Finding the Limit of Curvature for a Polar Curve

In summary, using the polar coordinate formula for radius of curvature, the curvature K for the given polar curve is 1/sqrt((a^2 + 1)e^(2ax)) and the limit of K as (a) theta approaches infinity and (b) as a approaches infinity is 0.
  • #1
WK95
139
1

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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  • #2
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.
 
  • #3
I'm getting 0 for the limit of x', x'', y', y'' as a approaches infinity and when theta approaches infinity.
 
  • #4
A curvature that approaches zero makes sense.
 
  • #5
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.
 
  • #6
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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  • #7
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.
 

What is the curvature of a polar curve?

The curvature of a polar curve is a measure of how much the curve deviates from a straight line at a given point. It is a value that describes the rate of change of the direction of the curve at that point.

How is the curvature of a polar curve calculated?

The curvature of a polar curve can be calculated using the formula K = |r'(θ)| / [1 + (r(θ))^2]^(3/2), where r(θ) is the polar equation of the curve and r'(θ) is its derivative with respect to θ. This formula takes into account the changing radius and angle of the curve as it is traced out.

What does a high curvature value indicate for a polar curve?

A high curvature value indicates that the curve is sharply bending or changing direction at that point. This could correspond to a sharp corner or cusp in the polar curve, or a very small radius of curvature.

How does the curvature of a polar curve relate to its graph?

The curvature of a polar curve is related to the shape of its graph. For example, a polar curve with a constant curvature will have a circular shape, while a curve with varying curvature may have a more complex shape with sharper or smoother turns.

Can the curvature of a polar curve be negative?

Yes, the curvature of a polar curve can be negative. This occurs when the curve is bending in the opposite direction of the positive curvature, such as in a concave shape. A negative curvature value indicates that the curve is curving inwards at that point.

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