Find the point on the curve

r(t)=(12sint)i-(12cost)j+5tk 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=t1 to t=t2 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 t1 = 0 to a negative value of t2. Thus evaluate the integral and solve

for t2.

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