Question

A particle is moving with a position function of r(t) = 〈3t2 −4t, 1 −5t, −3 + t3〉, with distance in meters, and

time in seconds.

(a) At what time is the particle at (4, −9, 5)?

(b) What are the velocity and acceleration vectors of the particle at (4, −9, 5)?

(c) Give a parametrization for the line tangent to the path of the particle at the point (4, −9, 5).

(d) Give a parametrization for the line tangent to the path of the particle at the point (7, 6, −4).

Answer 1

(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: