Trigonometry Arc Length Problem

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The discussion revolves around solving an arc length problem in a pre-calculus context while the user is simultaneously taking calculus. The user initially struggled with the lack of formal instruction on arc length and attempted to use the Pythagorean theorem to find the arc length, arriving at approximately √8 units. After confirming this with integration, the user sought a more formal trigonometric approach and calculated the central angle to be 16.3°. Ultimately, the arc length was consistently found to be around 2.8 units, aligning with both the Pythagorean and integration methods. The conversation highlights the challenges of navigating different mathematical concepts across courses.
Euler2718
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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.

a=\theta r

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

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 a=\theta r seems to be out of the question. This seemed to be an unusual problem, so I used Pythagoras after a while. I went

( \Delta x )^{2} + (\Delta y)^{2} = a^{2}

This yields the arc length to \sqrt{8} 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:

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

Which is pretty much \sqrt{8} 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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Can you figure out the angle AQB?
 
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phinds said:
Can you figure out the angle AQB?

Ahhh. Using the right triangles no doubt. I'll give it a go. Thanks.
 
Problems involving derivatives or integrals should be posted in the calculus section, not in the precalculus section. I have moved this thread.
 
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.
 
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.
 
Sounds good.
 
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