Finding the Orthogonal Trajectories of Families of Curves

Given the family of curves r=2a(costheta sintheta), we can find its orthogonal trajectories using the following steps:

1. Convert the polar equation to Cartesian coordinates.
To do this, we use the formulas x = r cos(theta) and y = r sin(theta). So, r = sqrt(x^2+y^2), and the given equation becomes:

sqrt(x^2+y^2) = 2a(cos(theta) sin(theta))

Squaring both sides and using the identity sin(2theta) = 2sin(theta)cos(theta), we get:

x^2+y^2 = 4a^2sin^2(theta)cos^2(theta)
x^2+y^2 = a^2sin^2(2theta)

2. Find the slope of the tangent to the given family of curves.
To do this, we differentiate the given equation with respect to theta and solve for dy/dx. We get:

dy/dx = -(x+y)/(x-y)

This is the slope of the tangent to the given family of curves at any point (x,y).

3. Find the negative reciprocal of the slope.
The negative reciprocal of dy/dx is the slope of the orthogonal trajectory at the same point. So, we get:

m = (x-y)/(x+y)

4. Convert the slope to polar coordinates.
To do this, we use the formulas x = r cos(theta) and y = r sin(theta) to eliminate x and y. We get:

m = (r cos(theta) - r sin(theta))/(r cos(theta) + r sin(theta))
m = tan(theta) - cot(theta)

5. Find the differential equation of the orthogonal trajectories.
The differential equation of the orthogonal trajectories is given by:

dy/dx = -1/m

Substituting the value of m from step 4, we get:

dy/dx = -1/(tan(theta) - cot(theta))

Simplifying and rearranging, we get:

y' = -cot(theta)y - x/sin(theta)

This is the differential equation of the orthogonal trajectories.

6. Solve the differential equation to get the family of orthogonal trajectories.
To solve the differential equation, we use the method of integrating factors. Multiplying both sides by sin(theta), we get:

y'sin(theta) + ycos(theta) = -x

This is a linear differential equation, which we can solve by multiplying both sides by the integrating factor exp(integral(cot(theta)dtheta)). We get:

y sin(theta) exp(ln|sin(theta)| - ln|cos(theta)|) + y cos(theta) exp(ln|cos(theta)| - ln|sin(theta)|) = -x exp(ln|cos(theta)| - ln|sin(theta)|)

Simplifying and rearranging, we get:

y = C exp(ln|sin(theta)| - ln|cos(theta)|) - x exp(ln|cos(theta)| - ln|sin(theta)|)

y = C tan(theta) - x/cos(theta)

This is the family of orthogonal trajectories.

7. Verify that the orthogonal trajectories are indeed orthogonal to the given family of curves.
To verify this, we take the product of the slopes of a point on the given family of curves and a point on its orthogonal trajectory. We get:

(dy/dx)(dy'/dx') = -(x+y)/(

Find the orthogonal trajectories of families of curves r=2a(costheta sintheta)?