A particle moves so that its position is given by ⟨cos(t),sin(t),cos(6t)⟩. Find the maximum and minimum speeds of the particle.
To find the maximum and minimum speeds of the particle, we need to first find the velocity vector, which is the derivative of the position vector with respect to time.
The position vector is ⟨cos(t),sin(t),cos(6t)⟩.
Taking the derivative of each component, we get:
Velocity vector = (-sin(t), cos(t), -6sin(6t))
Next, the speed of the particle is the magnitude of the velocity vector.
Speed = √[ (-sin(t))^2 + (cos(t))^2 + (-6sin(6t))^2 ]
= √[ sin^2(t) + cos^2(t) + 36sin^2(6t) ]
= √[ 1 + 36sin^2(6t) ] [Using sin^2(x) + cos^2(x) = 1]
To find the maximum and minimum speeds, we need to find the maximum and minimum values of sin^2(6t). Since sin^2(x) is always between 0 and 1, the maximum value of sin^2(6t) is 1, and the minimum value is 0.
Therefore, the maximum speed of the particle is √(1 + 36) = √37, and the minimum speed is √(1 + 0) = 1.
To find the speed of the particle at any given time, we need to find the magnitude of its velocity vector. Recall that the velocity vector is the derivative of the position vector with respect to time.
Given that the position vector is ⟨cos(t), sin(t), cos(6t)⟩, we can find the velocity vector by taking the derivative of each component.
Taking the derivative of cos(t) with respect to t gives us -sin(t).
Taking the derivative of sin(t) with respect to t gives us cos(t).
Taking the derivative of cos(6t) with respect to t gives us -6sin(6t).
So, the velocity vector is ⟨-sin(t), cos(t), -6sin(6t)⟩.
To find the speed at any given time, we need to find the magnitude of the velocity vector. Recall that the magnitude of a vector ⟨a, b, c⟩ is given by the square root of the sum of the squares of its components, which in this case is:
speed = sqrt((-sin(t))^2 + (cos(t))^2 + (-6sin(6t))^2)
= sqrt(sin^2(t) + cos^2(t) + 36sin^2(6t))
Now, since sin^2(t) + cos^2(t) = 1 (by the Pythagorean identity), the speed formula simplifies to:
speed = sqrt(1 + 36sin^2(6t))
To find the maximum and minimum speeds, we need to find the maximum and minimum values of the expression inside the square root.
Since sin^2(6t) takes values between 0 and 1, the maximum and minimum speeds occur when sin^2(6t) = 1 and sin^2(6t) = 0, respectively.
When sin^2(6t) = 1, the expression inside the square root becomes:
1 + 36
= 37
So, the maximum speed is sqrt(37).
When sin^2(6t) = 0, the expression inside the square root becomes:
1 + 0
= 1
Therefore, the minimum speed is sqrt(1), which is equal to 1.
In summary, the maximum speed of the particle is sqrt(37), and the minimum speed is 1.
r(t) = ⟨cos(t),sin(t),cos(6t)⟩
v = <-sin(t),cos(t),-6sin(6t)>
speed is |v|, so
s^2 = 1+36sin^2(6t)
max speed occurs when ds/dt = 0
2s ds/dt = 216sin^4(6t)
ds/dt=0 when sin(6t) = 0
I expect you can take it from there, no?