Final answer:
To find the parametric equations for the tangent line at the point where the curve intersects the xy-plane, we determine the point of intersection by setting z=0, then find the direction of the tangent by taking derivatives of the parametric equations, and finally, write the parametric equations for the tangent line using the intersection point and direction vector.
Step-by-step explanation:
To find the parametric equations for the tangent line to a given curve at a specific point, we must first locate the point where the curve intersects the xy-plane. The curve is defined by the parametric equations x = 2 - t^3, y = 2t - 1, and z = ln(t). The intersection with the xy-plane occurs when z=0, which implies t = 1 since ln(t) is zero only when t is 1. Knowing the value of t, we can find the point of intersection by substituting t back into the equations for x and y, yielding (x, y, z) = (1, 1, 0).
Next, we find the derivatives of the parametric equations with respect to t to get the direction of the tangent line: dx/dt = -3t^2, dy/dt = 2, and dz/dt = 1/t. Evaluating these derivatives at t = 1 gives us the direction of the tangent line: (-3, 2, 1).
Finally, the parametric equations for the tangent line can be formed using the point of intersection and the direction of the tangent: x = 1 - 3s, y = 1 + 2s, and z = s, where s is the parameter along the tangent line.\