(a) To find the time when the particle is at (4,-9,5), we need to set the position function equal to the given point and solve for t:
3t^2 - 4t = 4
1 - 5t = -9
t^3 - 3 = 5
Solving these equations, we get:
t = 2 or -1 or ∛8
Since time cannot be negative, the particle is at (4,-9,5) at t = 2 seconds.
(b) To find the velocity vector, we need to take the derivative of the position function:
r'(t) = 〈6t - 4, -5, 3t^2〉
Evaluating r'(2), we get:
r'(2) = 〈8, -5, 12〉
To find the acceleration vector, we need to take the second derivative of the position function:
r''(t) = 〈6, 0, 6t〉
Evaluating r''(2), we get:
r''(2) = 〈6, 0, 12〉
(c) The line tangent to the path of the particle at the point (4,-9,5) has direction vector equal to the velocity vector r'(2) that we found in part (b). A parametrization of this line is:
〈4, -9, 5〉 + t〈8, -5, 12〉
(d) The line tangent to the path of the particle at the point (7,6,-4) has direction vector equal to the velocity vector r'(t) evaluated at t such that r(t) = 〈7,6,-4〉. To find this value of t, we need to solve the system of equations:
3t^2 - 4t = 7
1 - 5t = 6
t^3 - 3 = -4
Solving these equations, we get:
t = -1 or 1 or ∛125
Since the particle is moving along a path, we need to choose the value of t that corresponds to the direction of motion. Since the particle is moving away from the point (7,6,-4), we choose t = 1. A parametrization of the line is: