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Math Help - Shortest distance between 2 curves

  1. #1
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    Question Shortest distance between 2 curves

    Find the shortest distance between,

    y = x^{2}-8x+15

    and

    2y + 7 + 2x^{2} = 0

    We can say that the vector perpendicular to the tangent vector of each curve, \vec{d}, is described as follows,

    \vec{d} = a\hat{i} + b\hat{j} + c \hat{k}

    Now if I could get my two curves in the form of a position vector, I can differentiate both position vectors to obtain their tangent vectors.

    Once I have their tangent vectors (Let's say vectors \vec{u} and \vec{v}) I can cross them with my vector \vec{d} and equate them to 0.

    But how do I get my curves in the form of a position vector?

    And even after all this, I've only got 2 equations, so I need another one! Where can I find that last equation?Shortest distance between 2 curves-graph.jpg

    EDIT:

    \vec{r_{1}(t)} = t\hat{i} + (t^{2} - 8t +15)\hat{j}

    \vec{r_{2}(t)} = t\hat{i} + (-t^{2} - \frac{7}{2})\hat{j}

    How do those look for position vectors of my curves?
    Last edited by jegues; September 23rd 2010 at 05:24 PM.
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  2. #2
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    If I were you, I would dot the tangent vectors with the vector \vec{d}. You want the normal vector \vec{d} to be orthogonal, not parallel, to the tangent vectors. If you do this, you'll only need a 2-dimensional vector \vec{d}. You'll have two dot products equalling zero, hence two equations with two unknowns. Turn the crank.

    Your position vectors seem correct.
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    If I were you, I would dot the tangent vectors with the vector \vec{d}. You want the normal vector \vec{d} to be orthogonal, not parallel, to the tangent vectors. If you do this, you'll only need a 2-dimensional vector \vec{d}. You'll have two dot products equalling zero, hence two equations with two unknowns. Turn the crank.
    Your position vectors seem correct.
    Okay so for my tangent vectors,

    \vec{v_{1}(t)} = \hat{i} + (2t -8)\hat{j}

    \vec{v_{2}(t)} = \hat{i} + (-2t)\hat{j}

    Now if I dot my first tangent vector with \vec{d},

    \vec{v_{1}(t)} \cdot \vec{d} = a + (2t - 8)b = 0

    \vec{v_{2}(t)} \cdot \vec{d} = a + -2tb = 0

    This gives me 2 equations with 3 unknowns (I don't know t).

    What can I do about this?
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  4. #4
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    Hmm. In working through things, I discovered that you actually have 4 unknowns. It turns out that you can't assume the parametrization of your two curves have the same x value. That is, you really ought to have

    \vec{r}_{1}(t) = t\hat{i} + (t^{2} - 8t +15)\hat{j} and
    \vec{r}_{2}(s) = s\hat{i} + (-s^{2} - \frac{7}{2})\hat{j}

    The two extra equations you get come from the fact that the two points corresponding to t and s must lie on their respective curves. Does that make sense?
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  5. #5
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    Yes. That makes perfect sense.

    But now I'm going to need 2 more equations, correct? Where can I get those?
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  6. #6
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    How about this: If I go from the origin to the point \vec{r}_{1}(t), and then go in the direction of \vec{d} a distance of \|\vec{d}\|, where should I end up?
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  7. #7
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    Quote Originally Posted by Ackbeet View Post
    How about this: If I go from the origin to the point \vec{r}_{1}(t), and then go in the direction of \vec{d} a distance of \|\vec{d}\|, where should I end up?
    at \vec{r_{2}(s)} no?

    Are you trying to say that,

    \vec{r_{1}(t)} + \vec{d} = \vec{r_{2}(s)} ?
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  8. #8
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    There you go. And since that's a vector equation with two components, you have two equations, right?
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    Alright, so now its down to solving 4 equations with 4 unknowns.

    There's no easy way to do this is there :S ?
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  10. #10
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    Hehe. I'm working through it now myself. Incidentally, I don't think I really understood how to solve this problem when I first started replying. I've been learning as I go.

    The system of equations is nonlinear, so doing ERO's is not going to work. I'd recommend substitution methods. Also, remember that the trivial solution a=b=0 is not allowed. That might be important in simplifying certain expressions.

    First step: write out all your equations.
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  11. #11
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    Quote Originally Posted by Ackbeet View Post
    Hehe. I'm working through it now myself. Incidentally, I don't think I really understood how to solve this problem when I first started replying. I've been learning as I go.

    The system of equations is nonlinear, so doing ERO's is not going to work. I'd recommend substitution methods. Also, remember that the trivial solution a=b=0 is not allowed. That might be important in simplifying certain expressions.

    First step: write out all your equations.
    Here are my equations in order

    a + (2t - 8)b = 0 Equation 1.

    a - 2sb = 0 Equation 2.

    s = a + t Equation 3.

    -s^{2} - \frac{7}{2} = b + t^{2} - 8t + 15 Equation 4.

    How's this look?
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  12. #12
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    Correct so far. What next?
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    I'm not really sure... I can't think of an efficient way of solving these. Any ideas?
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  14. #14
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    Try the substitution method: solve one equation (say, 2) for one variable. Once you've done that, plug in that expression in the other three equations.
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  15. #15
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    Okay from equation 2,

    a = 2sb

    then I'll have the equations,

    2sb + (2t-8)b = 0

    2sb + t = s

    -s^{2} - \frac{7}{2} = b + t^{2} - 8t + 15

    Now if I use the 2nd equation in this post,

    t = s - 2sb

    So then I'll have these 2 equations,

    2sb + (2s - 4sb -8)b = 0

    -s^{2} - \frac{7}{2} = b + (s-2sb)^{2} - 8s + 16sb + 15

    Seems to be going okay so far hopefully...

    Now solving for s in that first equation,

    s = \frac{2b}{b-b^{2}}

    Now I've plugged this into the last remaining equation so I have it in terms of b on paper, but its so messy I don't see how I can solve it.
    Last edited by jegues; September 23rd 2010 at 07:41 PM.
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