# Curvature in the plane 3 space

Find the value of $\alpha$ such that the curvature of $y=e^{\alpha\cdot x}$ at $x=0$ is as larger as possible. Please help

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Welcome to MSE! It might enhance your chances to get helpful feedback if you share with us what you have tried; where you got stuck; which context (book, class, ...) this is in; and similar such questions. – gnometorule Feb 28 '13 at 21:21

Given:

$$\tag 1 \displaystyle y (x) =e^{\alpha\cdot x}$$

Find the value of $\alpha$ such that the curvature of $(1)$ at $x=0$ is as large as possible.

The curvature, $k(x)$, of a curve is given by:

$$\displaystyle k(x) = \frac{|y''(x)|}{(1 + (y'(x))^2)^{3/2}}$$

You want to calculate $\displaystyle k(0)$ as a function of $\displaystyle \alpha$ and make this value as large as possible for the maximum.

Update

So, we get:

$$\displaystyle k(x) = \frac{|\alpha^2 e^{\alpha x}|}{(1 + (\alpha e^{\alpha x})^2)^{\Large \frac{3}{2}}}$$

At $x = 0$, we get:

$$\displaystyle k(0) = \frac{\alpha^2}{(1 + \alpha^2)^{\Large \frac{3}{2}}}$$

Let $\displaystyle f(\alpha) = k(0) = \frac{\alpha^2}{(1 + \alpha^2)^{\Large \frac{3}{2}}}.$

Now, how do you make this value as large as possible? Find the maximum based on $\alpha$, that is, find $\alpha$ when $f'(\alpha) = 0.$

You should get a maximum at $\alpha = \pm \sqrt{2}$, which has a maximum value $\displaystyle \frac{2}{3 \sqrt{3}}$

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Replace "calculate $k'(x)=0,\ \ldots$" by: Calculate $k(0)$ as a function of $\alpha$ and make this value as large as possible. – Christian Blatter Mar 1 '13 at 21:16
@ChristianBlatter: Thanks for correcting my error! Regards – Amzoti Mar 1 '13 at 21:25
+ Great start and nice update...no feedback from OP :-( – amWhy Apr 27 '13 at 0:12