• Nov 30th 2011, 07:54 AM
mami
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
http://i1182.photobucket.com/albums/...irasi/5625.jpg
MY SOLUTION...
http://i1182.photobucket.com/albums/.../5625_ms_s.jpg
• Dec 2nd 2011, 08:41 AM
Opalg
Quote:

Originally Posted by mami
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
http://i1182.photobucket.com/albums/...irasi/5625.jpg

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 $\displaystyle (x_0,y_0)$ on the graph of f, where the derivative is $\displaystyle y_0'$, the inequality $\displaystyle y_0^2+y'_0^2 \leqslant1$ must hold.

To start with, suppose that $\displaystyle y_0$ and $\displaystyle y_0'$ are both positive. If the derivative stays at the constant value $\displaystyle y_0'$ 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 $\displaystyle y'^2+y''^2\leqslant1$ shows that the most negative value y" can take is $\displaystyle -\sqrt{1-y'^2}.$ 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 $\displaystyle y'' = -\sqrt{1-y'^2}.$ The solution is $\displaystyle \arcsin y'=c-x$, for some constant c. So $\displaystyle y' = \sin(c-x).$ Integrating once more, we get $\displaystyle y = \cos(c-x)+d$, 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 $\displaystyle y\leqslant \cos(c-x) = \sqrt{1-\sin^2(c-x)} = \sqrt{1-y'^2}.$ Put $\displaystyle y=y_0$ to get the required condition $\displaystyle y_0^2+y'_0^2\leqslant1.$

That deals with the case when $\displaystyle y_0$ and $\displaystyle y'_0$ 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 $\displaystyle -x_0.$ So the previous argument will work again.
• Dec 22nd 2011, 08:59 AM
mami
Quote:

Originally Posted by Opalg
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 $\displaystyle (x_0,y_0)$ on the graph of f, where the derivative is $\displaystyle y_0'$, the inequality $\displaystyle y_0^2+y'_0^2 \leqslant1$ must hold.

To start with, suppose that $\displaystyle y_0$ and $\displaystyle y_0'$ are both positive. If the derivative stays at the constant value $\displaystyle y_0'$ 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 $\displaystyle y'^2+y''^2\leqslant1$ shows that the most negative value y" can take is $\displaystyle -\sqrt{1-y'^2}.$ 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 $\displaystyle y'' = -\sqrt{1-y'^2}.$ The solution is $\displaystyle \arcsin y'=c-x$, for some constant c. So $\displaystyle y' = \sin(c-x).$ Integrating once more, we get $\displaystyle y = \cos(c-x)+d$, 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 $\displaystyle y\leqslant \cos(c-x) = \sqrt{1-\sin^2(c-x)} = \sqrt{1-y'^2}.$ Put $\displaystyle y=y_0$ to get the required condition $\displaystyle y_0^2+y'_0^2\leqslant1.$

That deals with the case when $\displaystyle y_0$ and $\displaystyle y'_0$ 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 $\displaystyle -x_0.$ 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.
http://i1182.photobucket.com/albums/...asi/5625-1.jpg
and this is my solution.(i assume that when x goes infinity, the limit value of functions equals zero.)
part1
http://i1182.photobucket.com/albums/x443/mamirasi/1.jpg
part2
http://i1182.photobucket.com/albums/x443/mamirasi/2.jpg