Question

A particle has a position function r(t)=cos(3t)i+sin(3t)j+tk, where the arguments of the cosine and sine functions are in radians. What is the particle's velocity vector?

A) v(t)=−3sin(3t)i+3cos(3t)j+k
B) v(t)=−3sin(3t)i−3cos(3t)j+k
C) v(t)=3sin(3t)i+3cos(3t)j+k
D) v(t)=3sin(3t)i−3cos(3t)j+k

Answer 1

The velocity vector v(t) of the particle with the position function r(t) = cos(3t)i + sin(3t)j + tk is v(t) = -3sin(3t)i + 3cos(3t)j + k.

Step-by-step explanation:

The velocity vector of a particle is found by differentiating its position vector r(t). Given the position function r(t) = cos(3t)i + sin(3t)j + tk, the velocity vector v(t) is obtained by the derivative of each component of r(t) with respect to time t. So, the velocity vector is:

• Derivative of cos(3t)i with respect to time is -3sin(3t)i.
• Derivative of sin(3t)j with respect to time is 3cos(3t)j.
• Derivative of tk with respect to time is k since the derivative of t with respect to t is 1.

Thus, the velocity vector v(t) of the particle is v(t) = -3sin(3t)i + 3cos(3t)j + k.