Find the point on the curve

r(t)=(12sint)**i**-(12cost)**j**+5t**k**
at a distance 13pi units along the curve from the point
(0,-12,0) when t=0 in the direction opposite to the direction of increasing arc length.

Thanks vix

Hi Vix,
You have the curve expressed in vector form

r(t) = x(t)**i** + y(t)**j** + z(t)**k**
and hence you can write the arclength aoong this curve from t=t_{1} to t=t_{2} as
Calculate (x'(t))^{ 2}, (y'(t))^{ 2} and (z'(t))^{ 2} and simplify the square root of their sum. The result is a very simple expression.

Since the integrand (the square root) is a positive expression, the arclength increases as t increases. The given point, (0,-12,0) results when t=0 and hence to get a decreasing arclength you can integrate from t_{1} = 0 to a negative value of t_{2}. Thus evaluate the integral and solve

for t_{2}.
Harley