Let $T(x,y,z)=xy^{2}+2z-x^{2}z^{2}$ be the temperature at the point $(x,y,z)$.
I'm asked to find the unit vector belonging to the direction in which the temperature decreases most rapidly at $(1,0,-1)$.
I have calculated the unit gradient vector at point $(1,0,-1)$ to be $\left(\frac{-1}{\sqrt5},0,\frac{2}{\sqrt5}\right)$, but according to the solution manual the answer is $\left(\frac{1}{\sqrt5},0,\frac{-2}{\sqrt5}\right)$.
What have I done wrong?
The gradient (if it exists) at a point points to the direction of steepest ascent, therefore it's additive inverse points to the direction of the steepest descent.
Let $p = (1,0,-1)$, and let $\nabla T$ be the gradient of $T$.
Now, for any unit vector $u$, $(\nabla T)(p) \cdot u$ is going to tell you how fast $T$ is changing in the direction $u$ at the point $p$.
So you are asked to find the $u$ unit vector that minimizes the expression above, i.e. what's
$$\operatorname{argmin}_{u\in S^2}\left((\nabla T)(p) \cdot u\right)?$$
Well, if the angle betwen $u$ and $(\nabla T)(p)$ is $\alpha$, then $$(\nabla T)(p) \cdot u = |(\nabla T)(p)| \cdot 1 \cdot \cos(\alpha). $$
Therefore we have to minimize $\cos(\alpha)$, implying that $\alpha = {\pi} + 2k\pi$, and thus
$$ u = -\frac{(\nabla T)(p)}{\|(\nabla T)(p)\|}.$$