Arc Length Curve: Find Point at Distance 26pi

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SUMMARY

The discussion focuses on finding a point on the curve defined by the parametric equation r(t) = (5Sin(t))i + (5Cos(t))j + 12tk at a distance of 26π units from the initial point (0, 5, 0). The solution involves calculating the arc length using the formula L = ∫ |v| dt from 0 to T, where T is determined to be 2π. The final position at this distance is (0, 5, 24π). The initial position at t=0 is crucial as it establishes the starting point for the arc length calculation.

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Homework Statement



Find the point on the curve r(t) = (5Sint)i + (5Cost)j + 12tk
at a distance 26pi units along the curve from the point (0,5,0) in the direction of increasing arc length.

Homework Equations



L = int (|v|) from 0 to T.

The Attempt at a Solution



T comes to be 2pi and when substituted, the position comes to be (0,5,24pi). However, this calculation and answer though correct (according to the back of the book) does not involve the use of the fact that at time t=0, the particle is at (0,5,0).

What for is that information given then?
 
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It certainly uses that info, in a rather subtle way; namely indirectly telling you that the initial value of t should be zero.

Thus, if s is the arclength parameter, we have the relationship t=s/13.
 
Thanks!

Thanks a ton! I never knew that!
 

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