Question

Hi could someone help explain
- How to find the tangent vector to the level curve
- Find the vector that's orthogonal to the tangent vector

Thanks :)

Answer 1

tangent vector T = dr / dt, where r = position vector, t is parameter

normal/orthogonal N = dT/dt = d2r/dt2

Answer 2

Do you know how to explain it in terms of gradients and dot products, etc?

Answer 3

yes.

do you have a specific example?

Answer 4

Sure f(x,y)=x+y

Answer 5

I already found that the level curve is y(x)=c-x

Answer 6

ok. to use the del gradient (is that what you want), write it as z = x + y, W = z - x - y = 0 and find the directional derivative:
\[\nabla \psi = <-1, -1, 1>\]

that (ie the <-1, -1, 1>, or <1, 1, -1>) is the normal vector at any point on that surface,

which makes sense because what you have described is the plane -x -y +z =0 or x +y -z =0.

Answer 7

Okay, so how would be find the vector orthogonal to that? Use the gradient again? I'm sorry I was lost during this lecture.

Answer 8

there are countless vectors orthogonal to that one, and tangential to the surface, all lying on the plane itself. that is why you use the normal to describe surfaces.

another example, f(x,y) = z = x^2 + sin y

so S(x,y,z) = x^2 + sin y - z = 0
grad(S) = <∂S/∂x, ∂S/∂y, ∂S/∂z> = <2x, cosy, -1> = "directional derivative", which i assume is what you are learning.

the <2x, cosy, -1> vector is the "normal" to that surface at any point (x,y,z).

in this example, the tangents specifically in the x and y direction would be dz/dx = 2x, dz/dy = cos y

there are some detailed worked examples here, if that helps:
http://tutorial.math.lamar.edu/Classes/CalcIII/DirectionalDeriv.aspx

i hope i am following you correctly

Answer 9

So would you use the directional derivative to find the tangent vector?

Answer 10

Would the vector way be correct? Or is this completely different?

Answer 11

if you are operating in 3d, eg f(x,y) = x + y in your example, we are primarily looking at finding a "tangent plane", and that plane will be defined by a normal.

so lets take f(x,y) = x + y^2

we can do it in vectors and we can write that as position vector r = <x,y,x+y^2>

we can calculate tangent vectors in the x and y directions
dr/dx = <1,0,1>
dr/dy = <0,1,2y>

most interestingly, their vector/ cross product should give the ONE normal vector at any point (x,y,z). as opposed to the many tangent vectors

<0,1,2y> X <1,0,1> =
\[\left[\begin{matrix}i & j & k \\ 0 & 1 & 2y\\1 & 0 & 1\end{matrix}\right]\]

= <1, 2y, -1>

same result as prev approach, ie grad( x + y^2 - z) = <1, 2y, -1>

is that more useful?

Answer 12

let me read that.

i think i just happened to post something that looks more like how you have been shown it.

Answer 13

It's okay. I'm not really sure what my professor wants so I can just ask my TA and if I still don't get it I could ask again? But thanks anyway!

Answer 14

Wait I think I got it but I'm not completely sure. Would I find the gradient and derivative then multiply them using the dot product? Or no?

Answer 15

The word tangent vector is throwing me off

Answer 16

yes, your use of tangent vector is what is stumping me a little

you can find the tangents wrt x and y and then cross product them and you get the normal

the normal should then dot to zero with either tangent

Eg, with u and v as the parameters for position vector r:
r(u,v) = <u,v,f(u,v)>
dr/du = <1,0,∂f/∂u>
dr/dv = <1,0,∂f/∂v>

normal =<1,0,∂f/∂u> X <1,0,∂f/∂v>

= <-∂f/∂u, -∂f/∂v, 1>

normal • dr/du = <-∂f/∂u, -∂f/∂v, 1>• <1, 0, ∂f/∂u> = 0
normal • dr/dv = <-∂f/∂u, -∂f/∂v, 1>• <0, 1, df/∂v> = 0

however, that's not what this proof you have been given is really going at. it's very mathsy for my taste and not very intuitive, i much prefer the geometric approach to this stuff.

i hope it goes well with the TA and sorry i could not be of more help.

PS both Riley and Boas (Maths for Engineers and Physicists) treat this very well. but maybe again not mathsy enough for what your prof wants.

Answer 17

It's okay but thank you! Maybe you can help me tomorrow if I have a better explanation to what they mean by "tangent vector." And my professor does more of analytical approaches, that's why i don't really get a lot of problems and it confuses me. But thank you for helping! :)

Answer 18

cheers!

Answer 19

I figured it out. Just to let you know lol

Answer 20

cool!