Question about arc length and the condition dx/dt > 0

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The discussion centers on the condition dx/dt > 0 in the context of arc length calculations from James Stewart's Multivariable Calculus. Participants explore whether the formula for arc length can still be applied when dx/dt < 0, with some arguing that it can, while others emphasize that the condition ensures the curve is traversed in one direction. A specific example of the parametric curve x = sin(t), y = 1 is analyzed, illustrating that the arc length should remain consistent regardless of the direction of traversal. It is clarified that if dx/dt < 0, adjustments in the integral's limits are necessary to account for the curve being traversed in reverse. Overall, understanding the implications of the condition is crucial for accurate arc length calculations.
songoku
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That passage is from James Stewart (Multivariable Calculus). I want to ask about the condition dx/dt > 0. If dx / dt < 0, the formula can't be used?

Thanks
 
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Why not ? Can you find a counter-example ?
 
BvU said:
Why not ? Can you find a counter-example ?
In my opinion, it can because from the derivation I don't see the need for dx/dt to be positive.

I just don't understand why dx / dt > 0 is written there in the text.

Thanks
 
songoku said:
I just don't understand why dx / dt > 0 is written there in the text.
If dx/dt > 0 then then x is increasing. Conversely, if dx/dt < 0 then x is decreasing.

Consider the parametric curve ##x = \sin(t), y = 1## for ##t \in [0, \pi]##. Graph this simple "curve" and determine its arc length from your graph. What does the integral formula for arc length of this curve produce?
 
The length of a curve should be the same, whether you start measuring it from the left end (dx/dt &gt; 0) or the right end (dx/dt &lt; 0). Some details of the derivation will change if dx/dt &lt; 0; in particular the assumption that f(\alpha) = a &lt; b = f(\beta) must be replaced by f(\alpha) = b &gt; a = f(\beta).
 
Mark44 said:
If dx/dt > 0 then then x is increasing. Conversely, if dx/dt < 0 then x is decreasing.

Consider the parametric curve ##x = \sin(t), y = 1## for ##t \in [0, \pi]##. Graph this simple "curve" and determine its arc length from your graph. What does the integral formula for arc length of this curve produce?
The graph is horizontal line y = 1 and 0 ≤ x ≤ 1.

I think the arc length should be 2 because the curve is traversed twice, once from left to right for 0 ≤ t ≤ π/2 and then from right to left for π/2 ≤ t ≤ π.

From integration, I get zero

pasmith said:
The length of a curve should be the same, whether you start measuring it from the left end (dx/dt &gt; 0) or the right end (dx/dt &lt; 0). Some details of the derivation will change if dx/dt &lt; 0; in particular the assumption that f(\alpha) = a &lt; b = f(\beta) must be replaced by f(\alpha) = b &gt; a = f(\beta).
I think I understand more clearly now from your post and Mark's example. The condition dx / dt > 0 is to ensure the curve is traversed once, only from left to right. If I want to use integration to find the arc length, I need to divide it into two cases, for 0 ≤ t ≤ π/2 and π/2 ≤ t ≤ π then subtract.

Thank you very much for the help and explanation BvU, Mark44, pasmith
 
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