Finding the Limit of Curvature for a Polar Curve

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Homework Help Overview

The discussion revolves around finding the curvature of a polar curve defined by r=e^(a*theta) where a>0. Participants are tasked with determining the limits of curvature as theta and a approach infinity.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss converting the polar equation to Cartesian coordinates and calculating derivatives to apply the curvature formula. There are questions about the behavior of trigonometric functions as their arguments approach infinity, particularly regarding the limits of asin(x) and a*sin(x).

Discussion Status

Some participants express confusion about the limits leading to indeterminate forms, while others suggest alternative approaches using polar coordinate formulas. There is acknowledgment of the expectation that curvature approaches zero in both limit cases, but no consensus on how to resolve the indeterminate forms has been reached.

Contextual Notes

Participants note the complexity of the calculations involved and the potential for encountering zero in the denominator, which complicates the limit evaluation. The original problem setup and the imposed conditions on a and theta are under discussion.

WK95
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Homework Statement


Given the polar curve r=e^(a*theta), a>0, find the curvature K and determine the limit of K as (a) theta approaches infinity and (b) as a approaches infinity.

Homework Equations


x=r*cos(theta)
y=r*sin(theta)
K=|x'y''-y'x''|/[(x')^2 + (y')^2]^(3/2)

The Attempt at a Solution


I've tried converting the polar curve using the first equation, solving for their first and second derivatives, then plugging them into equation 3 but that gets very, very long.

So next, I apply the properties of limits to the relevant equations. However, I get stuck when I need to find the limit of asin(x), acos(x), sin(x), cos(x) as x approaches infinity. What now?
 
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You know sin(x) and cos(x) are limited. That should be sufficient.

If asin is the inverse sine, then asin(x) with x to infinity shouldn't occur. If it is a*sin(x), then see above.
 
I'm getting 0 for the limit of x', x'', y', y'' as a approaches infinity and when theta approaches infinity.
 
A curvature that approaches zero makes sense.
 
mfb said:
A curvature that approaches zero makes sense.

Yes, but then I'd end up with a 0 in the denominator in the third equation which wouldn't work. However, that would require me to find a work around and solve for the limit some other way so I don't get 0/0. I haven't been able to figure out what I have to do to get around that.

I am expecting the curvature to be 0 in both cases when a or theta approaches infinity so I'm expecting the numerator to be 0.
 
WK95 said:

Homework Statement


Given the polar curve r=e^(a*theta), a>0, find the curvature K and determine the limit of K as (a) theta approaches infinity and (b) as a approaches infinity.

Homework Equations


x=r*cos(theta)
y=r*sin(theta)
K=|x'y''-y'x''|/[(x')^2 + (y')^2]^(3/2)

The Attempt at a Solution


I've tried converting the polar curve using the first equation, solving for their first and second derivatives, then plugging them into equation 3 but that gets very, very long.

So next, I apply the properties of limits to the relevant equations. However, I get stuck when I need to find the limit of asin(x), acos(x), sin(x), cos(x) as x approaches infinity. What now?
Instead of using the radius of curvature formula for cartesian coordinates, why don't you try the equivalent formula for polar coordinates?

http://mathworld.wolfram.com/RadiusofCurvature.html
 
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SteamKing said:
Instead of using the radius of curvature formula for cartesian coordinates, why don't you try the equivalent formula for polar coordinates?

http://mathworld.wolfram.com/RadiusofCurvature.html
Thanks! That's just what I need.

I've ended up with R=sqrt((a^2 + 1)e^(2ax)) so K=1/sqrt((a^2 + 1)e^(2ax)). Taking the limit as a approaches infinity I get 0 and taking the limit as x approaches infinity, I also get 0.
 

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