Results 1 to 5 of 5

Math Help - Find the closest point on a curve?

  1. #1
    Super Member
    Joined
    Mar 2006
    Posts
    705
    Thanks
    2

    Find the closest point on a curve?

    Q: Find the point on the line -5 x + 3 y + 5 =0 which is closest to the point ( -3, 3 ).

    Solution: I know I have to use the distance formula, and I have to plug in the x and y with x and f(x);

    therefore I got to here d = (x+3)^2+[(5x-5)/3-3]^2;

    and f'(x) = 2(x+3) + (10/3)(5x-14)/3;

    then I have no idea what to do, as our prof didn't bother to explain it afterward...

    Any idea how to solve it?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    First thing is you're doing the right thing by differentiating the distance squared, because it will have the same critical points as the distance, but just as a note the distance between two points is D=\sqrt{(x-x_0)^2+(y-y_0)^2. You're working with D^2. But you've set up your problem correctly and differentiated correctly, so now set the derivative equal to zero and find your critical point(s)!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Grand Panjandrum
    Joined
    Nov 2005
    From
    someplace
    Posts
    14,972
    Thanks
    4
    Quote Originally Posted by tttcomrader
    Q: Find the point on the line -5 x + 3 y + 5 =0 which is closest to the point ( -3, 3 ).

    Solution: I know I have to use the distance formula, and I have to plug in the x and y with x and f(x);

    therefore I got to here d = (x+3)^2+[(5x-5)/3-3]^2;

    and f'(x) = 2(x+3) + (10/3)(5x-14)/3;

    then I have no idea what to do, as our prof didn't bother to explain it afterward...

    Any idea how to solve it?
    If a function f(x) is at a local minimum then f'(x)=0.

    (Note you have to be careful as f'(x)=0 does not guarantee that x
    is a minimum, it may be a maximum, or a point at which the tangent is horizontal
    for an otherwise locally increasing or decreasing function.

    Such problems don't arise with your d function as it goes to
    infinity as x \to \pm \infty, and has only a single root to the equation:

    d'(x)=0

    RonL


    RonL
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Mar 2006
    Posts
    705
    Thanks
    2
    I find the critical point as 43/34. So the minimum distance is f(43/34)? But how do I find the coodinate after I have found the minimum distance?
    Last edited by tttcomrader; May 1st 2006 at 10:09 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by tttcomrader
    I find the critical point as 43/34. So the minimum distance is f(43/34)? But how do I find the coodinate after I have found the minimum distance?
    No f(43/34) tells you the y-coordinate. Since you know the x-coordinate and y-coordinate that minimize distance you know use the distance formual
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: April 6th 2011, 07:44 AM
  2. Closest point on a curve
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 18th 2010, 02:20 PM
  3. Replies: 2
    Last Post: November 5th 2009, 01:05 PM
  4. Find the point on the line, closest to a point
    Posted in the Calculus Forum
    Replies: 3
    Last Post: July 14th 2009, 06:41 PM
  5. Replies: 1
    Last Post: April 27th 2009, 03:30 PM

Search Tags


/mathhelpforum @mathhelpforum