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I am learning how to find the tangent plane at a point to a surface.

I am trying to understand why the gradient vector is always normal to the surface's level set.

Reading this post helped quite a lot, it explains:

Now defining $\nabla F = \langle\dfrac{\partial F}{\partial x}, > \dfrac{\partial F}{\partial y},\dfrac{\partial F}{\partial z}\rangle$, we see that

$$dF\big|_p (\textbf{v})= \nabla F\cdot \textbf{v}$$

This one formula packages a lot of mathematics.

Hopefully this gives you some more intuition about what the gradient does. In words, to see how much $F(\textbf{p})$ changes when you move away a little bit to $\textbf{p}+\textbf{v}$, just dot product $\nabla > F\big|_p$ with $\textbf{v}$.

Armed with this intuitive understanding of the gradient, we can see why it must be perpendicular to the level curves of $F$ quite intuitively.

If $p$ is a point of the surface $F(x,y,z) = 0$, then the tangent vectors $\textbf{v}$ to the surface must satisfy $dF\big|_p(v) = 0$, because moving in the direction of the surface should not change the value of $F$ much since the value of $F$ is constant on the surface. Translating this into a gradient statement we see that $\nabla > F\big|_p \cdot \textbf{v} = 0$ for each tangent vector to the surface.

My understanding is this: if point $P$ lies on the level curve of a surface, then if I move a tiny amount away from $P$, the rate of change should be $0$ because the value of $F$ is constant on the surface at/around that point, because that point is on a level curve. And I understand that for $dF\big|_p(v) = 0$, $cos\theta$ must equal $0$ - or the angle between $ \nabla F$ and $\textbf{v}$ must be $\frac{\pi}{2}$ or $90$ degrees. (Please correct me if this is incorrect, I'm still trying to grasp it)

However, I'm struggling to understand: what does a level curve have to do with the vectors tangent to the surface?

maddie
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  • Slight confection to the beginning of your question: the gradient is always normal to the level set. And to attempt an answer to your question at the end: moving along a surface means moving along a vector tangent to the surface. – themathandlanguagetutor Oct 02 '17 at 05:51
  • I'm sorry, I meant correction. – themathandlanguagetutor Oct 02 '17 at 05:57
  • Thank you for the correction! Do you have a link or anything that could help explain why moving along a surface means moving along a vector tangent to the surface? Thanks! – maddie Oct 02 '17 at 05:58
  • Unfortunately this is something I find rather intuitive, so I cannot exactly identify a good source, but if I had to make a wager, I would imagine that the text I used for multivariate/vector calculus covers it decently. I used Marsden and Weinstein's text Calculus III which is supposedly available as a free PDF from the publishers. – themathandlanguagetutor Oct 02 '17 at 06:05
  • To follow up, do you understand that moving along a curve means moving along a vector tangent to the curve? And do you understand that a vector is said to be tangent to a surface if it is tangent to some curve on that surface? – themathandlanguagetutor Oct 02 '17 at 06:20
  • The title of your question is very strange... A surface has no level sets; that something that functions have. And a vector can't be tangent to a function, but it can be tangent to a surface. – Hans Lundmark Oct 02 '17 at 06:23
  • @themathandlanguagetutor no still not grasping that - I'm trying to find resources to explain that – maddie Oct 02 '17 at 07:22
  • This https://youtu.be/E9Q_Lc0g1xE seems to explain the first point decently well. I hope it helps – themathandlanguagetutor Oct 02 '17 at 07:29
  • @themathandlanguagetutor is the tangent vector to the level curve, the derivative of the level curve? – maddie Oct 02 '17 at 18:44
  • More precisely, the derivative of any curve with respect to some parameter is tangent to the curve (as there are infinitely many tangent vectors to a curve). – themathandlanguagetutor Oct 02 '17 at 18:46

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