Finding the Arc Length of a Polar Function

In summary, the conversation is about finding the arc length of the curve r=2/(1-cosθ) from π/2 to π using integration. The attempt at a solution involved finding the derivative of r and using a half-angle formula to simplify the integration. However, there was a mistake in identifying the half-angle formula, leading to an incorrect evaluation of the integral. After correcting the formula, the correct integral was found and evaluated, resulting in a solution for the arc length.
  • #1
jgiarrusso
7
0
Hi, I've been having some issues in solving this problem.

Homework Statement


Find the arc length of r=2/(1-cosθ) from π/2 to π

Homework Equations


L =(integrate) sqrt(r2+(dr/dθ)2)dθ

The Attempt at a Solution



I found (dr/dθ) = (-2sinθ)/(1-cosθ)2

so (dr/dθ)2 = (4sin2θ)/(1-cosθ)4

Then r2 = 4/(1-cosθ)2

both have a factor of 4/(1-cosθ)2, so I pulled that outside the sqrt to get

L=(integrate) 2/(1-cosθ) * sqrt(1+(sin2θ)/(1-cosθ)2)

then I multiplied the 1 by (1-cosθ)2/(1-cosθ)2 to give common denominators.

After multiplying it out, the numerator of the fraction was 1-2cosθ+cos2θ+sin2θ, so I got rid of the sin and cos and added a 1 to get 2 - 2cosθ

I pulled out a factor of sqrt2 and ended up with:

L = (integrate) 2sqrt2/(1-cosθ) * sqrt(1/(1-cosθ))
or
L= (integrate) 2sqrt2*(1-cosθ)-3/2

This is where I got stuck. I can't think of any way to integrate that problem using any of the means we have gone over so far.

Some direction in which way to go would be extremely helpful, thanks in advance.
 
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  • #2
Can you think of a way to express 1-cos(theta) as the square of something using a half-angle formula?
 
  • #3
Aha, thank you so much. That was just the push I needed!
 
  • #4
Hmm, that did allow me to solve the integration (at least, I think I did it properly). But now, when I go to plug in my evaluation, I'm getting sec(π/2) which is doesn't work out.

Using the half angle, I got (1-cosθ)=2cos2(θ/2).

Plugging that in canceled out my 2sqrt2, and made the integration into sec3(θ/2)

Solving that gave me: sec(θ/2)tan(θ/2) + ln|sec(θ/2)+tan(θ/2)|

Now in evaluating from π to π/2, plugging in for π is not a finite number. Did I go wrong in my integration of sec3?
 
  • #5
1-cosθ is not equal to 2cos2(θ/2); it's equal to 2sin2(θ/2).
 
  • #6
Thanks! That's what I get for trying to do my homework far too late at night and without my list of identities nearby ~_~
 

1. What is the definition of arc length in polar coordinates?

Arc length in polar coordinates is the distance along a curve from one point to another, measured along the curve. It is the polar equivalent of the linear distance in Cartesian coordinates.

2. How do you find the arc length of a polar function?

The arc length of a polar function can be found using the following formula: ∫√(r^2 + (dr/dθ)^2)dθ, where r is the polar function and dr/dθ is its derivative with respect to θ. This integral can be evaluated using integration techniques such as substitution or integration by parts.

3. Why is it important to find the arc length of a polar function?

Finding the arc length of a polar function is important in many real-world applications, such as calculating the length of a coastline or the distance traveled by an object in circular motion. It also helps in understanding the shape and behavior of polar curves.

4. Can the arc length of a polar function be negative?

No, the arc length of a polar function cannot be negative. It represents a distance along a curve, which is always a positive value.

5. Are there any limitations to finding the arc length of a polar function?

Yes, there are certain limitations to finding the arc length of a polar function. The function must be continuous and have a continuous derivative on the interval of interest. Additionally, the function must not have any vertical tangents or cusps within the interval, as these would result in an infinite arc length.

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