Final answer:
The parametric equations for the line tangent to the given curve at a given parameter value t₀ can be found by taking the derivative of the curve and evaluating it at t₀.
Step-by-step explanation:
The parametric equations for the line that is tangent to the curve can be found by taking the derivative of the curve at the given parameter value, t₀. Given that r(t) = (cost)i + (sint)j + (sin2t)k, the derivative of r(t) with respect to t is dr/dt = -sinti + costj + 2cos2tk. Evaluating the derivative at t = t₀ and using the point-slope form of a line, we can write the parametric equations for the tangent line as:
x = x₀ + (-sint₀)t
y = y₀ + (cost₀)t
z = z₀ + 2cos2(t₀)t
where (x₀, y₀, z₀) is the position vector of the given curve at t₀.