Spring 2008_MTH301_5_SOL by asimsaeedch

VIEWS: 2 PAGES: 9

									                                               Solution
                                       Assignment # 5
Note: Text in blue color is the information related to the given question and text in black
color gives the solution of the question.
   Q 1.     Let C be the curve with parametric equations
        x = t , y = t2 , z = t3            ; t≥0
       Find parametric equations for the tangent line to C at the point corresponding to
        t = 2.
       Solution:
       DEFINITION:
                                           →                       →
       If a vector-valued function r (t ) has a tangent vector r ′(t0 ) at any point t0 on its
                                               →
       graph, then the line parallel to r ′(t0 ) and passing through the tip of the radius
                 →                                                  →      →
       vector r (t0 ) is called the tangent line of the graph of r (t ) at r (t0 ) . Vector

       equation of the tangent line at t0 is
        →   →            →
        r = r (t0 ) + t r ′(t0 ) − − − − − − − − − − ( A)


       We will use the above definition for the solution of this question.


                                 →
       The vector equation r (t ) of the given curve C is
        →            ^       ^   ^     ^       ^      ^
        r (t ) = x i + y j + z k = t i + t 2 j + t 3 k       for t ≥ 0


       The tangent vector to C at the point corresponding to t is
        →        ^       ^       ^
        r ′(t ) = i + 2t j + 3t 2 k


       In particular, radius vector of the point corresponding to t = 2 is
        →            ^       ^   ^
        r (2) = 2 i + 4 j + 8 k
   And a tangent vector at the point whose radius vector is as above is
   →            ^       ^            ^
   r ′(2) = i + 4 j + 12 k


   For vector equation of tangent line at the point where t = 2, put values in Equation
   (A)
   →        ^       ^            ^       →           →
   r = x i + y j + z k = r (2) + t r ′(2)
     ^     ^     ^
                     ⎛ ^       ^     ^
                                       ⎞     ⎛^      ^      ^
                                                              ⎞
   x i + y j + z k = ⎜ 2 i + 4 j + 8 k ⎟ + t ⎜ i + 4 j + 12 k ⎟
                     ⎝                 ⎠     ⎝                ⎠
                                 ^           ^   ^           ^       ^           ^
                            = 2 i + 4 j + 8 k + t i + 4t j + 12t k
                                             ^                   ^               ^
                            = ( 2 + t ) i + ( 4 + 4t ) j + ( 8 + 12t ) k


   Equating the corresponding components gives the parametric equation of tangent
   line at the point where t = 2


   x = 2 + t , y = 4 + 4t , z = 8 + 12t


                                                                             →       ^   ^
Q 2.     Find parametric equations for the curve r (t ) = (3t − 2) i + (4t + 3) j using arc
   length s as a parameter. Use the point on the curve where t = 0 as the reference
   point.
   Solution:
                →            ^           ^               ^               ^
   Since r (t ) = x i + y j = (3t − 2) i + (4t + 3) j
   So in parametric form we can write
   x = 3t − 2 ⎫
              ⎬ − − − − − − − − − −(1)
   y = 4t + 3⎭
   Now rewrite these parametric equations with u in place of t
   x = 3u − 2
   y = 4u + 3
dx
   =3
du
dy
   =4
du
Theorem:
Let C be a curve in 2D-space given parametrically by
x = x(t) , y = y (t)
              /              /
where x (t) and y (t) are continuous functions. If an arc-length parameter s is
introduced with its reference point at (x(t0), y (t0)), then the parameters s and t are
related by
                        2            2
              ⎛ dx ⎞ ⎛ dy ⎞
         t
s=∫           ⎜ ⎟ + ⎜ ⎟ du − − − − − − − − − − − − − − − ( A)
     t0       ⎝ du ⎠ ⎝ du ⎠


Use this theorem for further procedure.


Since it is stated in the question that use the point on the curve where t = 0 as the
reference point, so here t0 = 0 .
Put values in equation (A)
                        2             2
              ⎛ dx ⎞ ⎛ dy ⎞
         t
s=∫           ⎜ ⎟ + ⎜ ⎟ du
     t0       ⎝ du ⎠ ⎝ du ⎠
         t
s =∫          ( 3) + ( 4 )
                    2        2
                                 du
         0
     t
 =∫           9 + 16 du
     0
         t
 =   ∫
     0
             5 du

 =5u0
              t


 = 5 (t − 0)
 = 5t
                                 s
So, s = 5t ⇒ t =
                                 5
Substituting the value of t in the parametric equations (1).
        ⎛s⎞
   x = 3⎜ ⎟ − 2
        ⎝5⎠
         ⎛s⎞
   y = 4⎜ ⎟ + 3
         ⎝5⎠
   Or
      3
   x = s−2
      5
       4
   y = s+3
       5


Q 3.          Determine whether the following differential is exact? If so, find z by
   integration of exact differentials.
       dz = ( y sec 2 x + sec x tan x) dx + (tan x + 2 y ) dy
   Solution:
   As we know that if z = f(x, y) then
            ∂z      ∂z
   dz =        dx +    dy
            ∂x      ∂y
   Let
          ∂z    ⎫
       P=       ⎪
          ∂x    ⎪
          ∂z    ⎬ − − − − − − − − − − − −( A)
       Q=       ⎪
          ∂y    ⎪
                ⎭
   Then above differential can be written as
   dz = P dx + Q dy


   Test for exact differential:
   Any differential
   dz = P dx + Q dy − − − − − − − −( B )
   where P and Q are functions of x and y, is an exact differential if
   ∂P ∂Q
      =
   ∂y   ∂x
Equate the given differential
dz = ( y sec2 x + sec x tan x) dx + (tan x + 2 y ) dy
with equation (B)
we get


P = y sec 2 x + sec x tan x
and
Q = tan x + 2 y


∂P
   = sec2 x
∂y
∂Q
   = sec2 x
∂x
         ∂P ∂Q
Since       =    , so the given differential is exact.
         ∂y   ∂x
Integration of Exact differential to determine z:
For an exact differential dz = P dx + Q dy where

   ∂z      ⎫
P=         ⎪
   ∂x      ⎪
   ∂z      ⎬ − − − − − − − − − − − − As in equation ( A)
Q=         ⎪
   ∂y      ⎪
           ⎭


z = ∫ P dx + f ( y ) − − − − − − − − − − − − − (i )
where f ( y ) is an arbitrary function of y only and is same as cons tan t of int egration
in a normal int egration

and also


z = ∫ Q dy + f ( x) − − − − − − − − − − − − − − − − − −(ii )
where f ( x) is an arbitrary function of x only and is same as cons tan t of int egration
 in a normal int egration
   Since expressions (i) and (ii) has same left hand sides so their right hand sides are
   also equal.
   Compare both expressions and find the value of f(y) and f(x), and show z.


   Since it is proved that the given differential is exact, so we will do the above
   stated procedure to determine z.


   P = y sec 2 x + sec x tan x
   and
   Q = tan x + 2 y


   z = ∫ P dx = ∫           ( y sec   2
                                          x + sec x tan x ) dx
                = y tan x + sec x + f ( y ) − − − − − − − (1)


   z = ∫ Q dy = ∫           ( tan x + 2 y ) dy
                      = y tan x + y 2 + f ( x) − − − − − − − − − − − −(2)


   Compare right hand side of both the expressions (1) and (2) we get
       f ( y) = y 2
   and
       f ( x) = sec x
   Thus
   z = y tan x + sec x + y 2


Q 4.      Evaluate      ∫
                        C
                            xy dx + x 2 dy if

          a. C consists of line segments from (2, 1) to (4, 1) and from ( 4, 1) to (4, 5) .
          b. C is the line segment from ( 2, 1) to (4, 5) .
   Solution:
   (a)
Let us sub-divide C into two parts c1 (line segment joining the points
(2, 1) to (4, 1) ) and c2 (line segment joining the points ( 2, 1) to (4, 5) ) , as shown
in the figure.


Line integral along C may be expressed as a sum of two line integrals, the first
along c1 and the second along c2. That is,

∫
C
     xy dx + x 2 dy = ∫ xy dx + x 2 dy + ∫ xy dx + x 2 dy − − − − − − − ( A)
                     c1                               c2




c1 is parallel to line y = 1. So dy = 0. To find the limits of integral we can see that
along c1 , x varies from 2 to 4. Put values in given line integral
                      4

∫
c1
     xy dx + x 2 dy = ∫ x (1) dx + x 2 (0)
                      2
                      4
                   = ∫ x dx
                      2
                                  4
                     x2
                   =                      = 8 − 2 = 6 − − − − − − − − − − − (1)
                     2        2




c2 is parallel to line x = 4. So dx = 0. To find the limits of integral we can see that
along c2 , y varies from 1 to 5. Put values in given line integral
                      5

∫
c2
     xy dx + x 2 dy = ∫ xy (0) + (4) 2 dy
                      1
                          5
                   = 16 ∫ dy
                          1

                   = 16 y 1
                                      5


                   = 16 (5 − 1)
                   = 16(4) = 64 − − − − − − − − − (2)
Put values from equations (1) and (2) in (A)

∫
C
     xy dx + x 2 dy = ∫ xy dx + x 2 dy + ∫ xy dx + x 2 dy − − − − − − − ( A)
                     c1                               c2

                   = 6 + 64 = 70
Thus line integral along C is 70.
(b)
Here, since C is the line segment
from ( 2, 1) to (4, 5) . So first we
will find the equation of C.
Let ( x1 , y1 ) = (2, 1) and

( x2 , y2 ) = (4, 5) then

               y2 − y1
Slope = m =
               x2 − x1
      5 −1 4
m=        = =2
      4−2 2
Use Point-Slope form of the line, to find equation of C.
y − y1 = m( x − x1 )


y − 1 = 2( x − 2)
y = 2x − 4 +1
  = 2x − 3
Now
dy = 2 dx
As we are converting given line integral to one variable x, so for limits of integral
we see how x varies along C, which is from 2 to 4.
Put values in given line integral
                     4

∫   xy dx + x dy =
            2
                     ∫   x(2 x − 3) dx + x 2 ( 2 dx )
C                    2
                     4
                 =∫      ( 2x   2
                                    − 3x + 2 x 2 ) dx
                     2
                     4
                 =∫      ( 4x   2
                                    − 3 x ) dx
                     2
                                           4
                   x3  x2
                = 4 −3
                   3   2               2

                  ⎛ 256 48 ⎞ ⎛ 32     ⎞
                =⎜      − ⎟ − ⎜ − 6⎟
                  ⎝ 3     2 ⎠ ⎝ 3     ⎠
                  512 − 144 − 64 + 36
                =
                           6
                  340 170
                =      =
                   6     3


Attention please: In the above question we get different values of the integral
while considering two separate paths joining the same two end points. Can
anyone of you give me the reason? Send me through mail.

								
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