Unit I - Matrices PDF

 SCHOOL OF SCIENCE AND HUMANITIES DEPARTMENT OF MATHEMATICS UNIT – 2 – GEOMETRICAL APPLICATIONS OF DIFFERENTIAL CALCULUS – SMTA1101  UNIT – II GEOMET...

 SCHOOL OF SCIENCE AND HUMANITIES DEPARTMENT OF MATHEMATICS UNIT – 2 – GEOMETRICAL APPLICATIONS OF DIFFERENTIAL CALCULUS – SMTA1101  UNIT – II GEOMETRICAL APPLICATIONS OF DIFFERENTIAL CALCULUS Curvature: At each point on a curve, with equation y=f(x), the tangent line turns at a certain rate. A measure of this rate of turning is the curvature Radius of curvature in Cartesian form: If the curve is given in Cartesian coordinates as y(x), then the radius of curvature is where y’= ,. Radius of curvature in Parametric form: If the curve is given parametrically by functions x(t) and y(t), then the radius of curvature is , Examples: 1. Find the radius of the curvature at the point on the curve. Solution: Differentiating w. r. t x ,we get. At ,  At , = 4. 2. Show that the radius of the curvature at any point of the curve is. Solution: Differentiating y w. r. t x we get. 3. Find the radius of the curvature of the curve y = x2(x-3) at the points where the tangent is parallel to the x – axis. Solution: y = x2(x-3) Differentiating y w. r. t x we get The points at which the tangent parallel to the x – axis can be found by equating y’ to zero. i.e., At x = 0,y” = -6. At x = 2, y” = 6. Therefore at x = 0 and x = 2,. 4. Prove that the radius of the curvature of the curve at any point of the cycloid is.  Solution: We have. Therefore , Now Also = Hence Centre and Circle of curvature: Let the equation of the curve be y = f(x). let P be the given point (x,y) on this curve and Q the point (x+Δx,y+Δy) in the neighborhood of P. let N be the point of intersection of the normals at P and Q. As Q→P, suppose N→C. Then C is the centre of curvature of P. The circle whose centre C and radius is called the circle of curvature. The co-ordinates of the centre of curvature is denoted as where ,. Equation of the circle of curvature: If be the coordinates of the centre of curvature and be the radius of curvature at any point (x,y) on a curve, then the equation of the circle of curvature at that point is Examples: 1. Find the centre of curvature of the curve. Solution:. Therefore the required centre of curvature is. 2. Find the centre of curvature of y = x2 at Solution: y’ = 2x, y” = 2. At , y’ = 1, y” = 2. Therefore. Therefore the required centre of curvature is 3. Find the centre of curvature of the curve xy = a2 at (a,a). Solution:. At (a,a) y’ = -1, y” = Therefore The required centre of curvature is (2a, 2a). 4. Find the circle of curvature of the curve at the point Solution: at is -1.. The circle of curvature is 5. Find the circle of curvature at the point (2,3) on. Solution: , , , The circle of curvature is Evolute and Involute Evolute: Evolute of the curve is defined as the locus of the centre of curvature for that curve. Involute : If C’ is the evolute of the curve C then C is called the involute of the curve C’. Procedure to find the evolute: Let the given curve be f(x,y,a,b) = 0. (1) Find y’ and y” at the point P. Find the centre of curvature. Using ,. (2) Eliminate x, y from (1), (2) we get f(. (3) Equation (3) is the required evolute. Examples: 1. Show that the evolute of the cycloid is another cycloid given by Solution: ,. The locus of is. 2. Prove that the evolute of the curve is a circle Solution:. , Eliminating , we get The evolute of the given curve is. ENVELOPE A curve which touches each member of a given family of curves is called envelope of that family. Procedure to find envelope for the given family of curves: Case 1: Envelope of one parameter family of curves Let us consider y = f(x,α) to be the given family of curves with ‘α’ as the parameter. Step 1: Differentiate w.r.t to the parameter α partially, and find the value of the parameter Step 2: By Substituting the value of parameter α in the given family of curves, we get the required envelope. Special Case: If the given equation of curve is quadratic in terms of parameter,i.e. Aα2+Bα+c=0, then envelope is given by discriminant = 0 i.e. B2- 4AC=0 Case 2: Envelope of two parameter family of curves. Let us consider y = f(x,α, β) to be the given family of curves, and a relation connecting the two parameters α and β, g(α, β) = 0 Step 1: Consider α as independent variable and β depends α. Differentiate y = f(x,α, β) and g(α, β) = 0, w.r. to the parameter α partially. Step 2: Eliminating the parameters α, β from the equations resulting from step 1 and g(α, β) = 0, we get the required envelope. Problems on envelope of one parameter family of curves : 1. Find the envelope of y  mx  am p where m is the parameter and a, p are constants Solution : Differentiate y  mx  am p (1) with respect to the parameter m, we get, 0  x  pam p 1 1  x  m    p  1 (2)  pa   Using (2) eliminate m from (1) 1 p  x  x y    p  1 x  a  p  1  pa   pa   p  x  p 1    x   y p  1    x  a p  1   pa   pa  i.e. ap p y p  1   x p p p  1  ( x) p which is the required equation of envelope of (1) 2. Determine the envelope of x sin   y cos  a , where θ being the parameter. Solution : Differentiate , x sin   y cos   a (1) with respect to θ , we get, x cos  y sin   a (2) As θ cannot be eliminated between (1) and (2) ,we solve (1) and (2) for x and y in terms of θ. For this, multiply (2) by sinθ and (1) by cosθ and then subtracting, we get, y  a(sin    cos ). Using similar simplification, we get, x  a( sin   cos ). 3. (Leibnitz’s problem) Calculate the envelope of family of circles whose centres lie on the x-axis and radii are proportional to the abscissa of the centre. Solution : Let (a,0) be the centre of any one of the member of family of curves with a as the parameter. Then the equation of family of circles with centres on x-axis and radius proportional to the abscissa of the centre is  ( x  a) 2  y 2  ka 2 (1) where k is the proportionality constant. Differentiating (1) with respect to a, we get,  2( x  a)  2ka x i.e. a . 1 k  x  2 k From (1) , x    y2  x2  1 k  (1  k ) 2  2  k  x 2  1  k 2 y 2  0, k 1 i.e.  k   4. Find the envelope of x sec 2   y cos ec 2  a , where θ is the parameter.  2   2  Solution : The given equation is rewritten as , x1  tan    y 1  cot    a i.e. x tan 4   ( x  y  a) tan 2   y  0 , which is a quadratic equation in t  tan 2 . Therefore the required envelope is given by the 2 discriminant equation : B -4AC = 0 i.e. ( x  y  a) 2  4 xy  0 i.e. x 2  y 2  2 xy  2ax  2ay  a 2  0. Envelope of Two parameter family of curves : 1. Find the envelope of family of straight lines ax+by=1, where a and b are parametersconnected by the relation ab = 1 Solution : ax  by  1 (1)  ab  1 (2) Differentiating (1) with respect to a ( considering ‘a’ as independent variable and ‘b’ depends on a ). db x y0 da db  x  (3) i.e. da y Differentiating (2) with respect to a db ba 0 da db  b i.e. da  a (4) From (3) and (4), we have x b  y a ax by ax  by 1 i.e.    1 1 2 2 1 and 1  a b (5) 2x 2y Using (5) in (2), we get the envelope as 4xy = 1 x y 2. Find the envelope of family of straight lines a  b  1 , where a and b are parameters connected by the relation a  b  1 Solution :  x y  1 (1) a b a b 1 (2) Differentiating (1) with respect to a x y db  0 32 3 2 da  2a  2b 32 db  x b i.e.  (3) da y a3 2 Differentiating (2) with respect to a 1 1 db  0 2 a 2 b da db  b i.e.  (4) da a From (3) and (4), we have xb 1 ya x y x y  i.e. a  b  a b 1 a b a b 1  a x and b y (5) 14 14 Using (5) in (2), we get the envelope as x  y 1 x y 3. Find the envelope of family of straight lines a  b  1 , where a and b are parameters 2 3 5 connected by the relation a b = c  x y  1 (1) a b 2 3 5 ab =c (2) Differentiating (1) with respect to a, x y db  0 a 2 b 2 da db  b 2 x i.e.  (3) da a2 y Differentiating (2) with respect to a db 2ab3  3a 2b 2 0 da db  2b i.e.  (4) da 3a From (3) and (4), we have 3x 2 y  a b x y x y  i.e. a  b  a b 1 3 2 5 5 5x 5y  a b 3 and 2 (5) 72 5 Using (5) in (2), we get the envelope as x2 y3  c 3125 x2 y2 4. Find the envelope of the family of circles whose centres lie on the ellipse a 2  b 2  1 and which pass through its centre.  Solution: Let (α,β) be the centre of arbitrary member of family of circles which lie on the x2 y2 ellipse a 2  b 2  1 , whose centre is (0,0). Therefore, equation of the circles passing through origin and having centreat (α,β) is x 2  y 2  2x  2 y  0 (1) with 2 2  2 1 a2 b (2) Differentiating (1) with respect to α ( ‘α’ as independent variable and ‘β’ depends on α ), d x y0 d d  x i.e. d  y (3) Differentiating (2) with respect to α 2 2 d  0 a 2 b 2 d d  b 2 i.e.  (4) d a 2 From (3) and (4), we have x b 2  y a2 x y x   y k    2 2 2 2 1 i.e.      , where k = αx+βy a2 b2 a2 b2  a2x b2 y   and  (5) k k 2 2 From (1), we have , x  y  2k (6)  x2  y2  2 Using (5) and (6) in (2), we get the envelope as    4 a 2 x 2  b 2 y 2      x2 y2 5. Determine the equation of the envelope of family of ellipses   1 where the a2 b2 a2 b2 parameters a and b are connected by the relation l 2  m 2  1 , l and m are non-zero constants. x2 y2  1 (1) a2 b2 a2 b2  1 (2) l2 m2 Differentiating (1) with respect to a,  2 x 2 2 y 2 db  0 a3 b 3 da db  b3 x 2 i.e.  (3) da a3 y 2 Differentiating (2) with respect to a 2a 2b db  0 l 2 m 2 da db  m 2 a i.e.  (4) da l 2b From (3) and (4), we have  b4 x 2 m2  a4 y2 l2 x2 y2 x2 y2  a2  b2  a2 b2  1 i.e. a2 b2 a2 b2 1  l2 m2 l2 m2  a 4  l 2 x 2 and b4  m2 y 2 i.e. a 2  lx and b 2  my (5) x y Using (5) in (2), we get the envelope as  1 l m Problems on Evolute as envelope of its normals : x2 y2 1. Determine the evolute of hyperbola a 2 b 2  1 by considering it as an envelope of its normal  Solution : Let P (a cosht, b sinht) be any point on the given hyperbola. Then dy dy dt b cosh t b    coth t dx dx a sinh t a dt Equation of normal line to the hyperbola is a ( y  b sinh t )  x  a cosh t  (1) b cosh t by ax    a2  b2 sinh t cosh t (2) Differentiating (2) partially with respect to t, we have,   by ax cosh t  sinh t  0 (sinh t ) 2 (cosh t ) 2  by  13  tanh t     ax  13 13  sinh t  ∓  by   and cosh t   ax  (3)  h   h  Where h  (ax) 2 3  (by) 2 3 Using (3) in (2) , we get, by ax h h  a2  b2 13 13  (by ) (ax) i.e. (ax) 23  (by ) 2 3 (ax) 23  (by ) 2 3  12  a 2  b2 2 3 i.e. (ax) 2 3  (by ) 2 3   a 2  b 2    2. By considering the evolute of a curve as the envelope of its normal, find the evolute of x  cos    sin  , y  sin    cos  Solution : dy dy  sin   d   tan  dx dx  cos . d Equation of normal line to the hyperbola is 1 ( y  (sin    cos  ))  x  cos    sin   tan    y sin   sin 2    sin  cos    x cos   cos 2    sin  cos  i.e. y sin   x cos   1 (1) Differentiating (1) with respect to the parameter θ, we have y cos   x sin   0 (2) Multiplying (1) by cosθ and (2) by sinθ and then subtracting, we have, x  cos  (3) Similarly we get, y  sin  (4) Eliminating θ between (3) and (4) we get the required evolute as x2  y 2 1  SCHOOL OF SCIENCE AND HUMANITIES DEPARTMENT OF MATHEMATICS UNIT – 3 - FUNCTIONS OF SEVERAL VARIABLES – SMTA1101  SCHOOL OF SCIENCE AND HUMANITIES DEPARTMENT OF MATHEMATICS UNIT – 4 - INTEGRAL CALCULUS I– SMTA1101  Definite integrals Properties of definite integrals and problems Beta and Gamma integrals Relation between them Properties of Beta and Gamma integrals with proofs Evaluation of definite integrals in terms of Beta and Gamma function. q  SCHOOL OF SCIENCE AND HUMANITIES DEPARTMENT OF MATHEMATICS UNIT – 5 - INTEGRAL CALCULUS II – SMTA1101