
#1
Mar2004, 06:21 AM

P: 45

how do you calculate the radius of curvature and the center p(h,k) of the circle with respect to the curve and how do you do improper integrals? kinda forgot improper integrals.




#2
Mar2004, 07:22 AM

P: 1,006

[tex]R = \frac{[1 + (\frac{dy}{dx})^2]^\frac{3}{2}}{\frac{d^2y}{dx^2}} = \frac{[1 + f'(x)^2]^\frac{3}{2}}{f''(x)}[/tex]




#3
Mar2004, 07:31 AM

P: 1,006

Do you mean like this?
[tex]\int _1 ^\infty \frac{1}{x^2}dx = \lim _{t \rightarrow \infty} \int _1 ^t \frac{1}{x^2}dx = \lim _{t \rightarrow \infty} 1  \frac{1}{t} = 1[/tex] 



#4
Mar2004, 01:54 PM

P: 45

radius of curvature
yes...i was just wondering for the radius of curvature, why is it the sercond derivative? i can compute the curvature but i walsy wondered why the second derivatie, after all, i believe a book said the first was acceptable..




#5
Mar2004, 04:33 PM

P: 1,006

That's not the part that concerns me, I don't understand why the square root of [tex]1 + f'(x)^2[/tex] is raised to the power of three.
I tried to find the radius myself. It is given by: [tex]R = \frac{\mbox{arc}}{\theta}[/tex] I found [tex]\theta[/tex] using the slopes of the tangent lines. The angle between them is given by the formula: [tex]\tan \theta = \lim_{dx \rightarrow 0}\frac{m_1  m_2}{1 + m_1m_2}[/tex] And since [tex]\theta[/tex] is very small you can say that [tex]\tan \theta = \sin \theta = \theta[/tex]: [tex]\theta = \lim_{dx \rightarrow 0}\frac{f'(x_1)  f'(x_2)}{1 + f'(x)^2}[/tex] To find the arc length I just used the normal [tex]d = \sqrt{(x_2  x_1)^2 + (y_2  y_1)^2}[/tex] formula: [tex]\mbox{arc} = \lim_{dx \rightarrow 0}\sqrt{(x_2  x_1)^2 + (f(x_2)  f(x_1))^2} = [/tex] [tex]\mbox{arc} = \lim_{dx \rightarrow 0}f(x_2)  f(x_1)[/tex] From this R is: [tex]R = \lim_{dx \rightarrow 0}\frac{1 + f'(x)^2}{\frac{f'(x_1)  f'(x_2)}{f(x_2)  f(x_1)}} = \frac{1 + f'(x)^2}{f''(x)}[/tex] So where's my mistake? 



#6
Mar2104, 11:14 AM

P: 1,006

Here's a little drawing that might help you understand what I did... still want to know where my mistake is.




#7
Mar2104, 12:08 PM

P: 657

[tex] ds=\sqrt{1+(f')^2}dx[/tex] 



#8
Mar2104, 12:17 PM

P: 1,006





#9
Mar2104, 12:22 PM

P: 657

but in calculus, we are not interested in zeroth order approximations, we are interested in first order approximations. 



#10
Mar2104, 01:28 PM

P: 1,006

Ahha! I see now how to get the correct formula for the curvature. Thank you very much.




#11
Mar2104, 04:54 PM

P: 45

thank you all always a great help! now i think i understand the concept of radius of curvature proving



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