Calculus 3 Lecture 13.7: Finding Tangent Planes and Normal Lines to Surfaces

Calculus 3 Lecture 13.7: Finding Tangent Planes and Normal Lines to Surfaces: How to find a tangent plane and/or a normal line to any surface (multivariable function) at a point. This is based on the ability to find the gradient of a function.

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0:00 Tangent planes to a surface at a point 29:20 Example 1 52:16 Example 2 1:02:22 Planes and Lines 1:06:40 Example 3 1:15:30 Example 4 1:24:35 example 5 Not completed 1:27:30 Example 6 1:35:25 Example 7

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[@GenuinePeacefulTimes]

I have been wondering how in the world the gradient which is just the partial derivatives produces a normal vector to the function for weeks now, and now I finally get it. thank you.

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[@Peter_1986]

I improvised a way to find normal lines a few years ago, which involved finding the partial derivatives along the x/y-coordinates, then creating lines through a common point, and then using the vector product to find the normal line - I guess that that method works as well, although the method in this video is certainly a lot faster.

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[@EeDymonNij]

27:38 if you're on a point of the level curve where the tangent line points to the center, and so gives you a gradient in a complete different direction than the center, how's that the fastest way to the center?

[@lujiabai]

exercise start at 41:33. for people want to do some difficult exercises, begin at 1:24:47

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[@_avenger9709]

Shouldn't r'(t) give you a tangent vector and not just the slope of the tangent vector? @19:38

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[@anshishkumarray0383]

1:34:23 Shouldn't be there a negative one at the denominator in that normal line equation?

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prof leonard im on a section in my calc 3 class called tangent planes and linear approximation is this video based on that section?

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[@omer8836]

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[@Peter_1986]

To me, the most intuitive way to find a normal plane is to simply find the 2 partial derivatives so that I get the 2 lines that touch the point, and then I use those lines to find the normal vector to that point by using the cross product.

[@nickvanderwater1088]

1:41:45 POV you now understand the math concept

[@manstuckinabox3679]

lol I got bored from all the easy exercises, so I made one of my own: f(x,y,z,u) = logbase(x^(y^(z(^||w||^v)))) where w is a vector function defined by and v = ||w(v)|| / the family of spheres (x^2+y^2+z^2). 1- kill me 2- find the tangent plane of f(x,y,z,u) = -5 at a point of your choice, and the tangent line as well. its actually quite easy, the implicit functions cancel out during the deriving process.

[@davidkrowa4153]

Professor, at 13:23 You said it is two things being multiplies and the added together, it looks like a dot product and used that, what about product rule for differentiation?

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Do you cover linear approximation?

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Laurelindo which is unfortunately a trashy outdated mentality to think about it

[@michaeljjan6344]

For the last two questions at 1:31, why do you move the z variable and make the gradient in terms of i,j,k? Wouldn't you just turn z variable into f(x,y) instead and treat the gradient in terms of i and j? This was done in the last word problem you did in section 13.6 (last question of this video). Seems like a contradiction.

[@hycao4437]

In that case, you would get the level curve in 2D and not the 3D surface itself.

[@anshishkumarray0383]

In the last question of video 13.6 , the function was in 3D in that case gradient must be in 2D . In the last problem of this lecture, it is an equation of a level surface which is in 3D but the actual function which we have to pretend is in 4D, as professor said that the gradient must be one dimesion lower than the function therefore gradient for this question must be in 3D. In 13.6 question, "function" was given. But in this problem "equation of level surface" was given we had to pretend that a function exists in 4D. thank you for your question I was confused myself after reading your comment but I figured it out after thinking for a while.

[@FtwChaotics]

I yawned right at 46:05. I felt like he was staring into my soul.

[@annabelleellis8098]

hi, I was wodering for example 2, is the gradient a normal to a bunch of level surfaces or level curves? What I got from it was that when in 4-D and the gradient is taken which gives a normal vector, it gives the normal to a bunch of level surfaces. When in 3-D and the gradient is taken, we would get normals to level curves. WLOG, that we were first given a 3-D surface the we cheated to get a function in 4-D and same for 2-D. Clarify, please.

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professor leonard, please Keep fighting the good fight!

[@yinkak3921]

Im not sure if I'm the one who made a mistake but for the last question isn't the gradient at p(0,3,1) supposed to be instead of

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are you going to go over differentiability in several variables, iam really having a hard time trying to grasp the idea, it would be awesome if you did !!

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Love you

[@thezeneris]

At 1:27:45, I found it easier to make it tan z - y/x = 0.

[@ton146]

If we know the two partials fx and fy and we could turn these into direction vectors we could do a cross product to get the normal. Therefore how do we get the direction vectors from the slopes which are scalars? How could on do this in 2D - ie turn dy/dx into a direction vector? Tony Johannebsurg

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Hello Professor many thanks for all your great efforts. Is there a way we can get the lectures notes beside the videos. Thanks again

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[@Buranku-go3wu]

so basically, is the tangent plane directed depending on the very close points those surround the original point that we want to find the tangent plane on it ?

[@estefaniakiara-elizabeth8538]

Thank you very much professor :)

[@yaredmamo7451]

why did we not do an example where we dot the gradient with the derivative of r(t)

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Waiting a lecture on differential equation

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just keep waiting

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Fredde He probably already mastered it, needn’t wait anymore.

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Well it needs its own course not a lecture

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makes so much sense now

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what would I do without you!!