CURVATURE by infojustwin

VIEWS: 8 PAGES: 6

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                                UNIT-2-CURVATURE

Def. Curvature. Curvature       means bending .That is the arc rate of turning of
       the tangent is called the curvature of the curve.
       Let P & Q be two neighbouring points on the curve C . Let δ      ψ be the
       angle between the tangents at P & Q and     δs is the arc length P . Q
       In moving from P to Q tangent has turned thro’ an angle δ     ψ and
       arc has gone thro’ distance δs .
                                                                            δψ
       Hence curvature of arc P   Q        ψ
                                       is δ . Average curvature of P = δ s .
                                                                       Q

                                             dψ          ψ
                                                         δ          ψ
                                                                    δ
               Curvature at P is equal to        = QLt P    = Lt        .
                                              ds     → δs     δ s→ δ s
                                                                  0


Def. Radius of curvature.
      Reciprocal of curvature at any point P of the curve C is called the radius
                                                                              ds
       of curvature at P and is denoted by ρ . Hence ρ = dψ .

Calculation of radius of curvature:
 (a) Equation of a curve in Cartesian form: y=f(x)
                                                   3
                                                2 2                        dy               d2y
      Radius of curvature ρ = (1 + y1            )       , where     y 1 = dx       &   y2= d x 2   ..(1)
                                  y2
                                                3
           dy                      (1 + x1
                                             2 2
                                              )                             dx       d 2x
      If      =∞   , then                               , where      x1 =      & x2 = 2 .
           dx                   ρ=                                          dy       dy
                                       x2
 (b) Equation of a curve in parametric form:                          x =φ(t ) , y =ψ(t )

                                                                            φ ′ψ ′ − ψ ′φ ′
                                             ψ′
            At points where φ′(t ) ≠ 0 , y1 = φ ′ ,                  y2 =
                                                                                  φ ′3
                                          3
                                      ′ 2)2
                             (φ ′ 2 + ψ ….
              Hence
                          ρ=                                        (2)

                             φ ′ψ ′ − ψ ′φ ′
 (c) Equation of a curve in polar form :                        r = f (θ)
                                 3
                      2       2 2
                    (r +    r1 )                               dr                  d 2r
                                     , where            r1 =         &      r2 =
              ρ=            2                                  dθ                  dθ 2
                                                                                        …(3)
             r 2 + 2r1 − r r2
(d) If the curve is in the implicit form f(x,y)=0 , then ρ is given by
             ρ = ( f x 2 + f y 2 ) 3 / 2 / ( f xx f y 2 − 2 f x f y f xy + f yy f x 2 ) …(4)
                                                         2


Radius of curvature at the origin :
 (a) x-axis is tangent to the curve at the origin :
                                       x2 
               ρ( 0,0) =      Lt              .                                                 (5)
                           x →0 , y →0 2 y 
                                           
  (b) y-axis is tangent to the curve at the origin :
                                       y2 
                ρ( 0,0) =      Lt                                                               (6)
                            x →0, y →0 2 x 
                                           
  (c) When curve passes thro’ the origin but neither axis is tangent to
      the curve at the origin : (Expansion Method is used )
      Let the curve be y= f (x) and it passes thro’ the origin .i.e. f
      (0)=0
                                                          x2                       x2
      Then         y= f ( x) = f (0) + xf ′(0) +               ′
                                                             f ′ (0) + ... = xy1 +    y 2 + ...
                                                          2!                       2!
      Put this value of y in the equation of the curve and then equate
      the coefficients of like powers of x from both sides of the Eq. of
      the curve .This way we can get two equations in y1 , y 2. which
      can be solved for y1 & y 2 . Hence ρ can be obtained.
  (d) Curvature at the origin when polar equation of the curve is given.
                                                     1 dr
           If initial line is the tangent , then ρ = 2 dθ .                                        (7)
  NOTE: Tangents at the origin to an algebraic curve are obtained by
             Equating to zero the lowest degree terms .
Ex. Find the radius of curvature of the following curves:
    (i) x 3 + y 3 =3axy at the po int P(3a / 2 , 3a / 2) .
   (ii) x= at 2 , y =2at , at point t
  (iii) r = a(1+cos θ ) , at any point θ
  (iv)     x + y − a =0 , at the po int P ( a / 4 , a / 4 )

  (v) y + x 4 + a( x 2 + y 2 ) − a 2 y = 0 , at the origin .
         4


  (vi) 2 x 3 + 4 x 2 y + xy 2 +5 y 3 − x 2 − 2 xy + y 2 + 4 x = 0 ,at the point P(0,0)
 (vii). y − x = x 2 + 2 xy + y 2 , at the point P (0,0)
             x2 y2
  (viii)       +   =1               ,               at the point P( a,0)
             a2 b2
                 ay − x 2 
Sol. (i)    y1=  y 2 − ax 
                                                = −1       , y 2 = 32 / 3a ; Hence    ρ = 3a / 8 2
                          x =3a / 2 , y =3a / 2


                                                          ( x′ 2 + y′ 2 )3/ 2 =2
     (ii) x′ = 2at , y ′ = 2a , x′′ = 2a , y ′′ = 0 ; ρ =                               a (1 + t 2 ) 3 / 2

                                                             x′ y ′ − y ′ x′
     (iii) r 1 = − a sin θ , r 2 = − a cos θ ,
                                                        3


                         [a 2 (1 + cos θ ) 2 + a 2 sin 2 θ ]3 / 2
        ρ=
              a 2 (1 + cos θ ) 2 + 2a 2 sin 2 θ + a 2 cos θ (1 + cos θ )
                                                  θ
                                     a 3 ( 4 cos 2 ) 3 / 2
  (2a 2 + 2a 2 cosθ ) 3 / 2                       2         4a       4a r   2
=                           =                              = cos θ =      =   2ar .
                                                  2θ
                                                                 2
     3a 2 (1 + cosθ )                        2
                                        6a cos
                                                             3        3 2a 3
                                                    2
                                                         1                    1
(iv)      f(x,y)=           x+     y − a ; fx =             =1 / a , f y =        =1 / a ,
                                                        2 x                 2 y
                                1        2                   1      2
                  f xx = −       3/ 2
                                      =     , f yy      = − 3/ 2 =      , f xy = 0 .
                              4x        a a                4y      a a
                                  [2 / a ]3 / 2 2 2
                             ρ=                =    a = a/ 2
              Hence                    4         4           .
                                       5/ 2
                                    a
(v)             Tangent at the origin is y=0 i.e. x-axis .Hence
                 x2
ρ=       Lt             .
       x→ , y → 2 y
         0     0

        Dividing the equation of the curve by 2y we get
                y3      x2     x2     y  a2
                   + x.    + a     + −
                              2y 2  2 =0                   , taking limits we get
                2       2y              
                   0+0 ⋅ ρ + a ⋅ ( ρ + 0) − a 2 / 2 = 0        ⇒ ρ = a/2   .
                                                                                                 y2
(vi) Tangent at the origin is x=0 i.e. y-axis. Hence ρ = x→0, y→0
                                                           Lt
                                                                                                 2x
        Dividing the equation of the curve by 2x we get
                      y2        y2 x         y2
        x 2 + 2 xy +     + 5y ⋅   − −y+         + 2 = 0. Taking limits we                  get
                       2        2x 2        2x
           0+0       + 0+ 5.0. ρ − 0 − 0 + ρ + 2 = 0 ⇒ ρ = 2 (in magnitude )

(vii) This curve passes thro’ the origin but no axis is the tangent .
         Tangent is the line y=x .We can do it by computing y1 &
    y 2 also . But we will do it by the method of expansion.
                                        x2
              Since         y = xy1 +      y 2 + ....   ,Putting it in eq. of the curve we get
                                        2!
                                                                                       2
             x2                                  x2                    x2           
      ( xy1 +   y 2 + ...) − x = x 2 + 2 x xy1 +
                                                    y1 + ...  +  xy1 +
                                                                           y 2 + ... 
                                                                                       
             2                                   2                     2            
           Comparing       coefficien tsof var ious powers of x from both sides ,
              we get
                                                                 2
                y1 − 1 = 0 ⇒ y1 = 1; y 2 / 2 = 1 + 2 y1 + y1 = 1 + 2 + 1 = 4 ⇒ y 2 = 8
                                                 2
                                (1 + y1 ) 3 / 2 2 3 / 2    1
                      Hence ρ =                =        =     .
                                     y2           8       2 2
(viii)     Differenti        ating                e
                                      w.r.t. x , w get
                                                4


              2 x 2 yy1          b2 x
              a2
                 + 2 = 0 ⇒ y1 = − 2
                   b             a y
                                                     which becomes                  ∞ at P( a,0)
                                                                                2
                                              (1 + x1 ) 3 / 2
       We will use the other formula i.e. ρ =
                                                   x2
                       dx      a2 y                                        a 2 x − yx1    a
       Now x1 =        dy
                           = − 2 = 0 at ( a,0)                  , x2 = −     2
                                                                               ⋅   2   =− 2
                               b x                                         b     x       b
                         2
                       b
       Hence        ρ=     (in magnitude )
                        a
Radius of curvature for pedal curve p = f(r) is given by
                            dr
                   ρ =r
                            dp
Ex. Find Pedal equation and hence the radius of curvature for the
    curve
                                 r 2 − a2
                         θ=               − cos −1 ( a / r )
                                    a
            dθ      r           a        r 2 − a2
    Sol.       =          −
            dr a r 2 − a 2 r r 2 − a 2
                                       =
                                            ar
                                                                  , since p = r sin φ
                                                                       2
                1       1      1 + cot 2 φ 1    1  1 dr                       tan φ =
                                                                                           r
            ∴       = 2      =            = 2 + 2                        (                   )
                p 2
                     r sin φ
                          2
                                   r 2
                                           r   r  r dθ                                   r1
                        1    1 a2r 2       1
                   =     2
                           + 4 2     2
                                       = 2   2
                                               ⇒ p 2 = r 2 − a 2 is pedal eq.
                       r    r r −a      r −a
                                                          dr    dr
     Differenti ating w.r.t. p we get 2 p = 2r               ⇒r    = p = r 2 − a2
                                                          dp    dp
                Hence     the radius     of curvature          ρ = r 2 −a 2 .
                                                                                             c
Radius of curvature if the equation of the curve is                                  r=
                                                                                           f (θ)
                                                                                       2
                                           u       uu − 2u
           We put r = 1 / u ; then r1 = − 1 & r2 = 2 3 1 , giving
                                             2
                                          u           u
                                         2     2
                                      (u + u1 )
           Radius of curvature ρ = 3                 (1)
                                      u (u + u 2 )
Ex. Find radius of curvature ρ at the point P(r,θ) for the curve
                            l
                       r=
                       1 + e cos θ
                1 + e cos θ
     Sol.    u=               , u1 = −e sin θ / l , u 2 = − e cos θ / l                ,
                     l
                     2  1 + e 2 + 2 cos θ                               (1 + e cos θ ) 3
            u 2 + u1 =                        , u + u2 = 1/ l , u 3 =                     ,
                                l2                                            l3
               (1 + e 2 + 2e cos θ ) 3 / 2         l4           l (1 + e 2 + 2e cos θ ) 3 / 2
            ρ=                             ⋅                  =                               .
                          l3                 (1 + e cos θ ) 3         (1 + e cos θ ) 3
                                         5



Centre Of Curvature & Circle of Curvature .

Def. Let PQ be a normal to the curve C at the point P on C .
     If PQ =ρ , the radius of curvature , then Q is called the
     centre of curvature of the curve C at the point P .
     The coordinates (α, β) of centre of curvature Q at the point
      P( x , y ) is given by
                                     2                    2
                            y1 (1 + y1 )         1 + y1
                   α =x−                 , β =y+
                                 y2                 y2

     Hence the circle of curvature is given by
            ( x − α) 2 + ( y − β ) 2 = ρ 2 .
 Def. Evolute . Locus of centre of curvature is called the evolute .

  Ex. Find the centre of curvature and evolute of parabola y 2 =4ax .
  Sol. General point. on parabola y 2 =4ax is given by ( at 2 ,2at )
                                                  1       1
                                                    (1 + 2 )
                                1
          y1 = 1 / t , y 2 = −     3 ;
                                       α = at 2 − t      t = at 2 + 2a(1 + t 2 )
                               2at                      1
                                                    −
                                                      2at 3
                                             1
                                        (1 + 2 )
            = 2a + 3at 2 & β = 2at +        t = 2at − 2at (1 + t 2 ) = −2at 3
                                            1
                                         −
                                           2at 3
        Hence Centre of curvature is (2a +3at 2 , − 2at 3 ) .
        Eliminating t from α = 2a + 3at 2 , β = −2at 3 , we get
                     3          2
           α − 2a   β 
                          ⇒ 4(α − 2a ) = 27 aβ
                                       3        2
                   =
           3a   − 2a 
        Hence locus of (α, β) is the evolute 4( x − 2a) 3 = 27 ay 2             .
Ex. Show that the centre of curvature at any point θ of the ellipse

      x2 y2           (a 2 − b 2 ) cos 3 θ (b 2 − a 2 ) sin 3 θ 
        + 2 = 1 , is 
                                          ,                     
                                                                     and its
      a2 b                    a                    b            
    evolute is given by (ax ) 2 / 3 + (by ) 2 / 3 = (a 2 − b 2 ) 2 / 3 . Also find
    circle of curvature at point θ = π / 2.

Sol. General point on the ellipse is P (a cos θ               , b sin θ) .
                                                 6

                b cos θ                   b
          y1 = −         , y2 = − 2                 ,
                a sin θ              a sin 3 θ
                        (a 2 sin 2 θ + b 2 cos 2 θ ) cos θ (a 2 − b 2 ) cos 3 θ
         α = a cos θ −                                    =
                                        a                          a
                        (a sin θ + b cos θ ) sin θ (b − a 2 ) sin 3 θ
                           2     2       2      2            2
         β = b sin θ −                                    =
                                        b                          b
      Eliminating θ from above two equations , we get
                2/3               2/3
      αa       βb 
                          = 1 ⇒ ( aα ) + (bβ ) = (a − b )
                                       2/3         2/3     2     2 2/3
      2  2
              + 2
     a −b      a − b2 
     Hence locus of (α, β) is (ax ) 2 / 3 + (by ) 2 / 3 = (a 2 − b 2 ) 2 / 3
                                           (a 2 sin 2 θ + b 2 cos 2 θ ) 3 / 2
    Radius of curvature at θ is ρ =
                                                         ab
    At   θ =π / 2 , ρ = a 2 / b   ,     α = 0 , β = (b 2 − a 2 ) / b        .
                                                                                              2
                                                                                (a 2 − b 2 )   a4
    Hence circle of curvature atθ = π / 2 is x 2 +  y +
                                                                                              = 2
                                                                                              
                                                                                     b         b
Ex. Show that the radius of curvature for the curve                                      r 2 cos 2θ = a 2   is
    r3 / a2 .
                                                                                          sin 2θ
   Sol. Putting r = 1/u ,we get u =                    cos 2θ / a       ;       u1 = −
                                                                                         a cos 2θ
             1 + cos 2 2θ                                  1                 1
      u2 = −                    , (u 2 + u12 ) 3 / 2 = ( 2        )3/ 2 = 3
             a(cos 2θ ) 3/ 2
                                                        a cos 2θ         a cos 3 / 2 2θ
      u + u2 =
                     1                                1        1          r       r3
                                 .Hence    ρ= 2 3 =                   =      = 2.
               a(cos 2θ ) 3 / 2                     a u    u cos 2θ cos 2θ a

								
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