Find the orthogonal trajectories of family of the circle x2+y2=a2 where a is a parameter
To find the orthogonal trajectories of a given family of curves, we need to find equations of curves that intersect the given curves at a right angle.
We start by considering the family of circles given by the equation x^2 + y^2 = a^2, where a is a parameter. Let's first differentiate this equation with respect to x:
2x + 2yy' = 0.
Now, we solve this equation for y':
y' = -x/y.
To find the equations of the orthogonal trajectories, we need to find curves with slopes that are negative reciprocals of -x/y. So, the slope of the orthogonal trajectories at any point (x, y) will be given by m = y/x.
Let's differentiate y = f(x) with respect to x to obtain y':
y' = f'(x).
We can now set y' = y/x:
f'(x) = y/x.
Next, we want to eliminate x and y from this equation. We can do that by multiplying both sides of the equation by x:
x * f'(x) = y.
This equation represents the orthogonal trajectories of the given family of circles x^2 + y^2 = a^2.
To summarize, the orthogonal trajectories of the family of circles x^2 + y^2 = a^2 are given by the equation x * f'(x) = y, where f'(x) represents the derivative of the function f(x) defining the orthogonal trajectory.
To find the orthogonal trajectories of the family of circles given by the equation x^2 + y^2 = a^2, we can use the concept of orthogonal gradients.
First, let's find the gradient of the circles' family.
Differentiating both sides of the equation x^2 + y^2 = a^2 with respect to x, we get:
2x + 2y(dy/dx) = 0
Now, let's solve this equation for dy/dx:
dy/dx = -x/y
The gradient of the family of circles is given by -x/y.
To find the orthogonal trajectories, we need to find the negative reciprocal of -x/y, which is y/x.
Therefore, the differential equation for the orthogonal trajectories is dy/dx = y/x.
To solve this differential equation, let's rewrite it in a separable form:
dy/y = dx/x
Integrating both sides, we get:
ln|y| = ln|x| + C
Using properties of logarithms, this can be simplified to:
ln|y| - ln|x| = ln|x/y| = C
Exponentiating both sides, we have:
|x/y| = e^C
Since e^C is a positive constant, we can remove the absolute value signs and simplify it to:
|x/y| = k
where k is the constant of integration.
To find the equation of the orthogonal trajectories, we can rearrange this equation as follows:
x^2 = k * y^2
This equation represents a family of hyperbolas centered at the origin, with the constant k determining the shape of the hyperbolas.
Therefore, the orthogonal trajectories of the family of circles x^2 + y^2 = a^2 are given by the equation x^2 = k * y^2, where k is a constant.
Haven't done this kind of question in over 55 years, had to look it up
try this:
http://math.stackexchange.com/questions/315317/orthogonal-trajectories-for-a-given-family-of-curves