Restricting a function to the span of a vector

Question

Let $ \mathbf v \in \Bbb R^2$ be a vector. Let $g$ be the one-variable function you get by restricting $ f(x,y)=3y^2+2xy $ to the line spanned by $\mathbf v$ and moving with constant speed 1 in direction $\mathbf v$. Find an expression for $g'(0)$.

Answer 1

You want to find the directional derivative at $0.$ Write $v=(a,b)$ and use the fact that $D_{u+w}(f) = D_u(f) + D_w(f).$

Answer 2

Since $$Df(x,y)=\begin{bmatrix}2y & 6y + 2x\end{bmatrix},$$ $Df(0,0)$ is just zero. We have $g(t)=f(tv),$ so by the chain rule, $$ g'(t) = Df(tv) v. $$ It should be pretty obvious what $g'(0)$ is.