Calculate the velocity and acceleration vectors, and speed for r(t)=⟨cos(6t),sin(t),sin(6t)⟩ when t=5π/6 .
r(t)=⟨cos(6t),sin(t),sin(6t)⟩
v = r' = <-6sin(6t),cos(t),6cos(6t)>
a = r" = <-36cos(6t),-sin(t),-36sin(6t)>
Now just plug in t=5π/6 (6t=5π)
and recall that speed is |v|
10 miles per hour south
To calculate the velocity and acceleration vectors, and speed for r(t) = ⟨cos(6t), sin(t), sin(6t)⟩ when t = 5π/6, we need to find the derivative of r(t) with respect to time.
1. Velocity vector:
The velocity vector is the derivative of the position vector r(t) with respect to time, given by v(t) = dr(t)/dt.
Taking the derivative of r(t) = ⟨cos(6t), sin(t), sin(6t)⟩ with respect to t, we get:
v(t) = ⟨-6sin(6t), cos(t), 6cos(6t)⟩
Substituting t = 5π/6, we get:
v(5π/6) = ⟨-6sin(6(5π/6)), cos(5π/6), 6cos(6(5π/6))⟩
= ⟨-6sin(5π), cos(5π/6), 6cos(5π)⟩
= ⟨0, cos(5π/6), 6(-1)⟩
= ⟨0, √3/2, -6⟩
So, the velocity vector at t = 5π/6 is ⟨0, √3/2, -6⟩.
2. Acceleration vector:
The acceleration vector is the derivative of the velocity vector with respect to time, given by a(t) = dv(t)/dt.
Taking the derivative of v(t) = ⟨-6sin(6t), cos(t), 6cos(6t)⟩ with respect to t, we get:
a(t) = ⟨-36cos(6t), -sin(t), -36sin(6t)⟩
Substituting t = 5π/6, we get:
a(5π/6) = ⟨-36cos(6(5π/6)), -sin(5π/6), -36sin(6(5π/6))⟩
= ⟨-36cos(5π), -sin(5π/6), -36sin(5π)⟩
= ⟨-36(-1), -sin(5π/6), -36(0)⟩
= ⟨36, -√3/2, 0⟩
So, the acceleration vector at t = 5π/6 is ⟨36, -√3/2, 0⟩.
3. Speed:
The speed is the magnitude of the velocity vector, given by ||v(t)||.
Substituting t = 5π/6 into v(t), we get:
||v(5π/6)|| = ||⟨0, √3/2, -6⟩||
= √(0^2 + (√3/2)^2 + (-6)^2)
= √(0 + 3/4 + 36)
= √(39/4 + 36)
= √(39/4 + 144/4)
= √(183/4)
= √183/2
So, the speed at t = 5π/6 is √183/2.
To calculate the velocity and acceleration vectors, and speed for the given position vector r(t), we need to take the derivatives of r(t) with respect to time (t).
Given: r(t) = ⟨cos(6t), sin(t), sin(6t)⟩
Step 1: Find the derivative of r(t) to obtain the velocity vector v(t):
v(t) = ⟨d/dt (cos(6t)), d/dt (sin(t)), d/dt (sin(6t))⟩
For each component, take the derivative of the trigonometric function:
d/dt (cos(6t)) = -6sin(6t)
d/dt (sin(t)) = cos(t)
d/dt (sin(6t)) = 6cos(6t)
Therefore, the velocity vector v(t) is:
v(t) = ⟨-6sin(6t), cos(t), 6cos(6t)⟩
Step 2: Find the derivative of v(t) to obtain the acceleration vector a(t):
a(t) = ⟨d/dt (-6sin(6t)), d/dt (cos(t)), d/dt (6cos(6t))⟩
Again, take the derivative of each component:
d/dt (-6sin(6t)) = -36cos(6t)
d/dt (cos(t)) = -sin(t)
d/dt (6cos(6t)) = -36sin(6t)
Therefore, the acceleration vector a(t) is:
a(t) = ⟨-36cos(6t), -sin(t), -36sin(6t)⟩
Step 3: Now that we have the velocity vector v(t) and acceleration vector a(t), we can calculate the speed at t = 5π/6 by substituting the value of t into the velocity vector.
Given: t = 5π/6
Evaluate the velocity vector v(t) at t = 5π/6:
v(5π/6) = ⟨-6sin(6(5π/6)), cos(5π/6), 6cos(6(5π/6))⟩
= ⟨-6sin(5π), cos(5π/6), 6cos(5π)⟩ (sin(6(5π/6)) = sin(5π))
= ⟨0, -1/2, 6⟩ (sin(5π) = 0, cos(5π) = -1)
Therefore, the velocity vector v(5π/6) is ⟨0, -1/2, 6⟩.
To calculate the speed, we need to find the magnitude of the velocity vector:
Speed = |v(5π/6)| = √(0^2 + (-1/2)^2 + 6^2) = √(1/4 + 36) = √(37/4) = √37/2
Hence, the velocity vector at t = 5π/6 is ⟨0, -1/2, 6⟩ and the speed is √37/2.