Results 1 to 3 of 3

Math Help - a problem about differentiability

  1. #1
    Newbie
    Joined
    Nov 2011
    Posts
    21

    a problem about differentiability

    i tried to solve this problem. i can do it a little. but i can't progress. as far as i'm concerned, it requires outstanding performance. thanks for now...
    PROBLEM

    MY SOLUTION...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7

    Re: a problem about differentiability

    Quote Originally Posted by mami View Post
    i tried to solve this problem. i can do it a little. but i can't progress. as far as i'm concerned, it requires outstanding performance. thanks for now...
    PROBLEM
    Your idea of involving an inverse sine function is good. I can use it to give an intuitive, geometric way to tackle the problem. But I think that this method is essentially rigorous, or at least can be made so.

    i am going to write y=f(x) and y', y" for the derivatives. I want to show that at a point (x_0,y_0) on the graph of f, where the derivative is y_0', the inequality y_0^2+y'_0^2 \leqslant1 must hold.

    To start with, suppose that y_0 and y_0' are both positive. If the derivative stays at the constant value y_0' then the function will increase steadily as shown by the line with the arrow in the diagram below, and it will soon exceed the maximum permitted value of 1. Therefore the derivative must decrease, so that the function increases more slowly. That means that the second derivative y" must be negative, and it must be sufficiently negative to stop the graph of the function crossing the line y=1 as it does in the diagram.

    The inequality y'^2+y''^2\leqslant1 shows that the most negative value y" can take is -\sqrt{1-y'^2}. So let's assume that y" takes this value, and we'll see whether this is sufficient to bend the graph of the function enough to stop it crossing the line y=1.

    Thus we need to solve the differential equation y'' = -\sqrt{1-y'^2}. The solution is  \arcsin y'=c-x, for some constant c. So y' = \sin(c-x). Integrating once more, we get y = \cos(c-x)+d, where d is another constant. When x=c, this has the value 1+d. But y must not exceed the value 1, so we conclude that d must be negative, and thus y\leqslant \cos(c-x) = \sqrt{1-\sin^2(c-x)} = \sqrt{1-y'^2}. Put y=y_0 to get the required condition y_0^2+y'_0^2\leqslant1.

    That deals with the case when y_0 and y'_0 are both positive. A similar argument deals with the case when they are both negative. If one of them is positive and the other one is negative, you can turn the whole diagram upside down and replace the function f(x) by the function f(x). This function will have the feature that it and its derivative have the same sign at -x_0. So the previous argument will work again.
    Attached Thumbnails Attached Thumbnails a problem about differentiability-graphic.png  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Nov 2011
    Posts
    21

    Re: a problem about differentiability

    Quote Originally Posted by Opalg View Post
    Your idea of involving an inverse sine function is good. I can use it to give an intuitive, geometric way to tackle the problem. But I think that this method is essentially rigorous, or at least can be made so.

    i am going to write y=f(x) and y', y" for the derivatives. I want to show that at a point (x_0,y_0) on the graph of f, where the derivative is y_0', the inequality y_0^2+y'_0^2 \leqslant1 must hold.

    To start with, suppose that y_0 and y_0' are both positive. If the derivative stays at the constant value y_0' then the function will increase steadily as shown by the line with the arrow in the diagram below, and it will soon exceed the maximum permitted value of 1. Therefore the derivative must decrease, so that the function increases more slowly. That means that the second derivative y" must be negative, and it must be sufficiently negative to stop the graph of the function crossing the line y=1 as it does in the diagram.

    The inequality y'^2+y''^2\leqslant1 shows that the most negative value y" can take is -\sqrt{1-y'^2}. So let's assume that y" takes this value, and we'll see whether this is sufficient to bend the graph of the function enough to stop it crossing the line y=1.

    Thus we need to solve the differential equation y'' = -\sqrt{1-y'^2}. The solution is  \arcsin y'=c-x, for some constant c. So y' = \sin(c-x). Integrating once more, we get y = \cos(c-x)+d, where d is another constant. When x=c, this has the value 1+d. But y must not exceed the value 1, so we conclude that d must be negative, and thus y\leqslant \cos(c-x) = \sqrt{1-\sin^2(c-x)} = \sqrt{1-y'^2}. Put y=y_0 to get the required condition y_0^2+y'_0^2\leqslant1.

    That deals with the case when y_0 and y'_0 are both positive. A similar argument deals with the case when they are both negative. If one of them is positive and the other one is negative, you can turn the whole diagram upside down and replace the function f(x) by the function f(x). This function will have the feature that it and its derivative have the same sign at -x_0. So the previous argument will work again.
    as you know i have been asked a question which no no way i couldn't tackle it. and its is about differentiabilty. at long last i found a solution. i want to share with you. could you check out please. thanks for now.this is the question.

    and this is my solution.(i assume that when x goes infinity, the limit value of functions equals zero.)
    part1

    part2
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. differentiability and continuity problem
    Posted in the Calculus Forum
    Replies: 10
    Last Post: December 13th 2011, 11:06 PM
  2. a problem about differentiability
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: November 29th 2011, 11:47 AM
  3. Differentiability problem
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: April 21st 2010, 10:28 PM
  4. continuity / differentiability problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 26th 2009, 06:52 AM
  5. A problem on continuity and differentiability
    Posted in the Calculus Forum
    Replies: 5
    Last Post: June 27th 2007, 11:42 AM

Search Tags


/mathhelpforum @mathhelpforum