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Given a point and a curve: how to find perpindicular POI?

  1. Dec 17, 2008 #1
    Hi all,

    This seems more like a basic level general math question, but since it was asked in my calc III/linear class on my final exam and possibly involves derivatives, I decided to post here.

    The original question:
    Given f(x,y) = x^2 + y^2, find the point on the surface that is closest to the point (3,0,0).

    I sketched the figure (a parabloid) relative to the point and saw that it was obvious that the point on the surface which was closest to (3,0,0) was on the xz-plane.

    I drew a projection of the xz-plane with the point, and then realized that the shortest line would be the one which was perpindicular to the curve z=x^2.

    It's probably got something to do with relating the derivative of the curve to the line, but I'm not sure how. I ended up saying that the slope of the curve = 2x (dy/dx), so the slope of the line from (3,0,0) to the point on z=x^2 must be 1/(2x).

    However, that's where I got stuck. It's really frustrating because it seems like it's so obvious, but the answer keeps eluding me!

    How could I go about solving this problem? Also, what's the more generalized way of doing this in three dimensions? I'm sure that mine is not the quickest or most general form, considering I had a special case where the point lied on an axis.
  2. jcsd
  3. Dec 17, 2008 #2
    I'm just a student (and it seems I'm at about the same level in math as you) but just something I noticed in reading your post. You said the slope perpendicular to 2x is 1/2x. Remember than if the slope is 2x, the slope perpendicular to that is -1/2x. I'll read your post more thoroughly and see if I can't help you out.

    edit: I gave it a shot but I'm no where near confident enough in my method to advise someone else. It seems though, that you have to take the distance formula d = sqrt((x-3)^2 + y^2 + z^2) and find the derivate, solve for d' = 0 to find critical values, figure out which of these is a minimum (there should only be one) and that should be your answer... I think.

    quick help with how to insert latex expressions, anyone? (not to hijack the thread or anything)
    Last edited: Dec 17, 2008
  4. Dec 17, 2008 #3
    Yeah, I had that equation written out on my paper, too... for some reason I didn't realize that I could just take the derivative of the equation to find a rel. min. It makes sense, I was just stuck on solving it my way because it's how I visualized it at first.

    LaTeX can be inserted by using:
    [ tex] Latex code here [/ tex]

    You can learn how to insert more complex things here: https://www.physicsforums.com/showthread.php?t=8997
  5. Dec 17, 2008 #4
    Thanks for the tip. I hope the admin will excuse this, but I'm going to try it out real quick:
    [tex]d\; =\; \sqrt{\left( x-3 \right)^{2}\; +\; y^{2}\; +\; z^{2}}[/tex]
    p.s. If you have a mac, write equations in grapher and then highlight them and right-click (you'll be able to copy an equation as a latex expression). alt+v is the sqrt symbol
    ...that would've been immensely useful to me a year ago. i'm going to make a thread on this in case anyone is unaware
  6. Dec 17, 2008 #5
    Blah! This makes me want to have a mac. :P

    Btw, taking the derivative of [tex]d\; =\; \sqrt{\left( x-3 \right)^{2}\; +\; y^{2}\; +\; z^{2}}[/tex] is pretty complicated (two partial derivatives)... so I guess it works out a lot better to simplify the problem to a projection of the xz-plane so that you only need to take one derivative with respect to x.

    [tex] d = sqrt{(x-3)^2+z^2}[/tex]
    z^2 = d^2 - (x-3)^2[/tex]
    z = sqrt{d^2 - (x-3)^2}[/tex]
    dz/dx = -(x-3)/(sqrt(-x+d+3)*sqrt(x+d-3))[/tex]

    Then what? What am I doing wrong? This was an extra credit question, so I don't feel that bad for being so clueless...

    Does this make sense? :/ UGH I'm so lost. I've been studying too much linear algebra stuff, I think, and I've lost my ability to do calculus.
  7. Dec 17, 2008 #6
    We are trying to pick a point (x,y,z) on our curve [tex]z = x^2 + y^2 [/tex] that will be closest to (3,0,0). Since that point is on our curve it will have the form [tex](x,y,x^2+y^2)[/tex]. So now we have our distance formula:

    [tex]D = \sqrt{(x-x_{0})^2 + (y-y_{0})^2 + (z-z_{0})^2} [/tex], plug our information in to get

    [tex]D = \sqrt{(x-3)^2 + y^2 + (x^2 + y^2)^2}[/tex]

    So now you differentiate D with respect to x, then differentiate with respect to y.

    Get your points, figure out if they are min, max or saddle.
    Last edited: Dec 17, 2008
  8. Dec 17, 2008 #7


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    let me think a bit more about this problem before I reply with nonesense
    Last edited: Dec 17, 2008
  9. Dec 17, 2008 #8
    I guess I understand how this works... but it seems like it gets really complicated. Normally our professor doesn't give us abnormally tedious calculations.

    [tex]D_x,y = (6*y^3+(-18*x^2+35*x+3)*y)*sqrt(y^4+(2*x^2+1)*y^2+x^4+x^2-6*x+9)/(y^8+(4*x^2+2)*y^6+(6*x^4+6*x^2-12*x+19)*y^4+(4*x^6+6*x^4-24*x^3+38*x^2-12*x+18)*y^2+x^8+2*x^6-12*x^5+19*x^4-12*x^3+54*x^2-108*x+81)[/tex]

    Then I find all of the roots, and there should be one minimum?
  10. Dec 17, 2008 #9
    I don't see the complication, if your point is a min/max of D it will also be min/max of D^2 so now let's say

    [tex] \bar{D} = D^{2} = (x-3)^{2} + y^{2} + (x^{2} + y^{2})^{2} [/tex]

    So now differentiate and you get

    [tex] \bar{D}_{x} = 2(x-3) + 4x(x^2 + y^2); \bar{D}_{y} = 2y + 4y(x^2 + y^2) [/tex]

    Each one must equal 0 therefore [tex] x^2 + y^2 = \frac{3-x}{2x} [/tex] and [tex] x^2 + y^2 = -\frac{1}{2} [/tex] (provided y <> 0 remember to treat this case separately)

    Can you proceed from here?
  11. Dec 17, 2008 #10
    AH! I understand now. I don't know what I was trying to do by trying to find [tex]D_x_y[/tex].

    Thanks for explaining.
  12. Dec 17, 2008 #11
    You will need [tex] D_{xy} [/tex] or my [tex] \bar{D}_{xy} [/tex] to test for min/max/saddle.
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