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 $\displaystyle \cos\theta$ is found by $\displaystyle \frac{ac+bd}{|f|*|g|}$ I think that gives me $\displaystyle \frac{3t}{{13t^2}+4t+1}$ Would $\displaystyle \frac{d\theta}{dt}$ be be $\displaystyle \frac{d}{dt}$ arcos u with $\displaystyle \frac{3t}{{13t^2}+4t+1}$ plugged in for u? I'd like to know I'm not barking up the wrong tree before that much algebra. Thanks