Results 1 to 8 of 8

Math Help - Orthogonal Trajectories

  1. #1
    Member
    Joined
    Jan 2011
    Posts
    156

    Orthogonal Trajectories

    Two curves are orthogonal if their tangent lines are perpendicular at each point of intersection. Are the given below families of curves orthogonal trajectories of each other? (In other words, is every curve in one family orthogonal to every curve in the other family?)

    x^2 + y^2 = ax
    x^2 + y^2 = by

    Then the problem goes on to give me five geometric graphs, one of which is presumably the graph of the equations shown above. Other than the fact that we are studying implicit differentiation, which I have attempted with the hope that I would see a negative reciprocal relationship between these two equations, I have no idea how to do this. I did not find a negative reciprocal relationship, so my instinct that these are not orthogonal, but I don't trust my calculations.

    Can anybody help?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    Quote Originally Posted by joatmon View Post
    Two curves are orthogonal if their tangent lines are perpendicular at each point of intersection. Are the given below families of curves orthogonal trajectories of each other? (In other words, is every curve in one family orthogonal to every curve in the other family?)

    x^2 + y^2 = ax
    x^2 + y^2 = by

    Then the problem goes on to give me five geometric graphs, one of which is presumably the graph of the equations shown above. Other than the fact that we are studying implicit differentiation, which I have attempted with the hope that I would see a negative reciprocal relationship between these two equations, I have no idea how to do this. I did not find a negative reciprocal relationship, so my instinct that these are not orthogonal, but I don't trust my calculations.

    Can anybody help?

    Thanks.
    Hint: Start by finding the tangents to those curves. (Use implicit differentiation.) If the two tangents are perpendicular at a point (x, y) on the curve, then the slopes of the tangents will be in the relationship of m and -1/m.

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2011
    Posts
    156
    I did that, but I don't know what to do with my answer.

    \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx} (ax) \rightarrow y' = \frac{a-2x}{2y}

    \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx} (by) \rightarrow y' = \frac{-2x}{2y-b}

    Then, I tried solving the original equations for y and plugging the results into the differentials, but nothing looks like anything that I can use. I'm totally stuck.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Jan 2011
    Posts
    156
    Let me post a little more on this question:

    First the derivatives:

    \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx} (ax) \rightarrow 2x + 2yy' = a \rightarrow 2yy' = a - 2x \rightarrow y' = \frac{a-2x}{2y}

    \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx} (by) \rightarrow 2x + 2yy' = by' \rightarrow y'(2y - b) = -2x \rightarrow y' = \frac{-2x}{2y-b}

    So far, so good (I think). But then, I get to the final step, which is to solve one of the original equations for y and then plug that value into the other equation. I should then come out with a negative reciprocal of the first derivative if they are orthogonal. Here goes:

    x^2 + y^2 = ax \rightarrow y^2 = ax - x^2 \rightarrow y = \sqrt{ax - x^2}

    And then, plugging into derivative #2:

    y' = \frac{-2x}{2\sqrt{ax-x^2}-b}

    This has no relationship (that I can see) to y' = \frac{a-2x}{2y}, which might just mean that the two functions are not orthogonal.

    Can somebody please take a look at this and confirm whether I have done this correctly or if I have missed the boat?

    Thanks.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    Hello,

    You have {y_1}^{'} = \dfrac{a-2x}{2y} ... (1) and {y_2}^{'} = \dfrac{-2x}{2y-b} ... (2)

    If {y_1}^{'} \cdot {y_2}^{'} = -1 ... (3) then the two curves are othogonal to each other.

    From (1) & (2) you may think that (3) is not true, but actually it is!

    The reason is that your derivatives shouldn't have parameters(a&b).

    In other words, we want {y_1}^{'} and {y_2}^{'} in terms of x & y only without a & b.

    Can you find them in terms of x&y only?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Jan 2011
    Posts
    156
    That's what has me so stumped. I can't get the a and b parameters to go away no matter what I do. With those still in the equations, I have no way to resolve the two derivatives since there is always an undefined a or b hanging around. Any advice?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Super Member General's Avatar
    Joined
    Jan 2010
    From
    Kuwait
    Posts
    562
    Well, you have two ways:

    First way:

    Rewrite your curves as :

    \dfrac{x^2+y^2}{x}=a ... (1)

    \dfrac{x^2+y^2}{y}=b ... (2)

    Now, differentiate (1) & (2) implicitly with respect to x. The parameters will vanish.


    Second way(Which I do not like) :

    From (1), you have a in terms of x&y, substitute it in your {y_1}^'. Similary for (2).
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,211
    Thanks
    419
    Awards
    1
    There may be a more elegant way to do this, but I see no reason to abandon the a's and the b's.

    These two equations are circles. The top one is the set of circles tangent to the y axis and passing through (0, 0). Similarly the second set of circles "expand" up the y-axis. this means that the two sets of circles have two points of intersection. It is at these points that the two equations must be orthogonal.

    So where do they intersect? I get that the circles always intersect at the origin and at the point (if I did it right)
    \displaystyle \left ( \frac{ab^2}{b^2 + a^2}, \frac{a^2b}{b^2 + a^2} \right )

    The first derivatives must be in the form \displaystyle y_1 \cdot \frac{1}{y_2} = -1 at those two points. Lots of algebra ahead doing it this way.

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Orthogonal Trajectories
    Posted in the Differential Equations Forum
    Replies: 4
    Last Post: June 26th 2011, 07:15 PM
  2. Orthogonal Trajectories
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 17th 2009, 07:57 AM
  3. Orthogonal trajectories?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 6th 2008, 08:33 AM
  4. Orthogonal Trajectories
    Posted in the Calculus Forum
    Replies: 1
    Last Post: June 2nd 2008, 07:18 PM
  5. orthogonal trajectories
    Posted in the Calculus Forum
    Replies: 9
    Last Post: October 22nd 2007, 12:29 AM

Search Tags


/mathhelpforum @mathhelpforum