Theorem 1: If two families of curves in a plane are orthogonal trajectories of each other, then the product of their slopes at any point is equal to -1.
Orthogonal trajectories are curves that intersect each curve of a given family at right angles.
Let’s consider two families of curves: y = f(x, C1) and y = g(x, C2).
C1 and C2 are constants that vary from curve to curve within their respective families.
The slope of a curve y = f(x, C1) is given by: dy/dx = f’(x, C1).
The slope of a curve y = g(x, C2) is given by: dy/dx = g’(x, C2).
According to Theorem 1, we have the equation f’(x, C1) * g’(x, C2) = -1.
This equation holds for every point of intersection between the two families of curves.
Find the orthogonal trajectory of the family of curves y = Cx, where C is a constant. Solution:
Differentiate y = Cx with respect to x to find the slope of the family of curves.
dy/dx = C
Now, let’s substitute the slope (C) into the equation f’(x, C1) * g’(x, C2) = -1.
C * g’(x, C2) = -1. We need to solve this equation to find the orthogonal trajectory.
Solving the equation, we get: g’(x, C2) = -1/C.
Integrating g’(x, C2) = -1/C with respect to x, we get g(x, C2) = -ln|C2| + K, where K is another constant.
Therefore, the final equation of the orthogonal trajectory is: y = -ln|C2| + K.
Find the orthogonal trajectories of the family of curves y = Cx^n, where C and n are constants. Solution:
Differentiate y = Cx^n with respect to x to find the slope of the family of curves.
dy/dx = nCx^(n-1)
Substituting the slope into the equation f’(x, C1) * g’(x, C2) = -1, we get: nCx^(n-1) * g’(x, C2) = -1.
Solving for g’(x, C2), we find g’(x, C2) = -1/(nCx^(n-1)).
Integrating g’(x, C2) = -1/(nCx^(n-1)) with respect to x, we get: g(x, C2) = -(x^n)/(nC) + K, where K is a constant.
Therefore, the orthogonal trajectory can be represented as: y = -(x^n)/(nC) + K.
That concludes our discussion on Theorem 1 for orthogonal trajectories. Next, we will explore some additional examples to further solidify our understanding.
Find the orthogonal trajectory of the family of curves y^2 = e^x. Solution:
Differentiate y^2 = e^x with respect to x to find the slope of the family of curves.
2yy’ = e^x
Solving for y’, we get y’ = (e^x) / (2y).
To find the orthogonal trajectory, we need to substitute the slope (y’) into the equation f’(x, C1) * g’(x, C2) = -1.
(e^x) / (2y) * g’(x, C2) = -1
Solving for g’(x, C2), we find g’(x, C2) = -2y / (e^x)
Integrating g’(x, C2) = -2y / (e^x) with respect to x, we get: g(x, C2) = -2y / (e^x) + K, where K is a constant.
We can rewrite this equation in terms of y as: y = -x / 2 + K(e^x), where K is a constant.