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Homework Help: Shortest distance between 2 curves

  1. Sep 23, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the shortest distance between,

    [tex]y = x^{2} - 8x + 15[/tex] and,

    [tex]2y + 7 + 2x^{2} = 0[/tex]


    2. Relevant equations



    3. The attempt at a solution

    Rearranging the 2nd function into a function of y in terms of x,

    [tex] y = -x^{2} - \frac{7}{2}[/tex]

    From here I was able to graph the two in the x-y plane. (See figure)

    Now the shortest distance between them will be a vector that runs from one curve to the other, and is perpendicular to both curves, correct?

    How can I go about finding this vector? If I can somehow create this vector, I can compute his magnitude and I'll have the shortest distance between the two curves.

    Any ideas?
     

    Attached Files:

  2. jcsd
  3. Sep 23, 2010 #2

    gabbagabbahey

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    You mean perpendicular to the tangent vector of each curve, right? :wink:


    Why not start by saying that this vector is [itex]\textbf{d}=a\textbf{i}+b\textbf{j}+c\textbf{k}[/itex] and then solving fo the 3 unkowns [itex]a[/itex], [itex]b[/itex], and [itex]c[/itex]? You will need 3 independent equations to solve for these variables, so try putting the condition that d is perpendicular to the tangent vector of each curve into equation form (giving you 2 equations) and then finding another equation that it must satisfy...
     
  4. Sep 23, 2010 #3
    If d is to be perpendicular to the tangent vector of each curve then,

    [tex] \vec{d} \cdot \vec{v} = \vec{0}[/tex] Where, [tex]\vec{v}[/tex] is the tangent vector of the given curve.

    But my curves aren't in vector notation, so how do I go about finding the tangent vector?

    If I can describe my two curves in the form of position vectors then I can differentiate them to find their tangent vectors, but I don't know how to describe my curves in terms of a position vector!?!?

    Can I get a few more nudges!? :wink:
     
    Last edited: Sep 23, 2010
  5. Sep 23, 2010 #4

    Office_Shredder

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    Once you have solved for y as a function of x, y=f(x), then you can easily parametrize your curve to be all points (t,f(t))
     
  6. Sep 23, 2010 #5
    So,

    [tex]\vec{r_{1}(t)} = t^{2} - 8t +15[/tex]

    and

    [tex]\vec{r_{2}(t)} = -t^{2} - \frac{7}{2}[/tex]

    How's that?

    EDIT: Wait,

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

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

    I think these 2 look better.

    Okay so for my tangent vectors,

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

    [tex]\vec{v_{2}(t)} = \hat{i} + (-2t)\hat{j}[/tex]

    So if I dot these with d,

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

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

    I'm stuck with 2 equations and 3 unknowns (I don't know t!), how can I fix this!? Can I get a third equation somehow!?

    EDIT: We can't assume that both curves will have the same parameter t, can we?

    So,

    [tex]\vec{r_{2}(s)} = s\hat{i} + (-s^{2} - \frac{7}{2})\hat{j}[/tex]

    and,

    [tex]\vec{v_{2}(s)} = \hat{i} + (-2s)\hat{j}[/tex]

    therefore,

    [tex]\vec{v_{2}(s)} \cdot \vec{d} = a + -2sb = 0[/tex]

    So it looks like I need another 2 equations hmmm...
     
    Last edited: Sep 23, 2010
  7. Sep 23, 2010 #6
    What could it be!??!? I'm so close!

    EDIT: Not so close anymore :wink: had to edit post above this

    I have it solved!
     
    Last edited: Sep 23, 2010
  8. Sep 24, 2010 #7

    gabbagabbahey

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    You also know that [itex]\textbf{d}[/itex] goes from [itex]\textbf{r}_1(t)[/itex] to [itex]\textbf{r}_2(s)[/itex], right?
     
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