − by) a
b ∂x + a ∂y = b e
+ {a[e(ax + by) bf(ax – by) + e(ax+
ab e(ax + by)
=
= 2abz–
f(ax – by) + ab
e(ax + by)
by) f'(ax
– by)(– b)]}
· f(ax – by)
Example 2: Let r2 = x2 + y2 + z2 and V = r m, prove that Vxx + Vyy + Vzz = m(m + 1)r m – 2
[Raipur, 2005; PTU, 2006]
Partial Derivatives and their Applications
265
m
Solution: Given V = rm = (r2 )m/2 = (x2 + y2 + z2 ) 2
…(1)
Here Vxx denotes 2nd order partial derivative of V(x, y, z) with respect to x keeping y and z
constant
...
Example 5: Prove that
∂2 u ∂2 u
+
= 0, if
∂x2 ∂y2
*(i) u = (tan–1a) [log(x2 + y2)] + b tan–1 (y/x),
*[NIT Kurukshetra, 2008]
2xy
...
= ( tan−1 a ) 2
+b 2
2
∂y
x + y2 x
(x + y )
Again,
2y
bx
= ( tan−1 a ) 2
2 +
2
(
x
y
)
+
x
+ y2 )
(
(x2 + y2 )2 − 2y ⋅ 2y b ⋅ x(−1)(2y)
∂2 u
−1
tan
a
=
(
)
2
+
(x2 + y2 )2
∂y2
( x2 + y2 )
…(2)
2(x2 − y2 )
2bxy
= ( tan −1 a ) 2
− 2
2 2
2 2
(x + y ) (x + y )
Adding (1) and (2), we get
(ii)
Let
∂2u ∂2u
+
=0
∂x2 ∂y2
}
x = r cos θ
so that x2 + y2 = r2 and
y = r sin θ
y
θ = tan−1
x
2xy
−1 2 sin θ cos θ
−1 sin 2 θ
u = tan−1 2
= 2θ
2 = tan
2
2 = tan
cos 2 θ
−
θ
−
θ
x
y
cos
sin
Therefore
and
Similarly
y
∂u
∂θ
∂
=2
= 2 tan−1 = 2
x
∂x
∂x
∂x
…(4)
−2y
1
y
− 2 = 2
2
2
y x x +y
1 +
x
…(5)
4xy
∂2u
−1
⋅ 2x =
2 = −2y
2
∂x
( x2 + y2 )
( x2 + y2 )2
…(6)
−4xy
∂2u
2 =
2
∂y
( x + y2 )2
…(7)
On adding (6) and (7),
∂2u ∂2u
+
= 0
...
2
Example 6: If u =
log[x3
+
y3
+
z3
∂
∂
∂
-2
– 3xyz], show that ∂x + ∂y + ∂ z- u = −9 (x + y + z)
Solution: Given u = log (x3 + y3 + z- 3 – 3xyz)
...
268
Engineering Mathematics through Applications
∂u ∂u ∂u
Therefore first find, ∂x , ∂y , ∂z
∂u ∂
=
log (x3 + y3 + z3 − 3xyz)
∂x ∂x
(i
...
partial derivative of u with respect to x keeping y and z– constant)
∂u
∂ 3
1
=
(x + y3 + z3 − 3xyz)
∂x x3 + y3 + z3 − 3xyz ∂x
3x2 − 3yz
∂u =
∂x y , z x3 + y3 + z3 − 3xyz
⇒
…(2)
Similarly we obtain
and
3y2 − 3xz
∂u
∂y = 3
x + y3 + z3 − 3xyz
x,z
…(3)
3z2 − 3xy
∂u =
∂z x, y x3 + y3 + z3 − 3xyz
…(4)
On adding (2), (3) and (4), we get
2
2
2
∂u ∂u ∂u 3 ( x + y + z − xy − yz − zx )
+
+
=
∂x ∂y ∂z
x3 + y3 + z3 − 3xyz
3 ( x2 + y2 + z2 − xy − yz − zx )
∂u ∂u ∂u
∂x + ∂y + ∂z =
( x + y + z) ( x2 + y2 + z2 − xy − yz − zx )
or
∂u ∂u ∂u
3
⇒ ∂x + ∂y + ∂z = x + y + z
(
)
or
…(5)
U = 3(x + y+ z)–1
Likewise, obtain partial derivatives of expression (5) with respect to x, y, z respectively
i
...
∂
∂
U=
3(x + y + z)−1
∂x
∂x
or
∂U
= − 3(x + y + z)−2
∂x
…(6)
∂U
= −3(x + y + z)−2
∂y
…(7)
∂U
= − 3(x + y + z)−2
∂z
…(8)
Similarly
and
Partial Derivatives and their Applications
269
Adding (6), (7), and (8), we get
∂U ∂U ∂U
+
+
= −9(x + y + z)−2
∂x ∂y ∂z
or
∂
∂ ∂
−2
∂x + ∂y + ∂z U = −9(x + y + z)
or
∂ ∂ ∂
−2
∂x + ∂y + ∂z u = −9(x + y + z)
2
Hence the result
...
e
...
∂2v ∂2v ∂2v
Example 9: If v = log (x2 + y2 + z2), prove that (x2 + y2 + z2 ) 2 + 2 + 2 = 2
∂y
∂z
∂x
Solution: Given v = log (x2 + y2 + z2)
1
∂v =
2x
∂x y , z (x2 + y2 + z2 )
⇒
…(1)
And
(x2 + y2 + z2 ) ⋅ 1 − x ⋅ 2x
(y2 + z2 − x2 )
∂2 v
∂ ∂v
= 2 2
2 = ∂x ∂x = 2
2
2
2 2
2
2 2
∂x
(x + y + z )
(x + y + z )
…(2)
Similarly
(x2 + z2 − y2 )
∂2 v
=
2
(x2 + y2 + z2 )2
∂y2
…(3)
(x2 + y2 − z2 )
∂2v
2 = 2 2
2
2 2
∂z
(x + y + z )
Adding (2), (3) and (4), we get
…(4)
and
∂2v ∂2v ∂2v
(x2 + y2 − x2 ) + (x2 + z2 − y2 ) + (x2 + y2 − z2 )
∂x2 + ∂y2 + ∂z2 = 2
(x2 + y2 + z2 )2
⇒
∂2 v ∂2 v ∂2 v
2
∂x2 + ∂y2 + ∂z2 = (x2 + y2 + z2 )
∂2v ∂2v ∂2v
(x2 + y2 + z2 ) 2 + 2 + 2 = 2
∂y
∂z
∂x
or
Hence the result
...
e
...
e
...
e
...
r
sinθ
∂r ∂r sinθ ∂θ
∂θ
∂V
n −1
2
Solution: Given V = rn (3 cos2θ – 1) so that ∂r = n r (3 cos θ − 1)
θ
272
Engineering Mathematics through Applications
r2
and
Now
…(1)
∂ 2 ∂V
= n(n + 1) rn(3 cos2 θ − 1) = n(n + 1)V
r
∂r ∂r
Further,
and
∂V = n +1(3 cos2 θ − 1)
nr
∂r
∂V = n −
r ( 6 cos θ sin θ)
∂θ r
so that
sin θ
…(2)
∂V
= − 6rn(cos θ sin2 θ)
∂θ
∂
∂V
= − 6rn (− sin θ sin2 θ + cos θ ⋅ 2 sin θ cos θ)
sin θ
∂θ
∂θ
= –6rnsinθ(2 cos2θ – sin2θ)
…(3)
…(4)
∂V
1 ∂
n
2
2
Implying sin θ ∂θ sin θ ∂θ = − 6r (2 cos θ − sin θ)
= –6rn(2cos2θ + cos2θ – 1)
= –6rn(3cos2θ –1) = – 6V, (using given relation)
…(5)
On adding expressions (2) and (5), we get
n(n + 1)V – 6V = 0 or [n(n + 1) – 6]V = 0
which implies either (n2 + n – 6) = 0 or V = 0
∴
(but V ≠ 0)
n2 + 3n + 2n – 6 = 0 or (n + 3) (n - 2) = 0 or n = 2, –3
...
Show that zxy = zyx if z = x3 + y3 – 3axy
...
If u = x2 tan−1 − y2 tan−1 , show that
∂x∂y x + y2
x
y
3
...
If v = (x2 + y2 + z2 ) 2 , prove that ∂x2 + ∂y2 + ∂z2 = 0
∂3 u
5
...
6
...
If z(–x + y) =
x2
+
y2,
∂z ∂z
∂z ∂z
show that ∂x − ∂y = 4 1 − ∂x − ∂y
8
...
If u = log
2
2
(x2 + y2 )
, verify ∂ u = ∂ u
...
If u = (1 − 2xy + y2 )− 2 , prove that
∂x
∂x ∂y ∂y
[MDU, 2006]
∂u
∂u
x
y
+y
=0
11
...
Verify that fxy = fyx when f(x, y) = (log x )tan–1 (x2 + y2)
13
...
−
14
...
3 WHICH VARIABLE TO BE KEPT CONSTANT
For
}
x = r cos θ
y = r sin θ
…(i),
1
we get r = (x2 + y2 )2
y
θ = tan −1
x
…(ii)
The above relations clearly indicate that x is a function of (r, θ) in x = r cos θ and x is a
function of (r, y) in r2 = x2 + y2 or x2 = r2 – y2
...
∴
∂x = cos θ and 2x ∂x = 2r
∂r θ
∂r y
⇒
∂x = r = r = 1
∂r y x r cos θ cos θ
and there is no reason for them to be equal since we have to be specific while finding the
partial derivative that which variable is to be kept constant
...
e
...
e
...
∂θ
∂θ r
Example 13: If u = lx + my, v = mx – ly, show that
∂u ∂x = l2 , ∂y ∂v = (l2 + m2 )
∂x y ∂u v l2 + m2 ∂v u ∂y x
l2
Solution: Given relations imply,
lu = l2x + lmy
mv =
m2x
– mly
…(i)
…(ii)
On adding the two (lu + mv) = (l2 + m2)x ⇒
x=
(lu + mv)
(l2 + m2 )
…(1)
Again given implies
mu = mlx + m2y …(iii)
lv = lmx – l2y …(iv)
On subtracting (iv) from (iii), (mu – lv) = (l2 + m2)y ⇒
y=
(mu − lv)
(l2 + m2 )
…(2)
Partial Derivatives and their Applications
275
Also from given relations,
∂ u = l ∂ v = −l
, ∂y
∂x y
x
…(3)
From (1) and (2),
l
∂x =
∂u v (l2 + m2 ) ,
∂y
l
∂v = − (l2 + m2 )
u
…(4)
From above, we can deduce the desired results
...
}
2
2
∂r
∂2r ∂2r 1 ∂r
+
=
+
(i)
∂y
∂ x2 ∂ y2 r ∂x
x = r cosθ
, prove that
Example 15: If
y = r sinθ
(ii)
∂2 θ ∂2 θ
+
=0
∂x2 ∂y2
Solution:
(x ≠ 0, y ≠ 0)
}
x = r cos θ
y = r sin θ
From (A),
2r
⇒
∂r
= 2x
∂x
[ KUK, 2004, 2008; NIT Kurukshetra, 2006]
r2 = x2 + y2
y
θ = tan−1
x
or
∂r x
=
∂x r
or
…(A)
…(B)
∂r
x
=
∂x (x2 + y2 )1
2
…(1)
276
Engineering Mathematics through Applications
Differentiating (1) again partially with respect to x, we get
x2
1
y2
(x + y2 ) 2 (x2 + y2 ) − x2
= 2
3 2 =
2
2
2
2
(x + y )
(x + y )
(x + y2 )3
(x2 + y2 ) −
∂r
=
∂x2
2
Similarly,
2
∂y y
=
∂r r
∂2 y
x2
2 =
∂r
(x2 + y2 )3
and
2
…(2)
…(3)
…(4)
2
Adding (2) and (4), we get
y2
∂2 r ∂2 r
2 +
2 =
2
∂x
∂y
(x + y2 )3
2
+
x2
(x + y2 )3
y2
x2
(x2 + y2 ) + (x2 + y2 )
=
1
2
(x + y2 )1
=
2
2
1 ∂r ∂r
+
r 2 ∂y ∂x ,
2
2
2
Using (1) and (3)
Hence the result
...
2
2 2
(x + y ) (x + y2 )2
∂x
∂y
…(6)
…(7)
…(8)
Partial Derivatives and their Applications
277
Example 16: If u = f(r), where r2 = (x2 + y2 + z2), show that
∂2 u ∂2 u ∂2 u
2
+
+
= f''(r ) + f'(r)
r
∂ x2 ∂ y2 ∂ z2
[UP Tech, 2005; Rajasthan, 2006]
Solution: Here u = f(r), where r2 = (x2 + y2 + z2)
Therefore
…(1)
∂u
∂r
= f'(r)
∂x
∂x
…(2)
∂2u ∂
∂r = f"(r) ∂r ⋅ ∂r + f'(r) ⋅ ∂2 r
=
f'(r)
2
∂x ∂x
∂ x2
∂x
∂x
∂x
and
2
∂r
∂2 r
= f"(r) + f'(r) 2
∂x
∂x
…(3)
2
Similarly
∂r
∂2 u
∂2 r
2 = f"(r ) ∂y + f'(r ) ⋅
∂y
∂y2
…(4)
2
∂2u
∂2r
∂r
2 = f"(r ) ∂z + f'(r )
∂z
∂z2
and
…(5)
Adding (3), (4) and (5), we get
∂r 2 ∂r 2 ∂r 2
∂2r ∂2r ∂2r
∂2u ∂2u ∂2u
+
+
=
f"
r
(
)
∂x + ∂y + ∂z + f'(r) ∂x2 + ∂y2 + ∂z2
∂x2 ∂y2 ∂z2
Now
1
∂r
∂ 2
x
1
=
⋅ 2x =
(x + y2 + z2 ) 2 =
r
∂x ∂x
2
2
2 12
2(x + y + z )
1
Further
∂r
=
∂x2
2
=
Precisely
…(6)
(x2 + y2 + z2 )
2
⋅1− x
1
2(x + y2 + z2 )
2
2
(x + y + z2 )
(x2 + y2 + z2 ) − x2
(x2 + y2 +
1
3
z2 ) 2
=
2
2
⋅ 2x
r2 − x 2
r3
∂r x
= ,
∂x r
∂2r r2 − x2
;
=
r3
∂x2
…(7)
∂r y
= ,
∂y r
∂2r r2 − y2
=
∂y2
r3 ;
…(8)
∂r z
= ,
∂z r
∂2 r r2 − z2
=
∂z2
r3
…(9)
278
Engineering Mathematics through Applications
On using results given by (7), (8), (9) into equation (6), we get
x 2 y 2 z 2
∂2u ∂2u ∂2u
r2 − x2 + r2 − y2 + r2 − z2
(
)
+
+
=
f"
r
r + r + r + f'(r)
∂x2 ∂y2 ∂z2
r3
x2 + y2 + z2
3r2 − r2
2
= f"(r)
+
f'
r
= f"(r) + f'(r)
(
)
2
3
r
r
r
ASSIGNMENT 2
2
2
r
r
x
x
r
1
...
r ∂θ
∂x ∂y
∂x ∂r
∂x
}
x = r cos θ
1
∂2 u ∂2 u
+
= f"(r) + f'(r )
...
If u = f (r) and y = r sin θ , prove that
r
∂x2 ∂y2
3
...
u
u
∂u θ ∂x y ∂u θ ∂y x
4
...
The above expression mathematically may be rewritten as
xn[a0 + a1(y/x) + a2(y/x)2 + … + an(y/x)n] or more precisely xnφ(y/x),
where φ(y/x) is a polynomial of degree n in (y/x)
...
E
...
(i) x3 tan (y/x) is a homogeneous function of degree 3 in x and y
...
In general, a function f(x1, x2, x3, …) is said to be a homogeneous function of degree n in
1
3
x
x x
(x1, x2, … , xn) if it is expressible in the form as x1n φ 2 , 3 , …, n
...
∂x
∂y
[KUK, 2004]
Partial Derivatives and their Applications
279
y
Proof: u is a homogeneous function of degree n in x, y, i
...
u = xn φ
...
Hence the result
...
x1
∂x1
∂x2
∂x3
∂xn
From (1) and (2), x
Example 17: If u = sin–1x/y + tan–1y/x, prove that x
∂u
∂u
+y
= 0
...
Hence Euler’s Theorem, x
∂u
∂u
+y
= nu = 0,
∂x
∂y
(as n = 0)
...
∂x
∂y
(
x2 1 + ( y / x)2
(x2 + y2 )
Solution: We have ψ(x, y) = log (x + y) = log
1
/
x + ( y x)
(
ψ (x , y )
So that e
1 + ( y / x )2
= x
1 + ( y / x )
(
)
)
)
comparable to xn φ (y / x), where n = 1
...
By Euler’s Theorem,
x⋅
⇒
∂ ψ( x , y )
∂
(e
) + y (eψ(x, y) ) = 1 ⋅ eψ(x , y)
∂x
∂y
x eψ(x , y)
∂ψ
∂ψ
+ y eψ(x , y )
= eψ(x, y)
∂x
∂y
or
x
(
1 + ( y / x )2
= log x 1 + ( y / x )
∂ψ
∂ψ
+y
=1
∂x
∂y
(
)
)
280
Engineering Mathematics through Applications
(x2 y2 )
∂u
∂u
, show that x
+y
= 3tan u
( x + y)
∂x
∂y
Example 19: If sin u =
y
y
x4
x
x
= x3 φ y is a homogeneous function of degree
= x3
Solution: sin u =
y
y
x
x 1 +
1 + x
x
3 in x and y
...
2
By Euler’s Theorem,
⇒
x
∂z
∂z
+y
= nz
∂x
∂y
x
∂u
∂u −1
+y
=
cot u
2
∂x
∂y
or
x(− sin u)
∂u
∂u 1
1
+ y(− sin u)
= z = cos u
2
∂x
∂y 2
1
1 2 + y1 2 2
−1 x
prove that
Example 21: If u = cosec 1
1 ,
x 3 + y 3
x2
2
tan u 13 tan2 u
∂2 u
∂2 u
2 ∂ u
+
2 + 2 xy ∂x ∂y + y
2 = 12
12
12
∂x
∂y
[MDU, 2006]
Partial Derivatives and their Applications
281
1
x1 2 + y1 2 2
Solution: Let 1
1 = z
x 3 +y 3
so that
cosec u = z
1
1
1 2
y2
12
x +
z= 1
1
3
x + y3
Now
1 2
1
2
x 2 1 + y
1 1
−
x
2 3
x
=
=
1
1 y 2
x2 1 +
x
( )
1
1 2
0
2
y
y
+
1
x
x
2
1
0
y + y2
x x
1
y
y
= x12 Function with all terms of the form = xn φ
x
x
Clearly a homogeneous function of degree n =
1
in x and y
...
2
∂xy
∂x
∂y
Note: As here in this example, z = z1 + z2, is a sum of two homogeneous functions of diffferent degree of
homoginity
...
Example 23: If u = tan−1
and x2
x3 + y3
∂u
∂u
+y
= sin 2x
, then show that x
∂x
∂y
x−y
2
∂2 u
∂2 u
+ y2 2 + 2xy ∂ u = sin 4u − sin 2u
2
∂x ∂y
∂x
∂y
[NIT Kurukshetra, 2004; VTU, 2005; SVTU, 2007; KUK, 2007-2009; PTU, 2009]
3
y
x3 1 +
x
(x3 + y3 )
(x3 + y3 )
= x2 φ y
Solution: u = tan−1
⇒ tan u =
=
x
x+y
(x + y)
y
x 1 +
x
Thus, tan u is a homogeneous function of degree 2 in x and y
...
Here
Z2 = y–nf2(x/y), clearly a homogeneous function of degree –n in x and y
...
∴ By Euler’s Theorem, x
∂u1
∂u
+ y 2 = 2 ⋅ u1
∂x
∂y
∂u2
∂u
+ y 2 = 2 ⋅ u2
∂x
∂y
Differentiating (1) partially with respect to x and y
and
x
x
and
x
∂2u1 ∂u1
∂2u1
∂u1
2 + ∂x + y ∂x∂y = 2 ∂x
∂x
∂2u1 ∂u1
∂2u
∂u
+
+ y 21 = 2 1
∂y∂x ∂y
∂y
∂y
Now multiply (3) by x, (4) by y and, add the two
2
2 ∂2u1
∂u1
∂2u1
∂ u1
∂u1
∂u1
2 ∂ u1
x ∂x2 + x ∂x + 2xy ∂x∂y + y ∂y + y ∂y2 = 2 x ∂x + y ∂y
…(1)
…(2)
…(3)
…(4)
286
Engineering Mathematics through Applications
2
2 ∂2 u1
∂2 u1
∂ u1
∂ u1
2 ∂ u1
x ∂x2 + 2xy ∂x∂y + y ∂y2 = x ∂x + y ∂y = 2u1 ,
⇒
2
∂2u2
∂2u2
2 ∂ u2
Similarly x ∂x2 + 2xy ∂x∂y + y ∂y2 = 2u2 ,
(using (1))
(using (2))
On adding the two, we get the desired result
...
x
∂x
∂y
∂z
Hence the proof
...
1
...
If u is a homogeneous function of nth degree in x, y, z,
y
x
z
∂u
∂u
∂u
+
+
, then x
+y
+z
= 0
...
3
...
If u = f (x, y) = log(x2 + y2 ) + tan−1 , prove that
+
= 0
...
Show that x
x3 + y3
∂u
∂u
...
If –z is a homogeneous function of degree
6
...
2
y
xy
∂x∂y
∂x2
∂y2
8
...
[UP Tech, 2004]
n in x and y, show that
[Kottayam 2005; UP Tech, 2006; VTU, 2007]
y−3x
, find x ∂u + y ∂u
...
9
...
5 TOTAL DERIVATIVES
Derivatives of Composite and Implicit Functions (Chain Rule)
Let –z = f(x, y) be a function of two independent variables x and y which are themselves
functions of one independent variable t, say x = φ (t), y = ψ(t) with an assumption that these
functions are differentiable
...
Without actually substituting the values of x and y in f(x, y), we can find
dz ∂ z dx ∂ z dy
=
+
dt ∂ x dt ∂ y dt
or
d z ∂ f dx ∂ f dy
=
+
dt ∂ x dt ∂ y dt
…(1)
Proof: If t is given an increment ∆t, then x, y and –z will have corresponding increments as ∆x,
∆y and ∆z, so that
∆z = f(x + ∆x, y + ∆y) – f (x, y)
= f(x + ∆x, y + ∆y) – f(x, y + ∆y) + f(x, y + ∆y) – f(x, y)
Dividing both sides by ∆t,
∆z f (x + ∆x, y + ∆y) − f (x, y + ∆y) ∆x f (x, y + ∆y) − f (x, y) ∆y
=
⋅
+
⋅
∆t
∆x
∆t
∆y
∆t
Taking limits as ∆t → 0, ∆x and ∆y also tends to zero
...
These rules can be extended to a function of n variables
z = φ(x1, x2, … , xn) with x1 = φ1(t), x2 = φ2(t), …, xn = φn(t)
such that
∂f dxn
dz ∂f dx1 ∂f dx2
=
+
+…+
dt ∂x1 dt ∂x2 dt
∂xn dt
…(3)
Differentiation of Implicit Functions
If the function –z = f(x, y) = 0 defines implicitly a function y = φ(x) of one independent variable
x, then equation (1) reduces to
df ∂f ∂f dy
0=
=
+
dx ∂x ∂y dx
or
∂f
dy
∂
=− x,
dx
∂f
∂y
provided
∂f
≠0
∂y
…(4)
Further, if the function f(x, y, z) = 0 defines one of the variables x, y, z implicitly in terms
of the other two variables, then
df =
∂f
∂f
∂f
dx +
dy +
dz = 0
∂x
∂y
∂z
…(5)
Partial Derivatives and their Applications
289
(i) If we take y = constant implying dy = 0, (5) gives
∂f
∂x
∂f
∂f
dz
=−
dx + dz = 0 or
∂x
∂z
dx y
∂f
∂z
…(5a)
(ii) If we take x = constant implying dx = 0, (5) gives
∂f
∂y
∂f
∂f
dz
dy +
dz = 0 or = −
∂y
∂z
dy x
∂f
∂z
…(5b)
(iii) If we take z = constant implying dz
– = 0, (5) gives
∂f
∂f
∂f
∂y
dx
dx +
dy = 0 or = −
∂x
∂y
∂f
dy z
∂x
Multiplying (5a), (5b), (5c), we get
…(5c)
dx dy dz
∂x ∂y ∂z
dy dz dx = −1 or ∂y ∂z ∂x = −1
y
y
z
x
z
x
Example 27: If u = sin–1(x – y), x = 3t and y = 4t3, show that
−1
du
= 3(1 − t2 ) 2
dt
dy
Solution: Given, u = sin –1(x – y), so that du = ∂u dx + ∂u
dt ∂x dt ∂y dt
∂u
1
1
=
=
Now,
2
2
2
∂x
1 − x − y + 2xy
1 − (3t) − (4t3 )2 + 2(3t)(4t3 )
=
1
1
=
,
2
2
(1 − 4t ) (1 − t ) (1 − 4t ) (1 − t2 )
…(5d)
[PTU, 2005]
…(1)
…(2)
2 2
∂y
1
1
=
=
,
2
2
2
∂y
1 − x − y + 2xy (1 − 4t ) 1 − t2
…(3)
dy
dx
= 3,
= 12 t2
...
1
2
2
dt (1 − 4t ) 1 − t
(1 − 4 t2 )(1 − t2 )2
…(4)
Example 28: Find the total differential coefficient of x2y with respect to x when x and y
[KUK, 2008; NIT Kurukshetra, 2008]
are connected by the relation x2 + xy + y2 = 1
...
dx
290
Engineering Mathematics through Applications
dy
dz ∂z dx ∂z dy
=
+
= 2xy + x2
…(1)
dx ∂x dx ∂y dx
dx
2
2
Further, variables x and y are connected by an implicit relation, φ = x + xy + y = 0
comparable to φ(x, y) = 0
∂φ
dy
2x + y
…(2)
so that
= − ∂x = −
2y + x
dx
∂φ
∂y
dy
from (2) in (1), we get
On substituting value of
dx
2x + y
dz
= 2 xy − x2
...
Solution: As f (x, y) = 0 is an implicit relation between x and y which defines a differential
coefficient of y with respect to x,
∂f
∂x
dy
=−
dx
∂f
∂y
…(1)
∂φ
∂y
dz
Likewise for φ(y, z) = 0, dy = −
∂φ
∂z
write,
∂f
∂φ
∂y
dz dz dx
dz ∂y
=
⋅
⇒ −
=−
dy dx dy
dx ∂f
∂φ
∂z
∂x
…(2)
⇒
∂f ∂φ dz ∂f ∂φ
...
5 ft/s and 0
...
dy
= 0
...
5 ft/s,
dt
dt
Here
z = f(x, y) = x · y (area of the rectangle)
so that
∂z
∂
∂z
dz ∂z dx ∂z dy
=
= x
(xy) = y and
=
+
= y (1
...
5) , as
dt ∂x dt ∂y dt
∂y
∂x ∂x
= 3(1
...
5) = 4
...
0 = 6
...
Example 31: If z = (2xy2 – 3x2y) and if x increases that the rate of 2 cm/s and it passes
through x = 3 cm, show that if y is passing through the value y = 1 cm, y must be decreasing
2
at the rate 2 15 cm/s in order that z shall remain constant
...
e
...
fy3
Example 33: xn + yn = an, find
…(7)
d2 y
...
2 = −
dx
q3
p = fx = nxn –1, q = fy = nyn –1,
r = fxx = n(n – 1)xn –2, s = fxy = 0, t = n(n – 1)yn – 2
...
y2n −1
Example 34: If x2 + y2 + z2 – 2xyz = 1, show that
dy
dx
dz
+
+
=0
2
2
1−x
1−y
1 − z2
[NIT Kurukshetra, 2009]
Partial Derivatives and their Applications
293
Solution: As the given relation x2 + y2 + z2 – 2xyz = 1 is an implicit function comparable to
f(x, y, z) = c, has its total differential coefficient equal to zero
...
e
...
∴
x=
2 2
2
2
−b ± b2 − 4ac 2yz ± 4y z + 4(1 − y − z )
=
2a
2
x=
2yz ± 2 (1 − y2 ) − z2(1 − y2 )
= yz ± (1 − y2 )(1 − z2 )
2
(x − yz) = (1 − y2 )(1 − z2 ) , (Taking the positive sign only)
or
Similarly, (y − zx) = (1 − z2 )(1 − x2 ) ,
and
…(2)
…(3)
(z − xy) = (1 − x2 )(1 − y2 )
…(4)
On using above results in equation (1),
(1 − y2 )(1 − z2 ) dx + (1 − x2 )(1 − z2 ) dy + (1 − x2 )(1 − y2 ) dz = 0
or
dy
dx
dz
+
+
=0
1 − x2
1 − y2
1 − z2
(Dividing throughout by
(1 − x2 )(1 − y2 )(1 − z2 )
)
Example 35: If the three thermodynamical variable P, V, T are connected by the relation
∂P ∂T ∂V
f(P, V, T) = 0, show that
⋅
⋅
= − 1
...
∴
df = fx dP + fy dV + fz dT = 0
∂f
∂P = − ∂T
(i) if V is kept constant, i
...
dV = 0, then
∂T V
…(1)
∂f
∂P
(i
...
change in one variable with respect to the other keeping the 3rd constant)
∂f
∂
T
= − ∂V
(ii) if P is kept constant, i
...
dP = 0, then
∂V P
∂f
∂T
…(2)
294
Engineering Mathematics through Applications
∂f
∂P
∂V
(iii) if T is kept constant, i
...
dT = 0, then
=−
∂P T
∂f
∂V
…(3)
∂P ∂T ∂V
On multiplying the three, ∂T ∂V ∂P = −1
V
P
T
Note: In case of implicit function with more than 2 variables, the differential coefficients involved are obviously
partial in nature instead of total
...
If Q is a function of the state of a gas such
∂Q
∂Q
∂Q
= p then show that
that
=
+ R
...
∂v v,T
∂ T p,T ∂ T v,T
∂p
p,T
Solution: We had the relation pv = RT,
…(1)
connecting pressure p, volume v and temperature T
...
∂Q
Here in the given problem, clearly the notation, ∂v
defines that Q is a function of
v,T
two independent variables v and T, and its partial derivative is taken with v keeping T
constant
...
ASSIGNMENT 4
du
find
and verify the result by direct substitution
...
If u = x2 + y2 + z2 with x = e2t ,
dt
y = e2t cos 3t,
z = e2t sin 3t
2
...
If u = sin(x2 + y2), where a2x2 + b2y2 = c2, find
dy
...
dx
du
...
If x increases at the rate of 2cm/sec at the instant when x = 3cm and y = 1cm, at what rate
must y be changing in order that the function (2xy – 3x2y) shall neither increasing nor
du
decreasing, i
...
= 0
...
If the curves f(x, y) = 0 and φ(x, y) = 0 touch, show that at the point of contact ∂y ∂x = ∂x ∂y
4
...
]
∂y ∂z
,
7
...
Hint : If (x, y, z) = 0, then ∂x = − f , ∂x = − f
y
z
y
z
8
...
6 CHANGE OF VARIABLES: FUNCTIONS DEFINED ON SURFACES
Suppose that u = f(x, y) is a function of two independent variables (x, y) and x, y are further
dependent on two independent variables (s, t) given by x = φ(s, t) and y = ψ(s, t), then by
chain rule,
∂ u ∂ u ∂ x ∂ u ∂y
=
+
∂s ∂x ∂s ∂y ∂s
…(1)
296
Engineering Mathematics through Applications
∂ u ∂ u ∂ x ∂ u ∂y
=
+
…(2)
∂t ∂x ∂t ∂y ∂t
Here ordinary derivatives have been replaced by partial derivatives, since x and y are
further functions of s and t (instead of one variable t as in the case of total derivatives, Article 4
...
∂u
∂u
and ∂y , we get their values in
On solving (1) and (2) as simultaneous equations in
∂x
∂u ∂ u
,
, s, t which precisely means that if s and t are given as functions s = ξ(x, y)
terms of
∂s ∂t
and t = η(x, y), then we get
and
and
∂u ∂u ∂s ∂u ∂t
=
+
∂x ∂s ∂x ∂t ∂x
…(3)
∂u ∂u ∂s ∂u ∂t
=
+
∂y ∂s ∂y ∂t ∂y
…(4)
∂u
∂x
as shown in chain rule Fig
...
1
...
4
...
4
...
∂u
∂x
∂u
∂y
∂u
∂z
∂u
∂x
y
X
Z
∂y
∂s
∂x
∂s
s
Fig
...
2 (i )
u = f (x , y , z )
Dependent
variable
u = f (x , y , z )
Intermediate
variable
∂u
∂y
y
X
Z
∂y
∂t
∂x
∂t
∂z
∂s
Independent
variable
∂u
∂z
L
Fig
...
2 ( ii )
∂z
∂t
Partial Derivatives and their Applications
297
∂z ∂z
∂z
∂z
Example 37: If z = f(x, y) and x = eu + e− v prove that ∂u − ∂v = x ∂x − y ∂y
...
Example 38: For
(i)
}
x = r cosθ
, z = f (x, y) has continuous partial derivatives, show that
y = r sinθ
∂z
1 ∂z
= fx cosθ + fy sinθ and
= − fx sinθ + fy cosθ
r ∂θ
∂r
2
2
1 ∂z
∂z
2
2
(ii) ∂r + r 2 ∂θ = fx + fy
...
Example 39: If u = f (2x – 3y, 3y – 4z
–, 4z
– – 2x), prove that
1 ∂u 1 ∂u 1 ∂u
+
+
=0
2 ∂x 3 ∂y 4 ∂z
[NIT Kurukshetra, 2002, 2004; Raipur, 2005; UP Tech, 2006]
Solution: Consider u = f (2x – 3y, 3y – 4z, 4z – 2x) = f (r, s, t)
where r = 2x – 3y, s = 3y – 4z, t = 4z – 2x
So that
Similarly,
∂u ∂u ∂r ∂u ∂s ∂u ∂t
=
+
+
∂x ∂r ∂x ∂s ∂x ∂t ∂x
∂u
∂u
∂u
∂u
∂u
2+
0+
=
× −2 = 2
−2
∂r
∂s
∂t
∂r
∂t
∂u ∂u ∂r ∂u ∂s ∂u ∂t
=
+
+
∂y ∂r ∂y ∂s ∂y ∂t ∂y
=
and
…(1)
∂u
∂u
∂u
∂u
∂u
× −3 +
+3
3+
0 = −3
∂r
∂s
∂t
∂r
∂s
…(2)
∂u ∂u ∂r ∂u ∂s ∂u ∂t
=
+
+
∂z ∂r ∂z ∂s ∂z ∂t ∂z
=
∂u
∂u
∂u
∂u
∂u
× −4 +
+4
0+
4 = −4
∂r
∂s
∂t
∂s
∂t
…(3)
1 ∂u 1 ∂u 1 ∂u
On using (1), (2) and (3), we see that 2 ∂x + 3 ∂y + 4 ∂z = 0
Example 40: If u = f (x2 + 2yz, y2 + 2zx), prove that
∂u
∂u
∂u
+ (x2 − yz)
+ (z2 − xy)
=0
(y2 − zx)
∂x
∂y
∂z
Solution: Consider u = f (x2 + 2yz, y2 + 2zx) = f (r, s), r = x2 + 2yz and s = y2 + 2zx
Differentiating u(x, y, z) partially with respect to x, y, z respectively as below:
∂ u ∂f ∂ r ∂f ∂ s
=
+
∂x ∂r ∂x ∂s ∂x
∂ u ∂f ∂ r ∂f ∂ s
=
+
∂y ∂r ∂y ∂s ∂y
∂ u ∂f ∂ r ∂f ∂ s
=
+
∂z ∂r ∂z ∂s ∂z
and
∂r
= 2x,
∂x
∂s
= 2z,
∂x
∂r
= 2z,
∂y
∂s
= 2y,
∂y
∂r
= 2y;
∂z
∂s
= 2x;
∂z
…(1)
…(2)
…(3)
Partial Derivatives and their Applications
299
∂r ∂r ∂r ∂s ∂s ∂s
∴ On substituting values of ∂x , ∂y , ∂z , ∂x , ∂y , ∂z in equation (2), we get
∂u
= ( fr ⋅ 2x + fs ⋅ 2z)
∂x
∂u
= ( fr ⋅ 2z + fs ⋅ 2y)
∂y
∂u
= ( fr ⋅ 2y + fs ⋅ 2x)
∂z
Using determinant method, eliminate fr and fs from (4),
2x
2z
2z 2y
2y 2x
or
…(4)
∂u
∂x
∂u
=0
∂y
∂u
∂z
∂u
∂u
∂u
∂u ∂u
2x 2y
− 2x + 2z 2y
− 2z +
( 2z ⋅ 2x − 2y ⋅ 2y ) = 0
∂y
∂z ∂x
∂z
∂y
On re-arrangement of terms, we get the desired result
...
∂x
∂y
∂z
Example 41: Transform the equation
Solution: For finding
∂2 u ∂2 u
+
= 0 into polar co-ordinates
...
2004; KUK, 2007]
∂2 u
∂u
∂2 u
∂u
2 and ∂y2 , we need to find first ∂x and ∂y
...
…(8)
Partial Derivatives and their Applications
301
}
Example 42: If u = f(x, y) and x = r cosθ prove that
,
y = r sinθ
2
2
2
2
∂u
1 ∂u
∂u
∂u
(i) ∂x + ∂y = ∂r + 2 ∂θ ,
r
∂2 u ∂2 u ∂2 u 1 ∂2 u 1 ∂u
(ii) ∂x2 + ∂y2 = ∂r 2 + r2 ∂θ2 + r ∂r
[Pb
...
In the
∂2 u 1 ∂2 u 1 ∂u
∂2u ∂2u
previous example we have just proved that 2 + 2 = 0 transforms to 2 + 2 2 + r ∂r = 0
∂r
r ∂θ
∂x
∂y
Means the expression,
∂2u ∂2u
+
which is in cartisian coordinate system has value
∂ x2 ∂ y 2
∂2u 1 ∂2u 1 ∂u
+
+
in (r, θ) system
...
∂θ
∂φ
[NIT Kurukshetra, 2008]
Solution: Here by change of variable, we have
∂u ∂u ∂x ∂u ∂y ∂u ∂z
=
+
+
∂θ ∂ x ∂θ ∂ y ∂θ ∂z ∂θ
Now
∂x
= −r sin θ cos φ,
∂θ
x = r cos θ cos φ,
∂y
y = r cos θ sin φ, ⇒
= −r sin θ sin φ,
∂θ
z = r sin θ
∂z
= r cos θ
∂θ
…(1)
…(2)
302
Engineering Mathematics through Applications
Using (2), we get
∂u
= f'(x2 + y2 + z2 ) ⋅ 2x( −r sin θ cos φ) + f'(x2 + y2 + z2 ) ⋅ 2y(−r sin θ sin φ)
∂θ
+ f'(x2 + y2 + z2 ) ⋅ 2 z(r cos θ)
…(3)
Simplifying the above expressions in one system of variables either express (r, θ, φ) or
(x, y, z), say
...
∂θ
We know that
∂ u ∂ u ∂ x ∂ u ∂y
=
+
∂θ ∂x ∂θ ∂y ∂θ
…(1)
and
∂ u ∂ u ∂ x ∂ u ∂y
=
+
∂φ ∂x ∂φ ∂y ∂φ
…(2)
Given
x + y = 2 eθ cos φ
x − y = 2 i eθ sin φ
implies
x = eθ+ i φ
y = eθ− i φ
…(3)
Partial Derivatives and their Applications
So that
∂x
= eθ + iφ ⋅ 1 = x;
∂θ
∂y
= eθ − iφ ⋅ 1 = y;
∂θ
303
∂x
= eθ + iφ ⋅ i = ix;
∂φ
∂y
= eθ − iφ ⋅ − i = −iy
∂φ
…(4)
On using (4), (1) and (2) becomes
and
∂u ∂u
∂u ∂
∂
= x
+ y = x
+y u
∂θ ∂ x
∂y ∂x
∂y
…(5)
∂u ∂u
∂u ∂
∂
= ix
− iy = ix
− iy u
∂φ ∂ x
∂y ∂ x
∂y
…(6)
Clearly from above two equations, the value of the operators
and
∂
∂
and
∂φ are
∂θ
∂ ∂
∂
= x
+y
∂θ ∂ x
∂y
∂
∂
∂
= ix
− iy
∂φ ∂ x
∂ y
…(7)
Therefore,
∂2u ∂ ∂u ∂
∂ ∂u
∂u
2 = ∂θ ∂θ = x ∂x + y ∂y x ∂x + y ∂y
∂θ
∂
∂u
∂ ∂u
∂
∂u
∂ ∂u
= x x + x y + y x + y y
∂x ∂x
∂x ∂y
∂y ∂x
∂y ∂y
14444
4244444
3 14444
4244444
3
∂u
∂2 u
∂2u
∂2u
∂u
∂2 u
=x
+ x2 2 + xy
+ yx
+y
+ y2 2
∂x
∂x∂y
∂y∂x
∂y
∂24444
x
∂3
y
14444
3 1444424444
∂2 u
∂2 u
∂2 u ∂u
∂u
= x2 2 + 2xy
+ y2 2 + x
+y
∂x∂y
∂y
∂y ∂x
∂x
…(8)
Likewise,
∂2u ∂ ∂u ∂
∂ ∂u
∂u
=
= ix
− iy ix
− iy
∂y ∂x
∂y
∂φ2 ∂φ ∂φ ∂x
∂2 u
∂2 u
∂2 u ∂u
∂u
= −x2 2 + 2xy
− y2 2 − x
+y
∂
∂
∂
∂
x
y
x
y
∂x
∂y
On adding (8) and (9),
∂2u ∂2u
∂2u
+ 2 = 4xy
2
∂ x∂ y
∂θ
∂φ
…(9)
304
Engineering Mathematics through Applications
Example 45: If by the substitution u = x2 – y 2, v = 2xy, f(x,y) = θ (u,v)
...
, 2003]
∂2 f
∂f
∂2 f
∂f
and
and
2 , we need to find first
2
∂y respectively
...
If u =f(r, s), r = x + at, s = y + bt, and x, y, t are independent variables, show that
∂u
∂u
∂u
=a
+b
...
If u = F(x – y, y – z, z – x), prove that ∂x + ∂y + ∂z = 0
y
x
z
∂u
∂u
∂u
+y
+z
= 0 [Karnataka, 2006]
3
...
If x = u + v + w, y = uw + vw + uv and F is a function of x, y, z, show that
u
∂F
∂F
∂F
∂F
∂F
∂F
+v
+w
=x
+ 2y
+ 3z
...
If x = r cosθ, y = r sinθ, z = f(x, y), prove that ∂x∂y (r cos n θ) = −n(n − 1)r sin(n − 2) θ
∂
∂ sin θ ∂ ∂
∂ cos θ ∂
Hint : Follow Ex
...
If x = ercosθ, y = ersinθ; prove that
2
∂2u ∂2u
∂2u
−2r ∂ u
+ 2
2 +
2 = e
2
∂r
∂x
∂y
∂θ
∂ ∂
∂ ∂
∂
∂
Hint: Use ∂r = x ∂x + y ∂y , ∂θ = −y ∂x + x ∂y
7
...
If w = u(x, y), where x = x(u, v), y = y(u, v), ∂x =
and
∂
v
∂
u
∂u ∂v
2
2
∂2 w ∂2 w ∂2 w ∂2 w ∂x ∂x
2 +
2 =
2 +
2 ∂u + ∂v
...
If u = f(x, y, z) and x = r sinθ cosφ, y = r sinθ sinφ, z = r cosθ, then show that
2
2
2
2
2
∂f ∂f ∂f
∂f
∂f
1 ∂f
1
+
+
= + 2 + 2
∂x ∂y ∂z
∂r
r ∂θ
r sin 2 θ ∂φ
[Hint: Follow Example 42 (ii), Section 4
...
By changing the independent variables u and v to x and y by means of the relation
x = (u cos α – v sin α), y = (u sin α + v cos α), show that
∂2 z ∂2 z
+
transforms to
∂ u2 ∂ v2
∂2 z ∂2 z
+
...
7 JACOBIANS: CO-EFFICIENTS OF TRANSFORMATION OF SYSTEM OF
VARIABLES
∂u ∂u
∂x ∂y
For u and v functions of two independent variables x and y, the determinant
is
∂v ∂v
∂x ∂y
∂(u, v)
called* Jacobian of u, v with respect to x, y and is denoted by either ∂(x, y) or J(u, v)
...
∂(u, v) ∂( x, y)
I
...
∂(x , y) ∂(u, v)
Let u = u(x, y) and v = v(x, y) which on solving for x and y, gives, x = x(u, v) and y = y(u, v)
...
d
...
, astronomy, elliptic functions and calculus of variations
...
Chain Rule: If u, v are functions of r, s and r, s are further functions of x, y then
∂(u, v) ∂(u, v) ∂(r, s)
=
⋅
∂(x , y) ∂(r, s) ∂(x , y)
∂u
∂(u, v) ∂(r , s)
⋅
= ∂r
∂(r, s) ∂(x, y) ∂ v
∂r
∂u ∂r
∂s ⋅ ∂x
∂v ∂s
∂s ∂x
(on
∂u ∂r ∂u ∂s
+
∂r ∂x ∂s ∂x
=
∂v ∂r ∂v ∂s
+
∂r ∂x ∂s ∂x
∂r
∂u ∂u
∂y
= ∂r ∂s
∂s
∂v ∂v
∂y
∂r ∂s
inter changing
∂r
∂x
∂r
∂y
∂s
∂x
∂s
∂y
rows and columns in 2nd determinant)
∂u ∂r ∂u ∂s
∂u
+
∂r ∂y ∂s ∂y
∂x
=
∂v
∂ v ∂y ∂ v ∂ s
+
∂x
∂r ∂y ∂s ∂y
∂u
∂y ∂(u, v)
=
∂v ∂(x, y)
∂y
∂(u, v)
≠0
∂(x, y)
(identically), otherwise, they are dependent and this property is extenable to system of any number of variables
...
Jacobian of Implicit Functions: If u, v, w instead of being given explicitly in terms of x,
y, z be connected with them by equations such as
f1(u, v, w, x, y, z) = 0, f2(u, v, w, x, y, z) = 0, f3(u, v, w, x, y, z) = 0, then
∂( f1 , f2 , f3 )
∂(u, v, w)
∂(x, y, z)
,
= (−1)3
( f1 , f2 , f3 )
∂
∂(x, y, z)
∂(u, v, w)
dy
f
Observations: This result can be generalized
...
IV
...
∂(r , θ)
∂ ( x , y , z)
= r
...
[NIT Kurukshetra, 2003]
∂(r , θ, φ)
Solution:
(i) For x = r cos θ,
∂x
∂x
= cosθ,
= −r sin θ
∂r
∂θ
308
Engineering Mathematics through Applications
∂y
∂y
= sinθ,
= r cos θ
∂r
∂θ
∂x ∂x
∂(x, y)
cos θ −r sin θ
= ∂r ∂θ =
= r
...
∂x
∂x
∂x
= sin θ cos φ,
= r cos θ cos φ,
= −r sin θ sin φ;
(iii) For x = r sinθ cosφ,
∂r
∂θ
∂φ
∂y
∂y
∂y
y = r sinθ sinφ,
= sin θ sin φ,
= r cos θ sin φ,
= r sin θ cos φ;
∂r
∂θ
∂φ
∂z
∂z
∂z
z = rcosθ, ∂r = cos θ, ∂θ = −r sin θ, ∂φ = 0;
y = r sin θ,
∴
∂x
∂r
∂(x, y, z) ∂ y
=
∂(r, θ, φ)
∂r
∂z
∂r
∂x
∂θ
∂y
∂θ
∂z
∂θ
∂x
∂φ
sin θ cos φ r cos θ cos φ −r sin θ sin φ
∂y
= sin θ sin φ r cos θ sin φ r sin θ cos φ
∂φ
−r sin θ
cos θ
0
∂z
∂φ
= r2[sinθcosφ (0 + sin2θcosφ) – cosθcosφ (0 – sinθcosθcosφ)
– sinθsinφ (–sin2θsinφ – cos2θ sinφ)]
= r2[sin3θ(cos2φ + sin2φ) – sinθcos2θ(cos2φ + sin2φ)]
= r2[sin3θ + sinθcos2θ] = r2sinθ(sin2θ + cos2θ) = r2sinθ
Example 47: If
}
x = u(1 − v)
, prove that J J’ = 1
...
v
u
∂(u, v)
309
…(1)
x + y = u(1 − v) + uv = u
u = (x + y)
y
y = uv = (x + y) ⋅ v
v=
x + y
−y
u
∂v
−uv
∂u
=
= 2 =− ;
= 1;
v
∂x (x + y)2
u
∂x
and ∂v
∂u
u(1 − v) 1 − v
x
=1
...
u
1
1− v v 1
1− v = u + u = u
u
…(2)
Hence the result
...
yz
xy
zx
,v =
,w =
,
Solution: For given u =
x
y
z we know that
=
∂u
∂x
∂(u, v, w) ∂v
=
∂(x, y, z)
∂x
∂w
∂x
∂u
∂y
∂v
∂y
∂w
∂y
∂u
−yz
∂z
x2
z
∂v
=
y
∂z
∂w
y
∂z
z
z
x
−zx
y2
x
z
y
x
x
y
−xy
z2
310
Engineering Mathematics through Applications
=
−yz zx xy
−1 1 1
yz zx xy
1
yz
zx
xy
1 −1 1 = 0 + 2 + 2 = 4
...
Are u and v functionally
...
[KUK, 2004, 2008]
Example 50: If u =
Solution: Take
Now
1
1
θ = tan−1 x
∂θ
x = tan θ
=
= cos2 θ
so that
2 =
2
−1 and ∂x
1
1
tan
+
x
+
θ
y = tan φ
φ = tan y
1
1
∂θ
2
=
=
= cos φ
...
∴
= sec2(θ + φ)
cos2 θ cos2 φ
= 0
...
∂(u, v, w)
[VTU, 2003; NIT Kurukshetra, 2005; KUK 2009; PTU, 2009]
Example 51: If u = x + y + z, uv = y + z, uvw = z, show that
Solution: Here
Now
u = x + y + z
uv = y + z
uvw = z
∂x
∂u
∂(x, y, z)
∂y
=
∂(u, v, w) ∂u
∂z
∂u
∂x
∂v
∂y
∂v
∂z
∂v
⇒
∂x
∂w
∂y
∂w
∂z
∂w
x = u − (y + z) = u(1 − v)
y = uv − z = uv(1 − w)
z = uvw
…(1)
…(2)
Partial Derivatives and their Applications
From (1),
∂x
= (1 − v),
∂u
∂x
= −u, ;
∂v
∂x
=0
∂w
∂y
= v(1 − w),
∂u
∂y
= u(1 − w), ;
∂v
∂y
= −uv
∂w
∂z
= vw,
∂u
∂z
= uw,
∂v
∂z
= uv
∂w
311
…(3)
On using (3), we get
−v
−u
0
∂(x, y, z)
= v(1 − w) u(1 − w) −uv
∂(u, v, w)
vw
uw
uv
= (1 – v) {u(1 – w) uv + uv uw} + u{(1 – w) uv + uv uw}
= (1 – v) {u2v – u2vw + u2uw} + u{uv2 – uv2w + uv2w}
= (1 – v) u2v + u2v2
= u2v
...
∂(u, v, w)
∂l
∂w
1
0
0
∂m
u
= v
0 = u2 v
∂w
vw uw uv
∂n
∂w
…(5)
…(6)
…(7)
…(8)
312
Engineering Mathematics through Applications
ASSIGNMENT 6
1
...
∂(u, v)
2
...
If u = 2xy, v = x2 – y2; x = r cos θ, y = r sinθ; find
∂(F, G, H )
...
∂(r, θ)
∂(u, v)
...
If u = sin– 1x + sin– 1y and v = x 1 − y2 + y 1 − x2 ; find
∂(x, y)
related? If so, find this relationship
...
If u = xy + yz + zx, v = x2 + y2 + z2 and w = x + y + z, determine whether there is a
functional relationship between u, v and w
...
6
...
Find the relationships between them
...
If u = x + y + z, v = xy + yz + zx, w = xyz; evalutate ∂(u, v, w)
...
If u =
∂(u, v, w)
y
x
z
,v=
,w=
, find the value of
∂(x, y, z)
y−z
z−x
x−y
[NIT Kurukshetra, 2007]
4
...
This change in value of u by δu is
seen as from three point of view:
Absolute change: δu
δu
Relative change:
u
δu
Percentage change:
× 100
u
In practice, percentage change (or error) and relative change (or error) are usually more
important than the absolute change (or error)
...
Example 52: How sensitive is the volume V = πr2h of a right circular cylinder to small
changes in its radius and height near the point (r0, h0) = (1, 3)?
Solution: By increment method, we get
∂V
∂V
∆V ≈
∆r +
∆h
∂r (r , h )
∂h (r , h )
0 0
0 0
= Vr(1, 3) ∆r + Vh(1, 3)∆h
= (2πrh)(1, 3) ∆r + (πr2)(1, 3) · ∆h
= 6π · ∆r + π · ∆h
The above result shows that a one-unit
change in r will change V nearly by 6 π units
and a one-unit change in h will change V nearly
by 6π units
...
4
...
In contrast, if value of r and h are reversed,
so that r = 3 and h = 1, then ∆V ≈ 6π · ∆r + 9π ·
∆h
...
Thus the
sensitivity to change depends not only on the
increment but also on the relative size of r and
h (See Fig
...
3 (ii))
...
3
...
3
...
4
...
2
or
Further,
∆V × 100 ≈ 2 ∆r × 100 + ∆h × 100
V
r
h
…(3)
indicates that the percentage change in V is the percentage change in h plus two times the
percentage change in r
...
5
percent, about how accurately can we calculate V from the formula V = πr2h?
Solution: For V = πr2h or logV = log π + 2logr + log h, increment approximation implies
∆V
∆r ∆h
…(1)
≈2
+
...
5 =
≤
2
h
200
h
∆V
∆r ∆h
∆r
∆h
2
1
∴
…(3)
≈ 2
+
≤2
+
=
+
= 0
...
∆V
× 100, due to possible
V
percentage error (or change) in measurement of r and A will be about 2
...
However, it is very difficult to answer such question for functions which involves two or
more independent variables because there could be several possible setting for errors in r
and h giving less than 2% error in V
...
Example 55: Find a reasonable square about the point (a, b) = (1, 3) in which the value of
V = πr2h will not vary more than ±0
...
Solution: By change estimation,
∆V ≈ (2πrh)(r0, h0)∆r + (πr2)(r0, h0)∆h
= (2π × 1 × 3) ∆r + (π 9) ∆h = 6π ∆r + 9π ∆h
…(1)
Since we have been looking for a square about the point (1, 3), we will take ∆h = ∆r so that
(1) becomes
∆V ¾ 6π ∆r + 9π ∆r = 15π ∆r
…(2)
In order to get,
|∆V| ≈ 15π |∆r| ≤ 0
...
1
≈ 2
...
12 × 10–3, |h – 3| ≤ 2
...
1
...
Find the percentage error in R due to an error of 1% in v and an
g
error of 0
...
[KUK, 2009]
Solution: Given R =
v2 sin2α
g
or
logR = 2 logv + log sin 2α – log g
...
e
...
5 = 2 + 2α cot 2α ×
1
2
= (2 + α cot 2α)
Hence, the percentage error in calculation of R due to errors of 1% in R and 0
...
2
Example 57: If the Kinetic energy K = wv , find approximately the change in the kinetic
2g
energy as w changes 49 to 49
...
Solution: Here K =
wv2
so that δK = 1 [v2δw + w2vδv] approx
...
p
...
units, w = 49 f
...
s units, δv = –10 f
...
s
...
5 f
...
s
...
δK =
1
0
...
p
...
)
2 × 32
Example 58: Find the percentage error in the area of an ellipse
...
Solution: Let x and y be the major and minor axes of the ellipse, then area A = πxy implying
logA = logπ + logx + logy
...
=
+
or
A
x
y
A
x
y
Hence the percentage error in measurement of the area of the ellipse with the percentage
error of 1% each in measurement of major and minor axes is 2 percent
...
= +
r r1 r2
1 1 1
Solution: By error approximation, r = r + r
1
2
1
1 1
d ≈ d +
r
r1 r2
or
−
gives
1
1
1
δr ≈ − 2 δr1 − 2 δr2
r2
r1
r2
[UPTech, 2004]
316
implying
Engineering Mathematics through Applications
1 δr 1 δr1 1 δr2
+
;
r r r1 r1 r2 r2
1 δr
1 δr
1 δr
× 100 ≈ 1 × 100 + 2 × 100
r r
r1 r1
r2 r2
or
δr1
δr2
δr × 100 ≈ r 1 ⋅ 2 + 1 ⋅ 2 = 2r 1 + 1 = 2,
as r × 100 = r × 100 = 2
r1
r1 r2
r
r2
1
2
Hence in the percentage error in r is 2 percent when r1 and r2 has possible percentage error
of 2 percent in each
...
If h and α are in error by small quantities δh and δα respectively, find the corresponding
π
error in the area
...
33° in α
...
Area of the curved surface = πrl = π(htanα)(h sec α)
= πh2 tanα secα
∴ Total area, A (h,α) = πh2 tan2α + πh2 tanα secα
l = h sec α
h
= πh2(tan2α + tanα secα)
Implying
δA =
∂A
∂A
δh +
δα approximately
δh
δα
B
r = h tanα
= 2πh(tan2α + tan α secα)δh
D
r
C
Fig
...
4
+ πh2(2tanα · sec2α + tan2α · secα)δα
which gives the error in area, A(h, α) corresponding to errors δh and δα in h and α respectively
...
4646) ⋅ δα
100
Since π and h are non zero, therefore,
If δA = 0 then 0 =
δα = −
...
02
radians = −
57
...
33°
...
4646
3
...
If l
t2 r 4
is decreased by 2%, r is increased by 2%, t is increased by 1
...
Example 61: The torsional rigidity of a length of a wire is from the formula N =
Partial Derivatives and their Applications
317
Solution: Given,
δl
× 100 = −2
l
δr
× 100 = 2
r is increased by 2% i
...
r
δt
3
× 100 = 1
...
5% i
...
t
2
l is decreased by 2%
Torsional rigidity, N(l, t, r) =
i
...
8πI l
t2 r4
implies
log N = log(8πI) + logl – 2logt – 4logr
Taking differentials on both sides,
…(1)
δt
δr
δN δ l
=
−2 −4
N
t
r
l
or
(
δr
δN × 100 = δ l × 100 − 2 δ t × 100
t
−4 r × 100
N
l
)
3
− 4 × 2 = −13
2
Hence N diminishes by 13% if with the above given percentage changes in l, r, t
...
If the possible errors in measuring the distance and elevation are 2 cm and 0
...
[NITK, 2002; UPTech, 2004]
Solution: The given problem with h as height of the top
(point A) from the bottom (point B) and α the elevation of
the top with the ground is α° as explained in Fig
...
5
...
e
...
4
...
05° =
radians,
100
100 180
2
δh = (tan 30 °) ⋅
2 5
2
5 π
1 2
π
+ 50 ⋅ sec2 30° ⋅
=
+ 50
⋅
3 100 180
100
100 180
3 100
= 0
...
0582 = 0
...
07 mts
...
Example 63: If the sides and angles of a triangle ABC vary in such a way that its circumda
db
dc
+
+
=0
radius remains constant
...
74603)
= 0
...
732050
= 0
...
00807 = 1
...
01%
...
From (1),
∂ g 4π2
= 2
∂l
T
and
∂g
8π2l
=− 3
∂T
T
…(2)
We know that, inversely to find errors in x1, x2, …, xn when X = f(x1, x2, …, xn) is to have
a desired accuracy
i
...
if ∆X is the error in X, we have to determine errors ∆x1, ∆x2, …, ∆xn in x1, x2, … , xn as
∆X =
∂X
∂X
∂X
∆ x1 +
∆ x2 + … +
∆ xn
∂x1
∂x2
∂xn
…(3)
However, on using the principal of equal effects, viz
...
1 1 ∆g
1 ∆g
∆l
× 100 =
× 100 =
× 100 , (using (2))
2l 4π2
l
l 2 ∂g
2
T
∂l
=
1
1 ∆g
1 ∆g
× 100 = × 0
...
005
2 × 100 =
2 4π l
2 g
2
T2
…(7)
and percentage error in T viz
...
01 = 0
...
01%, percentage errors in l and T must
not exceed by 0
...
0025, respectively
...
5 cm and h = 5
...
Solution: The percentage error in R =
∴
∆R =
=
∆R
× 100 = 0
...
2
0
...
2 (4
...
5
R=
+
=
+
100
100 2h 2 100 2 × 5
...
2 50
...
002 × 50
...
002 × 50
...
5 ×
= 0
...
5)2
11
…(2)
Partial Derivatives and their Applications
Percentage error in h =
=
=
=
321
∆h
× 100
h
100
∆R
1
h r2
2− 2 +
2h
2
100 ∆ R
r2
+ h
−
2h
100
50
...
002
×
= 5
...
Find the percentage error in the area of a rectangle when an error of 1% is made in
measuring its length and breadth
...
The time oscillations of a simple pendulum is given by T = 2π l
...
4
...
6
...
5% are possible in the values of l
and T respectively
...
[NIT Kurukshetra, 2008]
[Hint: Possible pairs of errors (±1, ±0
...
50 and 0
...
If the H
...
required to propel a steamer varies as the cube of the velocity and square of
the length
...
P
...
What is the increase in deflection
corresponding to p% increase in w, q% increase in l and r% increase in d ?
...
Find approximately the increase of work necessary when the
t2
displacement is increased by 1%, the time is diminished by 1% and the distance is
diminished by 2%
...
The indicated horse power l of an engine is calculated from the formula
2
l = PLAN/33,000, where A = π d
...
322
Engineering Mathematics through Applications
8
...
Show that the total error in c due to errors δa, δb, δC in a, b, C is given
by δacosB + δbcosA + aδCsinB
[Hint: Take c2 = a2 + b2 – 2ab cos C, Use b cos C + c cos B = a]
9
...
Show that the error δc
in the computed length of the 3rd side c due to small errors in the angle C is given by
aδC sin B ·
a2 + b2 − c2
Hint
:
Use
cos
with a, b both constant
C
=
2ab
10
...
4
Hint : ∆ = s(s − a)(s − b)(s − c) where s = a + b + c
2
11
...
Show that the acceleration due to gravity is reduced by nearly 1% at an altitude equal to
0
...
x
[Hint: Here a = g(r/x)2, with g and r constant]
4
...
Define a function
F(t) = f(x, y) = f(a + ht, b + kt)
…(4)
Then by chain rule on (1),
∂ f dx ∂ f dy ∂ f
∂f
d
F(t) = F'(t) =
+
= h
+k
∂ x dt ∂ y dt ∂ x
∂y
dt
∂
∂
+k f,
F'(t) = h
∂y
∂x
2
∂
∂
+ k f,
F"(t) = h
∂y
∂x
...
…(5)
n+1
∂
∂
n +1
+k
F (t) = h
f
∂y
∂x
On using Taylor’s Theorem for functions of one variable, when t = 1, a = 0, we obtain
Precisely,
F(1) = F(0) + F'(0) + … +
where
1 n
1
F (0) +
Fn+ 1(0)
n
n+1
…(6)
F(1) = f (a + h, b + k )
F(0) = f (a, b)
∂
∂
+ k f (a, b)
F'(0) = f h
∂y
∂x
2
∂
∂
+ k f (a, b)
F"(0) = h
…(7)
∂
∂
y
x
………………………………
………………………………
n+1
∂
∂
+k
Fn +1(0) = h
f (a + θh, b + θk ), 0 < θ < 1
∂y
∂x
(n
+
1)
(0), in (6), we get the Taylor’s Theorem for
With above values of F(1), F(0), F’(0), …, F
functions of two variables given by equations (1) and (2) in the statement
...
Further, if Rn → 0 as n → ∞, Taylor’s theorem becomes Taylor’s infinite series as below:
∂
∂
f (a + h, b + k) = f (a, b) + (x − a) + (y − b) f (a, b)
∂
x
∂
y
2
+
1
∂
∂
( x − a)
+ (y − b) f (a, b) + ……∞
2
∂x
∂y
…(3)
Note: 1
...
Expansion of f(x, y) as given in (3) is also known as Taylor’s expansion of f(x, y) in powers of (x
– a) and (y – b)
...
If we put x = 0, y = 0 and replace h = x, k = y in (3), we arrive at the extension of Maclaurin’s Theorem
for two independent variables
...
Solution: Expansion
f (x, y) = f (a, b) + (x − a) fx (a, b) + (y − b) fy (a, b)
1
[(x − a)2 fxx( a, b) +2(x − a)(y − b) fxy (a, b) + (y − b)2 fyy (a, b)] + …
2
is Taylor’s expansion of f(x, y) in powers of (x – a) and (y – b)
...
Hence Compute F(1
...
9) approximately
...
1, 0
...
1 = 1 + 0
...
9 = 1 – 0
...
1, 0
...
1;
k = – 0
...
1 − (−0
...
1)2 − (−0
...
1)3 + 3(0
...
1) − 3(0
...
1)2 − (−0
...
7854 – 0
...
6887
...
Solution: Expansion of ex sin y in powers of x and y means it is a Maclaurin’s expansion viz
...
1
f (x, y) = f (a, b) + h fx(a, b) + k fy(a, b) + h fxx(a, b) + 2hk fxy(a, b) + k2 fyy(a, b) +……∞
Write
2
when a = 0 = b and h = x, k = y
So that
f (x, y) = f (0, 0) + h fx(0, 0) + k fy(0, 0) +
1 2
h f (0, 0) + 2hk fxy(0, 0) + k2 fyy(0, 0) + ……∞
2 xx
Partial Derivatives and their Applications
Here
∴
f(x, y) = ex siny,
f(0, 0) = 0
fx(x, y) =
ex siny,
fx(0, 0) = 0
fy(x, y) =
ex cosy,
fy(0, 0) = 1
fxx(x, y) = ex siny,
fxx(0, 0) = 0
fxy(x, y) = ex cosy,
fxy(0, 0) = 1
fyy(x, y) = – ex siny,
fyy(0, 0) = 0
fxxx(x, y) = ex siny,
fxxx(0, 0) = 0
fxxy(x, y) =
ex cosy,
fxxy(0, 0) = 1
fxyy(x, y) =
–ex siny,
fxyy(0, 0) = 0
fyyy(x, y) =
–ex cosy,
fyyy(0, 0) = –1
f (x, y) = ex sin y = 0 + x ⋅ 0 + y ⋅ 1 +
= y + xy +
327
1 2
x ⋅ 0 + 2xy ⋅ 1 + y2 ⋅ 0
2
1
+ x3 ⋅ 0 + 3x2 y ⋅ 1 + 3xy2 ⋅ 0 + y3 ⋅ 1 + …
3
x2 y y3
−
(upto 3rd degree)
2
6
1
1
Example 70: Evaluate log (1
...
98)4 − 1 approximately
1
1
1
1
Solution: Let f (x, y) = log x 3 + y 4 − 1 which is comparable to log (1
...
98)4 − 1
...
03 = 1 + 0
...
98 = 1 − 0
...
03, –0
...
03)3 + (0
...
03 fx (1, 1) + (−0
...
03 − 0
...
005 approx
...
Expand excosy about the point 1, 4 by Taylor’s Theorem
...
Expand f(x, y) = sin xy in powers of (x – 1) and y − 2 upto the second degree term
...
If f(x, y) = tan–1xy, compute f(0
...
2) approximately
...
Expand sin x sin y in powers of x and y as far as terms of third degree
...
Expand ex·log(1 + y) in powers of x and y upto terms of third degree
...
10 MAXIMA-MINIMA OF TWO FUNCTIONS
Definition: A function f(x, y) of two variables is said to be maximum at (a, b) if f(a + h, b +
k) – f(a, b) < 0 for sufficiently small positive or negative values of h and k, and minimum if
f(a + h, b + k) – f(a, b) > 0
...
The points at which maximum or minimum values occur are also known as points of
extrema of critical points and the maximum and minimum values taken together are extreme
values of the function
...
A function f(x, y) may also attain its extreme values on the boundary
...
The maxima-minima so defined are local relative maxima or local relative minima
...
3
...
E
...
for a function z = f(x, y) say representing a dom, maximum value of z occurs at the top from where
surface descends in all directiosns
...
Otherwise, a maximum or minimum value
may form a ridge such that the surface ascends or descends in all directions
...
Necessary Conditions for Having Extremum: fx(a, b) = 0 = fy(a, b)
...
…(2)
From above, sign of ∆ depends on the sign of h fx(a, b) + k fy(a, b) which is a function of h
and k
...
Therefore, the
function cannot extreme unless fy = 0
...
Hence the necessary condition for f(x, y) to have a maximum or minimum at (a, b) is
fx(a, b) = 0 and fy(a, b) = 0
If fx(a, b) = 0 = fy(a, b), then f(a, b) is called the stationary value of f(x, y) at (a, b)
...
2
1 2 2
= sign of
h r + 2hkrs + k2rt
2r
sign of ∆ = sign of
= sign of
1
(h2r2 + 2hkrs + k2 s2 ) − k 2 s2 + k2rt
2r
1
(hr + ks)2 + k 2(rt − s2 )
…(3)
2r
In (3), (hr + ks)2 is always positive and k2(rt – s2) can be made positive if (rt – s2) > 0
...
Hence, if (rt – s2) > 0, f(x, y) has a Maximum or Minimum according as r < 0 or r > 0
...
e
...
If, rt – s2 = 0 or r = t = s = 0, no conclusion can be made and further investigation is required
under such circumstances
...
Example 71: Find the extreemum values of x3 + y3 – 3axy
OR
3
Determine the points where the function (x + y3 – 3axy) has a maximum or minimum
...
So that
y=
x2
x2
= 0 for x = 0 and y =
= a for x = a
...
r=
∂2 f
∂2 f
∂2 f
=
x
s
=
=
−
a
t
=
= 6y
6
,
3
,
∂x2
∂ x∂ y
∂y2
At (0, 0), (rt – s2) = 6x · 6y – 9a2 = 36xy – 9a2 = – 9a2 < 0 (for all values of a), while r = 0
...
At (a, a), (rt – s2) = 36xy – 9a2 = 36a2 – 9a2 = – 27a2 > 0 (for all values of a) and r = 6a
...
Example 72: Examine the function f(x, y) = x4 + y4 – 2x2 + 4xy – 2y2 for extreme values
...
For finding extreme values of f(x, y), necessary conditions is fx = 0 = fy
...
2
Hence the possible points for extreme values are (0, 0),
(
)(
2, − 2 − 2,
)
2
...
Otherwise also, f (x, y)
= x4 + y4 − 2x2 + 4xy − 2y2 = 0 and for points along x-axis, where
(0,0)
y = 0, i
...
f (x, y) x,0 = x4 − 2x2 = x2(x2 − 2)
...
Thus f(x, y) is neither maximum nor minimum at
(0, 0)
...
Discuss their nature
...
e
...
…(3)
Hence the stationary points are (0, 0), (3, 3), (–3, –3)
...
Thus total number of stationary points are (0, 0), (3, 3), (–3, –3), (1, –1), (–1, 1)
...
At (3, 3):
and
(rt – s2) = (2y2 – 10)(2x2 – 10) – (4xy – 8)2 = 64 – (28)2 < 0
r = 2y2 – 10 = 8 > 0
whence it is a point of minimum
...
At (1, –1): (rt – s2) = (2y2 – 10)(2x2 – 10) – (4xy – 8)2 < 0 (negative)
whence again at (–1,–1), function has neither maximum nor minimum
...
Therefore the function has neither maximum nor minimum at (–1, 1)
...
11 LAGRANGES METHOD OF UNDETERMINED MULIPLIERS: CONSTRAINED
MAXIMA-MINIMA
In many engineering and science problem it is desired to find extrema of a function of
several variables that are not all independent but are connected to one another by certain
conditions
...
Another name of it viz
...
Illustration: Say, if we want to find the Maximum and Minimum values of
u = f(x, y, z)
…(1)
Consider function of three variables x, y, z which are connected by an implicit relation,
φ(x, y, z) = 0
…(2)
For function u to possess stationary values, it is necessary that
∂u
∂u
∂u
= 0,
= 0,
=0
∂x
∂y
∂z
∴
∂u
∂u
∂u
dx +
dy +
dz = du(x, y, z) = 0
∂x
∂y
∂z
…(3)
…(4)
Also from (2),
∂φ
∂φ
∂φ
dx +
dy +
dz = dφ = 0
∂x
∂y
∂z
…(5)
We see that (3) + λ(4) results in
∂u
∂φ
∂φ
∂φ
∂u
∂u
+ λ dx +
+ λ dy +
+ λ dz = 0
∂x
∂x
∂y
∂z
∂z
∂y
…(6)
but this will hold true only if
∂φ
∂u
+λ
= 0,
(i)
∂x
∂x
∂φ
∂u
+λ
= 0, (ii)
∂y
∂y
∂φ
∂u
+λ
= 0, (iii)
∂z
∂z
…(7)
whence these three equation of (7) taken together with (2) will determine those x, y, z and λ
for which u is a stationary
...
Partial Derivatives and their Applications
333
Working Rule:
1
...
∂F
=0=
∂x
∂F
=0=
2
...
∂y
∂F
=0=
∂z
3
...
Example 74: Find the dimensions of a rectangle box, open at the top, of maximum capacity
whose surface is 432 sq
...
[MDU, 2004]
Solution: Let x, y, z be the three dimension of the box so that Volume, V(x, y, z) = xyz
Now our object is to find such values of x, y, z for which V(x, y, z) is maximum with the given
condition that
Surface area, S(x, y, z) = xy + 2yz + 2zx = 432 sq
...
e
...
e
...
e
...
e
...
y
334
Engineering Mathematics through Applications
72
, into (iv),
y
y
72
1−
y + 2 ⋅ = 0 or 288 – (y2 + 144) = 0 or y2 = 144 means y = 12 cm
288
y
Now putting x = y, z =
72 72
=
= 6 cm
y
12
Hence x = 12 cm, y = 12 cm, z = 6 cm gives the maximum volume
...
a
b
c
[UP Tech, 2004; KUK, 2007; PTU, 2008]
Solution: Consider
F(x, y, z) = u(x, y, z) + λ φ(x, y, x) = (a3x2 + b3y2 + c3z2) + λ(x–1 + y–1 + z–1)
Then for stationary values,
∂φ
∂F ∂u
=
+λ
=0
∂x ∂x
∂x
∂φ
∂F ∂u
=
+λ
=0
∂y ∂y
∂y
∂φ
∂F ∂u
=
+λ
=0
∂z ∂z
∂z
λ
= 0 or 2a3 x3 = λ,
x2
λ
2b3 y + 2 = 0 or 2b3 x3 = λ ,
y
λ
2c3 z + 2 = 0 or 2c3 z3 = λ,
z
2a3 x +
(i)
(ii)
(iii)
…(1)
…(2)
From (i), (ii), (iii),
2a3x3
=
2b3y3
k
a
k
y=
b
k
z=
c
x=
or
so that
=
2c3z3
= λ or
λ
ax = by = cz =
2
1
3
= k (say)
(i)
(ii)
(iii)
…(3)
a b c
1 1 1
+ + = 1 gives
+ + = 1 implying (a + b + c) = k
x y z
k k k
Therefore, from (3), x =
…(4)
k a + b + c ∑a
k ∑a
k ∑a
=
=
, y= =
, z= =
,
a
a
a
b
b
c
c
Example 76: Find the volume of the greatest rectangular parallelopiped that can be
inscribed inside the ellipsoid
x2 y2 z2
+
+
=1
a2 b2 c2
Partial Derivatives and their Applications
335
Solution: Let edges of the parallelopiped be 2x, 2y, 2z parallel to the coordinate axes so that
volume,
V = 8xyz
…(1)
Our object is to maximise V(x, y, z) subject to the condition,
x2 y2 z2
x2 y2 z2
+
+
=
1
or
φ
(
x
,
y
,
z
)
=
+
+
−1= 0
a2 b2 c2
a2 b2 c2
…(2)
Define function,
x2 y2 z2
F(x, y, z) = V + λφ = 8xyz + λ 2 + 2 + 2 − 1
a
b
c
…(3)
So that for stationary values,
∂F
2x
= 8yz + λ 2 = 0
∂x
a
2y
∂F
= 8xz + λ 2 = 0
∂y
b
∂F
2z
= 8xy + λ 2 = 0
∂z
c
…(i)
…(ii)
…(iii)
…(4)
Equating the values of λ from (i) and (ii); (i) and (iii), we get
x2 y2
=
a2 b2
implying thereby
or
and
x2 z2
=
a2 c2
2
x2 y2 z2 1 using x2 = y = z2
;
=
=
=
in (2)
a2 b2 c2
a2 b2 c2 3
x=
…(5)
a
b
c
, y=
, z=
3
3
3
When x = 0, the parallelopiped mearly becomes a rectangular sheet and in that case the
volume, V = 0
...
3 3
Example 77: Find the maximum value of the function cosA cosB cosC
...
e
...
e
...
2
cosA sinB cosC = cosA cosB sinC or cosA(sinB cosC – cosBsinC) = 0
Implying either
sin(B – C) = 0
i
...
B = C
…(6)
π
...
or
cosA = 0 i
...
A=
Here f = cosA cosB cosC is maximum at A = B = C =
π
and fMaxi = (cos 60°)3 = 1
...
e
...
3
2
Further find r = fAA, s = fAB , t = fBB and observe signs of (rt – s2) and r to ascertain,
Together, (11) and (12) gives A = B =
A=B=C=
π
...
Solution: Here we need to determine the extreme values of the function f(x, y) = x2 + y2
(or d =
x2 + y2
) subject to the condition φ(x, y) = 5x
2
+ 5y2 + 6xy – 8 = 0
Thus,
F(x, y) = f(x, y) + λ φ(x, y) = (x2 + y2) + λ(5x2 + 5y2 + 6xy – 8)
Now
∂F
= 2x + λ(10x + 6y) = 0
∂x
…(1)
…(2)
∂F
= 2y + λ(6x + 10y) = 0
…(3)
∂y
In order to solve (2) and (3) for x and y, multiply (2) by y and (3) by x and then substract
the later from the former,
6λ(y2 – x2) = 0 i
...
y = ± x
…(4)
2
2
On substituting y = ±x into the relation, 5x + 6xy + 5y – 8 = 0, we get,
2x2 = 1 …(i)
…(5)
and
and x2 = 2 …(ii)
Thus, the distance, f(x, y) = (x2 + y2) from the origin becomes
d2 = x2 + y2 =
1
1 1
+ = 1, using x2 = = y2
2
2 2
and
d2 = x2 + y2 = 2 + 2 = 4, (using x2 = 2 = y2)
Obviously, the first value is a minimum whereas second value is a maximum
...
Example 79: Prove that the rectangular solid of maximum volume that can be inscribed in
a given sphere is a cube
...
∴
x2 + y2 + z2 = d
or S(x, y, z) ≡ (x2 + y2 + z2 – d2)
…(2)
338
Engineering Mathematics through Applications
Now the condition for the volume of the rectangular solid to be maximum is that
∂V
∂S
∂V
∂S
∂V
∂S
+λ
= 0,
+λ
= 0,
+λ
=0
∂x
∂x
∂y
∂y
∂z
∂z
…(3)
yz + 2λx = 0 …(i)
zx + 2λy = 0 …(ii)
i
...
xy + 2λz = 0 …(iii)
Multiply (i) by x, (ii) by y, (iii) by z and equate them
From (i) and (ii), we get
x=y
i
...
x = y = z
From (i) and (iii), we get x = z
Hence the given rectangular solid would be of maximum volume if it is a cube
...
∂S
∂z
∂S
∂z
= 2y + 2z
= 2x + 2z
…(i)
∂y
∂y …(ii)
∂x
∂x
(³ y is independent of x)
∂S
∂S
=0=
implies
∂x
∂y
Here
∂z
x
=−
and
∂x
z
y
∂z
=−
∂y
z
…(6)
Vx (x, y, z) = 1 ⋅ yz + xy
x2 y
∂z
x2
= yz −
= yz − = 0
∂x
z
z
…(7)
Vy(x, y, z) = 1 ⋅ xz + xy
xy2
y2
∂z
= xz −
= x z − = 0
∂y
z
z
…(8)
Now
and
(³ x is independent of y)
(7) and (8) collectively implies x = y = z
...
From (7),
∂z
2 x
2zx − x2
x 2zx − x − z − xy(x2 + 3z2 )
∂z
x
∂
r = y
−
= −4x
=
= y − z −
z2
z2
z3
∂x
(at x = y = z) …(9)
From (7),
∂z x2 ∂z
y x2 y
x2
x2
+ 2 = z − + y − + 2 − = −2x ,
s = 1 z − + y
z z z
z
z
∂y z ∂y
(at x = y = z)
From (8),
t=
−xy(y2 + 3z2 )
= −4x, (at x = y = z)
z3
rt – s2 = (–4x)(–4x) – (–2x)2 = 12x2 > 0 and r = –2x < 0 for all x
...
e
...
…(10)
…(11)
Partial Derivatives and their Applications
339
Example 80: Find the stationary values of (x2 + y2 + z2) subject to ax2 + by2 + cz2 = 1 and
lx + my + nz = 0
Solution: Let f(x, y, z) = (x2 + y2 + z2)
and
φ(x, y, z) = (ax2 + by2 + cz2 – 1) = 0
…(1)
…(2)
ψ(x, y, z) = (lx + my + nz) = 0
Then by Lagranges Multiplier,
F(x, y, z) = (x2 + y2 + z2) + λ(ax2 + by2 + cz2) + 2µ(lx + my + nz)
which implies
dF = (2x + λ2ax + 2µl)dx + (2y + λ2by + 2µm)dy + (2z + λ2cz + 2µn)dz
Then for F(x, y, z) to possess stationary values,
…(3)
∂F
= x + λax + µl = 0 …(i)
∂x
∂F
= y + λby + µm = 0 …(ii)
…(4)
∂y
∂F
= z + λcz + µn = 0 …(iii)
∂z
In the above results, first multiply (i) by x, (ii) by y, (iii) by z and then add the three,
(x2 + y2 + z2) + λ(ax2 + by2 + cz2) + µ(lx + my + nz) = 0
On using (1), (2), (3), the above expression becomes
f + λ · 1 + µ · 0 = 0 i
...
λ = –f
…(5)
µl
,
(af − 1)
µm
y=
,
Similarly from, 4(ii) and 4(iii),
(bf − 1)
µn
z=
(cf − 1)
Substituting the values of x, y, z again in the relation (3), we get the condition
Thus from 4(i), x + (–f)ax + µl = 0 or
x=
…(6)
µl
µm
µn
l
+ m
+ n
=0
af − 1
bf − 1
cf − 1
l2
m2
n2
+
+
= 0,
…(7)
af − 1 bf − 1 cf − 1
This is a quadratic in f from which we can obtain its extreme values
...
Therefore point (x, y, z) satisfying both these equations lies on the conic of their
intersection and the expression (x2 + y2 + z2) gives us the square of the distance of (x, y, z)
from the origin
...
Hence the equation (7) gives the squares the lengths of the semiaxes of the conic of intersection
...
Find x and y in terms of h, if the canvas required for its
construction is to be minimised for the tent to have a given
capacity
...
4
...
Here V(x, y, h) = Volume of the box (cuboid) + Volume
due to elevated Portion (pyramid)
= x2 y +
y
D
C
x
…(1)
x
A
and S(x, y, h) = Area of the four faces of the cuboid + Area
of the four faces of the pyramid
...
4
...
e
...
Now for minimum surface area, Sx = 0 = Sh
...
2
ASSIGNMENT 9
1
...
Determine the maxima of the function given by u = (x + 1)(y + 1)(z + 1)
subject to the condition xaybzc = k
...
Given x + y + z = a, find the maximum value of xm yn zp
...
(i) xy +
2
...
4
...
6
...
342
Engineering Mathematics through Applications
8
...
Find the highest temperature
on the surface of the unit sphere x2 + y2 + z2 = 1
...
Find a point within a triangle such that the sum of the squares of its distances from the
three vertices is minimum
...
10
...
are
the
roots
of
the
equation
+
+
=
1
2
2
1 − a u 1 − b u 1 − c2u
a2 b2 c2
11
...
[NIT kurukshetra, 2008]
2
[Hint: f = ∆ = s(s – a)(s – b)(s – c), where a + b + c = 2s (s const
...
Find the triangle of maximum area inscribed in a circle
...
12 DIFFERENTIATION UNDER INTEGRAL SIGN
If f(x, α) a function of two variables with α as parameter, be integrated with respect to x
∫
between the limits a and b, then
b
a
f (x, α) dx will be a function of α: F(α), say
...
Such problems are handled by the following
rules:
Leibnitz’s Rule* 1: (When Limits of integration are constant)
∫
To find
Statement: If f(x, α) and
d
dα
∫
b
a
f (x, α)dx =
Poof: Let
∫
b
a
∂
∂
f (x, α) be continous function of x and a, then
∂α
∫ ∂α f (x, α) dx, where a and b are constants independent of α
...
*Named after the German mathematician Gottfried Wilhelm Leibnitz (1646–1716)
...
Example 82: Differentiating
∫
of
x
0
∫
x
0
1
1
x
dx = tan−1 under the integral sign, find the value
a
a
x2 + a2
1
dx
(x2 + a2 )2
1
dx
x
= tan−1
2
0 (x + a )
a
a
with respect to ’a’ (the parameter) by Leibnitz’s Rule 1,
Solution: Differentiating both sides of
∫
x
0
1
1 ∂
x ∂ 1
x
⋅ 2a dx =
tan−1 +
⋅ tan−1
(x2 + a2 )2
∂a a
a ∂a
a
a
∫
i
...
− 2a ⋅
0
2
∂ 1
∂ 1
−1 x
dx =
2
⋅ tan
2
∂a x + a
∂a a
a
implying
x
∫
x
−
∫
x
0
1
1
−x
x
− 2 tan−1
dx =
2 2
2
2
(x + a )
a(a + x ) a
a
2
Dividing throughout by –2a,
∫
x
0
1
1
x
x
dx = 3 tan −1 + 2 2
(x2 + a2 )2
2a
a 2a (a + x2 )
Example 83: Prove that if a > 0,
Solution: Let F(m) =
∫
∞ − ax
0
e
∫
∞
0
sin mx
dx
x
e− ax sinmx
m
dx = tan−1
...
x
a
∫
Example 84: Prove that
∞ −x
e
(1 − e−ax )dx = log(1 + a), (a > – 1)
x
0
∞ −x
Solution: Let I =
∫
so that
dI
=
da
0
e
(1 − e−ax ) dx
x
dI
=
da
∫
∞
0
∫
∞
0
…(1)
∂ e−x
(1 − e−ax ) dx =
∂a x
∫
∞ −x
0
e
⋅ e−ax (−x) dx =
x
∫
∞
0
e−x e−ax dx
∞
e−(1+ a)x
1
e−(1+ a)x dx =
= (1 + ) , a > − 1
−
+
(1
)
a
a
0
On integration,
I = log (1 + a) + C
…(2)
e
(1 − 1) dx = 0 thereby implying C = 0
0 x
from (2), I = log(1 + 0) + C = C
Now when a = 0 then from (1), I =
∴
∫
∞ −x
I = log (1 + a), a > – 1
...
2
− π = π 1 + y − 1
...
2 2
π
Solution: Let
I(a) =
∫ log(1 + a cos x) dx
so that
dI
=
da
∫
0
π
0
∂ log(1 + cos ) =
a
x dx
∂a
∫
π
0
cos x
dx
1 + a cos x
346
Engineering Mathematics through Applications
=
For
∫
π
0
∫
1
a
1 + a cos x − 1
1
dx =
1 + a cos x
a
π
0
∫
1
dx, use
1 + a cos x
π
0
=
π
0
dx −
1
a
∫
π
0
1
dx
1 + a cos x
β + α cos x
1
1
cos−1
taking α = 1, β = a
...
Integrating both sides with respect to a,
π
∫ a da − π∫ a
I=π
1
1
da + C
1 − a2
∫ sin t 1 − sin t cos t dt + C (Put a = sint, da = cost dt)
1
= π log a − π∫
dt + C = π log a − π∫ cosec t dt + C
sin t
1
= π log a − π
2
= π loga + π log (cosec t + cot t)
= π log a + π log
1 + cos t
1 + 1 − sin2 t
+ C = π log a + π log
+C
sin t
sin t
(
When a = 0, from above
from given integral,
1 + 1 − a2
I = π log 1 + 1 − a2 − π log 2 = π log
2
(
∴
Example 87: Prove that
Solution: Let
so that
)
1 + 1 − a2
+ C = π log 1 + 1 − a2 + C
a
I(0) = π log 2 + C
implying C = – π log2
...
e
...
Example 88: Show that
∫
∞
0
tan−1ax
π
dx = log(1 + a), a ≥ 0
2
2
x(1 + x )
[KUK, 2000, 2005; NIT Kurukshetra, 2004]
348
Engineering Mathematics through Applications
Solution: Let
I=
∫
∞
0
tan−1 ax
dx
x(1 + x2 )
…(1)
Differentiating both sides with respect to a, we have
∂
dI
=
da ∂ a
∫
∞
=
∫
∞
dI
=
da
∫
∞
=
0
0
0
∫
∞
0
tan−1 ax
dx
x(1 + x2 )
1
∂ tan −1
ax dx
x(1 + x2 ) ∂a
1
x
⋅
dx
2
x(1 + x ) 1 + a2 x2
1
dx
(1 + a2 x2 )(1 + x2 )
…(2)
Dealing the integrand by making partial fractions on taking x2 = t, we get
1
A
B
=
+
(1 + a2 t)(1 + t) (1 + t) (1 + a2 t)
⇒
1 = A(1 + a2t) + B(1 + t)
…(3)
2
1
When 1 + a2t = 0, we get B 1 − 2 = 1 i
...
B = − a 2
a
1− a
1
(1 − a2 )
Therefore on substituting values of (A) and (B) in (3), we get
When 1 + t = 0, we get 1 = A(1 – a2), i
...
A =
1
1
1
a2
2
,
=
−
2
2
2
(1 + a t) (1 + t) (1 − a )(1 + t) (1 + a t)(1 − a2 ) t = x
Using (4), integrand (2) becomes
dI
=
da
or
∫
∞
0
1 1
a2
dx
−
(1 − a2 ) (1 + x2 ) 1 + a2 x2
1
dI
=
da (1 − a2 )
=
∫
∞
0
∫
1
dx −
(1 + x2 )
∞
0
1
2
1 + 2
x
a
dx
∞
∞
1
tan−1 x ) − ( a tan−1 ax )
(
2
0
0
(1 − a )
dI
π
(1 − a) π
π
1 π
=
− 0 − a − 0 =
=
2
da (1 − a ) 2
2
(1 − a2 ) 2 2(1 + a)
…(4)
Partial Derivatives and their Applications
349
On integrating both sides with respect to a,
or
⇒
π
log(1 + a) + C
2
Now on using conditions, when a = 0, I = 0 (from (1))
I=
…(5)
π
log(1 + 0) + C , i
...
C = 0
2
With C = 0, (5) becomes
0=
π
log(1 + a)
2
I=
∫
Example 89: Evaluate
2
∞ − x2 + a2
x
e
0
dx by differentiating under the integral sign
...
t
t
d
F(a) =
da
∴
2
0 − a + t2
t2
∫e
∞
0
2
0 − t2 + a
t2
∫e
2
∞ − x2 + a
x2
∫
dF(a)
= −2
da
− 2at2 − a
2 ⋅ 2 ⋅ dt = 2
a
t
e
dx = −2F(a) or
∞
dt
dF(a)
= −2 da
F(a)
…(2)
On integration,
log F(a) = – 2a + logC or logF(a) = log e– 2a + logC
implying
When a = 0,
F(a) = Ce– 2a
from (3), F(0) = C
F(0) =
from (1),
Using (4),
…(3)
F(a) =
Example 90: Evaluate
∫
∫
2
∞ − x2 + a
x2
e
0
∫
α
0
∞
0
e
− x2
π giving C = π
dx =
2
2
dx = Ce−2 a =
…(4)
π −2 a
e
...
8
(1 + x2 )
350
Engineering Mathematics through Applications
Solution: Let F(α) =
∫
α
0
log(1 + αx)
dx
(1 + x2 )
…(1)
By Leibnitz’s Rule 2,
dF
=
dα
=
∫
α
0
∫
α
0
log(1 + α2 )
∂ log(1 + α2 )
d
α
+
dx
(
)
2
(1 + α2 )
dx
∂x 1 + α
log(1 + α2 )
x
dx
+
(1 + α x)(1 + x2 )
1 + α2
…(2)
Resolving the integrand into partial fractions:
x
A
Bx + C
=
+
2
(1 + αx)(1 + x ) 1 + αx 1 + x2
x = A(1 + x2) + (Bx + C)(1 + αx) or x = (A + Bα)x2 + (B + Cα)x + (A + C)
implying
1 + αx = 0
Putting
x=
i
...
−1
1
−1
, we get α = A 1 + 2
α
α
Comparing coefficients of x2, 0 = A + αB or
Comparing constant terms, 0 = A + C or
∫
∴
α
0
x
1
dx =
2
(1 + αx)(1 + x )
(1 + α2 )
=
∫
α
0
∫
α
0
A=
−α
1 + α2
1
−A
=
1 + α2
α
B=
C = −A =
−α
1
dx +
1 + αx
2
or
∫
α
0
α
1 + α2
∫
2x
dx + α
(1 + x2 )
α
0
1
dx
2
1+ x
α
1 −α log(1 + αx) 1
+ log(1 + x2 ) + α tan− 1 x
(1 + α2 )
2
α
0
1
1
x
dx =
− log(1 + α2 ) + log(1 + α2 ) + α tan− 1 α
(1 + αx)(1 + x2 )
(1 + α2 )
2
…(3)
With the above integral value, (2) on cancellation of some terms reduces to
α
dF 1 log(1 + α2 )
tan −1 α
=
+
2
2
(1 + α )
dα 2 (1 + α )
Integrating (4) with respect to a,
F(α) =
=
1
1
α
(tan−1 α) dα + C (Integration by parts)
⋅ log(1 + α2 ) dα +
1 + α2
2 (1 + α2 )
∫
∫
α
α
1
−1
⋅ 2α tan− 1 α d α +
log(1 + α2 )(tan−1 α) −
(tan α) d α + C
2
2
1+ α
1 + α2
∫
…(4)
Partial Derivatives and their Applications
1
log(1 + α2 ) ⋅ (tan−1 α) + C
2
F(α) =
At α = 0;
…(4)
from (1), F(0) = 0
implying C = 0
from(5), F(0) = 0 + C
∫
Hence
Putting
351
α
0
log(1 + αx)
1
dx = log(1 + α2 ) ⋅ (tan−1 α)
(1 + x2 )
2
∫
α = 1 in (5),
Solution: Let F(a) =
∫
a2
0
so that
d
F(a) =
da
∫
1 log(1 +
αx)
1
π
dx = log 2(tan−1 1) = loge 2
(1 + x )
2
8
2
0
d
da
Example 91: Show that
…(5)
∫
a2
0
tan−1
1
x
dx = 2a tan−1 a − log (a2 + 1)
...
)
=
∫
a2
−1
d x
1
dx + 2a tan− a =
2
da
a
1 + x
a
1
2
0
∫
a2
0
2x
dx + 2a tan−1a
a2 + x2
a2
1
1
= − log(a2 + x2 ) 0 + 2a tan−1 a = 2a tan−1 a − log(1 + a2 )
2
2
Verification: Put
∴
∫
a2
0
whereas
∫
x
= tan θ so that dx = a sec2θ dθ and limits for x = 0, θ = 0,
a
2
−1
for x = a , θ = tan a
∫
x
dx = a
a
tan−1
tan−1 a
0
∫
a2
0
tan −1 a
0
tan −1 a
θ ⋅ sec2 θ dθ = a (θ ⋅ tan θ)0
−
tan θ dθ = ( log cos θ )0
tan−1 a
=−
implying
…(3)
tan−1
1
(log(1 + tan2 θ))0
2
= ( − log sec θ )0
tan−1 a
∫
tan−1 a
tan −1 a
0
1 ⋅ tan θ dθ
(
= − log 1 + tan2 θ
1
= − log(1 + a2 )
2
x
a
dx = a2 tan−1 a − log(1 + a2 )
2
a
Differentiation of (6) with respect to ‘a’ gives the desired result
...
0
d2 y
+ k2 y = k f (x )
...
Evaluate
∫
1
0
xα − 1
dx, α ≥ 0
log x
[MDU 2004; KUK, 2004]
2
...
By differentiating under the integral sign, evaluate the integral
Hence show that
∫
∞
0
∫
∞ − ax
e
0
sin x
dx
x
sin x
π
dx =
2
x
4
...
By successive differentiation of
∫
1
0
xm dx =
∫
α2
0
tan−1 x dx
α
1
with respect to m, evaluate
m+1
∫ x (log x) dx
1
0
m
n
Partial Derivatives and their Applications
353
ANSWERS
Assignment 1
5
...
2
Assignment 3
x
8
...
– 7 tanu
Assignment 4
1
...
log(cos x) − x cot y
a2
2
2
3
...
1 + log xy − x
5
...
21
y
1
x2 + y
y2 + x
y
Assignment 6
1
...
– 4r3; r2 = x2 + y2
5
...
(x − y)(y − z)(z − x)
1
7
...
x(u – 1 – vy) + 2uv – z
4
...
w = 4 (u2 + 3v)
J' = (x − y)(y − z)(z − x)
Assignment 7
1
...
(p – 3q – 4r)%
4
6
...
5r
12
...
ex cos y = e + (x − 1) e + y − π − e
2
2
4
2
354
Engineering Mathematics through Applications
+
1
2
2
y − π − e + y − π − e + …
2 e
(
1)
2(
1)
x
−
+
x
−
2
4
2
4
2
Hint : For a = 1, b = π ; x − a = h = x − 1, y − b = k = y − π
4
4
2
...
2
...
xy
5
...
x = y = 2z
1
...
p2/(a2 + b2 + c2)
bc
ca
ab
log k 2
log k 2
log k 2
a
b
c
4
...
x = 12, y = 8, z = 4
8
...
Areamaxi =
9
...
x=
mm nn pp am + n + p
(m + n + p)m + n + p
y +y +y
x1 + x2 + x3
, y= 1 2 3
3
3
−2r2
π
at A = B = C =
3
3
Assignment 10
1
...
(−1)n n
(
n +1
m + 1)
7
Title: Partial derivatives
Description: All concepts related to Partial derivatives
Description: All concepts related to Partial derivatives