- #1

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$$ds = \sqrt{dx^2 + dy^2}$$

$$ds = \sqrt{dx^2(1 + {\frac{dy}{dx}}^2)}$$

$$ds = \sqrt{dx^2} \sqrt{1 + [f'(x)]^2}$$

$$ds = \sqrt{1 + [f'(x)]^2} dx$$

Isn't ##\sqrt{dx^2}## equal to ##|dx|##, and not ##dx##?

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- Thread starter PFuser1232
- Start date

- #1

- 479

- 20

$$ds = \sqrt{dx^2 + dy^2}$$

$$ds = \sqrt{dx^2(1 + {\frac{dy}{dx}}^2)}$$

$$ds = \sqrt{dx^2} \sqrt{1 + [f'(x)]^2}$$

$$ds = \sqrt{1 + [f'(x)]^2} dx$$

Isn't ##\sqrt{dx^2}## equal to ##|dx|##, and not ##dx##?

- #2

mfb

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- #3

- 479

- 20

I'm not very familiar with the notion of "going in the positive direction" while plotting a function, I had no idea it makes a difference. Could you please elaborate?

- #4

mfb

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If you want to know the arc length between x=2 and x=4 for example, you integrate x from 2 to 4 and not from 4 to 2.

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