# CURVATURE by infojustwin

VIEWS: 8 PAGES: 6

• pg 1
1

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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