Accelerated reference frames equation derivation question

[charliepebs]

my question comes from the portion of the derivation regarding evaluating the rate of change of the principle axis vectors. this begins by supposing a vector, Q, is rotating about axis n by δθ. Specifically, my question is how from step 4 to step 5 the approximation becomes an equality.

Q' can be approximated as:
1. Q' ≈ Q + (|Q|sinα)δθ(in direction of n cross Q)
2. Q' ≈ Q + |n cross Q|(in direction of n cross Q)δθ
3. Q' ≈ Q + (n cross Q)δθ
4. Q' ≈ Q + δθ cross Q

The derivation then states: "This means that we can express the derivative of this vector as:
5. (Q'-Q)/dt = (δθ cross Q)/dt
6. dQ/dt = δθ/dt cross Q
7. dQ/dt = Ω cross Q

 

[LightHero]

My question is how the approximation in line 4 becomes an equality in line 5. Can someone explain this to me? The answer to your question lies in understanding how derivatives work. The derivative of a function f(x) is defined as the limit of the difference quotient (f(x+h)-f(x))/h as h approaches 0. In the case of your equation, the function is Q(θ), and the derivative is dQ/dθ. The difference quotient in this case is (Q' - Q)/δθ. Since δθ is a small angle, we can replace it with its limit as δθ approaches 0, which is dθ/dt. This gives us the equation (Q' - Q)/dt = (δθ cross Q)/dt, which is equivalent to the equation dQ/dt = Ω cross Q.