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Trigonometry Arc Length Problem

  • Thread starter Euler2718
  • Start date
  • #1
90
3
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.
 

Answers and Replies

  • #2
phinds
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Can you figure out the angle AQB?
 
  • #3
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Can you figure out the angle AQB?
Ahhh. Using the right triangles no doubt. I'll give it a go. Thanks.
 
  • #4
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Problems involving derivatives or integrals should be posted in the calculus section, not in the precalculus section. I have moved this thread.
 
  • #5
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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
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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
phinds
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Sounds good.
 

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