Assume you are located at a point $\vec{a}=(a,b)$ on a function $f(x,y)$. If you want to travel in a direction $\vec{u}=(u_1,u_2)$ corresponding to an angle $\alpha$ measured from the positive x-axis, show that
$D_{\vec{u}}f(\vec{a})=f_x(a,b)cos\alpha + f_y(a,b)sin\alpha$

My attempt at the solution.
$D_{\vec{u}}f(\vec{a}) = \nabla f(\vec{a})*\vec{u} = |\nabla f(\vec{a})||\vec{u}|cos\alpha$
I'm not sure where to go from here, or if I'm even using the correct angle, I think the angle for dot product should be the one between the two vectors and $\alpha$ is from the x axis.
Some help would be appreciated.

Thanks

Originally Posted by nubshat
Assume you are located at a point $\vec{a}=(a,b)$ on a function $f(x,y)$. If you want to travel in a direction $\vec{u}=(u_1,u_2)$ corresponding to an angle $\alpha$ measured from the positive x-axis, show that
$D_{\vec{u}}f(\vec{a})=f_x(a,b)cos\alpha + f_y(a,b)sin\alpha$

My attempt at the solution.
$D_{\vec{u}}f(\vec{a}) = \nabla f(\vec{a})*\vec{u} = |\nabla f(\vec{a})||\vec{u}|cos\alpha$
I'm not sure where to go from here, or if I'm even using the correct angle, I think the angle for dot product should be the one between the two vectors and $\alpha$ is from the x axis.
Some help would be appreciated.

Thanks
Think about decomposing $\vec{u}=(u_1,u_2)$ into the Cartesian basis vectors. Hint: how else could you write $\nabla f(\vec{a})\cdot\vec{u}$