Find the directional derivative of $f(x,y,z)=3xy+z^2$ at the point $(5,1,−4)$ in the direction of a vector making an angle of $π/3$ with $∇f(5,1,−4)$.
$f_\vec u(5,1,−4)=D_\vec uf(5,1,−4)=?$
I know how to do directional derivative questions but I have no idea about this one. I'm guessing that I'm thinking about the question wrong.
$D_\vec uf=f_x\vec u_x+f_y\vec u_y+f_z\vec u_z =\nabla f \cdot \vec u$ where $\|\vec u\|=1$
So $\nabla f= \langle 3y, 3x, 2z \rangle$ and $\nabla f(5, 1, -4)= \langle3,15,-8\rangle$
Then it says $\vec u$ makes a $\pi/3$ angle with $\nabla f(5, 1, -4)$ which would mean
$\nabla f(5, 1, -4) \cdot \vec u = \|\nabla f(5, 1, -4)\| * \| \vec u\| * \cos(\pi/3)$
and knowing $\|\vec u\| = 1$ gives
$3\vec u_x+15\vec u_y-8\vec u_z=\sqrt {298} * \cos(\pi/3)$
and with $\vec u_x^2+\vec u_y^2+\vec u_z^2=1$ gives and unsolvable system of equations (I think)
So… i'm wondering where I went wrong.
Best Answer
You don't need to know $\;u\;$ . Since your function is differentiable everywhere, we get that
$$D_uf(5,1,-4)=\nabla f(5,1,-4)\cdot\frac u{||u||}$$
But you also know that
$$\frac12=\cos\frac\pi3=\frac{u\cdot\nabla f(5,1,-4)}{||u||\,||\nabla f(5,1,-4)||}\implies$$
$$\nabla f(5,1,-4)\cdot\frac u{||u||}=\frac12||\nabla f(5,1,-4)||=\ldots$$...and voila!