Saturday
April 19, 2014

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.

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 .....

Thank you!

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