Let $h(x, y)$ be some function from $\mathbb{R}^2$ to $\mathbb{R}$ which outputs height of a hill at a given point $(x, y)$. At a given moment, I am travelling up the hill with velocity $\mathbf{v}$ at angle $\theta$ to $\nabla h$. How do I prove that the rate at which my height is increasing is $\mathbf{v} \cdot \nabla h$?
My progress:
I know that for some 2D vector $\mathbf{a}$, the expression $\mathbf{a} \cdot \nabla h$ will give me the slope of hill in direction of $\mathbf{a}$. So if $\mathbf{\hat{v}}$ is the unit vector in direction of $\mathbf{v}$, then $\mathbf{\hat{v}} \cdot \nabla h$ will give me the slope of hill in the direction of the velocity. Since $\mathbf{v} \cdot \nabla h = (\left|\mathbf{v}\right|)(\mathbf{\hat{v}} \cdot \nabla h)$, the equation is essentially saying that the rate at which my height is increasing is speed multiplied by my slope. But this equation is clearly incorrect. Where am I going wrong?
