Why is the directional derivative zero along a level curve tangent vector?

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Ryuzaki
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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 [itex]z = f(x,y)[/itex] which is smooth, i.e., its partial derivatives with respect to [itex]x[/itex] and [itex]y[/itex] exist and are continuous. Let [itex]k[/itex] be a real number in the range of [itex]f[/itex] and let [itex]\mathbf v[/itex] be a unit vector in ℝ2, which is tangent to the level curve [itex]f(x,y) = k[/itex].

Now, I am told to understand that the rate of change of [itex]f[/itex] in the direction of [itex]\mathbf v[/itex] is [itex]0[/itex], i.e., Dvf [itex]= 0[/itex]. And the explanation for it, in almost all of the sources I've seen is, that it's because [itex]\mathbf v[/itex] is a tangent vector to this curve.

1. Can anyone tell me how [itex]\mathbf v[/itex] 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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We might as well do it for functions of ##n## variables since the argument is the exact same; note that the argument I'm giving here is not exactly rigorous but hopefully it is satisfying enough. Let ##f:\mathbb{R}^{n}\rightarrow \mathbb{R}## be a smooth scalar field and let ##M = f^{-1}(c), c\in f(\mathbb{R}^{n})## be a level set of ##f##. Let ##p\in M## and ##v## be any vector tangent to ##M## at ##p##, and choose a smooth curve ##\gamma :(-\epsilon,\epsilon)\rightarrow M## with ##\gamma(0) = p, \dot{\gamma}(0) = v##. Note that ##f(\gamma(t)) = c = \text{const.}## identically, since the image of the curve lies in the level set, so ##\frac{\mathrm{d} }{\mathrm{d} t}(f(\gamma(t)))|_{t=0} = \nabla f(\gamma(0))\cdot \dot{\gamma}(0) = \nabla f(p)\cdot v = 0## where I have used the chain rule in the very first step. Since the chosen point and tangent vector were entirely arbitrary, ##\nabla f## will be perpendicular to all vectors tangent to ##M## at each point of ##M##.
 
Thank you, WannabeNewton! I think I understand it now. :smile:
 
Ryuzaki said:
Thank you, WannabeNewton! I think I understand it now. :smile:
Let me know if you have more questions. Also, work it out specifically for the case where ##f(x,y,z) = x^2 + y^2 + z^2## i.e. when the level sets are spheres, ##x^2 + y^2 + z^2 = k##. Take the gradient and hopefully you can picture the outward (outward from the origin) radial vector field that results and see how it is perpendicular to all the spheres centered at the origin with different radii given by different values of ##k##.