1. ## Curvature

Calculate the curvature at the top of the curve of the parabola $\displaystyle y=ax^2+bx+c$

2. Originally Posted by lynch-mob
Calculate the curvature at the top of the curve of the parabola $\displaystyle y=ax^2+bx+c$
$\displaystyle \kappa = \frac{|y''|}{(1 + y'^2)^{3/2}}$

In your case, since the slope is zero at the maximum point, so the curvature is 2|a|.

-Dan

Edit: I forgot to add the factor of 2 due to taking the derivatives of y, not to mention the absolute value bars. Sorry about that.

3. Originally Posted by topsquark
$\displaystyle \kappa = \frac{|y''|}{(1 + y')^{3/2}}$

In your case, since the slope is zero at the maximum point, so the curvature is a.

-Dan

Before I start, shouldnīt the formula be $\displaystyle \kappa=\frac{|y''|}{(1+y'^2)^\frac{3}{2}}$

4. Originally Posted by lynch-mob
Before I start, shouldnīt the formula be $\displaystyle \kappa=\frac{|y''|}{(1+y'^2)^\frac{3}{2}}$
Yes of course, it was a typo by topsquark

5. I fixed it in my original post. Thanks for the catch, both of you.

-Dan

6. ## help

Im sorry guys but Iīm getting nowhere. First of all. All I got is the formula but no examples in the literature except for the curvature of a circle. I need some examples. Second is curvature the correct word for this area of calculus. I read somewhere about concavity of a curve, is it the same thing?

I tried to put the second derivative of the function in the formula but it takes me nowhere.

Could somebody also explain to me why the curvature at the top of the curve must be $\displaystyle a$?

HELP!!

7. ## Pathetic mathematician - Curvature

Please, I really need some help on this one. I canīt find anything useful anywhere on the internet or in my books.

What is the curvature at the top of the parabola $\displaystyle y=ax^2+bx+c$

And how do I calculate it?

Originally Posted by topsquark
$\displaystyle \kappa = \frac{|y''|}{(1 + y'^2)^{3/2}}$

In your case, since the slope is zero at the maximum point, so the curvature is a.

-Dan
How do I know that slope=0 means the curvature is a?

Hereīs my effort, which probably is totally useless!
$\displaystyle f'(x)=2ax+b$
$\displaystyle f''(x)=2a$
so we have:
$\displaystyle \kappa=\frac{|2a|}{(1+(2ax+b)^2)^{\frac{3}{2}}}$

uhhh....uuhhhh??? Is there a need for a calculation at all or do we only need to explain the obvious? (well not obvious to me, thats for sure!)

8. Originally Posted by lynch-mob
Please, I really need some help on this one. I canīt find anything useful anywhere on the internet or in my books.

What is the curvature at the top of the parabola $\displaystyle y=ax^2+bx+c$

And how do I calculate it?

How do I know that slope=0 means the curvature is a?

Hereīs my effort, which probably is totally useless!
$\displaystyle f'(x)=2ax+b$
$\displaystyle f''(x)=2a$
so we have:
$\displaystyle \kappa=\frac{|2a|}{(1+(2ax+b)^2)^{\frac{3}{2}}}$

uhhh....uuhhhh??? Is there a need for a calculation at all or do we only need to explain the obvious? (well not obvious to me, thats for sure!)

At the top of the parabola $\displaystyle y'(x)=0$

RonL

9. Originally Posted by topsquark
$\displaystyle \kappa = \frac{|y''|}{(1 + y'^2)^{3/2}}$

In your case, since the slope is zero at the maximum point, so the curvature is a.

-Dan
$\displaystyle |y''|=2|a|$ not $\displaystyle |a|$

RonL

10. Originally Posted by CaptainBlack
$\displaystyle |y''|=2|a|$ not $\displaystyle |a|$

RonL
(sigh) I hate it when that happens!

Thanks for the catch.

-Dan