posted by JDS .
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...
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 user-unfriendly program based on Newton's Methos that I made up myself back in the 80's in "Dos-Basic" 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 .....