Calc Radius of Curvature & Center P(h,k) + Improper Integrals

In summary, the radius of curvature is the distance between the center of a circle and the curve it is drawn around. To find the radius, you use the equation: R = \frac{\mbox{arc}}{\theta} and the angle between the tangent lines is given by: \tan \theta.
  • #1
hawaiidude
41
0
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.
 
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  • #2
[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
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
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
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
Here's a little drawing that might help you understand what I did... still want to know where my mistake is.
 

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  • #7
Originally posted by Chen
Here's a little drawing that might help you understand what I did... still want to know where my mistake is.

maybe here?

Originally posted by Chen

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]

i don't follow that last step, but that is not the normal formula for ds that i know. it should look like:

[tex]
ds=\sqrt{1+(f')^2}dx[/tex]
 
  • #8
Originally posted by lethe
i don't follow that last step, but that is not the normal formula for ds that i know. it should look like:

[tex]
ds=\sqrt{1+(f')^2}dx[/tex]
Well, I just used the formula of distance between two points. I took two points on the graph [[tex]x_1, f(x_1)[/tex]] and [[tex]x_2, f(x_2)[/tex]] with [tex]x_2 - x_1[/tex] tending to zero... how is that wrong, and how do you get to the formula you posted? :smile:
 
  • #9
Originally posted by Chen
Well, I just used the formula of distance between two points. I took two points on the graph [[itex]x_1, f(x_1)[/itex]] and [[itex]x_2, f(x_2)[/itex]] with [itex]x_2 - x_1[/itex] tending to zero... how is that wrong
yeah, dx tends to zero, but so does f(x2)-f(x1), so to zeroth order, ds is just zero. you can't lose one term but keep the other.

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

and how do you get to the formula you posted? :smile:
multiply and divide the whole thing by x2-x1. do your division inside the radical (which means you are actually dividing by (x2-x1)^2) and your multiplication outside the radical
 
  • #10
Ah-ha! I see now how to get the correct formula for the curvature. Thank you very much.
 
  • #11
thank you all always a great help! now i think i understand the concept of radius of curvature proving
 

1. What is the formula for calculating the radius of curvature and center of a curve?

The formula for calculating the radius of curvature and center of a curve is given by:
R = [(1 + (dy/dx)^2)^(3/2)] / |d^2y/dx^2|
Where dy/dx is the first derivative of the curve and d^2y/dx^2 is the second derivative.

2. How do you find the center of curvature for a given point on a curve?

To find the center of curvature for a given point on a curve, first calculate the radius of curvature using the formula mentioned above. Then, draw a perpendicular line from the given point to the curve. The intersection of this perpendicular line and the tangent line at the given point will be the center of curvature.

3. What is the significance of the radius of curvature and center of a curve?

The radius of curvature and center of a curve determine the degree of curvature at a specific point on the curve. A smaller radius of curvature indicates a sharper curve, while a larger radius of curvature indicates a more gradual curve. The center of curvature helps to understand the direction in which the curve is bending at a given point.

4. What are improper integrals and how are they different from regular integrals?

Improper integrals are integrals that do not have a finite value. This can happen when the limits of integration are infinite or if the integrand is undefined at some point within the limits. Unlike regular integrals, improper integrals cannot be evaluated using the fundamental theorem of calculus and require special techniques to solve.

5. How do you evaluate improper integrals?

To evaluate improper integrals, you can use one of the following methods:
1. Split the integral into smaller parts and evaluate each part separately.
2. Use a substitution to transform the integral into a form that can be evaluated.
3. Use a comparison test to determine if the integral converges or diverges.
4. Apply the limit definition of the integral to find the value of the integral.
It is important to check for convergence before attempting to evaluate an improper integral.

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