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Orthogonal-Trajectories-ODE.Pdf Orthogonal Trajectaries 141 Chapter 5 Orthogonal Trajectories ORTHOGONAL TRAJECTORY OF A GIVEN CURVE (CARTESIAN FORM) A curve of family or curves (x , y , c ) 0 which cuts every member of a given family of curves f( x , y , c ) 0 according to a fixed rule is called a trajectory of the family of curves. Orthogonal trajectory Figure-1 If we consider only the trajectories cutting each member of family of curves f( x , y , c ) 0 at a constant angle, then the curve which cuts every member of a given family of curves at right angle, is called an orthogonal trajectory of the family. In order to find out the orthogonal trajectories the following steps are taken. Step-1: Let f( x , y , c ) 0 be the equation where c is an arbitrary parameter.. Step-2: Differentiate the given equation w.r.t. x and eliminate c. dx dy Step-3: Substitute for in the equation obtained in step 2. dy dx Step-4: Solve the differential equation obtained from step 3. Y x2 + y 2 = r 2 y = mx X Orthogonal trajectory Figure-2 Example -1 Find the orthogonal trajectory of family of straight lines passing through the origin. Soln. Family of straight lines passing through the origin is y = mx. ...(i) where ‘m’ is an arbitrary constant. dy Differentiating w.r.t. x, we get m ...(ii) dx South Delhi : 28-A/11, Jia Sarai, Near-IIT Hauz Khas, New Delhi-16, Ph : 011-26851008, 26861009 North Delhi : 33-35, Mall Road, G.T.B. Nagar (Opp. Metro Gate No. 3), Delhi-09, Ph: 011-27653355, 27654455 142 Orthogonal Trajectaries dy Eliminating ‘m’ from (i) and (ii), we get y x dx dy dx dx Replacing by, we get y x x dx y dy 0 dx dy dy x2 y 2 Integrating both sides, we get c 2 2 x2 y 2 2 c Which is the required orthogonal trajectory. Example -2 Find the orthogonal trajectories of the curve y = ax2. Soln. Let y = ax2 ...(i) dy Differentiating equation (i), we get 2ax ...(ii) dx dy2 y Eliminating a from equations (i) and (ii), we get x ...(iii) dx x2 dx dy dx2 y Putting in equation (iii) in the place of , we get dy dx dy x x22 y 2 xdx 2 y dy 2 2 2y2 x 2 0 This is the family of required orthogonal trajectories. Example -3 Find the orthogonal trajectories of the hyperbola xy = c. Soln. The equation of the given family of curves is xy = c ...(i) dy Differentiating (i) w.r.t. x, we get x y 0 ...(ii) dx dx dy dx Substituting for in (ii), we get x y 0 ...(iii) dy dx dy This is the differential equation of the orthogonal trajectories of given family of hyperbolas. (iii) can be rewritten as xdx = ydy, which on integration given x2 y 2 c . This is the required family of orthogonal trajectories. Example -4 x2 y 2 Find the orthogonal trajectories of the curves 1, being the parameter of the family.. a2 b 2 x2 y 2 Soln. Let 1 ...(i) a2 b 2 x y dy Differentiating (i) w.r.t. x, we have 0 ...(ii) a2 b 2 dx South Delhi : 28-A/11, Jia Sarai, Near-IIT Hauz Khas, New Delhi-16, Ph : 011-26851008, 26861009 North Delhi : 33-35, Mall Road, G.T.B. Nagar (Opp. Metro Gate No. 3), Delhi-09, Ph: 011-27653355, 27654455 Orthogonal Trajectaries 143 From (i) and (ii) we have to eliminate . 2 2 dy b x a y Now (ii) gives, dx dy x y dx 2 2 dy ()a2 b 2 x ()a b y a2 ; b 2 dx dy dy x y x y dx dx dx dy 2 2 Substituting these values in (i), we get x y x y a b ...(iii) dy dx is the differential equation of the given family of curves. dy dx dy dx 2 2 Replacing to in (iii), we obtain x y x y a b ...(iv) dx dy dx dy which is the same as (iii). Thus we see that the family (i) is self orthogonal. i.e., every member of the family (i) cuts every other member of the same family orthogonally. Example -5 Find the orthogonal trajectories of the circles x2 y 2 ay 0 where ‘a’ is a parameter.. dy dy Soln. Here, x2 y 2 ay 0. Differentiating w.r.t. x, we get 2x 2 y a 0 dx dx 2 2 2 2 dy x y dy 2 2 x y 2x 2 y 0 x y ay 0 a dx y dx y y2 x 2 dy 2x 0. This is the differential equation of the given circles. y dx y2 x 2 dx The equation of orthogonal trajectories is 2x 0 y dy dx dy putting in place of dy dx 2xydy ( x2 y 2 ) dx 0 It is a homogeneous equation. dy d d 2 2 2 Put y x , then x 2x . x . x x x 0 dx dx dx d 2 2 d dx 2 or 2 x 1 0 1 2 x 0 2 d 0; dx dx x 1 Integrating, we get n x n(1 2 ) n c 2 2 y 2 2 x(1 ) c x1 2 c , i.e., x y cx x Which is the required family of orthogonal trajectories. South Delhi : 28-A/11, Jia Sarai, Near-IIT Hauz Khas, New Delhi-16, Ph : 011-26851008, 26861009 North Delhi : 33-35, Mall Road, G.T.B. Nagar (Opp. Metro Gate No. 3), Delhi-09, Ph: 011-27653355, 27654455 144 Orthogonal Trajectaries Previous Year Solved Problems Example -6 The trajectories of the system of differential equations dx/ dt y and dy/ dt x are [D.U. 2014] (a) ellipses (b) hyperbolas (c) circles (d) spirals dx dy Soln. y and x dt dt dy dy x dy x y2 x 2 c 2 dt ydy xdx 0 dxdx y dx y 2 2 2 dt x2 y 2 c 2 which is equation of circle Option (c) is Correct Example -7 Let be the curve which passes through (0, 1) and intersects each curve of the family y c x2 othogonally.. Then also passes through the point (a) 2, 0 (b) 0, 2 (c) (1, 1) (d) (–1, 1) [GATE-2016] y dy Soln. y cx2 c c.2 x x2 dx We need to remove arbitary constnat c dy y2 y dy2 y 2x dx x2 x dx x For othogonal trajectories, we replace dy dx dx2 y by 2ydy xdx 0 dx dy dy x 2y2 x 2 c 2 2y2 x 2 c 2 which is required equation of y. 2 2 2 Since, passes through (0, 1) 2×12 + 02 = c2 c2 = 2 So, 2y2 x 2 2 Now, x2, y 0 satisfies above equation of y.. Option (a) is Correct x0, y 2 doesnot satisfy the equation of y.. x1, y 1 doesnot satisfy the equation of y.. x 1, y 1 doesnot satisfy the equation of y.. Option (b), (c), (d) are not Correct. South Delhi : 28-A/11, Jia Sarai, Near-IIT Hauz Khas, New Delhi-16, Ph : 011-26851008, 26861009 North Delhi : 33-35, Mall Road, G.T.B. Nagar (Opp. Metro Gate No. 3), Delhi-09, Ph: 011-27653355, 27654455 Orthogonal Trajectaries 145 Example -8 The orthogonal trajectories of the one parameter family of curves y2 4 k ( k x ) where k is an arbitrary constant is [ISM-2015] (a) x2 2 c ( c y ) (b) y4 c ( c x ) (c) x2 4 c ( c y ) (d) y2 4 c ( c x ) Soln. y2 4 k k x ...(i) dy y dy 2y 4 k k dx 2 dx To remove arbitary Constant k Put k in equation (i) 2 y dy y dy 2 dy y dy y4 x y2 y x 2dx 2 dx dx2 dx 2 dy y dy dy dy y2 x y y 2 xc (iii) dx2 dx dx dx To get orthogonal trajectary, replace dy dx by dx dy 2 2 dx dx dx dx y y 2 x y y 2 x dy dy dy dy 2 2 dx dx dy dx dy dy y2 x y y2 x y y 2 x y ...(iv) dy dy dx dy dx dx equation (iv) is same as equation (iii) Given equation is self orthogonal samily of carves. The orthogonal tracjector will be y2 4 c c x which c is orthogonal constant Option (d) is Correct Example -9 6. The orthogonal trajectory of the hyperbola x2 y 2 c is [SAU-2015] (a) xy c (b) xy2 c (c) x2 y c (d) None of the above Soln. x2 y 2 c 2 dy 2x 2 y 0 dx dy dy To get orthogonal trajectory, we replace by dx dx South Delhi : 28-A/11, Jia Sarai, Near-IIT Hauz Khas, New Delhi-16, Ph : 011-26851008, 26861009 North Delhi : 33-35, Mall Road, G.T.B. Nagar (Opp. Metro Gate No. 3), Delhi-09, Ph: 011-27653355, 27654455 146 Orthogonal Trajectaries dx dx dx 2x 2 y 0 2x 2 y 0 2y 2 x dy dy dy dx dy dx dy 0 logx log y log c logxy log c xy c x y x y Option (a) is Correct Example -10 Let c be an arbitrary non-zero constant. Then the orthogonal family of curves to the family y(1 cx ) 1 cx is [H.C.U.-2015] (a) 3y y3 3 x 2 constant (b) 3y y3 3 x 2 constant (c) 3y y3 3 x 2 constant (d) 3y y3 3 x 2 constant Soln.
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