Posted by JDS on Monday, December 10, 2007 at 1:25am.
I'm desperate!
Find the point where the curve r(t)=(12sint)i  12(cost)j+ 5tk is at a distance 13pi units along the curve from the point (0,12,0) in the direction opposite to the direction of increasing arc length.
Thanks for any advice...

MATHPlease help!!  Reiny, Monday, December 10, 2007 at 10:05am
changing your equation to parametric form gave me
x = 12sint
y = 12cost
z = 5t
using the distance formula I got
(12sint)^2 + (12cost+12)^2 + (5t)^2 = (13pi)^2
which simplified to the nasty transcendental equation
25t^2  288cost  1379.9631 = 0
none of the online equation solvers I found were able to solve this equation, so I went to an old very userunfriendly program based on Newton's Methos that I made up myself back in the 80's in "DosBasic" to find a solution near t = ±7.612121.
(Perhaps you have a programmable calculator which gives you a better solution, nevertheless this one gave quite a significant error when I substituted.)
I then used the positive result to get the point (11.651,2.874,38.061)
That point is 40.837 units away from your given point, and the value of 13pi is appr. 40.84, so .....

MATHPlease help!!  JDS, Monday, December 10, 2007 at 10:56am
Thank you!
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