Note that the radius is 6 (the length of the minute hand) and is a constant. Next, we take the
derivative of the equation with respect to t: = r^2.
Finally, we substitute = into the derivative: in.^2 /s.
SOLUTIONS TO PRACTICE PROBLEM SET 13
- v(t) = 3t^2 − 18t + 24; a(t) = 6t − 18
In order to find the velocity function of the particle, we simply take the derivative of the
position function with respect to t: = v(t) = 3t^2 − 18t + 24. In order to find the acceleration
function of the particle, we simply take the derivative of the velocity function with respect to t:
= 6t − 18.
- v (t) = 2cos(2t) - sin(t); a (t) = -4 sin(2t) - cos(t)
In order to find the velocity function of the particle, we simply take the derivative of the
position function with respect to t: = v (t) = 2 cos(2t) − sin(t). In order to find the
acceleration function of the particle, we simply take the derivative of the velocity function with
respect to t: = a (t) = −4 sin(2t) − cos(t).
- t = π and t = 3π
In order to find where the particle is changing direction, we need to find where the velocity of
the particle changes sign. The velocity function of the particle is the derivative of the position
function: = v(t) = cos . Next, we set the velocity equal to zero. The solutions are: t
= π and t = 3π. Actually, there are an infinite number of solutions but remember that we are
restricted to 0 < t < 4π. Next, we check the sign of the velocity on the intervals 0 < t < π, π < t
