To find the velocity, acceleration, and distance traveled between times t1 and t2 for the given parametric equations x(t) = acos(wt), y(t) = asin(wt), and z(t) = bt, we need to differentiate these equations with respect to time.
- Velocity (v): The velocity vector is the derivative of the position vector with respect to time.
v(t) = (dx/dt, dy/dt, dz/dt)
For x(t) = acos(wt), differentiate it with respect to t: dx/dt = -aw*sin(wt)
For y(t) = asin(wt), differentiate it with respect to t: dy/dt = aw*cos(wt)
For z(t) = bt, differentiate it with respect to t: dz/dt = b
Therefore, the velocity vector is: v(t) = (-awsin(wt), awcos(wt), b)
- Acceleration (a): The acceleration vector is the derivative of the velocity vector with respect to time.
a(t) = (dv/dt)
Differentiating v(t) = (-awsin(wt), awcos(wt), b) with respect to t, we get:
dv/dt = (-aw²cos(wt), -aw²sin(wt), 0)
Therefore, the acceleration vector is: a(t) = (-aw²cos(wt), -aw²sin(wt), 0)
- Distance traveled (d): The distance traveled from t1 to t2 can be found by integrating the magnitude of the velocity vector over the time interval [t1, t2]:
d = ∫(|v(t)|) dt, from t1 to t2
The magnitude of the velocity vector can be calculated as:
|v(t)| = √(v₁² + v₂² + v₃²)
where v₁, v₂, and v₃ are the components of the velocity vector.
|v(t)| = √((-awsin(wt))² + (awcos(wt))² + b²) = √(a²w²sin²(wt) + a²w²cos²(wt) + b²) = √(a²w²(sin²(wt) + cos²(wt)) + b²) = √(a²*w² + b²)
Now, integrate the magnitude of the velocity over the time interval [t1, t2]:
d = ∫(√(a²*w² + b²)) dt, from t1 to t2
Integrating this expression will give you the distance traveled between times t1 and t2.
Please note that these calculations assume that a, w, and b are constants and do not vary with time.