Question

The position of a particle in motion in the plane at time t is
X(t) = −7 t I + sin(−3 t) J.
At time any t, determine the following:
(a) the speed of the particle is: ???
(b) the unit tangent vector to X(t) is: ??? I + ??? J
(Note that, a unit vector is a vector whose length is 1. Multiplying any non-zero vector V by 1/|V| produces a unit vector, and multiplying any unit vector by -1 shows that there are two unit vector which are multiples of any non-zero vector. The unit tangent vector to X(t) is defined as V(t)/|V(t)|. )

Answer 1

\[x(t) = (-7t) i + \sin(-3t)j\]

(a) The velocity of the particle is the derivative with respect to time of the position, so take the derivative of x(t):
\[v(t) = x'(t) = -7i + (-3)\cos(-3t)j\]
[perhaps by "speed" they mean the absolute value of the velocity? Lets calculate that]:
\[\sqrt{(-7)^2 + (-3 * \cos(-3t) )^2}\]
\[\sqrt{(-7)^2 + (9 * \cos^2(-3t) )}\]

(b) The tangent unit vector of the position is the velocity vector divided by the absolute value of the velocity vector, which is a mess:

\[v(t) / \left| v(t) \right|\]
\[v(t) = (-7/\sqrt{49+9\cos^2(-3t)}) i - 3\cos^2(-3t)/\sqrt{49+9\cos^2(-3t)} j\]

Are you sure they didn't specify a value of t for the tangent vector calculation?

Answer 2

says the J part of the equation is wrong !