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    21 mth301 assignment no 1 spring 2011 solution on vuhelp.net

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    Assignment # 1
    MTH301 (Spring 2011)
    Question 1: Marks 5
    Cylindrical coordinates of a point are given in the following diagram. Find the
    rectangular and spherical coordinates of that point.

    Solution:
     , ,  2, ,1
    4
    r z

       
     
    Rectangular Coordinates:
       
    cos 2cos 2 1 2
    4 2
    sin 2sin 2 1 2
    4 2
    1
    , , 2, 2,1
    x r
    y r
    z z
    x y z




           
     
           
     
     
     
    Spherical Coordinates:
     
     
    2 2 2 2
    1 1
    , , 2, ,1
    4
    (2) (1) 5
    4
    tan tan 2 1.107
    1
    , , 5, ,1.107
    4
    r z
    r z
    r
    z




     


      
     
        
     
        
     
              
       
         
     
    Question 2: Marks 05
    Check the continuity of given function at (0, 0)
    2 2 ( , ) (0,0)
    ( , )
    0 (, ) (0,0)
    xy if x y
    f x y x xy y
    if x y
          
      
    Solution:
    We need to check following three concepts:
    1. Function is defined at f (0,0) . It is defined because f (0,0)  0
    2.
    ( , ) (0,0)
    lim ( , )
    x y
    f x y

    exists
    Now, we compute the limit ( , ) (0,0) 2 2
    lim
    x y
    xy
     x  xy  y
    ( , ) (0,0) 2 2
    Limit along path 0 ; lim (0) 0
    x y (0) (0)
    y x
     x x
     
     
    ( , ) (0,0) 2 2
    Limit along path 0 ; lim (0) 0
    x y (0) (0)
    x y
     y y
     
     
    2 2
    ( , ) (0,0) 2 2 2 2 2 2
    Limit along path
    lim ( ) 1
    x y ( ) ( ) 3 3
    y x
    x x x x
     x x x x x x x x

      
       
    Since, we are getting different values along different paths, therefore,
    ( , ) (0,0) 2 2
    lim
    x y
    xy
     x  xy  y
    does not exist.
    There is a third condition too to check continuity, and it is that
    ( , ) (0,0)
    (0,0) lim ( , )
    x y
    f f x y

     ,
    but since
    ( , ) (0,0)
    lim ( , )
    x y
    f x y

    does not exist, therefore, there is no need to check third
    condition.
    Conclusion:
    Given function is discontinuous at (0,0)
    Question # 3 Marks 05
    Let f  ln(u2  v2  w2 ), Where u  x  2y, v  2x  y,w  2xy
    Use chain rule to find f
    x


    and express the resultant function in the form of x and y.
    Solution:
    f f u f v f w ................(1)
    x u x v x w x
          
      
          
     
     
     
    2 2 2
    2 2 2
    2 2 2
    2 2 2
    2 2 2
    2 2 2
    ln( ) 2
    2 1
    ln( ) 2
    2 2
    ln( ) 2
    2 2
    f u v w u
    u u u v w
    u x y
    x x
    f u v w v
    v v u v w
    v x y
    x x
    f u v w w
    w w u v w
    w xy y
    x x
                 
     
      
     
                 
     
      
     
                 
     
     
     
    Put all derivatives in Eq. (1), we get
     
    2 2 2 2 2 2 2 2 2
    2 2 2 2 2 2 2 2 2
    2 2 2
    2 (1) 2 (2) 2 (2 )
    2 4 4
    1 2 4 4
    f u v wy
    x u v w u v w u v w
    f u v wy
    x u v w u v w u v w
    f u v wy
    x u v w

      
          

      
          

      
      

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