By Brian H. Chirgwin and Charles Plumpton (Auth.)

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**Extra resources for A Course of Mathematics for Engineers and Scientists. Volume 2**

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1 6 . 1 7 . 1 8 . 19. 1:6 Orthogonal trajectories and geometrical applications The general solution of a first order differential equation involves one arbitrary constant and hence represents a family of curves. r. 32) at right angles is called an orthogonal trajectory of that family. Two families of curves are orthogonal trajectories when every curve of one family cuts every curve of the other family at right angles. Suppose a curve of thefirstfamily passes through the point (x, y) and has slope tan ip (= dy/dx) there.

We can integrate this to obtain an approximation to y at the neighbouring point thus, 52 A COURSE OF MATHEMATICS We now use this expression to give a second approximation : This process can be continued as long as the integrals can be found explicitly. 47) for y(x\ viz. 47), is a solution of eqn. 45) satisfying the initial condition. e. the problem as to whether this iterative process is or is not convergent requires examination. It is shown in books on analysis that this process is convergent if a domain D can be found in the x-y plane which includes the point (x0, yo) and in which/(x, y) is single-valued and continuous, and if for any two points (x, y{) and (x, y2) on the same arbitrary ordinate inside D the condition where K is a constant, is satisfied.

11. Find the orthogonal trajectories of the family of circles 12. A family of plane curves has the property that if the tangent at any point P(x, y) of any one of the curves intersects the *-axis in N, then the distance 07V is equal to ky% where O is the origin and A: is a positive constant. Find the equation of the particular curve of the family that passes through the point (0, 1) and sketch it. 13. Find the differential equation of first order satisfied by the curves y2(2o-x) = xs where a is a parameter.