Differential Calculus and Coordinate Geometry Fall, 2018-19

Chapter 7
Partial Differentiation
A derivative of a function of one variable expresses a change in that function relative to its
argument. For example Newton’s law for rectilinear motion is F = mass times the second
derivative of distance with respect to time. But the world involves functions that depend on
several variables.

For example the pressure of a gas depends on density and temperature. The speed of sound
(squared), it turns out, in a nebula in space (which is very nearly at constant temperature due to
radiative transport) is the partial derivative of the pressure with respect to density keeping
temperature fixed.

Your happiness H depends on how much money, m, you make and the number of hours, h, you
spend with your family. H = H(m, h). But how much money you make also depends on how
much on how much time you spend with your family. The more time you spend with them the
less money you will make. So m = m(h) and we must write H = H(m(h), h)

Now, we want to know how many hours "h" to work to maximize happiness, so we take the total
derivative of H with respect to h and set it equal to zero:

dH/dh = (partial H/partial m)*(partial m/partial h) + (partial H/partial h) = 0.

**Reference https://www.physicsforums.com/threads/what-are-some-basic-applications-for-
partial-derivatives.334972/
n
A mapping from n-dimensional space R into R is called a function of n real variables. For
simplicity we shall mainly deal with functions of two and three variables, but the methods used
and the results obtained may be extended to functions of more than three variables.

Domain: The domain of functions of two variables, z=f ( x , y ) , is region from two dimensional

space and consist of all the coordinate pairs  x, y  that we could plug into the function and get
back a real number.

Example 1:  Determine the domain of each of the followings

f  x, y   x  y
(a) ,
f  x, y   x  y
(b) ,   
f  x, y   ln  9  x  9 y 2 
2
(c) .
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Solution:(a) In this case we know that we can’t take the square root of a negative number so this
means that we must require,    x  y  0 .

Here is a sketch of the graph of this region.

Fig. 1

(b) This function is different from the function in the previous part.  Here we must require that,

                                               x  0, y  0  

and they really do need to be separate inequalities.  There is one for each square root in the
function.  Here is the sketch of this region.

Fig. 2

(c) In this final part we know that we can’t take the logarithm of a negative number or zero. 
Therefore we need to require that,

x2 y2
9  x  9 y  0 or, 
2 2
1
9 1                                   

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Differential Calculus and Coordinate Geometry Fall, 2018-19

 and upon rearranging we see that we need to stay interior to an ellipse for this function.  Here is
a sketch of this region.

Fig. 3

7.1 Definition

f
The first order partial derivative of f (x , y) with respect to x is denoted by x or fx is
∂f f ( x + Δx , y )−f ( x , y )
= lim
defined by the equation ∂ x Δx →0 Δx , provided the limit exists.

Note that here x varies while y is kept fixed.

Similarly, the first order partial derivative of f (x , y) with respect to y is defined by the
equation

∂f f (x , y+ Δy )−f ( x , y )
= lim
∂ y Δx→0 Δy , provided the limit exists.

The concepts of partial derivatives can be extended to functions of three or more variables.

The second partial derivatives are as follows:

2 2
∂ ∂ f = ∂ f =f ∂ ∂ f = ∂ f =f
( )
∂ x ∂ x ∂ x 2 xx ( )
∂ x ∂ y ∂ x ∂ y yx
2 2
∂ ∂ f = ∂ f =f ∂ ∂f = ∂ f =f
( )
∂ y ∂ x ∂ y ∂ x xy ( )
∂ y ∂ y ∂ y 2 yy
Higher order partial derivatives may similarly be defined.

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Example 2: Find partial derivatives

f x , f y , f xx and f yy of the function
f (x , y )= √ x 2 + y 2 .

∂f 1 x ∂f 1
=f x= ×2 x =f y = ×2 y
Solution:
∂x 2 √ x 2+ y 2 = √ x2+ y2 , ∂y 2 √ x2 + y 2 =
y
√ x 2+ y 2
x
2
√ x 2 + y 2 −x× 2
∂ f √x + y2 y2
=f xx = =
∂ x2 ( x2 + y 2 ) 2 2
3
2
(x + y )

x
x2  y 2  x 
2 f x  y2
2
x2
 f yy  
y 2 x 2
 y2  x 2
 y2 
3
2

Theorem: If f (x , y) is continuous and

f xy and f yx exist and are continuous then
f xy =f yx .

Example 3: Show that

f xy =f yx for the function f (x , y )= √ x 2 + y 2 .

Solution: f (x , y )= √ x 2 + y 2

∂f 1 x
∴ =f x = ×2 x=
∂x 2 √ x 2+ y 2 √ x 2+ y 2
2
∂ f x =−
xy
=f xy= ∂ 3
∂ x∂ y ∂y √ x 2+ y 2 2
(x +y ) 2 2

2 f xy
 fyx  
yx 3
( x2  y 2 ) 2
Similarly,

Therefore,
f xy =f yx .

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7.2 Differential

The total differential (differential) of a function w=f ( x, y , z) is defined by the equation

∂f ∂f ∂f
df = dx+ dy + dz
∂x ∂y ∂z

Whether or not x, y & z are independent of each other, provided only that the partial
derivatives involves are continuous.

1 1 1
= +
Example 4: Find the differential of the function R R1 R2 .

R 1 R2 ∂R R 22 ∂R R21
R= = =
Solution: We have R 1 + R 2 , ∂ R1 ( R1 + R2 ) 2 , ∂ R2 ( R1 + R2 ) 2

2 2
∂R ∂R R2 R1
dR= dR1 + dR2 = 2
dR1 + 2
dR 2
∂ R1 ∂ R2 ( R 1 + R2 ) ( R1 + R2 )
Total differential is

Example 5: The ideal gas law state that pV =RT , where p (pressure), V (volume) and T
(temperature), R (=10 say) is a constant. Find approximate change of V when p decreases from 1
to 0.99 and T decreases from 5 to 4.99.

10 T
V=
Solution: We have p

∂V ∂V
dV = dp+ dT =10 dT −50dp
Total differential of V at (1, 5) is ∂p ∂T .

Here dT =−0.01,dp=−0.01

Hence approximate change in V of dV= 10(−0 .01 )−50(−0 . 01)≈0 .4 .

7.3 Chain Rule

Chain Rule I: Suppose u=f ( x , y , z) has continuous partial derivatives and let x=x (t )
y= y(t ) and z=z(t ) be differentiable. Then we can consider f as a function of single
variable t, we have from above differential
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df f dx f dy f dz
   .
dt x dt y dt z dt

Chain Rule II: Suppose u=f ( x , y ) has continuous partial derivatives and let x=x (s ,t )
and y= y( s, t ) be differentiable with continuous partial derivatives. Then

∂f ∂ f ∂ x ∂ f ∂ y
= +
∂s ∂x ∂s ∂ y ∂s
∂f ∂ f ∂ x ∂ f ∂ y
= +
∂t ∂ x ∂t ∂ y ∂t .

∂V ∂V ∂V
+ + =0
Example 6: If V =f ( x− y , y−z, z−x ) , prove that ∂ x ∂ y ∂ z .

Solution: Let V =f ( p , q ,r ) where p=x− y , q= y−z , r=z−x .

∂p ∂p ∂p
=1 , =−1 , =0
Now ∂x ∂y ∂z ,

∂q ∂q ∂q
=0, =1, =−1
∂x ∂y ∂z
∂r ∂r ∂r
=−1, =0 , =1
and ∂x ∂y ∂z .

Using Chain rule , we can write

∂V ∂V ∂ p ∂V ∂ q ∂V ∂r
= + +
∂ x ∂ p ∂ x ∂q ∂ x ∂r ∂ x
∂V ∂V ∂V
= 1+ 0+ (−1 )
∂ p ∂ q ∂r
∂V ∂V
= −
∂ p ∂r
∂V ∂V ∂ V ∂V ∂V ∂ V
=− + =− +
Similarly, ∂y ∂ p ∂q and ∂z ∂ q ∂r .

Therefore,

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∂V ∂V ∂ V
+ +
∂x ∂ y ∂ z
∂ V ∂ V ∂V ∂V ∂ V ∂V
= − − + − +
∂ p ∂ r ∂ p ∂ q ∂ q ∂r
=0 .

 dy 
 
7.4 Implicit Differentiation  dx  using Partial Derivatives

Suppose that the equation F( x , y)=0 has a solution y=f ( x ) , where f (x ) has
continuous derivative. Then

dy ∂ F /∂ x F
=− =− x ,
dx ∂ F/∂ y F y where F y≠0 .

dy
Example 7: Find using partial differentiation for x 3 y + x y 3=2.
dx

Solution: Let, F ( x , y ) =x 3 y + x y 3 −2=0

F x =3 x 2 y + y 3

F y =x 3+ 3 x y 2

dy −3 x 2 y+ y 3
Therefore, = 3 .
dx x +3 x y 2

Partial differentiation for the system of equations is shown through examples:

Example 8: If 2u+ 3 v=sin x and u+2 v=x cos y , find

u x ,u y ,v x ,v y .

Solution: 2u+3 v=sin x

u+2 v=x cos y

Solving this two equations for u and v we get,

u=2 sin x−3 x cos y …………….(1) and

v =2 x cos y−sin x …………..(2)

Differentiating equation no (1) & (2) w.r.t. x , we get

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u x=2 cos x−3 cos y and v x=2cos y−cos x .

Differentiating equation no (1) & (2) w.r.t. y , we get

u y =3 xsin y and
v y=−2x sin y .

2 2 2 u x ,u y ,v x ,v y
Example 9: Given u +x + y =3 and u−v 3 +3 x=4 , find .

2 2 2
Solution: u +x + y =3 ……(1)
3
u−v +3 x=4 ………(2)

Differentiating equation no (1) & (2) w.r.t. x , we get

2
2uu x +2 x=0 and u x−3 v v x +3=0

Solving this two equations for

u x and v x by Crammer’s rule we get

−2 x 0 2 u −2 x
| | | |
−3 −3 v 2 6 xv 2 x 1 −3 −6 u+2 x 3 u−x
u x= = =− v x= = =
2u 0 −6 uv 2 u 2u 0 −6 uv 2
3 uv 2
| | | |
1 −3 v 2
and 1 −3 v 2 .
Differentiating equation no (1) & (2) w.r.t. y , we get

2uu y +2 y=0 and

u y −3 v2 v y =0

Solving this two equations for

uy and
vy by Crammer’s rule we get

−2 y 0 2 u −2 y
| | | |
0 −3 v 2 6 yv 2 y v y=
1 0 2y y
uy = = =− = =− 2
2u 0 −6 uv 2 u 2u 0 −6 uv 2
3 uv
| | | 2|
1 −3 v 2 and 1 −3 v .
7.5 Homogeneous Function

A function f (x , y) is said to homogeneous of degree n if

f (kx, ky )  k n f  x, y 

where k is a constant.

Alternative:A function f (x , y) is said to homogeneous of degree n if

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 y  x
f ( x, y )  x n f   f  x, y   y n f  
x  y
or,
A similar definition applies to a function of any number of independent variables. For example,
f  x, y , z   x 2  yz  z 2

is a homogeneous function of degree 2 in x, y and z since

f  kx, ky, kz    kx   ky  kz    kz   k 2  x 2  yz  z 2 
2 2

Or,

 y z z2   x2 z z 2 
f  x, y , z   x 2  1    2  f  x, y , z   y 2  2   2 
 x x x  y y y 
or, , or,

 x2 y 
f  x, y , z   z  2   1 
2

z z 

f  x, y 
7.6 Euler’s Theorem: If is a homogeneous function of two variables x and y of degree

∂f ∂f
x + y =nf
n, then ∂x ∂ y .
2 2
x +y
f=
Example 10: Test whether the function x+ y is homogeneous or not. If homogeneous
verify Euler’s theorem for this function.

Solution: Here,

( λ x )2 + ( λ y )2 x2 + y 2
f ( λx , λy ) =
λ x+ λ y
=λ ( )
x+ y
=λ f ( x , y )

Therefore f ( x , y ) is a homogeneous function of order 1 .

∂f ∂f
x + y =f
Applying Euler’s theorem to this function we have ∂x ∂ y .

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x 2+ y 2
f ( x , y )=
x+ y
2 2
∂ f ( x + y ) 2 x−( x + y ) x 2− y 2 +2 xy
⇒ = = .
∂ x ( x + y )2 ( x+ y ) 2

2 2
∂ f ( x + y ) 2 y−( x + y ) y 2−x 2 +2 xy
= 2
=
Similarly,
∂ y ( x + y ) ( x+ y )2 .

and
2 2 2 2
∂f ∂f x − y +2 xy y −x +2 xy
x +y =x 2
+y
∂x ∂y (x+ y) ( x+ y )2

x 2  y 2  2 xy y 2  x 2  2 xy
x y
 x  y  x  y
2 2

 x  y   x2  y 2  x 2
 y2   f  x, y  .
 
 x  y
2
 x  y
Hence the Euler’s theorem is verified.

Application of Euler’s theorem

x2  y2  z 2 ∂u ∂u ∂u
u  arc csc 3 x + y +z
Example11: If x 3  y 3  z 3 , find ∂x ∂ y ∂z .

x2  y 2  z 2
f  x, y, z   csc u  3

Solution: Let x3  y 3  z 3 .

x2  y2  z 2
f  x, y , z   3

1
Here x  y  z is a homogeneous function of degree 3 .
3 3 3

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According to Euler’s theorem,

f f f 1
x y z   f  x, y , z 
x y z 3
   1
 x  csc u   y  csc u   z  csc u    csc y
x y z 3
 u  u  u 1
 x  csc u   y  csc u   z  csc u    csc u
u x u y u z 3
u u u 1
 x   csc u cot u   y   csc u cot u   z   csc u cot u    csc u
x y z 3
 u u u  1
  csc u cot u  x  y  z    csc u
 x y z  3
u u u 1
 x y z  tan u.
x y z 3

7.7 Relative Extrema and Their Nature

We shall now consider the extreme values for functions of two variables.

Suppose f (x , y) and all its derivatives up to order two are continuous and
f x ( a ,b )=f y (a , b )=0 . Then

(a) (a,b) is a local maximum if f xx (a , b )<0 or f yy (a , b )<0 and D>0 .

(b) (a,b) is a local minimum if f xx (a , b )>0 or f yy (a , b )>0 and D>0 .

(c) (a , b) is a saddle point if D<0.

(d) no conclusion can be drawn if D=0.

where
D=f xx f yy −( f xy )2 evaluated at (a,b) .

f  x, y   4  x 3  y 3  3xy
Example12:  Find the critical points of .Also state their nature.

 Solution: We first need all the first order (to find the critical points) and second order (to
classify the critical points) partial derivatives so let’s get those.
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f x  3 x 2  3 y , f y  3 y 2  3 x,
f xx  6 x, f yy  6 y

Now, Critical points will be solutions to the system of equations,

f x =0 , f y =0 ,

⇒ 3 x 2−3 y=0 ,3 y 2−3 x=0 ,

Solving
 x, y    0, 0  ,  1,1 .

Nature: D( x, y )  36 xy  9 , D(0, 0)  9  0

So, for (0, 0) , D is negative and so this must be a saddle point.

D(1,1)  27  0 and f xx (1,1)  6  0

So, for (1,1) ,D is positive and f xx  is positive and so we must have a relative minimum.

The following figures show the graph of f(x,y).

Fig 4: f(x, y)

Example 13: Determine the point on the plane 4 x−2 y+ z =1  that is closest to the point
(−2 ,−1, 5). 

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Solution: Let’s suppose that ( x , y , z ) is any point on the plane.  The distance between this point
and the point in question, (−2 ,−1, 5) is given by the formula,

                                            
 
Given that, 4 x−2 y+ z =1 or z=1−4 x +2 y
 
Substituting this into the distance formula gives,

                  
Now, notice that finding the minimum value of d will be equivalent to finding the minimum
value of  .
 
So,                               

 
  
Now,

                              
and,                                 

 
so any critical points that we get will be relative minimums.
Now let’s find the critical point(s). 

                                       

 Solving these two equations we get we get a single critical point:  .

So,

                                               

 So, the point on the plane that is closest to   is .

7.8Lagrange Multipliers

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The problem of finding extrema subject to side condition called constraint may be obtained by
eliminating one of the variable using constraint equation.

For example to find extrema of f (x , y ,z) subject to g( x , y , z)=0 , we may proceed as

follows.

Define a new function

H ( x , y , λ)=f ( x , y )−λg ( x , y )

and considering the equations

∂H ∂H H
Hx= =0 H y= =0 H   0.
∂x , ∂y , 

To find the extrema of f (x , y ,z) subject to the constraints g1 ( x , y , z )=0 and

g2 ( x , y , z)=0 , define a function with two Lagrange multipliers λ1 and λ2 as

H ( x , y , z , λ1 , λ 2 )=f ( x , y , z )− λ1 g1 ( x , y , z )− λ2 g2 ( x , y , z )

and then solve the equations

H x =H y=H z =0 , H λ1=g1 ( x , y , z )=0 and H λ2=g2 ( x , y , z )=0

2 2 2
Example 14: Find the points on x + y /4+z =33 at which x−2 y+4 z attains its
maximum and minimum.

y2 2
Solution: Let,
H ( x , y , z , λ )=x−2 y +4 z− λ x + +z −33
4 ( 2
)
The equations for the conditional extrema gives

1
H x =1−2 λx=0 x=
or 2λ

λy 4
H y =−2− =0 y=−
2 or λ

2
H z =4−2 λz=0 z=
or z

Substituting those values in the equation

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y2 2
( 2
H λ=− x + + z −33 =0
4 )
1 4 4
2
+ 2 + 2 =33
we have 4λ λ λ

1
2 λ=±
132 λ =33 or 2

(1, −8, 4) and (−1, 8, −4) .

The maxima and minima then occur at

Since f (x , y , z)=x−2 y+4 z ,

f (1 , −8 , 4 )=1+16+16=33

and f (−1 , 8 , −4 )=−1−16−16=−33

Thus f (x , y , z)=x−2 y+4 z has a maximum at (1, −8, 4) , and a minimum at (−1, 8, −4) .

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Exercise 7.1

1. Describe the region (domain) on which the following functions are continuous.
x y2 x2 y2 x
(a)
√ x 2 + y 2−4
, (b) cos (
16−x 2 + y 2
, (c) √ x−2 )
y , (d)
√16−x 2− y 2
2 2 4
(e) 1− x − y ,
√ 16 9
(f) ln ( x + y−3 ) and (g) 2 2 ∙

2. Find fxx, fxy, fyx, fyy, fxxx, and fyyy where

x +y

2 (b) f ( x , y )=exp ( 2 xy )+ x ,
(a) f ( x , y )=3 x y + xy , y
(c) f ( x , y )=( x 3 + y 2 ) sin ( y −x ) and (d) f ( x , y )=exp ( cosh x ) +sin ( y 2 −x2 ).
3. Find fxz, fyz, fzx, fzy, fzz , fxyz, fxzy, fyzx and fzzz where
2 2 3 and (b)
(a) f ( x , y , z )=exp ( y ) ln x+ x y −z x f ( x , y , z )=x 3 yz 2 + xz− yz 2

4. Evaluate the indicated partial derivatives.

1
(a) ∂w ( )
π , ∧∂ w
2 1
w= y 2 cos( xy ),
∂x ∂y
π, .
2 ( )
3 2 ∂ z ( 2, 1 )∧∂ z
(b) z=√ x + 2 y , (2 ,1).
∂x ∂y
x ∂V ∂V ∂V ∂V
2 2
5. If V = y ln ( x + y )−2 x arctan
y ()
, find
∂x
and
∂y
, and verify that x
∂x
+y
∂y
=V +2 y ∙
2 2
∂u ∂u
2
+ 2 =0
6. Verify the Laplace equation ∂ x ∂ y , where
2 2
(a) u=x − y + 2 xy , (b) u=exp ( 2 y ) sin ( 2 x ) and (c) u=exp ( y ) sin x +exp ( x ) sin y
7. Show that the following functions satisfy one dimensional heat equation
∂ z 2 ∂2 z
=c (c> 0
∂t ∂ x2 , constant)
x x
(a)
z  exp  t  sin   and
c (b)
z=exp (−t ) cos
c . ()
8. The relationship between the pressure P , the volume V and the temperature T of one mole
of CO 2 is given by the Van der Waals equations of state,
RT a
P= − 2
V −b V
where R=8.31 J/mpl.K is the universal gas constant, a, b are proportionality constants.

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2
∂P ∂ P
.
Evaluate ∂V and ∂V 2
9. Show that a function of the form u=f ( x +ct )+g( x−ct ) , (c >0 , constant) satisfies
the
∂2 u 2 ∂ 2 u
2
=c .
Wave equation ∂t ∂ x2

10. (i) Show that u  x  3xy , and v  3x y  y satisfies Cauchy-Riemann equations in

3 2 2 3

rectangular form u x =v y ¿ u y =−v x .

(ii) Show that u=r 4 cos 4 θ , v=r 4 sin 4 θ , satisfies Cauchy-Riemann equations in polar form
1 1
ur  v vr   u .
r and r
11. Define homogeneous function. Test whether the following functions are homogeneous. If
homogeneous, what is the degree of these functions?
f  x, y , z   yz  xy  zx, x 2− y 2
(a) (b) f ( x , y )=√ x+ yx, (c) f ( x , y ) =
√5 x + √5 y
12. State Euler’s theorem.Verify Euler’s theorem for the following functions:
xy
(a) f ( x , y , z )=x 2− y 2−z 2, (b) f ( x , y )= ,
y−x
(c) f ( x , y )=√ x5 + y 5 , (d) f ( x , y , z )=x 2 z 2 + y 2 xz

13. (a) Show that the function f ( x , y )=exp ( yx ) satisfies the relation xf  yf
x y  0.

(b) Prove the relation in part (a) using Euler’s theorem.

x3 + y3 + z3
14. If u=arctan( x− y −z) , prove that x u x + y u y + z u z=sin 2 u.
2 2 2
x +y +z −3
15. If u=arccsc ( ) , prove that x u + y u + z u =
x y ztan u.
√ x− √ y− √ z 2
dy
16. Use partial derivatives to find dx , given that
(a) x 2−5 y 3=6 , (b) y−exp ( y ) cos ( xy ) , (c) xy− y 2 +3 xy−9=0 , (d) x y + y 2 x =6

and (e) e xy + x 2 e x =2.Also check your result using implicit differentiation.

17. Suppose thatw=exp ( xyz ) , x=3 u+ v , y=2u+ v and z=u v2 . Use appropriate forms of the
∂w ∂w
chain rule to evaluate and .
∂u ∂v
∂T ∂T
18. Given that T =x 3 y −x 2 y 3 +5 , x =r cos θ , y=r sin θ . Use chain rule to find and .
∂r ∂θ
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Differential Calculus and Coordinate Geometry Fall, 2018-19

dz 
z=√ xy + x , x=cos θ , y =sin θ . 
19. Given that Use chain rule to find d when 2.
du
u= y 2 sin x + x 2 cos y , x=cos 2 t , y=sin2 t
20. If then use chain rule to find dt .

Exercise 7.2

1. Find the total differential of the following functions:

(a) f ( x , y , z )=exp ( x 2 ) cos ( y + z ) , (b) g ( x , y )=sin ( y−z 2 ) and (c)
−3
h ( x , y , z )= y 2 x−3 z 2
2. The total resistance R of two resistances R1 and R2 connected in parallel is given by
1 1 1
= +
R R1 R2 .
2 2
R R
(a) Show that
dR=
R1( ) ( )
dR1 +
R2
dR 2 .

(b) Suppose that R1 and R2 are measured to be 300 ohms and 500 ohms, respectively,
with
error in R1 is 1% and in R2 is 2%. Find the maximum percentage error in the
calculated
value of R .
E2
3. The power consumed in an electric resistor is given by P= (in watts). If E=300 volts
R
and R=6 ohms, by how much does the power changes if E is decreased by 4 volts and Ris
decreases by 0.1 ohms?
4. For the formula R=E/C, find the maximum error and the percentage error if C=30 with a
possible error of 0.1 and E=150 with a possible error of 0.04?

5. The altitude of a right circular cone is 12 inches and is increasing at 0.4 in/min. the radius of
the base is 10 inches and is decreasing at 0.2 in/min. How fast is the volume changing?
6. Examine the following functions for maxima, minima and saddle points:
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Differential Calculus and Coordinate Geometry Fall, 2018-19

f  x, y   x 3  3 xy  y 3
(a) ,
f  x, y   12 x  6 y  x  xy  y 2 .
2
(b)
7. Find all the relative maxima, relative minima, and saddle points (if any) of the following
functions:
(a) z  x y  x  y , (b) z  2 x  6 xy  3 y ,
2 2 2 2 3 2

(c) z  xy  x  y , (d) z  x  y  3 x  3 y  8, (e) z  x  4 x y  2 x  2 y  1.

3 2 3 3 2 2 4 2 2 2 2

8. A closed rectangular box with a volume of 56 m 3 is made from two kinds of materials. The
top and bottom are made of material costing Tk.100 per square meter and the sides from
material costing Tk.50 per square meter. Find the dimensions of the box so that the cost of
materials is minimized.
9. The volume of a rectangular box without top is to be 256 m3. Find the minimum surface area.

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