Mth301 Collection of Old Papers
Mth301 Collection of Old Papers
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                             MTH301 Calculus
                       Final Term Examination – Spring 2005
                                 Time Allowed: 150 Minutes
      ∫∫ x
        R
              (1− x2 ) dA            R ={(x, y) :0 ≤ x ≤1,2 ≤ y ≤ 3}
 Question No. 2                                                                        Marks : 06
                                  D r f (−2, 0)                 2
                                                   f ( x, y) = x − 3xy + 4 in the direction
Find the directional derivative u               of
           r                                         3
           u = $i + 2 j                            y
       of               at the point (-2, 0).
                                               2
Find the centre and radius of the sphere x         y2 +z2 −8x+6y +12z = 3.
                                         +
          r  4
          a = i$ + $j r
    Let      3      and
                        b = 4 $i 3 ,then
                        +        $
                                 j
               r     r
          o   a and b
                          are orthogonal
                                                   π
          o   The angle between a and b is 4 .
          o   a and b are parallel.
                                r      r π
          o   The angle between a and b is 6 .
                1
                2        if s > 2
          o   s +4
                1
                         if s > 2
          o   s −4
                1
                            if    s > 2
          o   s +2
                1
                   if      s>2
          o   s −2
                2
     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)
The differential
                   dz = Mdx + Ndy is an exact differential if we have
               ∂M       ∂N
                      =
               ∂y       ∂y
         o
               ∂M       ∂N
                      =
               ∂y       ∂x
         o
               ∂M       ∂N
                      =
               ∂x       ∂y
         o
               ∂M       ∂N
                      =
               ∂x       ∂x
         o
                                                      ur    2$       3 $   2 $
Compute Curl and Divergence for the given vector field F = x i + 4 xy j + y xk .
The function
                y = x2/ 3 + 6 is
         o     Even function
         o     Odd function
         o     Neither even nor odd
         o     Constant
 Question No. 10                                                                   Marks : 06
                  s − 19
          L { s 2 + 3s − 10}
             -1
Determine                            by using the concept of partial fraction.
Locate all the relative maxima, relative minima and saddle points for function
f(x, y) = x 2 +y2 - x 2 y
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                             FINALTERM EXAMINATION
                                       SPRING 2006                           Marks: 55
                          MTH301 - CALCULUS II (Session - 2 )               Time: 120min
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      ►     1
                2
            s
      ► e− s
      ►             s
      ►                        y+2     z−3
            x−2= −                 =−
                               3     5
      ►                        y−3     z−5
                x − 1= −           =−
                               5     2
      ► x + 1= y + 3 =                 z−5
                           5       5
► True.
      ► False.
Question No: 4          ( Marks: 1 )       - Please choose one
                    z= r
 The equation              is written in
► Rectangular coordinates
► Cylindrical coordinates
► Spherical coordinates
               C
                   1+x                              x=t , y=                t ,0≤t≤3
                                                                   3
Evaluate
Find
                                                                                z=   x +y       , y= 4− x       ,
                                                                                       2    2               2
Set up a double integral for the volume bounded by surface                                                  and first
octant.
(Do not evaluate.)
u = 7 i +3 j +5k , v = − 8 i + 4 j +2k
Question No: 12    ( Marks: 4 )
      JG     ∧    ∧      ∧
                                          JG
      F = x 2 i − 2 j + yz k              F
Given                       .Find div(Curl ).
                                     a0 , an a n d bn
Determine the Fourier coefficients                      for a periodic function of period 2 π defined
by
                                MTH301 Calculus II
                              Mid Term Examination – Spring 2006
                                   Time Allowed: 90 Minutes
     1
∫ 4x    2
            +
                          2
                dx = ln 4 x + x + c
    x
 True.
ミ False
Find
∂z     ∂z
   and    , where z(s, t) = f (r(s, t), θ (s, t)) ,
∂s     ∂t
              2r
f (r, θ ) = e sin(3θ ) , r(s, t) = st − , θ (s, t) = s2 + t 2
  2
t
ミ True.
ミ False
f (x, y) = 3 − x 2 −    is continuous at
             2
           y
 x2 + y 2 = 3
  2       2
ミ x +y ≥3
   2     2
 x +y ≤3
 x2 + y 2 < 3
Find an equation of the plane through the origin that is parallel to the plane:
       4 x − 2 y + 7 z + 12 = 0
Question No. 8                                                                  Marks : 10
Use double integrals to compute the area of the region bounded by the curves:
     2
y = x and y = 8 − x 2
 True.
 False
    2 2 1
     , ,
    3 3 3
     2 −1 1
      ,    ,
      3 3 3
     1 1 2
      ,     ,
      6 6        6
 None of these.
                            MTh301-Calculus II
                            Midterm Special 2006
                                      www.
                                vujannat.ning.com
      ¾ At critical points function can assume
            o maximum value.
            o minimum value.
            o both maximum and minimum value.
            o zero value
      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 percentage
        error in the calculated area.
Solution:
Let
x = length
y = width
A = area
Then
          A = xy So
∂A
      =
y
∂x
 ∂A
      =x
 ∂y
Therefore,
                 ∂A ∂A
          ∂A ≈     +   ∂y
                 ∂x ∂y
= y∂x + x∂y
We desire percentage change in A, which is relative change multiplied by 100 so
let’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
             ∂x                    ∂y
−0.04 ≤         ≤ 0.04 and −0.03 ≤    ≤ 0.04
             x                     y
                                   ∂A
The maximum possible value for         given the above constraints. This happens
                                    A
       ∂x            ∂y                 ∂A
when      = 0.04 and    = 0.03 , giving    = 0.07 . This is relative error, so the
        x            y                 A
percentage error is 7%.
Solution:
Given:
x=1
y= 2
z = −2
We know that:
         2      2
r= x +y
r= 1+ 2
r= 3
          y
tan θ =
          x
          −1       y
θ = tan ( )
               x
          −1      2
θ = tan        ( )
                1
          −1
θ = tan        2
θ = 54.73
z = −2
                           1           1 −1
                               ,           ,
               o           3           3 3
                       1 1 −1
                        , ,
               o       3 3 3
                           1−1 −1
                              ,    ,
               o         2 2 2
                       1 1 −1
                        , ,
               o       2 2 2
Find equation of the tangent plane to the surface z = x 3 +        at the point (2,1, 3) .
Solution:
                                                       y3
f x = 3x − y
fy =3y−x
f z = −1
fx ( P) =6−1=5
fy ( P) =3−2=1
fz ( P) = −1
5 ( x − 2 ) + 1 ( y − 1) − 1 ( z − 3 ) = 0
5x −10 + y −1 − z + 3 = 0
Hence:
5x + y − z − 8 = 0
======================================================
                 G ^ ^ ^            G       ^   ^      ^
                 a = i + 2 j + k and b = 2 i + 3 j + 2 k
Dot product of                                           is
          o 8
          o 5
          o 10
          o 13
                                      ∫∫R x 2 dA ;
   ¾ Evaluate the double integral                 R is the region bounded
              16
          y=       , y = x , and x = 8
       by      x                      .
Solution:
                                  ∫∫
                                           2
Area of R =                                x dA
                                       R
     8               x
=∫           ∫   16      x2 dydx
     2
                 x
                              x
     8               x
=∫           ∫   1 6
                          2
                              16 dx
     2
x                              x
                 x
         8
= ∫ x 3 − 16 xdx
     2
               4
      8    x
           16x
                                2
=∫                 −
    2 4      2
        4
    8 x
=∫        − 8x
               2
    2 4
  ⎛ 84         ⎞ ⎛ 24 ⎞
= ⎜ − 8 × 8 ⎟− ⎜ − 8× 2⎟
  ⎝ 4          ⎠ ⎝ 4  ⎠
= 972
                   x+ y
      ¾ Domain of         is
               x+ y> 0
           o
                ≥     ≥
           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 x 2 + y 2 = 9 and the planes y + z = 9 and z = 0
Solution:
∫∫R
      (9 − y)da
           2           9− x 2
V=        ∫∫
           −2      − 9− x 2
                                (9 − y )   dydx
                                                  2
       2   ⎡                    1          9−x
=                  9y−              y
2⎤
  ∫   −2
                          2                           2
                                                       dx
                                        y= − 9−x
      2
                                2
=
  ∫        18 9 − x dx
      −2
      ⎛1 2 ⎞
= 18⎜ π 9 ⎟ = 729π
      2
    ⎝     ⎠