mth301 collection of old papers
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MTH301 CalculusFinal Term Examination – Spring 2005
Time Allowed: 150 Minutes
Please read the following instructions carefully beforeattempting any of the questions:
1-Duration of the paper is 150 minutes and use of calculator isallowed2. Attempt all questions. Marks are written adjacent to eachquestion.3. Do not ask any questions about the contents of thisexamination from anyone.
a. If you think that there is something wrong with any of the questions, attempt it to the best of your understanding.b. If you believe that some essential piece of information ismissing, make an appropriate assumption and use it tosolve the problem.c . Write all steps, missing steps may lead to deduction of marks.
4- Pasting the equations of math type from word file into software maycause some visibility problem, so please note that do not copy equationsof math type into software from word file. Paste the equations from mathtype directly into software.
Total Marks: 65 Total Questions: 12
Question No. 1 Marks : 08
Evaluate the iterated integral
∫∫ x R
(1− x2 ) dA R ={( x, y) :0 ≤ x ≤1,2 ≤ y ≤ 3}
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$
e
2
2
Question No. 2 Marks : 06
D r
Find the directional derivative u
r u = i + 2 j
f (−2, 0)of f ( x, y) = x
2 − 3 xy + 4
y3
in the direction
of at the point (-2, 0).
Question No. 3 Marks : 05
Find the centre and radius of the sphere x2
+
y 2 + z 2 − 8 x + 6 y + 1 2 z = 3 .
Question No. 4 Marks : 02
r $ r a =Let
4i +
3
$ jand b = 4
$i
+
3$
j
,then
r r
oa and b
are orthogonalπ
o The angle betweena and b
is 4 .
oa and b are parallel.
r r πo The angle between a and b is 6 .
Question No. 5 Marks : 02
2 t
The Laplace transform of is
1o s + 4
1
o s − 4
if s > 2
if s > 2
1
o s + 21
o s − 2
i f s > 2
if s > 2
Question No. 6 Marks : 02
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2 $ 3$ 2 $
y = x + 6
π
If (4, 3 ,- 4) is a point in the Cylindrical coordinates then the same point in Rectangular coordinates is given by
o (2,2 3 ,4)
o (2,2 3 ,4)
o (2,2 3 ,-4)
o (- 2,2 3 ,4)
o (2 3 ,2,-4)
Question No. 7 Marks : 02
The differential dz = Mdx + Ndy is an exact differential if we have
∂M =
∂ N
o∂ y ∂ y
∂M =
∂ N
o∂ y ∂ x
∂M =
∂ N
o∂ x ∂ y
∂M =
∂ N
o ∂ x ∂ x
Question No. 8 Marks : 10
ur F = x i + 4 xy j+ y xk Compute Curl and Divergence for the given vector field .
Question No. 9 Marks : 02
2/ 3
The function is
o Even function
o Odd function
o Neither even nor odd
o Constant
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Question No. 10 Marks : 06
Determine whether the differentialor not?
dz = ( xlny + xy) dx + ( ylnx + xy)dy
is exact
Question No. 11 Marks : 10
DetermineL-1{
s − 19}
s2 + 3 s − 10 by using the concept of partial fraction.
Question No. 12 Marks : 10
Locate all the relative maxima, relative minima and saddle points for function
f(x, y) = x2+y
2- x
2y
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BEST SITE TO HELPSTUDENTS
FINALTERM EXAMINATION
SPRING 2006
MTH301 - CALCULUS II (Session - 2 )
Marks: 55
Time: 120min
StudentID/LoginID:
Student Name:
Center Name/Code:
Exam Date: Thursday, August 17, 2006
Please read the following instructions carefully before attempting anyof the questions:
1. Attempt all questions. Marks are written adjacent to each question.
2. Do not ask any questions about the contents of this examinationfrom anyone.
a. If you think that there is something wrong with any of thequestions, attempt it to the best of your understanding.
b. If you believe that some essential piece of information ismissing, make an appropriate assumption and use it to solve theproblem.
c . Write all steps, missing steps may lead to deduction of marks.
3. Calculator is allowed.
**WARNING: Please note that Virtual University takes serious note of unfair means. Anyone found involved in cheating will get an `F` grade in thiscourse.
For Teacher's use onlyQuestion 1 2 3 4 5 6 7 8 9 10 Total
Marks
Question 11 12 13Marks
Question No: 1 ( Marks: 2 ) - Please choose one
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Laplace transform of ‘t ‘ is
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► 1
s
► 1
s2
► e− s
► s
Question No: 2 ( Marks: 2 ) - Please choose one
Symmetric equation for the line through (1,3,5) and (2,-2,3) is
► x − 2 = −
y + 2= −
z − 3
3 5
► x + 2 = − y + 3 = − z + 55 2
► x − 1= −
y − 3= −
z − 5
5 2
► x + 1=
y + 3=
z − 5
5 5
Question No: 3 ( Marks: 1 ) - Please choose one
The level curves of f(x, y) = y Cscx are parabolas.
► True.
► False.
Question No: 4 ( Marks: 1 ) - Please choose one
z = r The equation is written in
► Rectangular coordinates
► Cylindrical coordinates
► Spherical coordinates
► None of the above
Question No: 5 ( Marks: 5 )
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∫1
ds, 23
C 1 + x x = t , y =
3t , 0 ≤ t ≤ 3
Question No: 6 ( Marks: 5 )
∫ ∫∫ (3 x + y 2 − z )dV over the box 0 ≤ x ≤ 2 , − 1 ≤ y ≤ 1 , − 3 ≤ z ≤ 0Q
Evaluate
Question No: 7 ( Marks: 10 )
UsJeJGSt
Gokes theorem to evaluate the integral
v∫ F .d r ,C JG ∧ ∧ ∧
where F = 2 z i + 3 x j + 5 y k , C is the circle x2
+ y2
= 1
In the xy-plane with counter clockwise orientation looking down the positive z-axis.
Question No: 8 ( Marks: 3 )
f for function f (w, x, y, z ) = w2 xyz − 5ewz
2
wwx
Find
Question No: 9 ( Marks: 4 )
Locate critical points for the function
a3 b3
f ( x, y) = xy + + (a ≠ 0 , b ≠ 0) x y
Question No: 10 ( Marks: 4 )
z= x 2
+ y 2
, y = 4 − x 2
,
Set up a double integral for the volume bounded by surface and first
octant.(Do not evaluate.)
Question No: 11 ( Marks: 4 )
G G
u and vDetermine whether make an acute angle ,an obtuse angle , or are orthogonal.G ∧ ∧ ∧ G ∧ ∧ ∧
u = 7 i + 3 j + 5 k , v = − 8 i + 4 j + 2 k
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Question No: 12 ( Marks: 4 )
JG ∧ ∧ ∧
F = x2 i − 2 j + yz k JG
Given .Find div(Curl F ).
Question No: 13 ( Marks: 10 )
a0
, an
a n d bn πDetermine the Fourier coefficients for a periodic function of period 2 defined
by
⎧ − x − π ≤ x < 0f(x) = ⎨
⎩0 0 ≤ x < π
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MTH301 Calculus IIMid Term Examination – Spring 2006
Time Allowed: 90 Minutes
Please read the following instructions carefully beforeattempting any of the questions:
1. Attempt all questions. Marks are written adjacent to eachquestion.
2. Do not ask any questions about the contents of thisexamination from anyone.
a. If you think that there is something wrong with any of thequestions, attempt it to the best of your understanding.
b. If you believe that some essential piece of information ismissing, make an appropriate assumption and use it to solvethe problem.
c . Write all steps, missing steps may lead to deduction of
marks.
3. Calculator is allowed.
**WARNING: Please note that Virtual University takes serious note of unfair means. Anyone found involved in cheating will get an `F` grade in
this course.
Question No. 1 Marks : 5
Find the directional derivative D G
uG f (1 , − 1,1)of f ( x, y, z ) = x
2
z + y
3
z
2
− xyz in the
direction of u = (−1, 0 ,3)
at the point (1 , − 1,1) .
Question No. 2 Marks : 1
1∫ 4 x
2+
x
dx = ln 4 x2
+ x + c
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True.
ミ False
Question No. 3 Marks : 10
Find
∂zand
∂z
∂s ∂t, where z(s, t) = f (r(s, t),θ (s, t)) ,
f (r,θ ) = e2r
sin(3θ ) , r(s, t) = st −
t2
, θ (s, t) = s2 + t 2
Question No. 4 Marks : 10
Compute the volume of the solid bounded by the given surface
z =1 − y , z = 0 , y = 0 , x = 1, x = 2
Question No. 5 Marks : 1
Not all critical points need to be local extrema.
ミ True.
ミ False
Question No. 6 Marks : 1
f (x, y) = 3 − x 2
−
y2is continuous at
x2
+ y2 = 3
ミ x2
+ y2
≥ 3
x2 + y2 ≤ 3
x2
+ y2 < 3
Question No. 7 Marks : 5
Find an equation of the plane through the origin that is parallel to the plane:4 x − 2 y + 7 z + 12 = 0
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Question No. 8 Marks : 10
Use double integrals to compute the area of the region bounded by the curves:
y = x2
and y = 8 − x2
Question No. 9 Marks : 1
If f ′(a ) is +ve then the graph of function is concave up.
True.
False
Question No. 10 Marks : 1
Direction cosines of straight line through A(1,1,1) and B(0,0,-1) are
2
,2
,1
3 3 3
2
,−1
,1
3 3 3
1
,1
,2
6 6 6 None of these.
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MTh301-Calculus IIMidterm Special 2006
www.
vujannat.ning.com¾ At critical points function can assume
o maximum value.o minimum value.o both maximum and minimum value.o zero value
¾ A plane can be perfectly determined byo one point.o one point and normal vector.
o one point and parallel vector.o one normal vector.
¾ The length and width of a rectangle are measured with the errors of at most 4 %and 3%, respectively. Use differentials to approximate the maximum percentageerror in the calculated area.
Solution:
Letx = length
y = width
A = area
Then
∂ A=
y∂ x
A = xy So
∂ A= x
∂ y
Therefore,
∂ A ≈∂ A
+∂ A
∂ y∂ x ∂ y
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= y∂ x + x∂ y
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We desire percentage change in A, which is relative change multiplied by 100 solet’s work out relative change first. This is given by∂ A
≈ y∂ x
+ x∂ y
A A A
=
∂ x
+∂ y x y
Since
A = xy
−0.04 ≤∂ x
≤ 0.04 and −0.03 ≤∂ y
≤ 0.04 x y
The maximum possible value for ∂ A
given the above constraints. This happens A
when∂ x
= 0.04 and∂ y
= 0.03 , giving∂ A
= 0.07 . This is relative error, so the x y A
percentage error is 7%.
Convert the Cartesian coordinates(1, 2 , −2) into cylindrical coordinates.
Solution:
We are to convert (x,y,z) int (r, θ ,z)
Given:
x = 1
y = 2
z = −2
We know that:
r = x2 + y
2
Put value of x and y in the above equation:
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r = (1)2 +
(
2 )2
r = 1 + 2
r = 3
tan θ = y
x
θ = tan −1
( y )
x
θ = tan −1 ( 2 )1
θ = tan −1 2
θ = 54.73
z = −2
So the cylindrical coordinates are ( 3,54.73,−2)
======================================================G ^ ^ ^
¾ A unit vector in the direction of a = i + j − k
is
1,
1,
−1
o3 3 3
1,1
,−1
o3 3 3
1,
−1,
−1
o 2 2 21
,1
,−1
o2 2 2
Find equation of the tangent plane to the surface
Solution:
z = x3+
y3
at the point(2,1, 3)
.
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f x
= 3 x − y
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∫
f y
= 3 y − x
f z
= −1
f x ( P ) = 6 − 1 = 5
f y ( P ) = 3 − 2 = 1
f z ( P ) = −1
Equation of tangent line to the surface through P is:
5 ( x − 2 ) + 1( y − 1) − 1( z − 3) = 0
5 x −10 + y −1 − z + 3 = 0
Hence:
5 x + y − z − 8 = 0
======================================================G ^ ^ ^ G ^ ^ ^
a = i + 2 j + k and b = 2 i + 3 j + 2 k Dot product of is
o 8o
5o 10
o 13
¾ Evaluate the double integral
y =16
, y = x , and x = 8
∫∫ x2 dA ; R R is the region bounded
by x .
Solution:
Area of R = ∫ ∫R x2dA
8 x2
= ∫2 ∫16 x dydxx
x8 x
2= ∫2 ∫1 6
xx
16 dx
x
=8 x
3 − 16 xdx2
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2
8 x4
= ∫ 16 x2
−2 4 28 x
4
= ∫ − 8 x2
2 4
⎛ 84= ⎜
⎞ ⎛ 24− 8 × 8 ⎟ − ⎜
⎞− 8 × 2 ⎟⎝ 4
= 972
⎠ ⎝ 4 ⎠
¾ Domain of x + y
is
o x + y > 0
o x ≥ 0 , y ≥ 0
o
x > 0, y > 0
o x + y ≥ 0
¾ Use double integrals to compute the volume of the solid bounded by the
cylinder x2 + y2 = 9 and the planes y + z = 9 and z = 0
Solution:
∫ ∫ (9 − y)da
R
R as a type 1 region we find that:
2 9− x2
V = ∫ ∫ (9 − y ) dydx−2 − 9− x
2
=2 ⎡
9 y −1
y2 ⎤
9 − x2
∫−2 2 y = −
dx
9 − x2
2
=∫
−2
18 9 − x dx
= 18⎛ 1
π 92 ⎞
= 729π⎜2
⎟⎝ ⎠