The position of a particle in motion in the plane at time t is...?

Question

The position of a particle in motion in the
plane at time t is
X (t ) = −9t I + sin(−9t ) 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

Answer 1

(a) The speed of the particle is given by 'norm' of the velocity. First you need to find velocity. This is done by taking the derivative of the position vector x(t).
Lets call the velocity vector v(t):
v(t) = x'(t) = -9 I -9cos(-9t) J............(*)

Now you need the 'norm' of v(t). This is done by taking the square root of the 'sum of squares': Lets call the speed |v(t)|:
|v(t)| = root of (81+ 81(cos(-9t))^2)
Simplifying:
root of [81(1+(cos(-9t))^2)]
applying the identiy that 1+cos^2(x) = sin^2(x):
= root of [81(sin(-9t))^2]
= 9sin(-9t)....(**)

(b) The vector tangent to x(t) is v(t), which was already found above as (*).
A unit vector is found by simply diving each term by the norm, which was also found above as (**)

you get:
[-9 I -9cos(-9t) J] / [9sin(-9t)]
= [-1/sin(-9t)] I - [cot(-9t)] J

Answer 2

Differentiate it ,
X'(t)= (-9)I + (-9) cos(-9t)

(a) V(t)= {81+81[cos(-9t)]^2}^(1/2) = 9 (1+cos(-9t)^2)

(b) the unit tangent vector to X (t ) is:
(-1)/(1+cos(-9t)^2) I + (-1)cos(-9t) /(1+cos(-9t)^2)J