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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?

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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. – Brian M. Scott Sep 25 '12 at 23:14
up vote 3 down vote accepted

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?

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