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I'm trying to understand the concept of a Gradient vector, and it seems I'm having trouble visualizing certain stuff. So, I was hoping if someone could resolve some of the questions I'm having on my mind.

Okay, so I'm considering a real-valued function $z=f(x,y)$ which is smooth, i.e., its partial derivatives with respect to $x$ and $y$ exist and are continuous. Let $k$ be a real number in the range of $f$ and let $\vec{v}$ be a unit vector in $\mathbb{R}^2$, which is tangent to the level curve $f(x,y)=k$

Now, I am told to understand that the rate of change of $f$ in the direction of $\vec{v}$ is $0$, i.e., $D_{\vec{v}}f =0$. And the explanation for it, in almost all of the sources I've seen is, that it's because $\vec{v}$ is a tangent vector to this curve.

  1. Can anyone tell me how $\vec{v}$ being simply a tangent vector implies this?

  2. Is this concept simply extended to the case of a function in three variables, by level surfaces and tangent planes? In this case, wouldn't there be an infinite number of tangent vectors, and thus an infinite number of gradient vectors (since the gradient vector is perpendicular to the tangent vector, i.e., normal to the surface). Can anyone point out any sources that help in visualizing this?

Thank you!

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1 Answer 1

up vote 2 down vote accepted

Let $\vec{x}(t)$ be a parametrisation of the level set $f(\vec{x})=k$, i.e. $f(\vec{x}(t)) = k$ for all $t$, then $\vec{v} = \frac{\mathrm{d}\vec{x}}{\mathrm{d}t}$. It follows that $0=\frac{\mathrm{d}}{\mathrm{d}t}f(\vec{x}(t)) = \frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}t}+\frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}t} = (\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})\cdot \vec{v}$.

On the other hand $\frac{\mathrm{d}}{\mathrm{d}t}f(\vec{x}+t\vec{v}) = \frac{\partial f}{\partial x}\vec{v}_x+\frac{\partial f}{\partial y}\vec{v}_y = (\frac{\partial f}{\partial x},\frac{\partial f}{\partial y})\cdot \vec{v}$.

Thus, indeed $D_{\vec{v}}f = \frac{\mathrm{d}}{\mathrm{d}t}f(\vec{x}+t\vec{v}) = 0$.

The result indeed generalises in that for any vector $\vec{v}$ tangent to a level set of $f$, $D_{\vec{v}}f =0$.

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