Explore BrainMass

Explore BrainMass

    Mathematics - Calculus

    Not what you're looking for? Search our solutions OR ask your own Custom question.

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    I need to find the orthogonal trajectories of the family of curves, y = 1/(x+c) where k is an arbitrary constant.

    So far, I had figured on c = (1/y) - x

    m1 = -1/(x^2 + (1/y) - x)
    m2 = x^2 + (1/y) - x

    I don't know how to figure beyond that. Probably because those were calculated wrong. Please show me how it's done. Thank you.

    © BrainMass Inc. brainmass.com December 24, 2021, 8:13 pm ad1c9bdddf
    https://brainmass.com/math/basic-calculus/orthogonal-trajectories-family-curves-258003

    SOLUTION This solution is FREE courtesy of BrainMass!

    y = 1/(x + c)
    Differentiating with respect to x, we get dy/dx = -1/(x + c)^2
    The differential equation of the orthogonal trajectories is dy/dx = (x + c)^2
    dy = (x + c)^2 dx
    Integrating on both the sides, we get y = (1/3)(x + c)^3 + k are the orthogonal trajectories.
    Members of this family for k = 5 and c = -2, -1, 0, 1 and 2 are shown below ...

    This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!

    © BrainMass Inc. brainmass.com December 24, 2021, 8:13 pm ad1c9bdddf>
    https://brainmass.com/math/basic-calculus/orthogonal-trajectories-family-curves-258003

    Attachments

    ADVERTISEMENT