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1. Homework Statement

1. Homework Statement

2. Homework Equations

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

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.