Calculus Of Multivariable - Solved Assignments - Semester Fall 2007

1

Assignment Solution Fall 2007 MTH301 (1)

Question 1 Consider the point (1, 0, 0) convert it into cylindrical and spherical coordinates.

Solution

Here the po int is in the cartesian cordinates x =1 here

y=0

z=0

r = x 2 + y 2 = 12 + 02 = 1 y x

0 1

θ = tan −1 ( ) = tan −1 ( ) = tan −1 (0) = 0 z=0 so po int in the cyclinderical coordinates is (1,0,0) now in sphercal coordinates we have

ρ = x 2 + y 2 + z 2 = 12 + 02 + 02 = 1 = 1 0  y tan θ =   ⇒ θ = tan −1 ( ) = 0 1 x z ) φ = cos −1 ( x2 + y 2 + z 2 0 π  = cos −1 ( ) = cos −1 (0) =   1 2

π

so the po int is (1,0, ) 2

2

Question 2

By considering different path of approach, show that the function has no limit as

x2 f ( x, y ) = 2 x −y Solution

x2 f ( x, y ) = 2 x −y here we check along the path y = kx 2 x2 x2 1 lim ( x , y )→(0,0) ( 2 ) = lim x→0 ( 2 ) = x − kx 2 1 − k along y = kx 2 x − y k ≠1

here we will get different lim its for different values of k so lim it does not exist

Question 3

Find the partial derivative w.r.t x and y for the function

Solution

x3 − y 3 z = ln( 2 ) x + y2

( x, y ) → (0,0)

3

x −y ) x2 + y 2 here we can have z = ln(

3

3

z = ln( x 3 − y 3 ) − ln( x 2 + y 2 ) now taking partial derivative w.rt x we will have ∂z 3x 2 2x = 3 − ∂x x − y 3 x 2 + y 2 ∂z 3 y2 2y = 3 − ∂y x − y 3 x 2 + y 2

4

Assignment Solution Fall 2007 MTH301 (2)

Question 1

∂w ∂w and when r = 1 and s = −1 if w = ( x + y + z ) 2 ∂r ∂s where x = r − s , y = cos(r + s ) , z = sin(r + s )

Find

Solution

we have the following formulae ∂w ∂w ∂x ∂w ∂y ∂w ∂z = + + ∂r ∂x ∂r ∂y ∂r ∂z ∂r ∂w ∂w ∂x ∂w ∂y ∂w ∂z = + + ∂s ∂x ∂s ∂y ∂s ∂z ∂s w = ( x + y + z)2 ∂w = 2( x + y + z ) ∂x ∂w = 2( x + y + z ) ∂y ∂w = 2( x + y + z ) ∂z ∂x =1 ∂r ∂x = −1 ∂s

∂y = − sin(r + s ) ∂r ∂y = − sin(r + s ) ∂s

∂z = cos(r + s ) ∂r ∂z = cos(r + s ) ∂s

5

∂w ∂w ∂x ∂w ∂y ∂w ∂z = + + ∂r ∂x ∂r ∂y ∂r ∂z ∂r = 2[ x + y + z ](r − s ) + 2[ x + y + z ] − sin(r + s ) + 2[ x + y + z ]cos(r + s ) = 2[1 + 1 + sin(0) + cos(0)](1) + 2[1 + 1 + sin(0) + cos(0)]sin(0) + 2[1 + 1 + sin(0) + cos(0)]cos(0) = 6 + 0 + 6 = 12 ∂w ∂w ∂x ∂w ∂y ∂w ∂z = + + ∂s ∂x ∂s ∂y ∂s ∂z ∂s = 2[ x + y + z ](−1) + 2[ x + y + z ] − sin(r + s ) + 2[ x + y + z ]cos(r + s ) = 2[1 + 1 + sin(0) + cos(0)](−1) + 2[1 + 1 + sin(0) + cos(0)] − sin(0) + 2[1 + 1 + sin(0) + cos(0)]cos(0) = −6 + 0 + 6 = 0 Question 2










Verify that ( a × b ).c = (b × c ).a = (c × a ).b and find the volume of the parallelepiped













a , bandc where a = i + j − 2 k b = − i − k c = 2 i + 4 j − 2 k determined by

Solution















a , bandc where a = i + j − 2k b = − i − k c = 2i + 4 j − 2k here we have








(a × b ).c = (b × c ).a = (c × a ).b 1 1 −2


(a × b ).c = −1 0 −1 2 4 −2 =1

−1 −1 −1 0 0 −1 −1 −2 4 −2 2 −2 2 4

= 1(0 + 4) − 1(2 + 2) − 2(−4 − 0) = 4−4+8 =8 −1 0 −1


(b × c ).a = 2 4 −2 1 1 −2 = −1

4 −2 2 −2 2 4 −0 −1 1 −2 1 −2 1 1

= −1(−8 + 2) − 0 − 1(2 − 4) = 6−0+2 =8 2 4 −2


(c × a ).b = 1 1 −2 −1 0 −1 =2

1 −2 1 −2 1 1 −2 −4 0 −1 −1 0 −1 −1

= 2(−1 + 0) − 4() − 1(2 − 4) = 6−0+2 =8 the volume of the parallelopiped is volume = 8 Question 3

6

7 Find the derivative of





f ( x, y, z ) = x3 − xy 2 − z at P0 (1,1,0) in the direction of a = 2i − 3 j + 6k

Solution





f ( x, y, z ) = x − xy − z at P0 (1,1,0) in the direction of a = 2i − 3 j + 6k ; 3

2

f x ( x, y , z ) = 3 x 2 − y 2 f y ( x, y, z ) = −2 xy f z ( x , y , z ) = −1 at P0 (1,1,0) f x ( x, y , z ) = 3 − 1 = 2 f y ( x, y, z ) = −2 f z ( x , y , z ) = −1 now



a = 2i − 3 j + 6 k
| a |= 4 + 9 + 36 = 7


2i − 3 j + 6 k aɺ = 7


∆f = 2i − 2 j − k





2i − 3 j + 6 k Du f ( x, y , x) = ( ).(2i − 2 j − k ) 7 4+6−6 4 = = 7 7

8

Assignment Solution Fall 2007 MTH301 (3)

Question 1

If

f ( x) = x 2 + 2 x , x0 = 0, dx = 0.1 then find

(a) The change

∆f = f ( x0 + dx) − f ( x0 )

(b) The value of the estimate

(c) The approximate error

df = f '( x0 )dx

∆f − df

Solution

here f ( x) = x 2 + 2 x x0 = 0

dx = 0.1

(a ) ∆f = f ( x0 + dx) − f ( x0 ) = [(0.1)2 + 2(0.1)] − 0 = (0.1)2 + 0.2 = 0.21 df ( x) (b ) = x2 + 2 x dx

9

df = (2 x + 2)dx at x0 = [(2 × 0 + 2)][0.1] = 0.2 (c ) Now ∆f − df = 0.21 − 0.2 = 0.01

Question 2

If

f ( x, y ) = 2 xy − 5 x 2 − 2 y 2 + 4 x + 4 y − 4

find all the local maxima, local minima, and saddle

points of the function .

Solution

f ( x, y ) = 2 xy − 5 x 2 − 2 y 2 + 4 x + 4 y − 4 now we calculate the partial derivatives w.r.t x f x = −10 x + 2 y + 4

(1)

f y = 2 x − 10 y + 4

(2)

now by 1× (1) and 5× (2) −10 x + 2 y + 4 = 0 10 x − 20 y + 20 = 0 −−−−−−−−−−−−−−− −18 y + 24 = 0

10

4 3 now for x −10 x + 2 y + 4 = 0 y=

4 −10 x + 2( ) + 4 = 0 3 8 2 −10 x = −4 − = 3 3 2 4 so the po int is ( , ) 3 3 f xx = −10 f yy = −4 f xy = 2 now D = f xx f yy − f 2 xy = (−10)(−4) − 4 = 36 > 0 and f xx > 0 so local max ima

11

Assignment Solution Fall 2007 MTH301 (4)

Question 1 Find the volume of the prism whose base is the triangle in xy-plane bounded by the x-axis and the lines y=x and x=1 and whose top lies in the plane.

z = f ( x, y ) = 3 − x − y

Solution here 1 x

V = ∫ ∫ (3 − x − y )dydx 0 0 1

= ∫ [3 y − xy − 0

y2 x ]0 dx 2

1

x2 = ∫ ([3 x − x − ] − [0])dx 2 0 2

1

= ∫ [3 x − 0

3x 2 ]dx 2

3x 2 x3 1 − ]0 2 3 3 1 = [ − ] =1 2 2 =[

Question 2 2

Evaluate the integral

Solution

0

∫∫ 0 x −4 2

4

dydx + ∫

x

∫ dydx

0 0

12 2

0

∫∫

4

dydx + ∫

0 x 2 −4

x

∫ dydx

0 0

2

4

= ∫ [ y]

dx + ∫ [ y ]0 x

0 x 2 −4

0

0

2

4

= ∫ [0 − ( x − 4)] dx + ∫ [ x ]dx 2

0

0

2

4

0

0

= ∫ [(4 − x 2 )] dx + ∫ [ x ]dx = (4 x −

3

3 2

3

3 2

x 2 x 4 )0 + ( )0 3 3/ 2

x 2 x 4 )0 + ( )0 3 3/ 2 8 8 = (8 − ) + ( ) 3 3/ 2 8 16 = [8 − + ] 3 3 32 = 3 = (4 x −

13

Assignment Solution Fall 2007 MTH301 (5)

Question 1 1

Change the Cartesian integral into polar integral and solve

∫ ∫

−1 − 1− x 2

Solution 1

1− x 2

∫ ∫

dxdy

−1 − 1− x 2

Chnaging int o polar coordinates we will change dxdy = rdrdθ As it is circle of radius 1 so the lim its of the int egration will be r = 0,1 and θ = 0, 2π so the int egral changes int o =

2π 1

∫ ∫ rdrdθ 0 0

int egrating w.r.t r we will have =

2π

∫ 0

=

2π

∫ 0

=

2π

∫ 0

1

r2 dθ 2 0 12 02 − dθ 2 2

12 02 − dθ 2 2

1− x 2

dxdy

14

=

2π

∫ 0

1 dθ 2

int egrating w.r.t theta =

1 θ 2

2π

= 0

1 (2π ) − 0 = π 2

Question 2 a

Solve the integral by changing into polar coordinates

a 2 − x2

∫ ∫

− a − a2 − x2

Solution

a 2 − x2

a

∫ ∫

dxdy

− a − a 2 − x2

Chnaging int o polar coordinates we will change dxdy = rdrdθ As it is circle of radius 1 so the lim its of the int egration will be r = 0, a and θ = 0, 2π so the int egral changes int o =

2π a

∫ ∫ rdrdθ 0 0

int egrating w.r.t r we will have =

2π

∫ 0

=

2π

∫ 0

=

2π

∫ 0

a

r2 dθ 2 0 a 2 02 − dθ 2 2

a 2 02 − dθ 2 2

dxdy

15

=

2π

∫ 0

a dθ 2

int egrating w.r.t theta =

a θ 2

2π

= 0

a (2π ) − 0 = aπ 2

Question 3

Solve the initial value problem for r as a function of t







dr differential equation
= −ti − tj − tk initial condition r (0) = −ti + 2 j + 3k dt

Solution









dr differential equation
= −ti − tj − tk initial condition r (0) = −ti + 2 j + 3k dt



dr = [−ti − tj − tk ]dt taking int egral on both sides



r = ∫ [−ti − tj − tk ]dt
−t 2
−t 2
−t 2
r= i− j− k +c 2 2 2 now the given initial condition is



r (0) = −ti + 2 j + 3k


c = −ti + 2 j + 3k


−t 2
−t 2
−t 2

r= i− j− k − ti + 2 j + 3k 2 2 2 2 2

−t
−t 2
−t − t )i − ( − 2) j − ( − 3)k r =( 2 2 2

16

Assignment Solution Fall 2007 MTH301 (6)

Question 1 1 1 Ιf dz = 2 cos ydx + ( − 2 x sin y )dy + dz y z Show that it is an exact differential

Solution

1 1 dz = 2 cos ydx + ( − 2 x sin y )dy + dz y z 1 M = 2 cos y N = − 2 x sin y y ∂M ∂N = −2sin y = ∂y ∂y ∂N ∂P =0 = ∂z ∂y ∂P ∂M =0 = ∂x ∂z

P=

1 z

Hence all the above mentioned partial derivatives are equal so it is an exact differential.

Question 2

Use Line integral to evaluate I = ∫ {( x3 + 2 y 2 )dx + x 2 y 2 dy} from Ο(0, 0) toA (2, 0) along the line y = 0 and then form (2, 0) to (2, 4) along the line x = 2

Solution

17

I = ∫ {( x + 2 y )dx + x y dy} 3

2

2

2

now first along the line y = 0 , dy = 0 the lim its of x = 0, 2 so int egral becomes I = ∫ {( x3 + 2 y 2 )dx + x 2 y 2 dy} 2

I = ∫ {( x3 + 2.0)dx + x 2 y 2 0} 0 2

I = ∫ x3dx 0

x4 2 |0 4 16 = =4 4 now first along the line x = 2 , dx = 0 the lim its of x = 0, 4 so int egral becomes I =|

I = ∫ {( x3 + 2 y 2 )dx + x 2 y 2 dy} 4

I = ∫ {( x3 + 2 y 2 )0 + 4 y 2 dy} 0 4

I = ∫ 4 y 2 dy 0

4 y3 4 |0 3 4.43 256 = = 3 3

I =|

Question 3 I = ∫ {3 x3 ydx + 2 x3 y 2 dy} between O(0,0) and A(3,5) (a) along c1 i.e y = x 2

Evaluate (b) along c2 i.e y = 2 x (c) along c3 i.e x = 0 from (0, 0) to (0,5) y = 5 from (0,5) to (3,5)

18 Solution

I = ∫ {3 x3 ydx + 2 x3 y 2 dy} between O(0,0) and A(3,5) (a) along c1 i.e y = x 2 here y = x 2

, dy = 2 xdx lim its of x = 0,3

I = ∫ {3 x3 ydx + 2 x3 y 2 dy} 3

= ∫ {3 x3 .x 2 dx + 2 x 3 ( x 2 ) 2 2 xdx} 0 3

= ∫ {3 x5 dx + 2 x3 .x 4 2 xdx} 0 3

= ∫ {3 x5 dx + 4 x8 dx} 0 3

3

= ∫ 3x dx + ∫ 4 x8 dx 5

0

0 6

3x 3 4 x9 3 |0 + |0 6 9 729 19638 ] + 4[ ] = 9112.5 = 3[ 6 9 (b) along c2 i.e y = 2 x =

I = ∫ {3 x3 ydx + 2 x3 y 2 dy} 3

= ∫ {3 x3 .2 xdx + 2 x 3 (2 x) 2 2dx} 0 3

= ∫ {6 x 4 dx + 2 x3 .4 x 2 .2dx} 0 3

= ∫ {6 x 4 dx + 16 x5 dx} 0

19 3

3

= ∫ 6 x 4 dx + ∫ 16 x5 dx 0

0

6 x 3 16 x 6 3 |0 + |0 5 6 243 729 ] + 16[ ] = 243 + 8[243] = 2235.6 = 6[ 6 6 (c) along c3 i.e x = 0 from (0, 0) to (0,5) y = 5 from (0,5) to (3,5) =

5

Along x = 0 I =0 along y = 5 , dy = 0

∫ {3x ydx + 2 x y dy} 3

3

3 ∫ 3x 5dx = 15 0

3

2

x4 3 81 1215 |0 = 15. = = 303.5 4 4 4

20

Assignment Solution Fall 2007 MTH301 (7)

Question 1

2 2 2 2 If I = ∫ {( y − x )dx + ( x + y )dy} Use Green’s theorem to evaluate the integral where C: the triangle bounded by y=0, x=3 and y=x

Solution



F = ( y 2 − x 2 )i + ( x 2 + y 2 ) j here M = y 2 − x2 N = x2 + y 2 ∂M = 2 y ∂N = 2 x ∂y ∂x now green ' s theorem states that ∂N ∂M ∫ {Pdx + Qdy} = ∫∫R ( ∂x − ∂y )dydx 3 x

= ∫ ∫ (2 x − 2 y )dydx 0 0 3

= ∫ {2 xy − 2 0

3

y2 x }0 dx 2

= ∫ {2 xy − y 2 }0x dx 0

21 3

= ∫ {2 xy − y 2 }0x dx 0

3

= ∫ {2 x 2 − x 2 }dx 0

3

= ∫ x 2 dx 0

1 = [ x3 ]30 3 1 = [ x3 ] = 9 3

Question 2



xyz
−1 z F ( x , y , z ) = ln xi + e j + tan ( ) k Evaluate the curl F and div F x

Solution




F ( x, y, z ) = ln xi + e xyz j + tan −1 ( z )k x

∂
∂
∂
div F = ( i + i + i ).(ln xi + e xyz j + tan −1 ( z ) x ∂x ∂y ∂z 1 1 1 = + xze xyz + ( ) 2 x 1+ z 2 x x 2 1 1 x = + xze xyz + 2 ( ) 2 x x +z x 1 x2 1 = + xze xyz + 2 ( ) 2 x x +z x 1 x = + xze xyz + 2 x x + z2

22




F ( x, y, z ) = ln xi + e xyz j + tan −1 ( z )k x i ∂ curlF = ∂x

j ∂ ∂y

ln x e xyz ∂ = i ∂y e xyz

∂ ∂z

k ∂ ∂z tan −1 ( z ) x

∂ ∂ ∂ ∂ ∂z − j ∂x + k ∂x ∂y −1 z −1 z ln x tan ( ) tan ( ) ln x e xyz x x z ) + k ( yze xyz ) = i ( yze xyz ) − j ( 2 2 x +z

23

Assignment Solution Fall 2007 MTH301 (8)

Question 1



2
If F (r ) = xyi + x zj + 3 yzk evaluate

∫ F .dr c

between A (0, 0, 0) and B (5, 2, 1) along the curve c

2 having the parametric equation x = 5t , y = 2t , z = t

Solution




F (r ) = xyi + x 2 zj + 3 yzk x = 5t , y = 2t , z = t 2 dx = 5dt dy = 2dt dz = 2tdt dr = (5i + 2 j + 2t )dt


F (t ) = 10t 2i + 5t 4 j + 6t 3k


F .dr = (10t 2 i + 5t 4 j + 6t 3k ).(5i + 2 j + 2t ) F .dr = (50t 2 + 10t 4 + 12t 4 )dt 1

2 4 ∫ F .dr = ∫ (50t + 22t )dt c

0

t t5 + 22 |10 3 5 1 1 250 + 66 316 = 50 + 22 = = 3 5 15 15 = 50

3

Question 2

f ( x) = 2 x 0 ≤ x ≤ 2π Find the Fourier series for the function f ( x) = f ( x + 2π ) take the limits of the integration 0, 2π

24

Solution

f ( x) = 2 x 0 ≤ x ≤ 2π f ( x) = f ( x + 2π ) a0 = a0 = = a0 = an = =

2

π = =

bn = = = =

1

2π

∫

π 1

2π

∫ 2 xdx

π 1

π 1

0

[ x 2 ]02π = 4π 2π

∫ 2 x cos nxdx

π 2

f ( x)dx

0

0 2π

∫ x cos nxdx

π

0

2π

∫ x cos nxdx 0

2

x sin nx 2π 1 }0 − n n

π

[{

2

[0 − 0] = 0

π

2

π 2

π 2

π

∫ sin nxdx] 0

2π

∫ x sin nx dx 0

2π

∫ x cos nxdx 0

[{

−2

π

2π

− x cos nx 2π 1 }0 + n n

[2π − 0] = −1

2π

∫ cos nxdx] 0