Answer:
Step-by-step explanation:
(a) Velocity (v(t)):
To find the velocity, we'll differentiate the position function, r(t), with respect to time (t).
r(t) = tsin(t)i + tcos(t)j + 6t^2k
Differentiating each component with respect to time:
r'(t) = (d/dt)(tsin(t))i + (d/dt)(tcos(t))j + (d/dt)(6t^2)k
Using the product rule and the chain rule, we can simplify the derivatives:
r'(t) = (sin(t) + tcos(t))i + (-cos(t) + tsin(t))j + (12t)k
Therefore, the velocity of the particle is given by:
v(t) = (sin(t) + tcos(t))i + (-cos(t) + tsin(t))j + (12t)k
(b) Acceleration (a(t)):
To find the acceleration, we'll differentiate the velocity function, v(t), with respect to time (t).
v(t) = (sin(t) + tcos(t))i + (-cos(t) + tsin(t))j + (12t)k
Differentiating each component with respect to time:
v'(t) = (d/dt)(sin(t) + tcos(t))i + (d/dt)(-cos(t) + tsin(t))j + (d/dt)(12t)k
Using the product rule and the chain rule, we can simplify the derivatives:
v'(t) = (cos(t) - tsin(t))i + (-sin(t) + tcos(t))j + 12k
Therefore, the acceleration of the particle is given by:
a(t) = (cos(t) - tsin(t))i + (-sin(t) + tcos(t))j + 12k
(c) Speed:
The speed of a particle is the magnitude of its velocity vector.
Speed = |v(t)| = sqrt[(sin(t) + tcos(t))^2 + (-cos(t) + tsin(t))^2 + (12t)^2]