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.

  • $\begingroup$ Before anything else: why do you thing (or know) you are wrong?? $\endgroup$
    – DonAntonio
    Oct 13, 2013 at 4:12
  • 1
    $\begingroup$ Because there's no way to find u and WebWorK doesn't let me have a u in the answer $\endgroup$
    – user100508
    Oct 13, 2013 at 4:14

1 Answer 1


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!

  • $\begingroup$ Hmm.... I think it's time for me to take a break from doing calculus problems. $\endgroup$
    – user100508
    Oct 13, 2013 at 4:25
  • $\begingroup$ @Don Antonio, I am confused about the first one. Isnt it just the dot product? why is there $||u||$ in the denominator?? $\endgroup$
    – Vishesh
    Oct 13, 2013 at 4:46
  • 1
    $\begingroup$ Because one has to take a unitary vector in that formula, @Vishesh ...but, of course, you can assume from the start that the $\;u\;$ is unitary, just as the OP did in his question. If this is true then, of course, $\;||u||=1\;$ and everything remains the same, nothing's lost. $\endgroup$
    – DonAntonio
    Oct 13, 2013 at 10:24

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