Why does (dr/dt)*θ = 0 when deriving Velocity?

[LearninDaMath]

Homework Statement


A derivation of Instantaneous Velocity

Arc length = s = θ * r

ds/dt = d(θr)/dt = Instantaneous V

Product rule:

V$_{t}$ = (dθ/dt)*r + (dr/dt)*θ

V$_{t}$ = (dθ/dt)*r + 0

V$_{t}$ = ωr




Question:

Why does dr/dt*θ = 0 ?

 

[xlava]

Since r is constant, the rate of change of the radius with respect to time (or anything for that matter) is going to be 0.

 

[LearninDaMath]

ah makes perfect sense, thanks.

 

[xlava]

ah makes perfect sense, thanks.

No problem, sir. Good luck.

 

[asteriatic]

The reason why (dr/dt)*θ = 0 is because when deriving velocity, we are looking at the rate of change of the arc length (s) with respect to time (t). Since the arc length is equal to θ * r, we can use the product rule to derive the velocity. However, the term (dr/dt)*θ represents the rate of change of the radius (r) with respect to time (t), multiplied by the angle (θ). In this case, since the radius is constant, its rate of change is 0. Therefore, (dr/dt)*θ = 0, and we are left with the simplified equation V_{t} = (dθ/dt)*r, which represents the velocity as the product of the angular velocity (ω) and the radius (r). This is why (dr/dt)*θ = 0 when deriving velocity.