MODULE - II
Partial Differentiation
(FUNCTIONS OF TWO OR MORE VARIABLES)
3.1.
FUNCTIONS OF TWO VARIABLES
If three variables x, y, z are so related that the value of z depends upon the values of
x and y, then z is called a function of two variables x and y, and this is denoted by z = f(x, y).
z is called the dependent variable while x and y are called independent variables.
For example, the area of a triangle is determined wheb its base and altitude are
known. Thus area of triangle is a function of two variables, base and altitude.
(In a similar way, a function of more than two variables can be defined).
Geometrically. Let z = f(x, y) be a function of two independent variables x and y
defined for all pairs of values of x and y which belong to an area A of the xy - plane. Then to
each point (x, y) of this area corresponds a value of z given by the relation z = f(x, y). Representing all these values (x, y, z) by points in space, we get a surface.
Hence the function z f(x, y) represents a surface.
3.2.
PARTIAL DERIVATIVES OF FIRST ORDER
Let z = f(x, y) be a function of two independent variables x and y. If y is kept constant
and x alaone is allowed to vary, then z becomes a function of x only. The derivative of z,with
respect to x, treating y as constant, is called partial derivative of z w.r.t.x and is denoted by
or
∂f or fx
∂z
∂x
∂x
Thus
∂z =
∂x
Lt
f(x+h, y) - f(x, y)
h !0
h
Similarly the derivative of z, with respect to y, treating x as constant, is called partial
derivative of z w.r.t. y and is denoted by
∂z
or
∂f
or fy
y
y
∂
∂
∂z = Lt
f(x, y +k) - f(x, y)
∂y k ! 0
k
∂z are called first order partial derivatives of z.
∂y
Thus
∂z
∂x
[In general, if z is a function of two or more independant variables, then the partial
derivative of z w.r.t. any one of the independent variables is the ordinary derivative of z w.r.t
that variable, treating all other variables as constant].
[Geometrically, Let z = f(x, y) be a function of two variables x and y. Then by Art. 3,
1 it represents a surface S. If y = k, a constant, thewn y = k represents a plane parallel to the
zx - plane.
... z = f(x, y) and y = k
represent a plane curve C which is the section of S by y = k.
∂z represents the slope of tangent to C at (x, k, z)
∂x
∂z
Thus gives the slope of the tangent drawn to the curve of intersection of the
∂x
surface z = f(x, y) and a plane parallel to zx - plane.
∂z
Similarly
gives the slope of the tangent drawn to the curve of intersection of the
∂y
surface z = f(x, y) and a a plane parallel to yz - plane.
3.3.
PARTIAL DERIVATIVES OF HIGHER ORDER
∂z
∂z
Since the first order partial derivatives∂x and ∂yare themselves functions of x and y,
they can be further differentiated partially w.r.t.x as well as y. These are called second order
partial derivatives of z. The usual notations for these second order partial derivatives are :
2
∂ ∂z
∂x ∂x = ∂∂xz2 or fxx
∂
∂
( )
∂
∂y
∂
∂x
∂
∂y
(∂∂zy)= ∂∂z2y2 or fyy
2
(∂∂zy)= ∂zx∂y
or fxy
∂
(∂z∂x)= ∂∂z2y xor fyx
∂
2
In general, ∂ z =
∂x∂y
∂2z
∂y∂x
or fxy = fyx
Note 1.If z = f(x), a function of single independent variable x, we get dz
dx
If z = f (x, y), a function of two independent variables x and y, we get ∂z and ∂z
∂y
∂x
Similarly, for a function of more than two independent variables x1, x2, ......., xn we get
∂z
∂x1,
∂z
∂z
∂x2, ......., ∂xn
x
Note 2.
(i) If z = u +v, where u = f(x, y), v = φ (x, y), then z is a function of x and y.
...
∂z = ∂u ∂v
+
;
∂x ∂x ∂x
∂z ∂u ∂v
=
+
∂y ∂y ∂y
(ii) If z = uv, where u = f(x, y), v = φ (x, y),
then
∂z ∂
∂v
∂u
=
(uv) = u x + v x
∂x ∂x
∂
∂
∂v
∂z ∂
= y (uv) = u y + v ∂u
y
∂
∂
∂
∂y
(iii) If z =u where u = f(x, y), v = φ (x, y),
v,
∂z ∂ u
=
∂x ∂x v
( ) = u ∂∂ux - u ∂v∂x
v2
∂z ∂ u
∂u
∂v
=
=v y -u y
∂y ∂y (v,)
∂
∂
then
v2
(iv) If z = f (u), where u = φ (x, y),
∂z = ∂z ∂u
,
;
∂x ∂u ∂x
∂z ∂u ∂u
=
,
∂y ∂u ∂y
ILLUSTRATIVE EXAMPLES
Example 1. Find the first order partial derivatives of the following:
(ii) u = tan -1 x2+y2
(i) u = yx
x+y
( ).
(iii) u = cos -1 x
y
Sol . (i) u = yx
[Treating y as constant, u is of the form ax]
∂u = yx log y
∂x
[Treating x as constant, u is of the form yn]
∂u = xyx -1
∂y
(ii)
u = tan -1 x2+y2
x+y
∂u =
1
2
x
2+y2
∂
1+ x
x+y
( )
=
, ∂ x2 + y2
∂x
x+y
(
)
2 + y2 ) - (x2 + y2 ) ∂ (x + y)
∂
(x
∂x
(x+y)
∂x
(x+y)2
2
(x+y)2+ (x2 + y2 )
=
(x+y)2
(x+y). 2x - (x2 + y2). 1
(x+y)2+ (x2 + y2 )
!
2
∂u = x2 + 2xy - y2
2
∂x (x+y)2 + (x2 + y2 )
[Since u remains the same if we interchange x and y, u is symmetrical w. r.t.x and y. Interchanging x and y in (1), we have]
Similarly,
(iii)
u = cos -1
∂u = y2 + 2xy-x2
∂y (x+y)2 + (x2 + y2)2
x
()
y
∂u =
∂x
=
-1
2
()
1- x
y
-y
y2 - x2
-1
∂u =
2
∂y
1- x
y
()
=
-y
y2 - x2
∂
∂x
(xy)
1
-1
y
y2 - x2
∂
∂y
(xy)
(- xy 2)=
x
y y2 - x2
y
Example 2. If z (x+y) = x2+y2, show that
2
∂z - ∂z = 4 1∂x
∂y
∂z - ∂z
∂x
∂y
Sol. z = x2+y2
x+y
(z is symmetrical w.r.t.x and y)
∂z
=
∂x
(x+y) ∂ (x2+y2) x2+y2) - ( x2+y2)∂ (x+y)
∂x
∂x
(x+y)2
(x+y) . 2x- (x2+y2) .1
=
Similarly,
∂z =
∂y
Now
∂z = ∂z
∂x ∂y
=
(x+y)2
x2+2xy-y2
(x+y)2
y2 + 2xy-x2
(x+y)2
2
=
2x2 - 2y2
(x+y)2
=
4(x+y)2 (x-y)2
(x+y)4
= 4(x-y)2
(x+y)2
4(1-∂z = ∂z = 4 x2+2xy-y2)- y2+2xy-x2
∂x ∂y
(x+y)2
(x+y)2
2
2 2
2- 2
2
= 4 1- x +2xy+y - x -2xy+y y -2xy+x
(x+y)2
=
4(x2-2xy+y2) 4(x-y)2
(x+y)2
...
∂z - ∂z
∂x ∂y
2
∂z
= 4 1- ∂z ∂y
∂x
(x+y)2
Example 3. If u = log (tan x + tan y), prove that
Sin 2x ∂u + Sin 2y ∂u = 2
∂x
∂y
Sol.
u = log (tan x + tan y) (u is symmetrical w.r.t.x and y)
∂u
1
=
∂x
tan x + tan y.
∂ (tan x + tan y) =
∂x
sec2 x
tan x + tan y
∂u
sec2 y
=
∂x
tan x + tan y
Sin 2x ∂u + sin 2y ∂u
∂x
∂y
Similarly
=
=
1
2
2
tan x + tan y. (sin 2x sec x + sin 2y sec y)
1
1
1
tan x + tan y. (2 sin x cos x cos2x + 2 sin y cos y.
cos2y
= 2 (tan x + tan y)
tan x + tan y
= 2.
x2
Example 4. If f (x, y, z) = x
1
fx+fy+fz = 0.
Sol.
f (x, y, z) =
x2
x
1
y2
y
1
z2
z , prove that
1
y2
y
1
z2
z
1
Expanding by first row = x2 (y-z ) - y2 (x-z )+ z2 (x-y)
= x2y-xy2 +y2z-yz2+ z2x-zx2
...
fx
fy
= 2xy-y2 +z2-2zx
= x2 -2xy+2yz -z2
fz
= y2 -2yz+2zx -z2
fx+fy+fz = 0
2
Example 5. Verify that
2
∂u = ∂u
for the following functions:
∂x∂y ∂y∂x
ax
(i) u = e sin by
(ii) u = log
(x+y)2
xy
(iii) u = tan-1 x
y
Sol. (i) u = eax sin by
∂u = sin by. ax
e a = aeax sin by
∂x
∂u
= eax cos by. b = beax cos by
∂y
2
∂ u = ∂ ∂u
∂x∂y
∂x ∂y = b cos by.
eax . ∂ = abeax cos by
2
∂ u = ∂ ∂u
ax
ax cos by.
∂y∂x
∂y ∂x = ae cos by. b = abe
2
∂ u = ∂2u
∂x∂y
∂y∂x
...
(ii)
u = log
x2 + y 2
xy
= log (x2 +y2) - log x - log y
∂u
=
∂x
2x
-1
x2 + y 2 x
∂u
=
∂y
x2 + y2
2y
-1
y
2
∂ u = ∂ ∂u
= ∂ 2y (x2 + y2)-1-1
∂x∂y ∂x ∂y
∂x
y
-2
= -2y (x2 + y2) , ∂ (x2 + y2) = ∂x
=
-
2y
(x2 + y2) 2. 2x
4xy
2
( x2 + y 2 )
2.
∂ u = ∂ ∂u
2
2 -1
= ∂ 2x (x + y ) -1
∂y∂x ∂y ∂x
∂y
x
=
-2x (x2 + y2)
= - 4xy
2
(x2 + y2)
2
... ∂2.u = ∂ u
∂x∂y
∂y∂x
-2.
∂
2
2
2x
. 2y
∂y (x + y ) = 2
2
2
(x + y )
(iii)
-1
u = tan
x
y
y
∂u
y2
x
= 1
1
∂
2
2
2
∂x
1+ x ∂x y = (x2 + y2) y = x + y
y
y
∂u
y2 - x
x
= 1
∂
2
2
2
∂y
1+ x ∂y y = (x2 + y2) y2 = x + y
y
2
(x2 + y2) ∂ (x)-x ∂ (x2 + y2)
∂ u = ∂ ∂u
∂x
∂x
=
∂x∂y ∂x ∂y
2
2
(x + y )2
x2 - y2
2
2
- (x + y ) 1-x. 2x
=
=
(x2 + y2)2
(x2 + y2)2
2
(x2 + y2) ∂ (y)-y ∂ (x2 + y2)
∂ u = ∂ ∂u
∂y
∂y
=
∂y∂x ∂y ∂x
(x2 + y2)2
=
x2 - y2
(x2 + y2) 1-y. 2y
=
(x2 + y2)2
...
(x2 + y2)2
∂2u
∂2u
=
∂x∂y ∂y∂x
Example 6. If z = log (ex + ey), show that rt - s2 = 0
2
2
Where r ∂ z2 , t = ∂ z , s = ∂2z
∂x
∂y2
∂x∂y
⇒
Sol . z = log (ex + ey)
1
∂z
= x y ∂ (ex + ey), ex
∂x
e + e ∂x
ex + ey
1
∂
∂z
= x y ∂y(ex + ey)= ey
∂y
e +e
ex + ey
... r =
∂
∂2z
=
∂x
∂x2
ex
x
e + ey
∂ (ex ) - ex ∂ (ex +ey )
∂x
∂x
x
y
(e + e )2
x y
(ex + ey) ex - ex .ex
= e +
(ex + ey)2
(ex + ey)2
(ex + ey).
By symmetry,
∂2z =
ex + y = r
∂y2
(ex + ey)2
t=
s=
2
∂
∂ z = ∂ ∂z
= ∂x [ey (ex + ey)-1]
∂x
∂y
∂x∂y
= ey . (-1) (ex + ey)-2
∂ (ex)
∂x
x y
= - e +
= -r
(ex + ey)2
... rt - s2 = r. r - (-r)2 = 0
Example 7. If u = xy, show that
∂ 3u
∂x2∂y
Sol.
∂ 3u
∂x∂y∂x
u = xy
∂u
= xy log x
∂y
∂ ∂u
∂2u
y-1
= ∂x
= yx
∂x∂y
∂y
log x + xy . 1
x
= xy-1 (y log x +1)
∂3u = ∂
∂x2∂y ∂x
∂2x
∂
[xy-1 (y log x +1)]
=
∂x∂y
∂x
.
....(1)
∂u
= yxy-1
∂x
∂2u = ∂ ∂u
∂y∂x ∂y ∂x
= xy-1 + yxy-1log x
= xy-1 (y log x +1)
∂3u = ∂ ∂2u = ∂
y-1
∂x [x (y log x +1)]
∂x∂y∂x ∂x ∂y∂x
From (1) and (2),
∂ 3u
∂x2∂y =
∂3u
∂x∂y∂x
....(2)
r cos 0
Example 8. If x = e
cos (r sin 0) and
r cos 0
∂x
1
y=e
r sin 0 prove that
=
∂r
r,
∂y ∂y
1
∂ 0 ∂r = - r
∂x
∂0
2
Hence deduce that ∂ x + 1 ∂x + 1 ∂2x
= 0
∂r2 r ∂r
r2, ∂ 02
Sol. x = e r cos 0 . cos (r sin 0 )
... ∂x = e r. cos 0 . cos .cos (r sin -er cos 0
0)
0)
∂r
sin (r sin 0 ) . sin 0
= e r cos 0 [cos 0 cos (r sin 0) - sin 0 sin (r sin
0)
= e r cos 0 cos ( 0 + r sin 0)
∂x =
e r cos 0 (- r sin 0) cos (r sin 0)
∂0
-e r cos 0sin (r sin 0) r cos 0)
= -re r cos 0[sin 0
cos (r sin 0) + cos 0 sin (r sin 0)]
r cos 0
-re
sin (0 + r sin 0)
r
cos
0
Also y = e
sin (r sin 0)
=
... ∂y = e r cos .cos
0
0. sin (r sin 0) + er cos 0 . cos (r sin 0 ) sin 0
∂r
0
= e r cos (sin
0 cos (r sin 0) + cos 0 sin (r sin 0 )
0
= e r cos sin(
0 + r sin 0 )
∂y
r cos 0
=
e
(-r sin 0 ) sin (r sin 0 )
∂0
r cos 0
+e
cos (r sin
0 ) . r cos 0
r cos 0
= re
[cos 0 cos (r sin 0 ) - sin 0 sin (r sin
0 )]
r cos 0
cos( 0 +r sin 0 )
∂x 1
∂y
From (1) and (4), ∂r =
r
∂0
∂y
1
∂x
From (2) and (3), =
=∂r
r ∂0
From (5),
∂2x - 1 ∂y
1
=
∂r2 r2 ∂0 + r
= re
∂2y
∂r∂0
From (6)
...
...
3.4.
∂x
∂y
= -r
∂0
∂r
∂ 2x
∂2y = - r ∂2y
=
r
∂ 02
∂ 0∂r
∂r∂ 0
2
∂ 2x
∂x
1
+ 1.
+ 2.∂ x
∂r2
r ∂r
r ∂ 02
2
2
= - 1 . ∂y + 1 . ∂ y+ 1 . ∂y - 1 . ∂ y = 0
∂0
∂r∂0 r2
∂r∂ 0
∂0 r
r2
r
HOMOGENEOUS FUNCTIONS
A function f(x, y) is said to be homogeneous of degree (or order) n in the variables x
and y if it can be expressed in the form xn O y or yn O x
x
y
An alternative test for a function f(x, y) to be homogeneous of degree (or order) n is
that f(tx, ty) = tn f(x, y)
For example, if f (x, y) =
x+y
x+ y , then
x 1+y
y
x
= x1/2 O
(i) f (x, y) =
x
x 1+ y
x
"
(ii)
"
(iii)
"
f(x, y) is a homogeneous function of degree 1/2 in x and y.
x
y y +1
f(x, y) =
= y1/2 O x
y x
y
y +1
f(x, y) is a homogeneous function of degree 1/2 in x and y.
f(tx, ty) =
tx+ty
tx+ ty
=
t(x+y)
t ( x+ y)
= t1/2f(x, y)
f(x, y) is a homogeneous function of degree 1/2 in x and y.
Similarly, a function f(x, y, z)is said to be homogeneous of degree (or order) n in the
variables x, y, z if
x y
x z
y z
n O
f(x, y, z) = xn O
or zn O z z
or
y
y
y
x x
Alternative test is f(tx, ty, tz) = tn f(x, y, z)
3.5.
EULER’S THEOREM HOMOGENEOUS FUNCTIONS
If u is a homogeneous function of degree n in x and y, then
∂u
∂u
∂x + y ∂x = nu
Since u is a homogeneous function of degree n in x and y, it can be expressed as
x
...
u = xn f y
x
y
y
∂u
n
-y 2
= nxn-1 f x + x f x
x
∂x
∂u
" x
= nxn f
∂x
y n -1
x -x yf
y
x
Also ∂u = xn f y . 1 = xn-1f y
∂y
x
x
x
∂u
"
y
= xn-1 y f y
∂y
x
Adding (1) and (2), we get
x ∂u + y ∂u = nxn f y = nu
∂x
∂x
x
Note : Euler’s theorem can be extended to a homogeneous function of any number of
variables. Thus, if u is a homogeneous function of degree n in x, y and z then
x
3.6.
∂u
∂u
∂u
= nu
∂x + y ∂x + z ∂x
If u is a homogeneous function of degree n in x and y, then
x2
∂2u
2
∂ 2u
+ y2 ∂ u
+
2xy
2
∂x
∂x∂y
∂y2
= n(n-1) u
Since u is a homogeneous function of degree n in x and y
...
By Euler’s theorem, we have
x
∂u
∂u = nu
+
y
∂x
∂y
Differentiating (1) partially w.r.t.x, we have
∂u
∂ 2u
∂2u = n ∂u
1. ∂x + x. ∂x2 + y.
∂x
∂x∂y
Differentiating (1) partially w.r.t.y, we have
∂2u
. ∂u
x
∂2u
∂u
∂y∂x + 1 ∂y + y 2 =n
∂y
∂y
But
.
..
x
∂2u
∂2u
=
∂y∂x
∂x∂y
∂2u + ∂u
∂u
∂ 2u
∂x∂y
∂y + y ∂y2 = n ∂y
Multiplying (2) by x, (3) by y and adding
2
2
x2 ∂ 2u +2xy ∂ u + y2∂2u + x ∂u + y ∂u
∂x
∂x∂y ∂u2
∂x
∂y
∂u
=n + x
+ y ∂u
∂x
∂y
or
"
x2
2
∂2u
∂ 2u
2 ∂ u + nu = n.nu
+
2xy
+
y
∂y2
∂x∂y
∂x2
2
∂2u
∂ 2u
x2 ∂ u2 + 2xy
+ y2 2
∂y
∂x∂y
∂x
= n2u - nu = n (n-1) u
Note . If f (x, y) is homogeneous n, show that
x2f11+xyf12+xyf21+y2f22 = n (n-1)f
ILLUSTRATIVE EXAMPLES
Example 1.
(i)
Verify Euler’s theorem for the functions:
u = (x½ +y½) (xn+yn)
Sol. i. u = (x½ +y½) (xn+yn)
n
= (x½ (1+y½ ) xn (1+ y )
xn
x½
½
n
= xn½ 1+( y )
1+ ( y )
x
x
-1 y
ii u = sin-1 x + tan
x
y
= xn+½ f (y)
x
[OR f(tx, ty) = tn+½ f(x, y)]
"
u is a homogeneous function of degree (n+½) in x and y.
...
By Euler’s theorem, we should have
x ∂u + y ∂u = (n+½ )u
∂x
∂y
From (1), ∂u = ½ x -½ (xn +yn) + nxn-1 (x½+y½)
∂x
x ∂u = ½ x (xn +yn) + nxn (x½+y½)
∂x
∂u
n-1 (x½+y½)
= ½ y -½ (xn +yn) + ny
∂y
y ∂u = ½ y ½
∂x
(xn +yn) + nyn (x½+y½)
∂u
∂u
... x ∂x +y ∂y = ½ (x½+y½) (xn +yn) + n(xn +yn) (x½+y½)
= ½ u +nu = (n+½)u
Which is the same as (2). Hence the verification.
-1 y
u = sin-1 x
+ tan
y
x
(ii)
-1 y
y
= cosec-1 y + tan
= x0f
x
x
x
[OR f (tx, ty) = f(x, y) = t0f (x, y)
" u is a homogeneous function of degree 0 in x and y.
... By Euler’s theorem, we should have
x
From (1),
=
x
∂u + y ∂u
∂y = 0 x u = 0
∂x
∂u
∂x =
1
1
y2-x2
∂u
∂x =
∂u
∂y =
-y
.
2
1- x
y2
-
1
+
y
1
y2
1+ x2
-
y
x2 + y2
x
y2-x2
1
2
1- x
y2
x +
2y x2
-.
xy
x2+y2
-. x +
y2
x
x2+y2
1
2
1+y
x2
.1
x
y
x2
y ∂u =
∂y
x
+
2y x2
xy
x2+y2
... x ∂u + y ∂u = 0
∂x
∂y
Which is the same as (2). Hence the verification.
3.7
COMPOSITE FUNCTIONS
(i)
If u = f (x,y)where x = φ(t), y = ψ(t)
du
Then u is called a composite function of (the single variable) t and we can find
dt
(ii)
If z = ƒ(x, y) where x = φ (u, v), y = ψ (u, v)
Then z is called a composite function of (two variables) u and v so that we can find
∂z
∂z
and
∂v
∂u
3.8
DIFFERENTIATION OF COMPOSITE FUNCTIONS
If u is a composite function of t, defined by the relations
u = f(x,y); x = φ (t), y = ψ(t), then
∂u
dx
dy
∂u
du
=
+
.
.
∂x
dt
dt
∂y
dt
∂u
and ∂u
∂x
∂y
... u is a function of two variables x and y
dx
and dy
dt
dt
... x and y are function of a single variable t
ILLUSTRATIVE EXAMPLES
Example 1 . Find du when u = xy2 +x2y, x = at2, y = 2at.
dt
Also verify by direct substitution
Sol. the given equations define u as a composite function of t.
∂u .dx
du
∂u . dy
=
...
∂x dt + ∂y
dt
dt
= (y2+2xy). 2at +(2xy+x2).2a
= (4a2t2+2at2.2at) 2at +(2at2.2at+a2t4).2a
= 8a3t3+8a3t4 +8a3t3+2a3t4
= 2a3t3(5t+8)
Also u = xy2+x2y = at2.4a2t2+a2t4.2at
= 4a3t4 + 2a3t5
du
= 16a3t3 + 10a3t4 = 2a3t3 (5t+8)
dt
Hence the verification
d2 y
If ax2 +2hxy+by2 = 1
dx2
Sol. Let f(x, y) = ax2 +2hxy+by2 -1
fx = 2ax+2hy, fy = 2hx +2by
fxx = 2a, fxy = 2h,
fyy = 2b
Example 2.
...
Find
d2y = fxxfy2- 2fxfyfxy+ fyyfx2
dx2
fy3
= - 2a(2hx+2by)2 - 2(2ax+2hy) (2hx+2by) (2h) +2b(2ax+2hy)2
(2hx+2by)3
= - a(hx+by)2 - 2h(ax+hy) (hx+by) +b(ax+hy)2
(hx+by)3
= (ah2 - 2ah2 +ba2 )x2+(2abh-2abh-2h3+2abh)xy+(ab2-2bh2+bh2)y2
(hx+by)3
= - a(ab - h2 )x2+2h(ab-h2)xy+b(ab-h2)y2
(hx+by)3
=
(h2 -ab) (ax2 +2xhy+by2)
(hx+by)3
=
h2-ab
(hx+by)3
[ ... ax2+2hxy+by2= 1]
H.W
1.
Find du when u = x2+y2, x=at2, y =2at.
dt
Also verify by direct substitution.
Ans: 4a2t(t2+2)]
2.
If u = sin
x
y,
x = et, y = t2 find du
dt
Ans: et(t-2)
et
cos
t2
t3
3.
If u = x3+y3, where x = a cos t, y= b sin t, find
du and verify the result.
dt
Ans: -3a3cos 2t sin t +3b3 sin2 t cos t
If z = u2 + v2, u = r cos φ, vr sin φ , find ∂z and
∂r
Ans: 2r, 0
4.
5.
Given v = f(x, y, z), x = r cos φ, y = r sin φ , z = t, obtain expressions for
∂v ∂v ∂v
in terms of
∂r , ∂φ , ∂t
Ans:
6.
∂v
∂v ∂v
∂x , ∂y , ∂z
∂v ∂v
∂v
∂v
∂v ∂v
∂v
cos φ + ∂v sin φ ;
=
=
.r cos φ;
φ +
=
.r
sin
∂r
∂φ
∂x
∂t
∂y
∂x
∂y
∂z
If u = f(r, s), r = x + y, s = x-y, show that
∂u
∂u + ∂u
=2
∂x
∂x
∂y
7.
8.
∂z
∂φ
If x = u+v, y = uv and z is a function of x, y show that
∂z
∂z
+ v ∂z = x ∂z + 2y
u
∂y
∂u
∂v
∂x
If u = f (r, s), r = x+at, s = y+bt, show that
∂u
∂u
∂u
=a
+b
∂x
∂t
∂y
9.
If u = emx (y-z), y = m sin x and z = cos x, find
Ans:
emx (m2 +1) sin x
10.
If u = x log (xy), where x3+ y3+3xy = 1, find
Ans: 1+log (xy) - x(x2+y)
y(y2+x)
du
dx
du
dx
11.
(i)
If u = f(r, s, t) and r =
x
z
y
y, s = z, t = x, prove that
x ∂u + y ∂u + z ∂u = 0
∂y
∂x
∂z
(ii)
If x = vw
, y = uw ,
x ∂φ + y∂φ + z ∂φ
∂x
∂y
∂z
12.
z = uv and φ is a function of x, y, z, then
= u ∂φ + v ∂φ + w ∂φ
∂v
∂u
∂w
If z = x2y and x2+ xy+y2 = 1, show that
dz
= 2xy - x2(2x+y)
dx
x +2y
13.
If x3+ 3x2 y+6xy2 +y3 = 1, find dy
dx
Ans: - x2+ 2xy+2y2
x2+ 4xy+y2
14.
If xy = yx show that dy = y(y- x log y)
dx
x (x-y log x)
using partial derivative method
15.
Find
Ans: (i)
16.
17.
(ii) xy = ex2 +y2
y(x+y)
log (x+y) xy (ii)
x2 2
- y-2xe x2+y 2
x (x+y) log (x+y) +xy ,
x-2ye +y
2
2 2
Prove that if y3-3ax2 +x3 = 0, then d y + 2a x = 0
dx2
y5
By changing the independent variables x and y to u and v by means of the relations
u = x -ay, v = x +ay, show that
a2
18.
dy from (i) xy log (x+y) = 1
dx
∂2z ∂2z
transforms in to 4a2
∂x2 ∂y2
∂ 2z
∂u∂v
If z is a function of x and y, and u and v be other variables sucha that u = lx+my,
v = ly -mx, show that
2
∂2z
+ ∂ z = (l2+m2)
2
∂x
∂y2
∂2z + ∂2z
∂u2 ∂v2
19.
If u = u (x, y) and x = er cosφ , y = er sinφ , show that
2
2
(i)
2
∂u
+ ∂u = e -2r
∂x
∂y
∂u
+ ∂u
∂r
∂φ
2
(ii)
20.
∂2u + ∂2u
∂r2
∂φ2
∂v
∂v ∂v
+
+
∂z = 0
∂x ∂y
1 ∂u + 1 ∂u + 1 ∂u
If u = f(x2-y2 , y2-z2 , z2-x2 ), prove that x ∂x y ∂y z ∂z = 0
(i) If v = f(x-y, y-z, z-x), show that
If x = ñ cosφ, y = ñ sin φ, show that
2
22.
2
∂2u ∂2u
-2r
+
∂x2 ∂y2 = e
(ii)
21.
2
2
2
2
∂v
+ ∂v =
∂x
∂y
∂v
+ 1 2 ∂v
∂ñ
ñ ∂φ
Prove that
∂2u ∂2u
∂2u + ∂2u
+
=
2
∂x
∂y2
∂s2 ∂t2
Where x = s cos a -t sin a. and y = s sin a + t cos a
23.
If z = uv, u2+v2-x-y=0, u2 -v2+3x+y=0, find ∂z
∂x
24.
If x = u2-v2, y = 2uv, find
Ans:
∂u ∂v ∂v
∂u
∂x , ∂y , ∂x , ∂y
u
∂u
∂u
∂x = 2(u2+v2) , ∂y =
2(u2+v2)
-v
∂v
∂v
∂x = 2(u2+v2) = ∂y =
u
2(u2+v2)
v
Ans.
2u2-v2
2uv
3.9.
JACOBIANS
If u and v are functions of two independent variables x and y, then the determinant
∂u
∂x
∂v
∂x
∂u
∂y
∂v
∂y
is called the Jacobian of u, v with respect to x, y and is denoted by the symbol
J
u, v
or
x, y
∂ (u, v)
∂ (x, y)
Similarly, if u, v, w be functions of x, y, z then the Jacobian of u, v, w with respect to x, y, z is
J
u, v, w
or ∂ (u, v, w)
x, y, z
∂ (x, y, z)
=
∂u
∂x
∂v
∂x
∂u
∂y
∂v
∂y
∂w
∂x
∂w
∂y
3.10.
PROPERTIES OF JACOBIANS
I.
If u, v are functions of r, s where r, s are functions of x, y, then
∂ (u, v)
∂ (u, v) ∂ (r, s)
x
=
∂ (x, y)
∂ (r, s)
∂ (x, y)
...
Now
Proof. Since u, v are composite functions of x, y
∂u . ∂s
∂u = ∂u
= urrx+ussx
. ∂r +
∂s
∂x
∂x
∂r
∂x
∂u = ∂u
.
∂y
∂r
∂r
∂y
+
∂u
∂s
. ∂s = urry+ussy
∂y
....A
∂v = ∂v
.
∂x
∂r
∂r
∂x
+
∂v
∂s
. ∂s = vrrx+ussx
∂x
∂v = ∂v
.
∂y
∂r
∂r
∂y
+
∂v
∂s
. ∂s = vrry+ussy
∂y
∂ (u, v) ∂ (r, s)
∂ (r, s) x ∂ (x, y) =
ur, us
vr, vs
.
rx ry
sx sy
∂u
∂z
∂v
∂z
∂w
∂z
Interchanging rows and columns in the second determinant
=
urrx+ ussx urry+ ussy
vrrx+ vssx vrry+ vssy
=
∂u
∂x
∂v
∂x
=
∂u
∂y
∂v
∂y
∂ (u, v)
∂ (x, y)
=
II.
us . rx sx
vs ry sy
ur,
vr,
If J1 is the Jacobian of u, v with respect to x, y and J2 is the Jacobian of x, y with
respect to u, v then J1, J2 = 1
∂ (u, v) . ∂ (x, y)
=1
∂ (x, y)
∂ (u, v)
i.e,
Proof. Let u = u(x, y) and v = v(x, y), so that u and v are functions of x, y.
Differentiating partially w.e.t.u and v, we get
1=
∂u .
∂x
∂x
+ ∂u
∂u
∂y
.
∂y
∂u = uxxu+uyyu
0=
∂u .
∂x
∂x
+ ∂u
∂v
∂y
.
∂y
= uxxv+uyyv
∂v
0 = ∂v .
∂x
∂x
+ ∂v
∂u
∂y
.
∂y
= vxxu+vyyu
∂u
∂v .
∂x
∂x
+ ∂v
∂v
∂y
.
∂y
∂v = vxxv+vyyv
1=
∂ (u, v) . ∂ (x, y)
=
∂ (x, y)
∂ (u, v)
Now
ux,
vx,
uy x
.
vy y u
u
....A
xv
yv
Interchanging rows and columns in the second determinant
=
ux, uy . xu
vx, vy xv
yu
yv
=
uxxu + uyyu uxxv +uyyv
vxxu + vyyu vxxv+vyyv
=
1
0
0
= 1
1
[Using (A)]
ILLUSTRATIVE EXAMPLES
Example 1.
J=
If x = r cos φ, y = r sin φ , prove that the Jacobians
∂ (x, y) = r and J =
∂ (r, φ)
[A.M.I.E (S) 1993; (S) 94; (S) 95; (W) 97]
y = r sin φ
x = r cos φ,
Sol.
...
∂ (r, φ) = 1
J
∂ (x, y)
∂x
= cos φ
∂r
∂x
∂φ
∂y
= sin φ ,
∂r
∂y
= r cos φ
∂φ
∂x
∂r
... J = ∂ (x, y) =
∂ (r, φ)
= -r cos φ
∂x
∂φ
cos φ
=
sin φ
∂y
∂φ
∂y
∂r
-r sin φ
r cos φ
= r cos 2 φ + r sin 2 φ = r
Now x2 + y2 = r2 and
" r =
and φ = tan -1
x 2 + y2
∂r
∂x
∂r =
y
,
∂
y
x2 + y2
x2 + y2
1
1+ y2
x2
(-y)
x2
( x1 ) =
... J = ∂ (r, φ) =
∂ (x, y)
x2
=
=
y
x
x
1
∂φ
= 1+ y2
∂x
x2
∂φ
∂y
y = tan φ
x
3/2
(x2 + y2)
1
x2 + y2
∂r
∂x
∂φ
∂x
+
=-
y
x2 + y2
x
x2 + y2
x
∂r
∂y
=
∂φ
∂y
x2 + y2
-y
x2 + y2
x
x2 + y2
y2
x2 + y2
x2+ y2
3/2
(x2 + y2)
1
1
= r =
J
y
=
3/2
(x2 + y2)
3.11.
TAYLOR’S THEOREM FOR A FUNCTION OF TWO VARIABLES
We know that by Taylor’s theorem for a function f(x) of ssingle variable x,
h3 f’’’(x) + .....
h2
f(x+h)=f(x) + hf’(x)+ 2! f”(x) + 3!
Now let f(x, y) be a function of two independent variables x and y. If y is kept constant, then by Taylor’s theorem for a function of a single variable x, we have
3
2
3
2 ∂
∂
f(x, y+k) + h ∂ 3 f (x, y+k)+... ..(1)
f(x, y+k) + h
∂x
3! ∂x
2! ∂x 2
Now keeping x constant and applying Taylor;s Theorem for a function of a single
variable y, we have
f(x+h, y+k) = f(x, y+k) +h
2
f(x, y+k) = f(x, y) +k
3
2 ∂
k3 ∂
∂
k
f(x,
y)
+
f (x, y)+... .
f(x, y) +
3! ∂y 3
∂y
2! ∂y 2
Using (2), we can write (1) as
3
k2 ∂ 2 f(x, y) +
f(x+h, y+k) = [f(x, y) +k ∂ f(x, y) +
.2
2! ∂y
∂y .
2
2
∂
∂ f(x, y) + ....]
+ h [f(x, y) +k ∂ f(x, y) + k
∂x
2
2!
∂y .
. ∂y
k3 ∂ f(x+y) + ......]
3! ∂y3
2
2
+ h ∂ 2 [f(x, y) +k ∂ f(x, y) + ....]
2! ∂x
∂y
+
h3
3!
∂3 [f(x, y) + ..........] + .......
∂x3
= [f(x, y) + h ∂f + k ∂f
∂x
∂y
(
+
(
3
)+ (
2
2
h2 ∂f + hk ∂f + k2 ∂f
2
2! ∂x
∂x∂y 2! ∂y2
)
3
h3 ∂f + h2k ∂f + hk2 ∂3f + k3 ∂3f
3
2! ∂x2∂ y 2! ∂x∂y2 3! ∂y3
3! ∂x
(
= f(x, y) + h
∂f + k ∂f
∂x
∂y
) + .......
2
)
) (
3
2
∂3f
+ 3hk ∂3f + k3
+ 1 h3 ∂f 3 + 3h2k
2
∂x ∂y
3!
∂x
∂x∂y2
(
(
= f(x, y) + h
+ 1
3!
(
2
2
∂f + 2 ∂f
1
∂f
2
k 2
+ h
2 + 2hk ∂x∂y
∂y
2!
∂x
∂ +k ∂
∂x
∂y
)
+1
2!
) + .......
2
∂ +k ∂ f
h
∂x
∂y
(
3
∂ +k ∂
f + .......
h
∂y
∂x
)
∂3f
∂y3
)
.(2)
Cor. 1. Putting x = a and y = b, we have
f(a+h, b+k) = f(a, b)+[hfx(a, b) +kfy(a, b)]
2
+ 1 [ h2 fxx (a, b) + 2hkfxy (a, b) +k fyy (a, b)]
2!
+
1 [h3 fxxx (a, b) + 3h2kfxxy (a, b)
3!
2
3
+3hk fxyy (a, b) + k fyyy (a, b)]+.......
Cor.2. In Cor 1 Putting a+h = x and b+k = y so that h = x - a and k = y - b, we have
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)
2!
+ (y-b)2 fyy (a, b) ]+.....
Cor. 3. Putting a = 0, b = 0 in Core 2, we have
f(x, y)= f(0, 0)+[xfx(0, 0) +yfy(0, 0)]+ 1 [x2fxx (0, 0)
2!
+ 2xy fxy (0, 0) + y2fyy (0, 0)]+.......
This is called Maclaurin’s theorem for two variables.
[Note. Cor. 3 is used to expand f(x, y) in powers of x and y (or to expand f(x, y) in the
neighbourhood of origin (0, 0)].
Cor 2 is used to expand f(x, y) in the neighbourhood of (a, b)].
ILLUSTRATIVE EXAMPLES
Example 1.
Expand ex sin y in powers of x and y as far as terms of the third degree.
Sol. Here
...
f(x, y) = ex sin y,
f(0, 0) = 0
x
fx(x, y) = e sin y,
fx(0, 0) = 0
fy(x, y) = ex cos y,
fy(0, 0) = 1
x
fxx(x, y) = e sin y,
fxx(0, 0) = 0
fxy(x, y) = ex cos y,
fxy(0, 0) = 1
x
fyy(x, y) = -e sin y,
fyy(0, 0) = 0
fxxx(x, y) = ex sin y,
fxxx(0, 0) = 0
x
fxxy(x, y) = e cos y,
fxxy(0, 0) = 0
fxyy(x, y) = -ex sin y,
fxyy(0, 0) = 0
x
fyyy(x, y) = -e cos y,
fyyy(0, 0) = -1
.........................................................................................
ex sin y= f(x, y)
1
= f (0, 0) + [xfx (0, 0) + y fy (0, 0)] + 2! [x2 fxx (0, 0)
1
+ 2xyfxy (0, 0) +y2fyy (0, 0)] + 3! [x3 fxxx (0, 0)
+ 3x2yfxxy(0, 0) + 3xy2 fxyy(0, 0) +y3 fyyy (0, 0)] + ....
1
1
= 0 +[x.0+y.1] + 2! [x2.0+2xy.1+y2.0]+ 3! [x3.0
+3x2 y.1 +3xy2 .0+y3 (-1)]+........
1 2 1 3
= y + xy + 2! x y - 6 y + ......
H.W.
2
1.
Show that ex log (1+x) = x+xy - x approximately.
2
[Hint. Find the expansion at (0, 0)]
2.
Expand ex cos y in powers of x and y as far as the terms of third degree.
[Ans. 1+x+1 (x2-y2)+ 1 (x3-3xy2)+.....]
3!
2!
3.
Expand eax sin by in powers of x and y as far as the terms of third degree.
4.
[Ans. by + abxy + 1 (3a2 bx2 y - b3 y3 ) +....]
3!
Expand exy at (1, 1).
1
[Ans. e{1+(x-1)+(y-1) + 2! (x-1)2 +4(x-1) (y-1)+(y-1)2 +....}]
(
5.
)
( )
( ) + (x-1)2 - (x-1) (y- 4ð )- -12 (y- ð4 )
2
ð
Expand ex cos y at 1,
4
e
ð
[Ans.
{1+(x-1)- y2
4
6.
Expand (1+x+y)2 1/2 at (1, 0)
7.
Expand sin (x+h) (y+k) by Taylor’s Theorem
8.
1
2
Ans. sin xy +(hy +kx) cos xy +hk cos xy 2 (hy +kx) sin xy + ....}]
y
Obtain the expansion of tan-1 about (1, 1) up to the third degree terms.
x
ð
- 1 (x-1) + 1 (y-1) +1 (x-1)2 -1 (y-1)2
2
4
4
4
2
Ans.
- 1 (x-1)3 - 1 (x-1)2 (y-1) + 1
12
4
4
(x-1) (y-1)2 + 1 (y-1)3 +....]
12
9.
Expand f (x, y) = yx in the neighbourhood of (1, 1) up to the terms of second degree.
Ans. 1+(y-1) + (x-1) (y-1) + .........]
10.
Expand (x2y+siny+ex ) in powers of (x-1) and (y-ð) through quadratic terms
by using Taylor’s series.
1
Ans. (ð + e) + (2ð+e) (x-1) +
(x-1)2 (2ð + e) + 2(x-1) (y-ð) + .................. ]
2
4.1.
MAXIMA AND MINIMA OF FUNCTIONS OF TWO VARIABLES
A function f(x, y) is said to have a maximum value at x=a, y=b if f(a, b) > f (a+h, b+k)
for small values of h and k, positive or negative.
A function f(x, y) is said to have a minimum value at x=a, y=b if f(a, b) < f (a+h, b+k)
for small values of h and k, positive or negative.
A maximum or a minimum value of a function is called an extreme value.
4.2.
RULE TO FIND THE EXTREME VALUES OF A FUNCTION
z = f (x, y)
(i)
Find ∂z and ∂z
∂y
∂x
(ii)
Solve ∂z = 0 and ∂z = 0 and simultaneously.
∂y
∂x
Let (x1 , y1); (x2, y2); .......... be the solutions of these equations.
(iii)
For each solution in step (ii), find
2
r = ∂ 2f s = ∂2 f
∂2 f
∂x ,
∂x∂y , t = ∂y2
(iv) (a) If rt - s2>0 and r <0 for a particular solution (x1, y1) of step (ii), then z has a
maximum value at (x1, y1)
(b) If rt - s2>0 and r > 0 for a particular solution (x1, y1) of step (ii), then z has a
minimum value at (x1, y1)
(c) If rt - s2<0 for a particular solution (x1, y1) of step (ii), then z has no extreme
value at (x1, y1)
(d) If rt - s2=0, the case is doubtful and requires further investigation.
Note: The points (x1, y1); (x2, y2); ..................... are called stationary points and the values of
f (x, y) at these points are called stationary values.
ILLUSTRATIVE EXAMPLES
Example 1.
Examine the function x3 + y3 - 3axy for maxima and minima.
Sol. Here f(x, y) = x3 +y3 - 3axy
fx = 3x2 - 3axy
fy = 3y2 - 3ax
r = fxx = 6x, s = fxy = -3a,
t = fyy = 6y
Now fx = 0 and fy = 0
"
x2 - ay = 0
y2 - ax = 0
From (1),
y = x2
a
... From (2), x4
3 3
2 - ax = 0 or x(x -a ) = 0
a
x = 0, a
or
When x = 0, y = 0, when x = a, y = a
...
There are two stationary points (0, 0) and (a, a).
Now
rt - s2 = 36xy - 9a2
At (0, 0),
"
rt - s2 = - 9a2 < 0
There is no extreme value at (0, 0)
At (a, a), rt - s2 = 36a2- 9a2 =27a2 > 0
"
f(x, y) has extreme value at (a, a)
Now
r = 6a
If a > 0, r >0 so that f (x, y) has a minimum value at (a, a)
Minimum value = f (a, a) = a3+a3-3a3= -a3
If a<0, r <0 so that f(x, y) has a maximum value at (a, a)
Maximum value = f (a, a) = a3+a3-3a3= -a3
HW
1.
Examine for extreme values:
(i)
x2+y2+3x+12
(ii)
x3+y3+3xy
(iii)
3x2-y2+x3
(iv)
x2y2-5x2-8xy-5y2
(v)
x3+3xy2-15x2-51y2+72x
Ans:
(i)
Min. value = 3at (-3, 0) (ii) Max value = 1 at (-1, -1)
(iii)
Max. value = 4at (-2, 0) (iv) Max value = 0 at (0, 0)
(v)
Max. value = 112at (4, 0) (ii) Min. value = 108 at (6, 0)
2.
Find the stationary points of
f(x, y) = y2+4xy+3x2+x3
Examine them for the extreme values of the function.
Ans. (0, 0), 2 -4 ; Main value = - 4 at
2 -4
27
3, 3
3, 3
(
)
(
)
3.
(i) Investigate f(x) = x5 - 5x4 for maxima and minima.
Ans. Min. value = - 256 at x = 4
(ii) Find the minimum value of the function
f(x, y) = x2+y2+xy+ax+by
where a and b arae constants
Ans. - 1 ( 3a2+3b2-2ab)
9
4.
When travelling x km per hour a truck burns diesel oil att eh rate of
1
300
( 900x +x )
litres per km. If diesel oil costs 40P per litre and the driver is paid Rs. 1.50per hour,
find the steady speed that will minimise the total cost of a trip of 500km.
Hint. Total cost of trip is C = 500
300
( 900x + x ) x 0.40 +900x x 1.50
Ans. 45km/hour
5.
Determine the points where the function
x2y+xy2-axy
has a maximum or a minimum
Ans. Min. value = - a3
a
a when a > 0
at
3,
3
27
3
a
a at
a when a < 0
Max. value =
3,
3
27
(
(
)
)
Find the stationary points of the function z = x3y3 (12-x-y) satisfying the condition
x>0, y>0 and examine their nature.
Ans. Maximum at (6, 4)
6.
Find the minimum value of x2+y2+z2 when
(i) x+y+z = 3a
(ii)
xyz = a3
Ans. (i) 3a2
(ii) 3a2
7.
Find the points of the surface z = xy+1 nearest to the origin
8.
Ans. (0, 0, +
- 1)
9.
Ans.
10.
(
Which point of the sphere x2+y2+z2 = 1 is farthest from the point (2, 1, 3)
-2
-1
-3
14,
14,
14
5xyz
Given f(x, y, z) = x+2y+4z find the values of x, y, z for which f(x, y, z) is a maximum,
)
subject to the condition xyz = 8
Ans. (4, 2, 1)
11.
A rectangular box, open at the top is to have a volume of 32 c.c. Find the dimensions
of the box requiing least material for its construction.
Ans. 4cmx4cmx2cm
12.
Find the dimensionn of the rectangular box without a top, of maximim capacity
whose surface is 108sq cm.
Ans. Length = Breadth = 6m, Height = 3cm
13.
The sum of three numbers is constant. Prove that their product is a maximum when
they are equal.
14.
Find the point on the plane ax+by+cz = p at which the functionso = x2+y2+z2 has a
minimum value and find this minimum o
Ans.
(∑aap2,
bp
∑a2,
p2
cp
∑a2; ∑a2
)
15.
If x increases at the rate of 2cm/sec at the instant when x = 3 and y = 1, at what rate
must y be changing in order that the function 2xy - 3x2y shall be neither increasing
nor decreasing?
32 cm/sec
Ans. Decreasing at
21
Hint. Let f(x, y) = 2xy-3x2y, then
df ∂f dx ∂f dy
=
.
.
+
dt ∂x dt ∂f dt
Put x = 3, y = 1, dx = + 2 cm/sec
dt
f is neither increasing nor decreasing when
df
= 0
dt