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?
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?
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.
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.