# Difficulty understanding the gradient vector

Assume I want to decide the sign of $\frac{dz}{dt}|_{t=1}$ (and not t=2). As far as I can see: At $t=1$ i think that the derivative is positive... this is because the $x'(1)=0 ,y'(1)>0$ and $z_y(3,0) > 0$ . In one of my previous posts, Babak S. said he thinks I'm right.

On the other hand, at the point $(3,0)$ , the gradient must be perpendicular to the corresponding level set, so it must point towards the positive region of the x-axis, which means that the y'th-coordinate term of the gradient vector must be $0$ and thus $\frac{dz}{dt}|_{t=1} = 0$ ...

Can someone please explain to me what am I getting wrong here?

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Why do you say $z_y(3,0) > 0$? As you observed correctly in your second paragraph, the gradient is in the $x$-direction at $(3,0)$, so $z_y(3,0) = 0$.
Because when walking from (3,0) to the positive direction of the $y$-axis, we ascend and not stay at the same place... – czash Jun 6 '13 at 7:27
At $(3,0)$ the contour line is also in the $y$ direction, so (to first order) you are walking along the contour line, i.e. $dz/dt=0$. – Anthony Carapetis Jun 6 '13 at 7:36