I am given two vector functions: f=(2t+1)i + (3t)j g=0i + (4t)j The problem wants the differential of the angle between the vectors. I think that \cos\theta is found by \frac{ac+bd}{|f|*|g|} I think that gives me \frac{3t}{{13t^2}+4t+1} Would \frac{d\theta}{dt} be be \frac{d}{dt} arcos u with \frac{3t}{{13t^2}+4t+1} plugged in for u? I'd like to know I'm not barking up the wrong tree before that much algebra. Thanks