i) The orthogonal trajectories to the family of curves y=Ax^3 are the family of curves x^2+3y^2=C, where C is a constant. ii) The solution finds the derivative of y=Ax^3 to get the slope dy/dx=3Ax^2. It then sets the product of the slopes of the original and orthogonal families equal to -1 and solves for the slope dx/dy of the orthogonal trajectories. iii) Integrating the resulting separable differential equation gives the equation x^2+3y^2=C for the orthogonal family of curves.

1. Q1
i) Find the othrogonal trajections for the following family of curves.
ii) Sketch on the same set of axes at least three members of each of the two families.
y= A x3
Solution
y= A x3
dy/dx = A 3x^2
now put dy/dx = -dx/dy
since othgonal trajecotries are nothign but which are orthogonal
becuase dy/dx * -dx/dy = -1
multiplication slopes of orthogonal trajectories is -1
-dx/dy = A 3x^2
-dx/dy = 3x^2 y/x^3
-dx/dy = 3y/x
3y dy + x dx = 0
take integration
3y^2/2 + x^2/2 = C
x^2 + 3y^2 = C C is constatn