Trigonometry Arc Length Problem

In summary: Thanks for helping.In summary, the lecture on arc length was not helpful, and the attempt at a solution used the Pythagorean approximation and integration.
  • #1
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First I'd just like to point out that I'm taking calculus and advance pre-calculus simultaneously (kind of a stupid system) and this is a problem in the pre-calc.

1. Homework Statement


IMG_3151.JPG


2. Homework Equations


Let 'a' be arc length.

[tex] a=\theta r [/tex]

[tex] a = \int_{a}^{b} \sqrt{1+[f'(x)]^{2}} dx [/tex]

3. The Attempt at a Solution

The lecture on arc length wasn't even a lecture. Merely a hand out of the chapter summery, not to be touched again. I saw no central angle so [tex] a=\theta r [/tex] seems to be out of the question. This seemed to be an unusual problem, so I used Pythagoras after a while. I went

[tex] ( \Delta x )^{2} + (\Delta y)^{2} = a^{2} [/tex]

This yields the arc length to [tex] \sqrt{8} [/tex] units.

I confirmed this with the integration for arc length. But I'm really not at the level of arc length integration yet. I can do basic integration only (I skipped ahead in the course a bit). Via wolfram alpha:

[tex] \int_{-8}^{-6} 10\sqrt{\frac{1}{100-x^{2}}} dx = 10\sin ^{-1}\left(\frac{7}{25}\right) \approx 2.83794 [/tex]

Which is pretty much [tex]\sqrt{8}[/tex] So really, my question is if there is another more formal way to approach this problem using trigonometric concepts to arrive at an answer.

Thank you in advance.
 
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  • #2
Can you figure out the angle AQB?
 
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  • #3
phinds said:
Can you figure out the angle AQB?

Ahhh. Using the right triangles no doubt. I'll give it a go. Thanks.
 
  • #4
Problems involving derivatives or integrals should be posted in the calculus section, not in the precalculus section. I have moved this thread.
 
  • #5
Mark44 said:
Problems involving derivatives or integrals should be posted in the calculus section, not in the precalculus section. I have moved this thread.

Sorry. I merely included the calculus to show what I used to compute the answer, as it was technically part of my attempt.
 
  • #6
Pretty sure I got it now! Thank you phinds for the tip. I calculated the central angle to be 16.3°, and used the arc length equation from there. It comes out to 2.8 units, which is what the Pythagorean approximation and the integration came to.
 
  • #7
Sounds good.
 
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1. What is the formula for finding the arc length in trigonometry?

The formula for finding the arc length in trigonometry is s = rθ, where s represents the arc length, r is the radius of the circle, and θ is the central angle in radians.

2. How do you convert from degrees to radians?

To convert from degrees to radians, you can use the formula θ (in radians) = θ (in degrees) x π/180. For example, if the central angle is 60 degrees, the equivalent in radians would be 60 x π/180 = π/3 radians.

3. Can you use the arc length formula for any shape?

No, the arc length formula (s = rθ) is only applicable to circles or circular arcs. For other shapes, the arc length must be calculated using a different formula or method.

4. How do you find the arc length if the central angle is given in degrees?

If the central angle is given in degrees, you must first convert it to radians using the formula mentioned in question 2. Then, you can use the formula s = rθ to calculate the arc length.

5. Is it possible for the arc length to be larger than the circumference of the circle?

No, the arc length can never be larger than the circumference of the circle. The arc length is a portion of the circumference, so it will always be smaller or equal to the circumference.

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