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Spring 2008_MTH301_2_SOLved

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                                                Solution
                                        Assignment No: 2


     Q 1.       Consider a point P(1, 2, 2) in rectangular co-ordinate system.
             a. What is the position vector of this point?
             b. Find the unit vector in the direction of this position vector?
        Solution:
                                          →
        a)       The position vector r of the point (1, 2, 2) is




                                                                    OM
                 →
                 r = i + 2 j + 2k




                                                              C
        b)       Since any vector can be expressed as product of its magnitude and unit




                                                           S.
                vector so we can write
                 →         ^
                 r= r r
                       →
                                                   N
                 ^     r
                 r=
                                                IA
                       r

                 r = (1) 2 + (2) 2 + (2) 2
                                      AL


                     = 9
                     =3
                          U



                           ^    i + 2 j + 2k
                 So,       r=
                       RT




                                      3
                                1     2     2
                               = i+ j+ k
                                3 3         3
             VI




                                                →              →
     Q 2.       Find whether the vectors a = i − 3 j + 7k ,    b = 8i − 2 j − 2k are
        orthogonal to each other or not?
        Solution:
        As we know, if two vectors are orthogonal their dot product is zero. So let us find
        the dot product of given vectors.




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        →       →
        a . b = (i − 3 j + 7 k ) . (8i − 2 j − 2k )
              = 8(i . i ) + 6( j. j ) − 14(k .k )
              = 8 + 6 − 14
              =0
        Thus, the two vectors are orthogonal or perpendicular to each other.


     Q 3.            Find the area of the parallelogram determined by the vectors
        →                               →
        a = i − j + 2k and b = 3 j + k




                                                                              OM
        Solution:
                                                          →   →       →   →
        Area of a parallelogram = magnitude of a × b = a × b




                                                                 C
                                                              S.
        →       →
        a × b = (i − j + 2k ) × (0i + 3 j + k )
                        i     j     k               N
                     = 1 −1 2
                       0 3 1
                                                 IA
                     = (−1 − 6)i + (0 − 1) j + (3 − 0)k
                     = − 7i − j + 3k
                                            AL


            →    →
            a × b = (−7) 2 + (−1) 2 + (3) 2
                               U



                     = 49 + 1 + 9
                     = 59
                            RT




                     = 7.68
        Thus area of parallelogram is 7.68 square unit
                VI




     Q 4.            Find the directional derivative of
            f ( x, y ) = y 2 ln x
                                                                  →
        at the point (1, 4) in the direction of the vector u = − 3i + 3 j
        Solution:




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            f ( x, y ) = y 2 ln x
                          y2
            f x ( x, y ) =   , f y ( x, y ) = 2 y ln x
                           x
            f x (1, 4) = 16 , f y (1, 4) = 0
        →
        u = − 3i + 3 j
        →              ^
        u= uu
                   →
        ^          u
        u=
                   u




                                                                           OM
        u = (−3) 2 + (3) 2




                                                                        C
               =       9+9




                                                                     S.
               = 18 = 3 2
        ^          −3i + 3 j
        u=
                  3 2
                                                           N
                                                        IA
                   1     1
              = −     i+    j
                    2     2
                                 ⎡                  ^
                                                      ⎤              ⎡                  ^
                                                                                          ⎤
                                            AL


        Du f (1, 4) = f x (1, 4) ⎢ x − component of u ⎥ + f y (1, 4) ⎢ y − component of u ⎥
                                 ⎣                    ⎦              ⎣                    ⎦
                         ⎛ −1 ⎞ ⎛ 1 ⎞
                   = 16 ⎜      ⎟ + 0⎜     ⎟
                                 U



                         ⎝ 2⎠ ⎝ 2⎠
                     −16
                              RT




                  =
                         2
     Q 5.              Find a unit vector in the direction in which the function
                   VI




            f ( x, y ) = 4 x3 y 2 increases most rapidly at the point P(-1, 1).
        Solution:
            f ( x, y ) = 4 x 3 y 2
            f x ( x, y ) = 12 x 2 y 2 ,     f y ( x, y ) = 8 x 3 y
        ∇f ( x, y ) = f x ( x, y )i + f y ( x, y ) j
                           = 12 x 2 y 2 i + 8 x3 y j
            f x (−1, 1) =12 ,          f y (−1, 1) = −8




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        ∇f (−1, 1) = f x (−1, 1)i + f y (−1, 1) j
                      = 12 i − 8 j
                             ^
        A unit vector u in the direction of ∇f (−1, 1) is
        ^       ∇f (−1, 1)
        u=
                ∇f (−1, 1)

        ∇f (−1, 1) = (12) 2 + (−8) 2
                      = 144 + 64
                      = 4 13




                                                                            OM
        ^       ∇f (−1, 1)   12 i − 8 j
        u=                 =
                ∇f (−1, 1)    4 13




                                                                 C
                  3     2
            =        i−    j



                                                              S.
                  13    13


     Q 6.
                                                   N
                   Write down the general equation in parametric form for the following.
                                                IA
                a. Straight line in two dimensional space.
                   Let (x0, y0) is any fixed point on the line and the line is parallel to the
                   vector ai + bj then parametric form of straight line in two dimensional
                                       AL


                   space in terms of parameter t is
                    x = x0 +at , y = y0 + bt
                            U



                b. Straight line in three dimensional space.
                         RT




                   Parametric equation of a line in three dimensional space passing through
                   the point (x0, y0, z0) and parallel to the vector ai + bj + ck is given by
                VI




                   x = x0 +at , y = y0 + bt         ,   z = z0 +ct
                c. Parametric form of curve in two dimensional space is
                     x = f(t) , y = g(t) , where t is parameter


     Q 7.          Find the equation of
                a. Tangent plane
                b. Normal line
                   of the surface f ( x, y, z ) = x 2 y − 4 z 2 + 7 at the point (-3, 1, -2)




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       Solution:
       f ( x, y , z ) = x 2 y − 4 z 2 + 7
       f x ( x, y, z ) = 2 xy
       f y ( x, y , z ) = x 2
       f z ( x, y, z ) = −8 z

       f x (−3, 1, − 2) = 2(−3)(1) = − 6
       f y (−3, 1, − 2) = (−3) 2 = 9
       f z (−3, 1, − 2) = − 8(−2) = 16




                                                                                       OM
       Equation of the tangent plane to the surface at (x0, y0, z0) = (-3, 1, -2) is
       f x ( x0 , y0 , z0 ) ( x − x0 ) + f y ( x0 , y0 , z0 ) ( y − y0 ) + f z ( x0 , y0 , z0 ) ( z − z0 ) = 0




                                                                         C
       −6( x + 3) + 9( y − 1) + 16( z + 2) = 0




                                                                      S.
       −6 x − 18 + 9 y − 9 + 16 z + 32 = 0
       −6 x + 9 y + 16 z + 5 = 0
                                                        N
       Equation of the normal line of the surface through (x0, y0, z0) = (-3, 1, -2) in
                                                     IA
      SYMMETRIC FORM is
               x − x0               y − y0               z − z0
                            =                    =
        f x ( x0 , y0 , z0 ) f y ( x0 , y0 , z0 ) f z ( x0 , y0 , z0 )
                                         AL


       x + 3 y −1 z + 2
            =    =
                           U



        −6    9    16
                        RT




       Equation of the normal line of the surface through (x0, y0, z0) = (-3, 1, -2) in
      PARAMETRIC FORM is
       x = x0 + f x ( x0 , y0 , z0 )
            VI




       y = y0 + f y ( x0 , y0 , z0 )
       z = z0 + f z ( x0 , y0 , z0 )
       Thus
       x = − 3 − 6t
       y = 1 + 9t
       z = −2 + 16t




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     Q 8.            Show that the sphere x 2 + y 2 + z 2 = a 2 and the cone z 2 = x 2 + y 2 are
        orthogonal at every point of intersection.
        Solution:
            f ( x, y, z ) = x 2 + y 2 + z 2 − a 2 = 0 − − − − − − − − − − − − − −(1)
        g ( x, y, z ) = x 2 + y 2 − z 2 = 0 − − − − − − − − − − − − − − − −(2)
            f x ( x, y , z ) = 2 x
            f y ( x, y , z ) = 2 y
            f z ( x, y , z ) = 2 z




                                                                                      OM
        g x ( x, y , z ) = 2 x
        g y ( x, y , z ) = 2 y
        g z ( x, y , z ) = − 2 z




                                                                          C
            f x g x + f y g y + f z g z = (2 x)(2 x) + (2 y )(2 y ) + (2 z )(−2 z )



                                                                       S.
                                       = 4 x2 + 4 y 2 − 4 z 2
                                      = 4( x 2 + y 2 − z 2 )
                                                          N
        Since from equation (2) , x 2 + y 2 − z 2 = 0 So
                                                       IA
            f x g x + f y g y + f z g z = 4( x 2 + y 2 − z 2 ) = 0

        Thus, the sphere x 2 + y 2 + z 2 = a 2 and the cone z 2 = x 2 + y 2 are orthogonal at
                                            AL


        every point of intersection.
                               U



     Q 9.            For evaluating RELATIVE EXTREME VALUES of a function at some
                            RT




        point in the domain, what we consider as a neighborhood of a point if it lies in
                a. Two dimensional space
                 VI




                b. Three dimensional space
        Solution:
                a. For evaluating RELATIVE EXTREME VALUES of a function at some
                     point (x0, y0) in the domain D, neighborhood of that point is an open disc.
                     Open disc K ∈ D, centered at point (x0, y0) and of radius r is defined as

                      K = {( x, y ) ∈ R 2 : ( x − x0 ) 2 + ( y − y0 ) 2 < r 2 }

                b. For evaluating RELATIVE EXTREME VALUES of a function at some
                     point (x0, y0, z0) in the domain D, neighborhood of that point is an open




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                 sphere. Open sphere S ∈ D, centered at point (x0, y0, Z0) and of radius r is

                 defined as S = {( x, y , z ) ∈ R 3 : ( x − x0 ) 2 + ( y − y0 ) 2 + ( z − z0 ) 2 < r 2 }


     Q 10.       Locate all relative maxima, relative minima and saddle point of
         f ( x, y ) = x 2 + 2 y 2 − x 2 y
        Solution:
         f ( x, y ) = x 2 + 2 y 2 − x 2 y
         f x ( x, y ) = 2 x − 2 xy , f y ( x, y ) = 4 y − x 2




                                                                                OM
        The critical points of f satisfy the equations
        2 x − 2 xy = 0 − − − − − − − − − −(1)




                                                                   C
        4 y − x 2 = 0 − − − − − − − − − −(2)




                                                                S.
        From equation (1)
        2 x(1 − y ) = 0                              N
        x(1 − y ) = 0
        x = 0 , 1− y = 0 ⇒ y = 1
                                                  IA
        Use this value of x and y in equation (2),
        If x = 0 then
                                        AL


        4 y − x2 = 0 ⇒ 4 y = 0
        ⇒y=0
                           U



        If y = 1 then
                        RT




        4 y − x2 = 0 ⇒ 4 − x2 = 0
        x2 = 4
              VI




        x = ±2
        So (0, 0), (2, 1) and (-2, 1) are the critical points
        Now find second-order partial derivatives
         f xx ( x, y ) = 2 − 2 y , f yy ( x, y ) = 4 , f xy ( x, y ) = − 2 x

        For critical point (0, 0)
         f xx (0, 0) = 2 , f yy (0, 0) = 4 , f xy (0, 0) = 0
        D = f xx (0, 0) f yy (0, 0) − f xy 2 (0, 0)
             = (2)(4) − 0 = 8




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       D > 0 and f xx (0, 0) > 0 So f has relative minimum at (0, 0)
       For critical point (2, 1)
       f xx (2, 1) = 0 , f yy (2, 1) = 4 , f xy (2, 1) = − 4
       D = f xx (2, 1) f yy (2, 1) − f xy 2 (2, 1)
          = (0)(4) − (−4) 2
          = −16
       D < 0 so f has a saddle point at (2, 1)
       For critical point (-2, 1)




                                                                             OM
       f xx ( x, y ) = 2 − 2 y , f yy ( x, y ) = 4 , f xy ( x, y ) = − 2 x
       f xx (−2, 1) = 0 , f yy (−2, 1) = 4 , f xy (−2, 1) = 4




                                                               C
       D = f xx (−2, 1) f yy (−2, 1) − f xy 2 (−2, 1)




                                                            S.
          = (0)(4) − (4) 2
          = −16
       D < 0 so f has a saddle point at (-2, 1)
                                                        N
                                                     IA
                          U            AL
                       RT
            VI




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