dt' role="presentation" tabindex="0">∫t0√∂x∂t2+∂y∂t2+∂z∂t2dtSo for instance, on our function above, we would have:
∫t0√(−2πsin(2πt))2+(2πcos(2πt))2+1dtNow considering that a negative squared is a positive, we can rewrite as follows:
∫t0√(2πsin(2πt))2+(2πcos2(2πt))2+1dtAnd remembering that sin2(θ)+cos2(θ)=1 we can simplify:
∫t0√4π2sin2(2πt)+4π2cos2(2πt)+1dt∫t0√4π2(sin2(2πt)+cos2(2πt))+1dt∫t0√4π2+1dt=6.36