Let $f(x,y)=x+y^2$ and $P = (1,1)$. Find a unit vector $u$ such that the directional derivative $D_uf(x,y)$ is zero.

$$ \nabla{f(x,y)} = \left\langle1, 2y\right\rangle\\ \nabla{f(1,1)} = \left\langle11, 2\right\rangle\\ D_uf(1,1) = \left\langle11,2\right\rangle\cdot u\\ $$

$u$ must be a unit vector so $u = \sqrt{x^2+y^2} = 1$. So we must solve the system of equations:

$$ \sqrt{x^2+y^2} = 1\\ x + 2y = 0 $$

Then we simply solve the system of equations. Is this the correct direction for this problem?

  • 3
    $\begingroup$ I wouldn’t think of it as solving a system. Like @nullUser, I would just use $x+2y=0$ to find a vector in the right direction, like $\langle -2,1\rangle$, and then normalize it by dividing each component by the length of the vector. $\endgroup$ Sep 25 '12 at 23:14

You are correct that that is what you must do, but you need not solve the equations as a system (well, this is what you are doing, but I find the connotation to be different in this particular situation).

If $\langle 1, 2 \rangle \cdot \langle x, y \rangle = x+2y=0$ then this result will hold for any multiple of $\langle x, y \rangle$. For instance, take $x=2, y=-1$. Then for any $\alpha \in \mathbb{R}$ when we multiply we always get $\langle 1, 2 \rangle \cdot (\alpha \langle 2, -1 \rangle) = 0$. In particular, we could take $\alpha = 1/\sqrt{2^2+(-1)^2}$, i.e. the reciprocal of the norm of $\langle x, y \rangle$, and the result still holds.

Do you see why this solves the problem?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.