Trigonometry Arc Length Problem

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Homework Help Overview

The discussion revolves around a problem related to arc length in trigonometry, specifically within the context of pre-calculus and calculus. The original poster describes their attempts to solve the problem using both Pythagorean theorem and integration methods, while expressing uncertainty about the appropriate approach.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to use the Pythagorean theorem and integration to find the arc length, questioning the applicability of trigonometric concepts. Some participants inquire about the angle AQB and suggest using right triangles to find it.

Discussion Status

The discussion is active, with participants exploring various interpretations of the problem. Some guidance has been offered regarding the calculation of the central angle, and the original poster has reported progress in their understanding and calculations.

Contextual Notes

There are indications of confusion regarding the appropriate mathematical level for the problem, as it involves both pre-calculus and calculus concepts. The original poster has noted a lack of formal instruction on arc length in their course materials.

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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