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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 $g'(0)$.

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

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Thanks for the response! Could you elaborate on your notation on the last part of your answer? –  L1meta Feb 18 '13 at 15:43
    
$D_v(f)$ is the derivative of $f$ in the direction of $v.$ –  lyj Feb 18 '13 at 21:11

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.

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