Final answer:
The parametrization of the tangent line at the point where t = 1 for the curve r(t) = (ln t)i + t^{-4}j + 3tk is L(t) = ti + (1 - 4t)j + (3 + 3t)k.
Step-by-step explanation:
To find a parametrization of the tangent line at a given point on the curve r(t), we first need to find the derivative of the vector function r(t) to obtain the tangent vector. The vector function given is r(t) = (ln t)i + t^{-4}j + 3tk, and we are interested in the tangent at t = 1.
The derivative of r(t), denoted as r'(t), gives us the components of the tangent vector at any point t. The derivative of each component is as follows:
- d(ln t)/dt = 1/t
- d(t^{-4})/dt = -4t^{-5}
- d(3t)/dt = 3
Thus, r'(t) = (1/t)i - 4t^{-5}j + 3k. At t = 1, this simplifies to r'(1) = i - 4j + 3k.
The point on the curve at t = 1 is given by r(1) = (ln 1)i + (1^{-4})j + 3(1)k = 0i + 1j + 3k.
Hence, the parametrization of the tangent line L(t) at t = 1 can be given by the point on the curve at t = 1 plus a scalar multiple of the tangent vector:
L(t) = r(1) + t * r'(1) = (0i + 1j + 3k) + t(i - 4j + 3k)
Therefore, the parametrization of the tangent line at t = 1 is L(t) = ti + (1 - 4t)j + (3 + 3t)k.