VIEWS: 8 PAGES: 6 CATEGORY: College POSTED ON: 10/27/2012
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