Math Problem Statement

Find the orthogonal trajectories of the family of the curves passing through the points (0,2 ) and (0,-2)

Solution

To find the orthogonal trajectories of a family of curves, we need to follow a systematic approach. Orthogonal trajectories are curves that intersect a given family of curves at right angles. Let's start by determining the family of curves passing through the points (0,2)(0,2) and (0,2)(0,-2).

Step 1: Identify the Family of Curves

The given points suggest a family of curves symmetric with respect to the y-axis. A suitable family of such curves could be circles centered at the origin. The general equation for this family is:

x2+y2=R2x^2 + y^2 = R^2

where RR is a constant representing the radius of the circle.

Step 2: Find the Differential Equation of the Family of Curves

Differentiating the equation x2+y2=R2x^2 + y^2 = R^2 with respect to xx gives:

2x+2ydydx=02x + 2y \frac{dy}{dx} = 0

This simplifies to:

x+ydydx=0x + y \frac{dy}{dx} = 0

or

dydx=xy\frac{dy}{dx} = -\frac{x}{y}

Step 3: Find the Differential Equation of the Orthogonal Trajectories

For orthogonal trajectories, the slope of the trajectory at any point must be the negative reciprocal of the slope of the original family of curves. Therefore, we need to replace dydx\frac{dy}{dx} with its negative reciprocal:

dydx=yx\frac{dy}{dx} = \frac{y}{x}

Step 4: Solve the Differential Equation for the Orthogonal Trajectories

This is a separable differential equation. To solve it, we can separate the variables and integrate:

dydx=yx\frac{dy}{dx} = \frac{y}{x}

Separating the variables:

dyy=dxx\frac{dy}{y} = \frac{dx}{x}

Integrating both sides:

dyy=dxx\int \frac{dy}{y} = \int \frac{dx}{x}

lny=lnx+C\ln |y| = \ln |x| + C

where CC is the constant of integration. Exponentiating both sides to solve for yy:

y=eCx|y| = e^C |x|

Let eC=ke^C = k, where kk is a new constant. We get:

y=kx|y| = k |x|

This implies two sets of straight lines:

y=kxory=kxy = kx \quad \text{or} \quad y = -kx

Conclusion

The orthogonal trajectories of the family of circles x2+y2=R2x^2 + y^2 = R^2 are the family of straight lines y=kxy = kx and y=kxy = -kx, which pass through the origin and intersect the circles at right angles.

Would you like any further details or have any questions about this solution?

Here are 8 related questions for further exploration:

  1. How do you find orthogonal trajectories for a family of parabolas?
  2. What are the orthogonal trajectories for the family of hyperbolas?
  3. Can orthogonal trajectories be found for families of curves not centered at the origin?
  4. How do you apply this method to non-Cartesian coordinate systems?
  5. What is the significance of orthogonal trajectories in physics?
  6. How would you visualize orthogonal trajectories using software tools?
  7. Are there numerical methods to find orthogonal trajectories when analytical solutions are difficult?
  8. What role do orthogonal trajectories play in potential theory?

Tip: When solving differential equations, always check if they are separable, as this can significantly simplify the solution process.

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Math Problem Analysis

Mathematical Concepts

Differential Equations
Orthogonal Trajectories

Formulas

General equation of a circle: x^2 + y^2 = R^2
Slope of a line: y = mx + c

Theorems

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Suitable Grade Level

Advanced High School