1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Newtons method and distance help

  1. Nov 26, 2008 #1
    1. The problem statement, all variables and given/known data

    Given: Let d be the distance function of a point on a parabola y=9+3x-x^2 and point (5,11)

    Questions:
    a) express f(x) = d^2 in terms of x
    b) show that there is only one critical point of f
    c) approximate the critical point by newton's method with the initial guess x sub zero = 2.5 (accurate to 3 decimal places). Is this a local maximum, local minimum or neither?
    d) find the furthest and the closest pings on the parabola to the ping (5,11) where 0 <= x <= 5


    2. Relevant equations



    3. The attempt at a solution

    i get a,b,and part of c. but im stuck in part c where they ask me if the point is a local maximum,minimum,or neither.

    part c what i did/need help) i did newtons method with x sub n -f'(x)/f''(x)
    and got that the zero is about 2.635999161. and im confused on how to find if it is a max,min, or neither.

    part d what i need did/ need help) i am totally stuck on this one i have no clue where to start it. i drew a few pictures and tried a few thing but it always got me to a dead end. could you help me on this too.



    Thanks Before hand.
     
  2. jcsd
  3. Nov 26, 2008 #2

    Mark44

    Staff: Mentor

    I found f(x) = d^2 = x^4 - 6x^3 + 14x^2 -22x + 29,
    from which I found f'(x) = 4x^3 - 18x^2 + 28x - 22

    I wasn't able to factor the cubic, so I wasn't able to find critical points using the derivative.

    It's clear geometrically, though, that there is just one critical point. By graphing the parabola, I see that it passes through (0, 9) and (3, 9) and that the vertex of the parabola (which opens downward) is at (3/2, 45/4). The point (5, 11) is to the right of the parabola and at a lower level than the vertex so there will be one point on the parabola that is closest, hence f'(x) = 0 for just one value. The critical point will be for the minimum distance. We can make f(x) arbitrarily large by taking larger or more negative values of x.

    I'm curious as to what equation you ran Newton's method on. I hope you ran it on the equation f'(x) = 0, since it's clear to me that the equation f(x) = 0 doesn't have any solutions.
     
  4. Nov 26, 2008 #3

    Mark44

    Staff: Mentor

    Take another look at the problem the OP posted. It was to find the minimum distance between the parabola and the point (5, 11), not to find the high point on the parabola.
     
  5. Nov 26, 2008 #4
    Yeah, I just saw that. My mistake. What is a ping though?
     
  6. Nov 27, 2008 #5

    Mark44

    Staff: Mentor

    I think he might have meant point, but not sure. Otherwise, I don't have any idea.
     
  7. Nov 27, 2008 #6

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    Perhaps he is getting a sonar picture of the parabola!:rofl:
     
  8. Nov 27, 2008 #7
    opps that should say point.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Newtons method and distance help
  1. Newtons method? (Replies: 2)

  2. Newton's method (Replies: 3)

  3. Newton's Method (Replies: 3)

  4. Newton method (Replies: 3)

Loading...