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Math Help - orthogonal trajectories

  1. #1
    Senior Member polymerase's Avatar
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    orthogonal trajectories

    Could someone explain what this question really means and how to start solving it....

    If m is a fixed constant such that the two families of curves y=\frac{1}{x+c} and y=m(x+k)^\frac{1}{3} are orthogonal trajectories of each others, what must be the value of m?
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  2. #2
    TD!
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    The functions y = g(x) and y = h(x) are orthogonal at x = a, if g'(a)h'(a) = -1.
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  3. #3
    Senior Member polymerase's Avatar
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    Quote Originally Posted by TD! View Post
    The functions y = g(x) and y = h(x) are orthogonal at x = a, if g'(a)h'(a) = -1.
    y'=\frac{-1}{(x+c)^2} and y'=\frac{m}{3(x+k)^\frac{2}{3}} therefore when i multiple and set them equal to -1 I get, \frac{m}{3(x+c)^2(x+k)^\frac{2}{3}}=1 but how do i solve for m?
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polymerase View Post
    y'=\frac{-1}{(x+c)^2} and y'=\frac{m}{3(x+k)^\frac{2}{3}} therefore when i multiple and set them equal to -1 I get, \frac{m}{3(x+c)^2(x+k)^\frac{2}{3}}=1 but how do i solve for m?
    Obviously for a given c and k the two functions will intersect at a set of points \{ a_i \}, so we have
    \frac{m}{3(a_i + c)^2(a_i + k)^\frac{2}{3}}=1
    for each such a_i.

    So your solution for m depends on the c, k, and set \{ a_i \}.

    -Dan
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    Senior Member polymerase's Avatar
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    Quote Originally Posted by topsquark View Post
    Obviously for a given c and k the two functions will intersect at a set of points \{ a_i \}, so we have
    \frac{m}{3(a_i + c)^2(a_i + k)^\frac{2}{3}}=1
    for each such a_i.

    So your solution for m depends on the c, k, and set \{ a_i \}.

    -Dan
    So are you saying there is no defined answer.....? Because it says the answer is \sqrt[3]{3}
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by polymerase View Post
    So are you saying there is no defined answer.....? Because it says the answer is \sqrt[3]{3}
    I haven't actually solved it, so I couldn't tell you. But start with where the two curves will intersect and go from there.

    -Dan
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    Senior Member polymerase's Avatar
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    Quote Originally Posted by topsquark View Post
    I haven't actually solved it, so I couldn't tell you. But start with where the two curves will intersect and go from there.

    -Dan
    What you mean? start where the original curves will intersect?
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  8. #8
    TD!
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    Quote Originally Posted by polymerase View Post
    y'=\frac{-1}{(x+c)^2} and y'=\frac{m}{3(x+k)^\frac{2}{3}} therefore when i multiple and set them equal to -1 I get, \frac{m}{3(x+c)^2(x+k)^\frac{2}{3}}=1 but how do i solve for m?
    Ok, so far so good! You know have m/(...) = 1.
    A fraction is 1 if numerator and denominator are equal.
    So you can easily solve this for m. We have:

    <br />
m = 3\left( {x + c} \right)^2 \left( {x + k} \right)^{\frac{2}{3}} <br />

    Now, this m doesn't seem "constant", but dependent of x, c and k!
    But, where do we have to look for ortogonality? At their intersection!
    Let's try to find something useful, equalling both equations:

    <br />
\frac{1}{{x + c}} = m\left( {x + k} \right)^{\frac{1}{3}}  \Leftrightarrow \frac{1}{m} = \left( {x + c} \right)\left( {x + k} \right)^{\frac{1}{3}}  \Rightarrow \frac{1}{{m^2 }} = \left( {x + c} \right)^2 \left( {x + k} \right)^{\frac{2}{3}} <br />

    You recognise this? Filling in, in what we had for m earlier:

    <br />
m = 3\left( {x + c} \right)^2 \left( {x + k} \right)^{\frac{2}{3}}  \Leftrightarrow m = \frac{3}{{m^2 }} \Leftrightarrow m^3  = 3 \Leftrightarrow m = \sqrt[3]{3}<br />
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  9. #9
    Senior Member polymerase's Avatar
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    Quote Originally Posted by TD! View Post
    Ok, so far so good! You know have m/(...) = 1.
    A fraction is 1 if numerator and denominator are equal.
    So you can easily solve this for m. We have:

    <br />
m = 3\left( {x + c} \right)^2 \left( {x + k} \right)^{\frac{2}{3}} <br />

    Now, this m doesn't seem "constant", but dependent of x, c and k!
    But, where do we have to look for ortogonality? At their intersection!
    Let's try to find something useful, equalling both equations:

    <br />
\frac{1}{{x + c}} = m\left( {x + k} \right)^{\frac{1}{3}} \Leftrightarrow \frac{1}{m} = \left( {x + c} \right)\left( {x + k} \right)^{\frac{1}{3}} \Rightarrow \frac{1}{{m^2 }} = \left( {x + c} \right)^2 \left( {x + k} \right)^{\frac{2}{3}} <br />

    You recognise this? Filling in, in what we had for m earlier:

    <br />
m = 3\left( {x + c} \right)^2 \left( {x + k} \right)^{\frac{2}{3}} \Leftrightarrow m = \frac{3}{{m^2 }} \Leftrightarrow m^3 = 3 \Leftrightarrow m = \sqrt[3]{3}<br />
    Thanks! I understand now
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  10. #10
    TD!
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    You're welcome
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