# This is a straight line?

• June 9th 2011, 07:44 AM
rainer
This is a straight line?
Can someone please point out to me how the following reduces to y=mx+b:

$\frac{y''(x)}{(1+y'(x))^{\frac{3}{2}}}=0$

Thanks
• June 9th 2011, 07:59 AM
Ackbeet
The only way for a fraction to be zero is when?
• June 9th 2011, 08:02 AM
Prove It
Let $\displaystyle Y = \frac{dy}{dx}$, which means $\displaystyle \frac{dY}{dx} = \frac{d^2y}{dx^2}$ and the DE becomes

\displaystyle \begin{align*} (1 + Y)^{-\frac{3}{2}}\,\frac{dY}{dx} &= 0 \\ \int{(1 + Y)^{-\frac{3}{2}}\,\frac{dY}{dx}\,dx} &= \int{0\,dx} \\ \int{(1 + Y)^{-\frac{3}{2}}\,dY} &= C_1 \\ \frac{(1 + Y)^{-\frac{1}{2}}}{-\frac{1}{2}} + C_2 &= C_1 \\ -2(1 + Y)^{-\frac{3}{2}} &= C_1 - C_2 \\ (1 + Y)^{-\frac{3}{2}} &= -\frac{1}{2}(C_1 - C_2) \\ 1 + Y &= \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} \\ Y &= \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} - 1 \end{align*}

So $\displaystyle \frac{dy}{dx} &= C$ where $\displaystyle C = \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} - 1$.

Can you go from here to solve for $\displaystyle y$?
• June 9th 2011, 09:01 AM
HallsofIvy
Quote:

Originally Posted by Prove It
Let $\displaystyle Y = \frac{dy}{dx}$, which means $\displaystyle \frac{dY}{dx} = \frac{d^2y}{dx^2}$ and the DE becomes

\displaystyle \begin{align*} (1 + Y)^{-\frac{3}{2}}\,\frac{dY}{dx} &= 0

So either [tex]\frac{dY}{dx}= 0[//tex] or $(1+ Y)^{-\frac{3}{2}}= 0$. But that second equation is impossible.

That's essentially what Ackbeat said.

Quote:

\begin{align*} \int{(1 + Y)^{-\frac{3}{2}}\,\frac{dY}{dx}\,dx} &= \int{0\,dx} \\ \int{(1 + Y)^{-\frac{3}{2}}\,dY} &= C_1 \\ \frac{(1 + Y)^{-\frac{1}{2}}}{-\frac{1}{2}} + C_2 &= C_1 \\ -2(1 + Y)^{-\frac{3}{2}} &= C_1 - C_2 \\ (1 + Y)^{-\frac{3}{2}} &= -\frac{1}{2}(C_1 - C_2) \\ 1 + Y &= \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} \\ Y &= \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} - 1 \end{align*}

So $\displaystyle \frac{dy}{dx} &= C$ where $\displaystyle C = \left[-\frac{1}{2}(C_1 - C_2)\right]^{-\frac{2}{3}} - 1$.

Can you go from here to solve for $\displaystyle y$?
• June 9th 2011, 10:33 AM
zg12
Is there not a way to delete a post?
• June 9th 2011, 10:48 AM
Jester
Quote:

Originally Posted by rainer
Can someone please point out to me how the following reduces to y=mx+b:

$\frac{y''(x)}{(1+y'(x))^{\frac{3}{2}}}=0$

Thanks

A question for the OP. Is it

$\frac{y''(x)}{(1+y'(x))^{\frac{3}{2}}}=0$ or $\frac{y''(x)}{(1+y'^2(x))^{\frac{3}{2}}}=0$

The first looks odd. The second - standard curvature. Just curious :-)
• June 9th 2011, 11:11 AM
Ackbeet
Quote:

Originally Posted by zg12
Is there not a way to delete a post?

Just report your own post and ask to delete. If it is appropriate, a moderator will delete your post. The report post button is in the lower left corner of your post - looks like a triangle with an exclamation mark inside it.
• June 9th 2011, 05:25 PM
rainer
Quote:

Originally Posted by Danny
A question for the OP. Is it

$\frac{y''(x)}{(1+y'(x))^{\frac{3}{2}}}=0$ or $\frac{y''(x)}{(1+y'^2(x))^{\frac{3}{2}}}=0$

The first looks odd. The second - standard curvature. Just curious :-)

Unless the author has made a type-o, it's the first one, which is the outcome of plugging the arclength formula into the Euler-Lagrange equation.
• June 10th 2011, 03:41 AM
Prove It
Quote:

Originally Posted by zg12
Is there not a way to delete a post?

You can click "Edit" and then "Delete Message".
• June 10th 2011, 04:50 AM
Ackbeet
Quote:

Originally Posted by Prove It
You can click "Edit" and then "Delete Message".

This works for you, because you are have the MHF Expert badge. There is no way for zg12 directly to delete a post. A moderator must do it for him.
• June 10th 2011, 05:24 AM
Prove It
Quote:

Originally Posted by Ackbeet
This works for you, because you are have the MHF Expert badge. There is no way for zg12 directly to delete a post. A moderator must do it for him.

Oops - didn't realise that was only made possible by the badge. Thanks :)
• June 10th 2011, 06:46 AM
Jester
Quote:

Originally Posted by rainer
Unless the author has made a type-o, it's the first one, which is the outcome of plugging the arclength formula into the Euler-Lagrange equation.

You mean $L = \sqrt{1+y'^2}$. If so, then it would be the second, wouldn't it?