Interval for the Length of an Arc

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ineedhelpnow
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find the length of an arc of a helix r(t)=(sint,cost,t) from the point (0,2,0) to (0,5,2pi)

would the interval when integrating be from 0 to 2pi because t in the case is (z=t)? please say yes. please say yes.
 
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ineedhelpnow said:
find the length of an arc of a helix r(t)=(sin2t,cos2t,t) from the point (0,2,0) to (0,5,2pi)

would the interval when integrating be from 0 to 2pi because t in the case is (z=t)? please say yes. please say yes.

Yes.
 

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ok ILS ima let you in on a little secret (i don't really remember the points. all i remember are the z coordinates 0 and 2pi) but it was a helix for sure. it was probably 5 instead of 2. i don't know i can't remember. i don't want to say i made them up but I am going to saaaaay improvised.
 
ineedhelpnow said:
ok ILS ima let you in on a little secret (i don't really remember the points. all i remember are the z coordinates 0 and 2pi)

Okay...
How about $r(t)=\left(\frac{3t}{2\pi}\sin(t),\ 2+\frac{3t}{2\pi}\cos(t),\ t\right)$?
 
what about it?
you would differentiate and pull out a common factor and simplify the sin and cos to 1. and then add the extra 1 and take the square root.
 
ineedhelpnow said:
what about it?
you would differentiate and pull out a common factor and simplify the sin and cos to 1. and then add the extra 1 and take the square root.

Did you really differentiate it?
How about the factor $t$ that is in both the x-coordinate and the y-coordinate (which is an integral part of a helix)?
 
factor it out before you differentiate.
 
why can't it be factored out? :confused:
 
ineedhelpnow said:
what about it?
you would differentiate and pull out a common factor and simplify the sin and cos to 1. and then add the extra 1 and take the square root.
:D soooo you can just do what i said earlier. by differentiating. pulling out a common factor of 3/2pi and---- oh i see what i did wrong. (Giggle) made a mistake with the differentiating part.